CONTENTS - Department of Mathematics and Statistics - University ...

CONTENTS
Introduction, General Information and Administration, Overview
SECTION 1
This covers an introduction to the package R-cmdr, presents an overview of biostatistics and
research methodology.
Biostatistics and Research Methodology; R-cmdr
Types of Data
Numerical Data and Histograms
Measures of Centre: Mean and Median
Measures of Variability: Standard Deviation, Variance and Interquartile range
Box-and-Whisker Plots
SECTION 2
This covers the measures of disease frequency and disease association with several examples
looking at prevalence, incidence, relative risks, attributable risk and odds ratios.
Prevalence and Incidence
Cumulative Incidence
Incidence Rate
Disease Association
Relative Risk
Attributable Risk
Odds Ratio
SECTION 3
This section covers a brief introduction to probability definitions, notation, rules and random
variables with examples, several involving tree diagram use.
Definitions including mutually exclusive and independent events
The Addition Rule for combining probabilities
The Multiplication Rule for probabilities
Tree diagrams with examples
Screening test terminology
Probability Distributions and Random Variables
Rules for combining Random Variables
SECTION 4
This section introduces both the Binomial and Normal Distributions which model many
phenomena arising in the real world. Consequently the distributions allow us to answer
some important and relevant questions.
The Binomial Distribution: Definition, mean and variance
The Binomial Table: Examples
The Normal Distribution: Definition
Standard Normal Distribution and Table
General Normal Distribution
Normal Approximation to the Binomial
Transforming Data to Normal
Contents

SECTION 5
This section defines sampling distributions, establishes the standard deviations of these
distributions called standard errors, and set up confidence intervals for population means,
differences between the means of two populations, proportions and difference between
proportions based on random samples drawn from the populations.
An outline of the Research Process
The Distribution of Sample Means
The Standard Error of the Mean
Confidence Interval for a Mean
The t-distribution and Its Use
Comparison of Two Independent Groups
The Standard Error of the Difference Between Two means
Pooled Estimate for the Common Variance
Comparison of Two Dependent Groups (Paired Data)
Confidence Interval for a Proportion
Confidence Interval for Difference Between Two Proportions
Summary of Distributions and Confidence Intervals
SECTION 6
This section reviews hypothesis testing, type 1 and type 2 errors, conclusive and inconclusive
results and the power of a study.
Null and Alternative Hypotheses
Study Based and Data Driven Hypotheses
One and Two Sided Tests
Four Steps in the Hypothesis Testing Procedure
Examples
Pooled proportion estimate
Clinical and Ecological Importance
Conclusive and Inconclusive Results
Errors in Hypothesis Testing
Power of a Study
Examples
SECTION 7
One factor analysis of variance
Post analysis of variance tests on means
Multiple comparison procedures
SECTION 8
This section covers the analysis of count data including the Chi-square test for contingency,
the chi-square test for trend as well as relative risks, attributable risks and odds ratios along
with their confidence intervals. The analysis of a three way table and Simpson’s paradox are
investigated as a way of introducing the concept of a confounding variable in the lead up to
regression analyses.
Categorical Data Examples
Relative Risk and its Confidence Interval
Attributable Risk and its Confidence Interval
Odds Ratio and its Confidence Interval
Chi-square Test for Contingency
Chi-square Test for Trend
Interpretation of Confidence Intervals
Simpson’s Paradox and Confounder Control
Contents

SECTION 9
This section introduces the topic of Simple Linear Regression which sets out to fit a straight
line through what is called a scatter diagram. One purpose of this analysis is to establish
whether one predictor variable is influencing the outcomes of a response variable and also
measuring the magnitude of the effect of this predictor variable on the outcome. It is possible
to use the fitted straight line to make predictions.
Simple linear regression is also the first step in controlling for a confounder variable. This
occurs with the extension to multiple regression which will be considered in the next section.
Scatter Diagrams and Examples
Equation of Fitted Straight Line
Analysis of Variance for Regression Model
Confidence Interval for Slope
Confidence Interval for Prediction
Correlation as Measure of Linear Association
Review Exercises
SECTION 10
Multiple regression models and logistic regression models are introduced in this section. In
the case of ordinary multiple regression the response or outcome variable is on a continuous
scale whereas in the case of a logistic regression the outcome measure is binary taking
therefore only two possible values interpreted as success versus failure.
The models allow us to identify those variables which have an effect on the outcomes and
those variables which do not.
Adding additional variables leads to adjusted values for estimated parameters and it is this
that allows us to control for confounding.
The Multiple Regression Model
R-cmdr Printout for Multiple Regression
Dummy Variables
Checking Model Fit
Parallel Regression Lines and Analysis of Covariance
Binary Outcomes and Logistic Regression
Study Design principles
Critical appraisal
Confounding analysis
Sources of bias
SECTION 11
Appendix 1: The Basics – mathematical rules and statistical concepts
Appendix 2: Some summaries
Appendix 3: Formulae
Contents

STAT115 INTRODUCTION TO BIOSTATISTICS 2012
Advances in our understanding of factors which affect health and wellbeing come through
research in the health sciences. Examples of such research include surveys to describe
patterns of disease in a community or risk factors for disease such as diet and smoking; studies
trying to find out whether a newly developed treatment works; studies of factors which may
prevent disease such as physical activity; studies of barriers to improving health such as
reasons for declining vaccination rates in children, prevention of smoking. Biostatistics
(statistics applied in the health sciences) is a vital tool in our mission to improve health and
wellbeing for all people.
STAT115 provides an introduction to the core principles and methods of biostatistics. In this
course you will gain an understanding of how statistics is used to answer research questions:
how to look for patterns in data, how to test hypotheses about disease causation and prevention
and improvement in well-being. The understanding and skills gained in STAT115 can be a
starting point for a career in biostatistics or can be used to assist understanding of research in
other disciplines including physiology, anatomy, human nutrition, sports science, and
psychology.
Lecturers
GENERAL INFORMATION AND ADMINISTRATION
Dr Katrina Sharples, Dept of Preventive and Social Medicine, Adams Building
Dr Janine Wright, Room 237, Science III building
Mr Daniel Turek, Room 231, Science III building
Dr David Bryant, Room 514, Science III building
Lectures
Lectures are held as follows: Monday, Tuesday, Thursday and Friday at 11.00 am,
commencing Monday 9 July. Although these notes are extensive, experience shows that
students who miss lectures have a severe disadvantage.
Help Sessions and Tutorials
These will be held in 539 Castle St Laboratory which has 36 computers. Tutorials are
cafeteria style which means that you can attend at any scheduled time when tutors are
available to help with weekly exercises. Times can be found on the STAT115 paper page on
the Mathematics and Statistics Department website. In addition, you may access the
computers to complete assignments outside of scheduled sessions. Attend early in the week
to avoid the inevitable rush before submission day.
STAT 115 Web Page and Resource Area
The STAT 115 web page: www.maths.otago.ac.nz/stat115 will contain course resource
material. Answers to weekly exercises, notices, old exam papers with solutions and any other
useful information will be posted here. You can access such information by clicking on the
Resources button. You are strongly advised to read through the solutions to weekly exercises
as students who fail to do this are at a severe disadvantage.
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Introduction & overview

Support Classes
There is also a Wednesday evening support class for students worried about their mathematics
background for this course. This class will be held in 539 Castle St at 6pm on Wednesday
evening. If you wish to attend the support class you will need to register using the form
which is available on the resource page or from the Maths and Statistics Reception, Science
III, 2 nd floor. Our experience is that only a small number of students will need to use the
support class. Note, there is no mathematics pre-requisite for this course. If you have
difficulty in carrying out the calculations in the Basics Booklet of Appendix 1 of these notes
you may find it helpful to attend the support class. In addition, you can access Mathercize by
going to the web page mathercize.otago.ac.nz, log-in password line. The options
STAT115 Exercises for Biostatistics
STAT115 Revision mathematics
will take you through background material for this course in an easy to use self-testing
environment.
Study Centre
A Study Centre will operate in a room at the back of 539 Castle St. This is an area where you
can go to work with fellow students. There will also be statistics help available at times as
shown on the door.
References
There is no set text for the course as this course booklet contains all material necessary. The
book: Harraway, J. Introductory Statistical Methods for Biological, Health and Social
Sciences. (University of Otago Press) has multiple copies on reserve in the Science Library at
the Loans Desk. The first 17 chapters are relevant for this course. A second book on close
reserve: Clark, M.J. and Randal, J.A. A First Course in Applied Statistics (Pearson).
Computing
The R-commander (R-cmdr) package will be used in tutorials. No prior knowledge of the
package is needed as a handout and full instructions will be available in the tutorials. All
students will have their own User Name and Password. The User Name is the name on your
student ID card and the Password is your student ID number.
Time Commitment
STAT 115 is a one semester course worth 18 points. It is expected that students should spend
an average of 12 hours per week on this course. After allowing four hours per week attending
lectures, this leaves eight hours for other course related activities such as assignments, reading
notes and revising.
Calculators
There is no restriction on the type of calculator that can be used except that no device with
communication capability shall be accepted as a calculator.
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Introduction & overview

Course content (in approximate lecture order)
Introduction: research methods and study design; designed experiments versus
observational studies; case control, cohort and intervention studies.
Data description and presentation: the use of R-commander; histograms, boxand-whisker
plots, measures of centre and spread of data, measures of disease
frequency and association.
Probability: the nature of random variation; diagnostic tests; probability
distributions including the binomial and normal distributions.
Estimation: sampling distributions; confidence intervals for means, differences
proportions.
Hypothesis testing: classical procedures for means, proportions, and differences;
the p-value; statistical vs clinical significance; power and sample size.
Analysis of variance: completely randomised design; bonferroni procedure for
multiple comparisons.
Categorical data: tests for association; rates, relative risk and risk differences,
odds ratios; confidence intervals for relative risk and odds ratio.
Regression and correlation: the simple linear regression model; tests on the slope;
predictions; confidence intervals for predictions; correlation.
Multiple regression: tests on the estimated parameters; dummy variables for
qualitative predictors; parallel regressions and control of confounding.
Ethics and Study design: Ethical issues, bias and confounding.
2 lectures
6 lectures
8 lectures
5 lectures
3 lectures
3 lectures
4 lectures
5 lectures
4 lectures
7 lectures
Internal Assessment
There will be eight assignments and three mastery tests. Each assessment will have a mark
recorded out of 20. These assessments will be admininstered on-line. The assignments can
be completed anywhere you have an internet connection. The mastery tests will be conducted
in the Castle St Computer Laboratory. A booking system for half-hour slots in which to
attempt the tests will operate. Cutoff times for each assignment will be announced in lectures.
Exam format
A three-hour exam will produce a mark out of 100.
Final mark
In your overall mark we will count your exam mark for 2/3 of the total and the internal
assessment for 1/3. However, if your final exam mark taken out of 100 is greater than this,
we will use just the final exam mark. That is, the final mark F will be calculated as:
F = {E, (2E + A)/3}
where E (exam mark) is out of 100 and A (internal assessment) is out of 100. The internal
assessment marks will be made up 1/3 from the eight assignments and 2/3 from the three
mastery tests.
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Introduction & overview

Email Contact with Students
From time to time lecturers may wish to email students taking STAT 115. This will be done
by contacting you using your Student email address. You should check your student address
regularly. If you have another address then you might like to arrange that emails sent to your
student address are forwarded automatically.
Disability and Impairment Support
The Department of Mathematics and Statistics encourages students to seek support if they
find they are having difficulty with their studies due to a disability, temporary or permanent
impairment, injury, chronic illness or deafness.
Contact either The Course Convenor,
or Disability Information and Support
Telephone 479 8235
Email: disabilities@otago.ac.nz
Website: http://www.otago.ac.nz/disabilities
Plagiarism
Students should make sure that all submitted work is their own. “Plagiarism is a form of
dishonest practice. Plagiarism is defined as copying or paraphrasing another’s work and
presenting it as one’s own” (University Council, December 2004). In practice this means that
plagiarism includes any attempt in any piece of submitted work (e.g. an assignment or test) to
present as one’s own work the work of another (whether of another student or a published
authority). Any student found to be responsible for plagiarism in any piece of work submitted
for assessment shall be subject to the University’s dishonest practice regulations which may
result in various penalties, including forfeiture of marks for the piece of work submitted, a
zero grade for the paper or in extreme cases exclusion from the University.
SURV 102 Computational Methods for Surveyors
Students enrolled for SURV102 will attend lectures in STAT115 for four weeks beginning on
Monday 23 July.
A separate notice about assessment in SURV102 will be made in the Surveying Department.
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Introduction & overview

Biostatistics and Health Research - An Overview
1 Health Research
Billions of dollars are spent every year in a quest to improve human health and well-being.
The broad goal of this quest is to acquire new knowledge to help prevent, detect, diagnose and
treat disease.
What sort of knowledge do we look for
What causes a disease
Once you have a disease, what happens
Who has the disease
What is the best strategy for treatment or prevention
How do societal factors affect health
What causes a disease
Understanding the factors which lead to the development of disease gives ideas about how to
prevent disease. For example:
• Drinking water is treated to kill bacteria, virus and other contaminants like giardia.
• Our ability to prevent heart disease has improved with our understanding of specific
dietary components which increase risk, and with our understanding of how exercise
works to reduce risk.
• The realization that the cause of AIDS was a virus (HIV) which could be transmitted
through sexual intercourse and blood transfusions led to prevention strategies to
reduce transmission. These included use of condoms, screening of blood products
and drugs to reduce of transmission from mother to baby.
• Understanding how and when sports injuries occur helps to develop rules of play and
training schedules which reduce injury burden.
Once you have a disease, what happens
Understanding how a disease progresses gives ideas about how to cure disease, or to prolong
survival or to improve quality of life. For example:
• Understanding how HIV affects the immune system has led to the development of
drugs such as zidovudine which prevent the virus from reproducing and seem to
slow the destruction of the immune system.
• Understanding how bacteria work allowed the development of different types of
antibiotics with different actions.
• Cancer develops when cells in a part of the body begin to grow out of control.
Knowledge of the cell cycle was important in developing cancer drugs
(chemotherapy) which work only on actively reproducing cells.
Who has the disease
Detecting who has a disease and diagnosing disease are the first steps in delivering effective
treatments. For example:
• Development of non-invasive technologies for looking inside the body (such as
ultrasound, CT scans, MRI) provided techniques for making the initial diagnosis of
cancer, or for identifying the form of damage to a knee following injury.
• Tests which look at cells from biopsies or blood can give more accurate diagnosis of
cancer than the non-invasive technologies.
• We identify people with HIV infection though a blood test which detects antibodies
to the virus.
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Introduction & overview

What is the best strategy for treatment or prevention
Once we have developed a new treatment or approach to prevention we need to evaluate the
risks and benefits of that treatment before it is made available for use. For example:
• Exercise and balance programmes have been demonstrated to reduce the risk of
falling in the elderly
• The statin family of drugs have been demonstrated to reduce risk of death from
cardiovascular disease
• Evaluations of the use of beta-carotene (which the body converts to vitamin A)
found that contrary to expectations, it did not prevent lung cancer; in fact it increased
the risk of lung cancer.
How do societal factors affect health
Working with individuals can lead to significant improvements in health, but societal factors
can also have an impact.
• Societal attitudes to alcohol and smoking can make it difficult for individuals to
change behaviour
• Understanding how societal factors operate is important for developing systems of
health care.
Where does knowledge come from
During the last century we have gained an enormous amount of knowledge, but there are still
many gaps.
• Cancer and cardiovascular disease still end many people’s lives prematurely.
• Back pain is very common. We still are not very good at treating it or preventing it.
• Diabetes is becoming increasingly common, particularly among Maori and Pacific
Island populations. It has many serious health consequences.
• New diseases provide additional challenges. HIV/AIDS, a disease thought to have
jumped the species barrier into humans, has had an enormous impact. Avian
influenza is common in birds in Asia, and can cause severe disease in humans, but
doesn’t currently spread directly from human to human. But it would only take a
small change in the genome of the virus to make it highly infectious amongst
humans.
Knowledge can come from ‘experience’ or ‘research’
Experience is a very unreliable way of obtaining knowledge. Humans are not objective; our
recall is very selective. The history of medicine is littered with treatments which doctors were
convinced, through their own experience, worked, but time has shown to be ineffective or
harmful in many of the settings where they were used: bloodletting, ground woodlice,
mercury, arsenic, and so on. These treatments were widely used centuries ago, but there are
more modern examples.
• An early treatment for heart attack, where blood flow to part of the heart muscle is
blocked, involved sprinkling powdered asbestos on to the heart to increase blood flow
to the affected areas. It was never truly shown to work, but thousands of these
operations were done.
• Hormone replacement therapy was widely used initially for treatment of the symptoms
of menopause, but was also believed to reduce risk of heart disease in postmenopausal
women. The results of a study published recently found in fact it
increased the risk of heart disease.
That leaves research.
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Introduction & overview

2 The Research Process and Biostatistics
What is research
Research is a systematic process for providing answers to questions
Examples of research questions:
• What are the causes of meningococcal meningitis
• What is the best treatment strategy for chronic back pain
• What are the genetic events that lead to childhood cancer
• Can this new drug improve survival in people with colon cancer
• What is the role of selenium as an antioxidant in the protection against risk factors for
cardiovascular disease
• To what extent do western diet and exercise habits need to change in order to reduce
insulin resistance
• Does this conditioning programme reduce serious knee injury in team sports
Biostatistics is the field of development and application of statistical methods to research in
health-related fields, including medicine, public health, and biology. Since early in the
twentieth century, biostatistics has become an indispensable tool for health research.
Statistics is often defined as the art and science of collecting, summarising, presenting and
interpreting data. Statistics is a set of techniques which formally implement the fundamental
principles of the scientific method. The scientific method underlies the research process:
observation and theories lead to the development of hypotheses. We work out the best test of
the hypothesis, then collect data and determine to what extent the data are consistent with the
hypothesis.
The research process
When we carry out research we often collect data on a sample or subgroup from a population.
Our goal is to use the information collected on that sample to draw inferences about a larger
population.
Underlying populations
Inference
Sample
Statistics
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Introduction & overview

Examples
• We use the frequency with which diabetes occurs in a sample to estimate the
frequency with which diabetes occurs in the population the sample came from.
• We study a new treatment in a subgroup of patients in order to be able to make claims
about the effects of the treatment in all such patients.
Steps in the research process
Development of the research questions
Design of the study
Collection of information
Data description and analysis
Interpretation of results
Ideas for research come from many places – from reading the literature, observation and
clinical experience, from talking to colleagues and from just sitting and thinking.
The first step is to refine the idea into a question, or series of questions, which can be
answered in a single study; that is, we need to be able to design a study to answer the
question. The question may be framed as a hypothesis. For example, we might wish to
answer the question “Does a low fat diet reduce risk of diabetes” The hypothesis would be
“Low fat diet reduces the risk of diabetes”. We then need to work out how best to test the
hypothesis.
The study design specifies the methods for selecting people (or other units) for the study and
for collecting the information that will be used to answer the questions. It needs to be feasible
and ethical. We need to identify which study designs can give us appropriate data, and how to
maximize our chance of being able to distinguish a true relationship from random noise.
Once we have collected the data we use statistical methods to describe and analyse the data
and interpret the results. The analysis and the interpretation of the results will depend on the
study design.
Biostatisticians work with scientists to identify and implement the correct statistical methods
for designing studies and analyzing and interpreting the results.
3. Introduction to study design
Understanding where data come from is vital for making sensible choices about statistical
analysis. At this stage in the course we will give an overview of some of the study designs
that are commonly used in epidemiology and clinical research. We will return to this material
in the second half of the course.
There are several different ways to classify study designs, and several specific ‘named’ study
designs. It can be confusing since different epidemiology books use the terms differently. The
classifications and definitions exist to help us think about the strengths and weaknesses of a
particular study for addressing the research questions. The differences in the ways the
definitions are used arise where textbooks emphasize the relative strengths and weaknesses a
little differently.
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2.1 Classifications of Study Designs
1. Descriptive versus analytic
This classification relates to the primary aims or objectives of the study. Where the study aims
to test an hypothesis we say the study is analytic. For example, does this vaccine reduce the
risk of meningococcal disease Here we hypothesize a relationship between vaccine and risk
of meningococcal disease (we hypothesize that vaccine reduces risk) and aim to test that
hypothesis. Analytic studies are studies which test hypotheses.
Descriptive studies are used where the aims are simply to describe something, with no prespecified
hypothesis. For example, if we wish to describe trends in incidence of
meningococcal disease over time we carry out a descriptive study. Here there are no prespecified
hypotheses about the reasons for a change over time.
Many descriptive studies in epidemiology describe patterns of disease in populations. This can
provide clues about causes of disease and lead on to further studies. The standard approach is to
examine the characteristics of disease according to time, place, and person:
TIME A descriptive study can be repeated in order to examine trends over time
examples: epidemics, seasonality eg: influenza
PLACE Many diseases vary according to country, or even within countries
examples: breast cancer incidence by country, multiple sclerosis and latitude
PERSON Characteristics of people with the disease can be studied, for instance age, sex,
ethnic group, socioeconomic group, occupation
example: heart disease in New Zealand according to age and sex and ethnic group
2. Experimental versus observational
In experimental studies the investigators intervene in the natural order (hence the alternative
name intervention study). The investigator decides the exact nature of the intervention,
chooses a control strategy, and decides who will receive the intervention under study and who
will be part of the control group. The goal is to control the conditions so that the effect of
interest can be isolated and studied. For example, if investigators want to know whether a
drug (nevirapine) reduces maternal-infant transmission of HIV they can construct an
experiment which isolates the effect of the drug from any other factors which might affect risk
of transmission. The extent to which we can isolate the effect of the intervention (eg drug)
determines how good the experiment is. Of course ethics are a fundamental consideration.
In observational studies we simply observe a naturally occurring process without intervening.
It is much harder to test a hypothesis in an observational study, but for many research
questions in the health sciences it is not ethical or feasible to conduct an experiment. We aim
to design our observational studies to get as close as possible to the information we would
have got if the experiment could have been done.
3. Randomised versus non-randomised (applies to experiments only)
Experiments always (should) have a control group as well as a group (or groups) which gets
the intervention(s) under study. Randomisation is a process we can use to allocate people to
either the intervention group or the control group – the simplest version of randomisation is
like flipping a coin: each person has a 50% chance of being in the intervention group. Careful
use of randomisation gives the best test of an hypothesis.
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In some experiments the investigators use a method other than randomisation to decide who
will be in the intervention group and who will be in the control group. For example in a
community intervention study the investigators might choose a set of communities to get the
intervention (often those interested or with structures in place to take part), and then choose a
matched set of control communities. Experiments like this which are non-randomised are
sometimes referred to as quasi-experiments. Sometimes they are the only practical alternative,
but they never provide the same strength of evidence as a randomised trial.
Note that the process of randomisation is not the same as random sampling. The purpose of
random sampling is to select a single group which is representative of a population (see
below).
4. Cross-sectional versus longitudinal
This classification refers to the data themselves and the (calendar) time points or periods
about which the information is collected. For example, we might do a study looking at the
relationship between oral contraceptive use and coronary heart disease. Fully cross-sectional
data would refer to one point in (calendar) time. For example, in a survey we might ask, do
you have coronary heart disease today Are you taking oral contraceptives today Note that if
we are collecting data on existing disease we are working with prevalence of coronary heart
disease rather than incidence of coronary heart, and so cross-sectional data is not very good
for testing hypotheses about the causes of disease. (The exposures may have changed after
disease was diagnosed.)
Longitudinal data have some time course present. The ideal for testing hypotheses about
disease causation is to get information about things that occurred before the disease
developed. Often the best we can do is collect information about exposures that occurred
before diagnosis of disease since the time between developing disease and diagnosis is often
unclear. Longitudinal studies collect information over a period of time, eg exposures which
occur before disease is diagnosed.
5. Study unit
The majority of studies in epidemiology collect data on individuals. However, there are some
where the ‘unit’ under study is something bigger – such as a family, a community or a
country. In some studies it is the group that is of interest, not the individual, and we might
want to test a hypothesis relating to the group (an analytic study). For example, the COMMIT
study asked, does a community prevention programme reduce the prevalence of smoking in
the community The intervention is carried out at the community level, and we can evaluate
by examining whether the prevalence of smoking in the community changes. Note the
outcome data are collected on the individual (whether someone smokes or not), to measure the
effect of the intervention in a community.
2.1 Common study designs in epidemiology and clinical research
1. Case report
Usually describes the occurrence of disease in one person. The purpose is to alert others to the
fact that this combination of factors can occur, and to encourage people to keep a look out for
other similar cases. Such case reports (to a central registry) led to the initial recognition of
AIDS. Case reports are always descriptive and observational. The cross-sectional longitudinal
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classification doesn’t really apply, but they could be considered ‘longitudinal’ in the sense
that they may collect data on the person’s experience over time.
2. Case series
A case series takes a group of people with a recognised disease and describes patterns among
them. A study of the initial case series of men diagnosed with AIDS recognised a common
dysfunction of the immune system and that the disease occurred in gay men, injecting drug
users and blood product recipients. This led to the hypothesis that it was caused by a
transmissible agent, and gave clues as to the modes of transmission. Case series are always
descriptive, observational, and are generally cross-sectional, but could be longitudinal if they
describe changes in individuals over time.
3. Descriptive study using population data
Many descriptive epidemiological studies make use of data that is collected routinely on a
population. This includes census data, death certificates, data reported to cancer registries,
hospital morbidity and mortality data, and infectious disease data reported as ‘notifiable’
diseases. Provided the data sources are reliable this can provide valuable descriptions of the
disease (or risk factor) experience in a population. These studies are descriptive and
observational.
4. Sample survey
Where data are collected specifically for a research study, they generally involve collecting
data for only a sample (subset) of the population of interest. This will give the opportunity to
collect more information about each person, at the cost of the random variation that comes
with sampling from a population. There are many way to go about selecting a sample. In
quantitative research we generally choose random samples. In a random sample everyone has
a known chance of being selected for the study; this allows us to use statistical methods to
accurately determine the influence of random error (through use of confidence intervals).
And hence, to make valid inferences regarding the population the sample came from. Random
sampling gives us the best chance of getting a sample which is representative of the
population.
The simplest type of random sample is a simple random sample, where everyone has the same
chance of being chosen. We can also draw stratified samples or cluster samples. In stratified
sampling we divide the population into groups (or strata) – for example ethnic groups. We
then choose to sample a fixed number from each stratum to ensure all groups are adequately
represented in the study. For example, we might wish to choose the same number of people
from each ethnic group to ensure we have enough data for reliable estimates in each group.
Cluster sampling is used where we can’t easily select a sample of individuals. For example, if
we wish to study children, we can’t carry select a simple random simple because we have no
list of children from which to select the sample. One approach commonly used is to select
schools at random, classrooms within a school at random, and children from a class at
random.
A true survey generally means getting people to fill in a questionnaire. However people have
extended the idea to include other forms of data collection: we may take measurements of
height and weight, fitness tests, blood tests and so on.
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These studies are most often descriptive, but can be analytic, are observational, and can be
cross-sectional or longitudinal.
5. Cross-sectional study
In epidemiology the term cross-sectional study often refers to a survey. The data are often not
fully cross-sectional according to the definition above. For example we might carry out a
survey of use of hormone replacement therapy (HRT) among New Zealand women.
Such a survey would generally ask about past life experiences and past use of HRT, rather
than just current use, which gives a longitudinal element to the data. When the study collects
information about disease status, it is generally prevalent disease. So while cross-sectional
studies can be used to test hypotheses they are not very good for testing hypotheses about
disease causation.
6. Case-control study
Two groups
Group with disease (cases)
Group free from disease (controls)
In a case-control study, people are selected for the study according to whether they have the
disease of interest (cases) or not (controls). Generally case-control studies identify incident cases
and collect information about experiences before diagnosis of disease of the cases, and for an
equivalent time period for the controls. Case-control studies are sometimes called retrospective
studies because information is collected about exposures that occurred in the past. For example, a
case-control study of cervical cancer selected a group of women with cervical cancer and a
control group of women who did not have cervical cancer. Information was collected about past
experiences which were hypothesised to be related to risk of cervical cancer including number of
sexual partners. Case-control studies are analytic, observational and longitudinal.
7. Cohort Study
A group of people is observed over a period of time in order to measure the frequency of the
disease being investigated. A cohort study starts by documenting exposures and then measuring
the subsequent risk of developing disease, according to exposure. Cohort studies aim to identify
associations between exposure to suspected causal agents and the development of disease. The
cohort may be selected by taking a random sample from a population (eg the Scottish Heart
Study); by selecting some geographical areas (eg Framingham study) or taking a particular group
(eg British Doctors study, Nurses Health Study). Researchers may also identify an exposed group
of interest (eg people working in a particular industry) and find an appropriate control group who
are not exposed to the substance under study. Exposure can be measured at the beginning of the
study (baseline) and also periodically during the follow-up period. The entire cohort of people is
followed up to determine if and when disease develops.
8. Randomised controlled trial (RCT)
In a randomised controlled trial a group of study participants are selected and then randomly
allocated to an intervention group (s) (who get the intervention under study) and a control
group. Since group allocation is entirely by chance, this is the best approach for getting two
groups who are comparable is all respects. This means that if there is a difference in outcome
xii
Introduction & overview

etween the two groups it can be attributed to the intervention (provided other aspects of the
study are well carried out).
9. Clinical trial
This the term used for an experiment which evaluates a treatment. They are often, but not
always, randomised controlled trials.
10. Prevention trial
This is the term used for an experiment used to evaluate a prevention strategy. They can be
randomised controlled trials.
11. Community intervention study
This is the term used for a study to evaluate a community intervention. They are usually
experiments, but often not randomised, and may not involve a control group.
4. Content of STAT115
Learning aims and objectives
By the end of the course students should
• be aware of the appropriate use of common study designs and their strengths and
weaknesses
• be able to describe the information contained in a data set
• be able to carry out common statistical data analyses
• be able to interpret the results of common statistical analyses in the context of the
particular study design used
• be aware of ethical issues relating to research involving humans
• be able to critically evaluate selected research articles published in health sciences
journals.
The material in this course will provide skills for interpreting research in your chosen field of
study, as well as some basic skills for analysing data that you collect through course projects
or labs using a computer and a statistical software package. If you have mathematical skills,
and are stimulated by the idea of being involved in health research, you may wish to pursue a
career in biostatistics. There are many jobs available for biostatisticians, in New Zealand and
overseas. Most are employed in research groups at universities or government or in
pharmaceutical or biotech companies.
Types of research questions covered in STAT 115
There are many types of research question in the health sciences:
• Laboratory studies: research involves understanding how cells and cell components
work, identifying compounds which can be used to treat disease and how they affect
cells.
• Animal studies: used as models for humans
xiii
Introduction & overview

• Human studies:
– anatomy and physiology consider the structure and function of the human body
– clinical research asks questions relating to patient care including evaluation of
new treatments
– epidemiology is the study of the distribution and causes of disease
• Studies of public health: the science and art of promoting health, preventing disease
and prolonging life through organised efforts of society
• Studies of society:
– medical sociology examines topics such as the social aspects of physical and
mental illness, physician-patient relationships, the organization and structure of
health organizations and the socio-economic basis of the health care system.
In STAT 115 we will focus on research questions involving humans, mainly clinical research
and epidemiology. There are many research questions in these areas which can be understood
without specialised knowledge. In the other areas, particularly laboratory studies, an in-depth
understanding of the field (eg biochemistry, molecular biology, anatomy or physiology) is
needed to understand the research questions.
Studying humans brings particular challenges, and it is these challenges which have driven the
specialised development of biostatistics from it statistical basis. The challenges arise from the
more complex ethical issues in research involving humans, as well as the complexities of the
biological system and the consequential research questions we wish to answer.
xiv
Introduction & overview

SECTION 1
This covers an introduction to the package R-cmdr, presents an overview of biostatistics and
research methodology.
Biostatistics and Research Methodology; R-cmdr
Types of Data
Numerical Data and Histograms
Measures of Centre: Mean and Median
Measures of Variability: Standard Deviation, Variance and Interquartile range
Box-and-Whisker Plots
1
Section 1

Biostatistics and research: an overview
Course aim:
An introduction to the core biostatistical methods
essential to the health sciences
• scientific method
• design of research studies
• description and analysis of data
The scientific method underpins the design of
research studies. Sound research design is vital
for obtaining reliable information. A major part
of this course is about techniques for describing
data and understanding the analysis principles.
This enables us to make sense of the mass of
information collected in a research study.
2
Section 1

Learning aims and objectives
By the end of the course students should
• be aware of the appropriate use of
common study designs and their strengths
and weaknesses
• be able to describe the information
contained in a data set
• be able to carry out common statistical
data analyses
• be able to interpret the results of common
statistical analyses in the context of the
particular study design used
• be aware of ethical issues relating to
research involving humans
• be able to critically evaluate selected
research articles published in health
sciences journals
3
Section 1

Goal of health sciences professions
To improve the health and well-being of
individuals and communities
This involves
• treatment of disease
• prevention of disease
• promotion of health
In order to do this we need knowledge about
• causes of disease
• diagnosis
• disease processes
• effectiveness of treatments
• societal factors which affect health
4
Section 1

Examples of current gaps in knowledge
• causes of meningococcal meningitis
How to prevent Vaccine
• SARS, avian influenza
New diseases
• back pain
Not good at treating
• cancer
Nasty treatments for child cancer
• diabetes
Common in Pacific communities
• cardiovascular disease
Common cause of death
• prevention of overweight and obesity
• effective promotion of behaviour change
Prevention of smoking
Knowledge may come from
• teaching
• experience
• research
5
Section 1

Research
A process for providing answers to questions for
which the answer is not immediately available
General research areas
What are the causes of meningococcal
meningitis
Can we develop a vaccine to prevent SARS
What are the genetic events which lead to
childhood cancer
Can a new drug improve survival in people with
colorectal cancer
How can we prevent childhood overweight and
obesity
What are the main factors affecting quality of life
of people with a chronic illness
Research provides a systematic process for
answering these questions
6
Section 1

Iron Deficiency – Should NZ parents be
Concerned
[Dr Elaine Ferguson, Dept of Human
Nutrition]
A survey randomly selecting 323 children
aged 6-24 months in Dunedin, Christchurch
and Invercargill.
To assess prevalence of iron deficiency.
To explore factors associated with low body
iron store. Possible Factors are:
Categorical:
Continuous:
• Sex
• Ethnicity
• Maternal Education
• Household Income
• Breast feeding
• Age
• Meat intake
Regression methods are used as well as
procedures for summarising data.
7
Section 1

Does early childhood circumcision reduce the
risk of acquiring genital herpes
[Dr Nigel Dickson, Dept of Preventive and
Social Medicine]
• Cohort of over 1000 births in 1972 in
Dunedin.
• Called the Dunedin Multidisciplinary
Health and Development study.
• Does early circumcision reduce the risk
of genital herpes.
• Initially appears to be the case but it is
an observational study.
• Number of sexual partners is a
confounder.
• When confounder allowed for early
circumcision appears not to be
protected.
• Designed experiments (or clinical trials)
set up in Africa to investigate effect of
circumcision on HIV.
8
Section 1

The research process
The objective for most studies is to use data from
a sample to draw inference about a larger
population:
Underlying population
Inference
Sample
Statistics
Examples:
• we use the frequency with which a disease
occurs in a sample to estimate the
frequency with which disease occurs in the
population
• we study a new treatment in a group of
patients in order to be able to make claims
about the effects of the treatment in all such
patients
9
Section 1

Steps in the research process:
Development of the research question
Design of the study
Collection of information
Data description and analysis
Interpretation of results
• the research question
- needs to be framed very carefully
- must be specific enough to be
answerable by a research study
• the study design
- is determined by the research
question
- describes the methods used to collect
the information
• analysis and interpretation
- depends on the study design
10
Section 1

Research questions relevant to this course:
Epidemiology:
the study of distribution
and determinants of disease
frequency
Clinical research: the study of questions
relating to care of patients
Descriptive questions:
What is the distribution of a disease
What is the natural history of a disease
Analytic questions:
What are the causes of a disease
Will this approach prevent disease
Does this treatment improve outcome
11
Section 1

Data Analysis and Computer Software
Easy to use software is essential for data
management and data analysis. In this course R-
cmdr (Statistical Package for the Social Sciences)
will be used. This package is widely available on
campus, used in most Departments which specify
first year statistics as a pre-requisite, and widely
available internationally.
At school you may have used EXCEL. Possibly
at University you have used EXCEL. EXCEL
is excellent for data management and reporting
but is poor for statistical analyses and clumsy for
graphical procedures.
R-cmdr is easy to use with good pull down menu
options. There are three windows in R-cmdr
• Data Editor (where data being analysed are
located)
• Output Window (where results appear)
• Syntax Window (not used in this course)
12
Section 1

Introduction to study design
1. Descriptive studies
2. Analytic studies
Experimental studies
Observational studies
Examples of analytic study types
3. Summary
Classification of research designs
Classification of common study types
There are two types of research questions.
Descriptive – describing things
Analytic – testing hypotheses
Strengths and weaknesses of the different designs
will be discussed.
13
Section 1

1. Descriptive studies
Aim: to describe, for example:
• the characteristics of people with a disease
(person, place, time)
• lifestyle patterns of a population
• attitudes to health care
• etc
Descriptive studies are often called surveys or
cross-sectional studies
Descriptive studies generally use a sample from a
population
14
Section 1

Example: What are the serum cholesterol levels
of New Zealanders
Method:
Select a subgroup (sample) of people
and measure their serum cholesterol
levels
Random sampling
• choose the sample in such a way that
every individual in the population has a
known chance of being selected
• in a simple random sample, everyone has
an equal chance of being chosen
• this method is the best way of obtaining a
sample which is representative of the
population
Suppose we want to estimate mean cholesterol in
the population:
15
Section 1

Sample average = true mean + error
unknown
random error:
systematic
error
random
error
• due to natural biological variability
• increasing the sample size will reduce
the random fluctuations in the sample
mean
systematic error (=bias)
• due to aspects of the design or
conduct of the study which
systematically distort the results
• occurs if a sample is not representative of
the population
• cannot be reduced by increasing the
sample size
16
Section 1

2. Analytic studies
Purpose: to test hypotheses, about, for
example:
• causes of disease
• methods for prevention of disease
• the effects of treatments
Experimental studies
• the researcher intervenes and records the
result of their intervention
• the aim is to control all other factors to
isolate the effects of the intervention
• best way to study causation
Observational studies
• the investigator does not intervene, simply
observes a naturally occurring process,
and collects information
• ideal is to get as close as possible to the
information that would have been
obtained if the experimental study could
have been done
17
Section 1

Example: Options for studying the
relationship between smoking and lung cancer
Experimental study
Randomly assign people
Follow for 20 years
Check lung cancer rates
Clearly unethical
Smokers (start)
Non-smokers
Observational study
Cohort
Smokers (known) 20yrs % with lung CA
Compare
Non-smokers (known) 20yrs % with lung
CA
Problem: groups may differ in other ways that are
related to CA risk – confounding.
Case control
% smokers 20 yrs lung cancer (now) (cases)
% smokers 20 yrs no lung cancer (now)(controls)
No long term follow up needed. Smaller
samples. Could be recall bias from 20 years ago.
Also confounding.
18
Section 1

Examples of analytic study types
Randomised controlled trial (RCT)
• a “Gold standard” analytic study (best)
• experimental
Characteristics of a RCT:
• select a group of people
• randomly allocate them to either an
intervention or a control group
• follow participants up over time, and
measure outcome
A control group is used to isolate the effects of
the intervention
Random allocation, or randomisation means
every person has the same chance of being in
each. This gives the best chance of getting two
groups which are comparable in all respects
Used to evaluate new treatments
Often not ethical in studies of disease causation
19
Section 1

Example RCT: LIPID study (NEJM, 1998)
Does treatment with pravastatin reduce the risk
of death in patients with coronary heart disease
Study participants:
9014 patients
age 31-75
coronary heart disease
cholesterol 155 - 271mg/decilitre
participants (selected)
control
(n=4502)
randomisation
randomly
allocated
intervention
pravastatin
(n=4512)
6 yrs
8.3% mortality 6.4%
20
Section 1

Advantages of RCT:
• experiment – the best way to test an
hypothesis
• differences in outcome can be attributed to
the exposure
Disadvantages of RCT:
• may not be ethical
Cohort study
Observational study, generally carried out to test
hypotheses
Characteristics:
• participants are selected before
disease has developed
• followed over time to determine
development of disease
• information is collected about
exposures at baseline and during
follow-up
• longitudinal
21
Section 1

Example of cohort study:
Study to investigate the relationship between
smoking and lung cancer (eg British Doctors
study)
Group of people
without lung cancer
smokers
non-smokers
20 years
% develop % develop
lung cancer lung cancer
Compare
22
Section 1

Case-control study
Observational study, generally carried out to test
hypotheses
Characteristics
• participants are chosen on the basis of
their disease status: a group with disease
(cases) and a group without (controls)
• information is collected from people with
and without disease about exposures that
occurred in the past
• longitudinal (retrospective)
23
Section 1

Example of case-control study
Study to investigate the relationship between
smoking and lung cancer
Known at start
group of people
with lung
cancer
group of people
without
lung cancer
document smoking history
% smokers %smokers
in the past
in the past
24
Section 1

Cohort vs case-control studies
Cohort study
Advantages:
• closest observational study to
randomised controlled trial
• good for examining common outcomes
• can evaluate the effect of exposure on
multiple outcomes
Disadvantages:
• long duration needed if the disease takes
a long time to develop after exposure
• if the disease is rare, the number of
participants needs to be very large
Case-control study
Advantages
• relatively quick
• smaller than cohort studies, particularly
for rare diseases
• can examine the effects of multiple
exposures
Disadvantages
• events have already occurred so the
potential for bias is higher
25
Section 1

3. Summary
Classification of research designs
Note: these provide a useful framework for
thinking about the strengths and weaknesses of
different study designs, but they will not always
work.
i) Classification by purpose of the study
descriptive (describe things)
versus
analytic (testing hypotheses)
ii) Classification by form of the design
experimental (researcher intervenes)
versus
observational (researcher observes)
iii) Classification by time
cross-sectional
(information collected about one point in time)
versus
longitudinal
26
Section 1

Classification of common study types
Randomised controlled trial
• analytic
• experimental
• longitudinal
(prospective)
Cohort study
• analytic
• observational
• longitudinal (usually prospective)
Case-control studies
• analytic
• observational
• longitudinal (retrospective)
27
Section 1

Types of data and graphical summaries
[A] Data and variables
There are two types of measurement of interest in
many scientific studies.
• First, the outcomes measured on each
experimental unit (plant, animal, person)
provide values of what is called a response
variable.
• Second the characteristics or levels of exposure
that explain at least some of the differences in
the observed values of the response variable
are called explanatory variables.
e.g. iron levels in new born children is the
outcome or response – what are the
explanatory variables
e.g. diabetes presence is outcome – what are
the explanatory variables
Data forming the response and exposure
variables can be either categorical or numerical
(otherwise known as qualitative and
quantitative).
28
Section 1

1. Categorical data:
The simplest case involves two categories.
For example a person could be
• male/female
• smoker/non-smoker
• diabetic/non-diabetic
Such data have other names such as binary
data, dichotomous data, yes/no data and 0 – 1
data (the last is particularly important, for
example 0 represents non-diabetic and 1
represents diabetic).
A problem could be to establish the chance
(or probability) that a woman with a certain
profile (defining the explanatory variables)
may drink alcohol during pregnancy (the
response) or equivalently to find the
proportion of pregnant women who will drink
alcohol. Ultimately, we are interested in who
will do this.
More than two categories can occur.
• blood group: A/B/AB/O
• Maori/Pacific Island/Caucasian/Asian.
29
Section 1

In these examples the data are said to be
nominal. But this type of data is said to be
ordinal if the categories are in some order.
For example, “degree of pain” may be
minimal/moderate/severe/unbearable
If more than two ordinal categories it is not
possible to use 0/1/2/3 to identify the
classes since “unbearable” is not three
times “moderate” even though the data are
ordered. Consequences of this will be
important in the second half of the
semester.
2. Numerical data:
(a) Discrete Here observations take only
certain numerical values. Usually they are
counts of events. For example,
• number of possums caught in traps
• number of children in a family
(0/1/2/3/4)
30
Section 1

These are not like categorical data as 3
children is three times as many as one.
This type of data can be treated as though it
is categorical but this discards information
about the magnitude of the relationships
between successive outcomes. Ordinal
categorical data is important.
(b) Continuous quantitative measures. Here
recorded values or observations result from
some form of measurement [e.g. height,
age, blood pressure, serum cholesterol,
oxygen levels in a lake].
• Often no restriction on values other than
that caused by accuracy of equipment
for recording values.
• Often the values show pattern similar to
what is called the bell-shaped normal
curve with many values clustered
around a central point and few values in
the tails.
31
Section 1

3. Rates, Ratios and Proportions
These are constructed from categorical data
and include for example measures of disease
frequency and disease association. Examples
of disease frequency are
• prevalence or proportion (concerned with
existing cases)
• incidence rate (concerned with new cases)
e.g. the prevalence of obesity in the New
Zealand population
(Gives indication of burden on the country
by identifying proportion affected)
e.g. the incidence rate of HIV in New Zealand
in 2008.
(This deals with number of new cases and
is useful if looking at causes)
Examples of disease association are
• absolute (or attributable) risk
• relative risk
• odds ratio
32
Section 1

e.g. the relative risk of melanoma for a
farmer compared with an office worker.
Here, the prevalence of melanoma
among farmers is divided by the
prevalence among office workers. This
will show if there is any association
between prevalence of melanoma and
occupation after an appropriate analysis
by essentially comparing the two
groups.
4. Other types of response data
• Scores (direct measurement not possible;
instead a patient is assessed on several
subjective scales then the values on each
are added to give a score for a patient)
e.g. 30 questions on a health survey. A
respondent gives values 0 to 3 on each
question then score out of 90 given. This
total has convenient properties whereas
individual values may not.
• Patients assess their degree of low back
pain after treatment on scale 1 (no pain) to
5 (unbearable pain).
33
Section 1

Two treatments may be assessed from the
two sets of values for patients in a new
treatment compared with a standard. The
data may be viewed as categorical or
continuous but there are problems as the
difference between 1 and 2 is not
necessarily the same as the distance
between 4 and 5. The data are certainly
ordinal.
• In social sciences, data are often ordinal.
e.g. In a questionnaire people are asked to
respond by checking the category that best
describes their level of agreement with a
statement from
a great deal somewhat not much not at all
usually coded as 4, 3, 2, 1.
Such data can be regarded as continuous or
categorical (ordinal). If ordinal then a
question is how many categories should be
chosen e.g. 4 (as here) or 5 or 7 or 9, and is
the distance between 1 and 2 the same as that
between 2 and 3 etc
34
Section 1

[B] Describing Numerical Data
Graphs can be used to summarise data but many
graphs can be highly misleading especially if too
much information is presented. We shall
summarise numerical data graphically using
• histograms
• box-whisker plots
Particular values which summarise numerical
data are:
• mean; median; mode
• standard deviation; interquartile range
These approximate the centre and the variability
of the data collected respectively.
35
Section 1

Example for Continuous Data: In a
hypertension study 56 men who are heavy
smokers (smoked for 25 years) have blood
pressures measured (in mm of Hg). Summarise
the outcomes.
Blood pressures are classified into intervals to
form a frequency table and interval frequencies
(f j ) are obtained as shown below.
Frequency Table
Pressure(mm of Hg) Frequency (f j )
59.5 – (69.5) 2
69.5 – (79.5) 7
79.5 – (84.5) 9
84.5 – (89.5) 10
89.5 – (94.5) 11
94.5 – (99.5) 7
99.5 – (109.5) 8
109.5 – (119.5) 2
Total
56 (sample size)
Although the readings are likely to be recorded to
the nearest mm and hence appear to be discrete,
the data are actually continuous and for this
reason the intervals are recorded as 59.5 – (69.5)
which is 59.5 up to but not including 69.5.
36
Section 1

Relative frequency: this is f j /n in the j th interval
where n is the sample size.
Pressure
Relative
(mm of Hg) Freq(f j ) Freq(f j /n)
59.5 – (69.5) 2 0.036
69.5 – (79.5) 7 0.125
79.5 – (84.5) 9 0.161
84.5 – (89.5) 10 0.179
89.5 – (94.5) 11 0.196
94.5 – (99.5) 7 0.125
99.5 – (109.5) 8 0.143
109.5 – (119.5) 2 0.036
Total 56 1.00
Here, 2/56 = 0.036 (rounded to 3 d.p.)
7/56 = 0.125
Percentage frequency: the relative frequency
multiplied by 100.
e.g. 0.036 = 3.6% (or 3.6 per 100) meaning that
3.6% of the values are in 59.5 – (69.5)
Note: Relative (or percentage) frequencies allow
comparison of samples when samples are of
unequal size. Absolute frequencies f j will not
allow this since all f j will be large for a large
sample of outcomes but small for a small sample.
37
Section 1

Histograms: These are simple pictures of the
data. The base of a rectangle is interval length
and area of a rectangle is proportional to class
frequency (or relative frequency). When class
intervals are all equal, rectangle heights are
proportional to the frequencies as well.
Example: Return to the blood pressure readings.
Pressure (mm) (f j ) (f j /n)
59.5 – (69.5) 2 0.036
69.5 – (79.5) 7 0.125
79.5 – (84.5) 9 0.161
84.5 – (89.5) 10 0.179
89.5 – (94.5) 11 0.196
94.5 – (99.5) 7 0.125
99.5 – (109.5) 8 0.143
109.5 – (119.5) 2 0.036
Total 56 1.00
12
10
8
6
4
2
0
FREQ PER
5mm Interval
59.5
69.5 79.5 89.5 99.5 109.5 119.5
38
Bl pr
Section 1

N.B. (1) The heights of the first two and last
two rectangles are halved but their bases are
doubled from 5 to 10mm. (Area therefore
remains proportional to frequency in these
intervals if 5mm is regarded as the horizontal
“unit”)
(2) The label on the vertical axis is given as
“Freq. Per unit interval” where “unit” = five.
(3) The relative frequency histogram follows:
0.20
0.15
REL FREQ PER
5mm Interval
0.196
0.125
0.10
0.063
0.072
0.05
0.00
0.018
59.5 69.5 79.5 89.5 99.5 109.5 119.5
Bl pr
(4) The frequency and relative frequency
histograms have the same shape. Only the
scales on vertical axis differ. Both give some
idea of the data centre, the extent of the
variability in the data and the distribution of
the data.
39
Section 1

(5) The relative (or percentage) frequency
histogram is used if comparing two (or more)
samples of data, one sample of values from a
control group and the other from a treated
group of experiment units.
(6) Notice how a histogram with rectangle
heights proportional to class frequencies
would give a misleading picture of the data.
(7) You will find that most of the histograms
produced by statistical packages like R-cmdr
have class intervals of equal length and you
can decide the number of intervals you want
in the graph. Usually between 5 and 20
intervals of equal length are chosen for a
good summary of the data.
40
Section 1

Measures of Central Tendency.
The mean is “typical” of the majority of data in a
sample.
Example: Six patients lived the following years
after diagnosis of HIV.
Datum (Outcome) Symbol
1.8 x 1
3.2 x 2
6.8 x 3
4.6 x 4
2.8 x 5
7.9 x 6
Mean = 6
1 (1.8 + 3.2 + 6.8 + 4.6 + 2.8 + 7.9)
= 27.1/6
= 4.52 years
Notation: mean
or x ∑ = 1
x1 + x2
+ x3
+ x4
+ x5
+
n
1
= n i
xi
n
x =
6
x
41
Section 1

Note The mean need not be one of the outcome
values and i is a suffix taking values i = 1 to i = n
(or 6 here). Any symbol can be used for this
suffix.
Example: The 56 blood pressure readings just
considered have a mean of 89.54mm of Hg. This
value is “typical” of the data in the sense that it is
near the centre of the region where most values
are located.
The Median is a second measure “typical” of data
in a sample and is the “middle value” of the data
after arranging the numbers in order from
smallest to largest.
Example: Data: 95 86 78 90 62 73 89
Rearrange: 62 73 78 86 89 90 95
Median = 86 (the middle value)
Note:
1. If 62 replaced by 5, the median is unchanged
(the mean would be much smaller). This
indicates that in general the median is not
affected by a few very extreme values
whereas the mean is.
42
Section 1

2. If even number of values, halve the two
centre values.
Example: For the 56 blood pressure readings,
the median turns out to be 89.30 (compare mean
of 89.54)
The mode is another measure of centre. It is the
commonest value in the data. This only makes
sense for discrete data. For continuous grouped
data it coincides with the peak in the histogram.
The histogram is bimodal if there is more than
one peak.
Further Notes
(1) The mean (89.54) and median (89.30) for the
blood pressure readings are close because the
data are almost “symmetrical.”
(2) For “non-symmetrical” data mean and
median are different since the mean is pulled
in the direction of the extreme values. The
data are said to be skew.
43
Section 1

0
Median
Mean
The mean may be unsuitable as a measure of
centre while the median is more “typical” of
most values.
(3) For measurements which cannot be negative
it is quite common to have many values close
to zero thus presenting a skew distribution.
This is called positive skewness. (The
histogram above represents positively skewed
data.)
(4)The opposite phenomenon with an extended
left hand tail is called negative skewness and
is rare.
(5) A trimmed mean is the mean with the lower
5% and upper 5% of values removed.
44
Section 1

Measures of Variability
“Looking at the world using data is like looking
through a window with ripples in the glass”.
(Professor Chris Wild, Auckland University)
Statistics is about variability. Variability reflects
differences in the values collected for different
units being measured, for example people, or
animals or plants or companies or readings on
different days etc. Two sets of values can have
the same mean and median yet show quite
different patterns.
Variability can be random or caused by different
treatments or “factors” acting on the experiment
units in a study in different ways. The hope is
that the random variation will be relatively small
or controlled by choice of appropriate study
designs. This will result in the identification of
important treatment effects explaining key
aspects of the variation.
45
Section 1

If data are highly variable there are problems
analysing the data and it will be necessary to
select larger samples.
The first measure of variation is the range (the
distance between the lowest and highest values).
It is sensitive to any extreme values and hence
not useful. But reduced ranges (encompassing
the central 95% say of the data) are useful as
extreme values (outliers) are excluded.
Note: In clinical chemistry (e.g. cholesterol
measures) a reference range encompassing the
central 95% of values describes variability in
normal people and allows test results for other
individuals to be assessed to see if corrective
action is needed.
A second measure is the (sample) variance defined
by s 2 2
= 1
s ∑ −
−
= 1(
x
n 1
n i i x
Although the divisor is (n – 1) in this equation, we
can see that s 2 is effectively the “average” of the
squared deviations of the individual data values
46
)
2
Section 1

(x i ) from their mean x . For technical reasons do
not divide by n.
Notes: 1. The variance is an overall measure of
the extent to which values x i differ from their
mean x .
2. Squaring is essential. If the deviations from x
are added, the value 0 is obtained always.
A third convenient measure is the standard
deviation (s) given by
1
= var iance = ∑i n = ( xi − x)
n −1
s 1
Note: The standard deviation s is measured in
the same units as the original data (taking the
square root cancels the squaring).
2
47
Section 1

Example: Find the standard deviation of 11, 18, 14,
15, 12
x i x i – x (x i – x ) 2
11 11 –14 = – 3 9
18 18 – 14 = 4 16
14 14 – 14 = 0 0
15 15 – 14 = 1 1
12 12 – 14 = – 2 4
70 0 30
x = 70/5 = 14 s = 30 / 4 = 2. 74
Note that 2.74 is a “typical” or “average”
deviation from the mean x = 14.
Example: Return to the 56 blood pressure
readings
Pressure Interval f j
59.5 – (69.5) 2
69.5 – (79.5) 7
79.5 – (84.5) 9
84.5 – (89.5) 10
89.5 – (94.5) 11
94.5 – (99.5) 7
99.5 – (109.5) 8
109.5 – (119.5) 2
Total 56
The standard deviation is s = 11.21. This value is
“typical” of deviations from x = 89. 54.
48
Section 1

The Interquartile Range is another measure of
variability.
25%
data
25%
data
25%
data
Q L median Q U
Interquartile range
Range
25%
data
The lower quartile Q L is the value below which a
quarter of data lie. The upper quartile Q U has 4
3
of data below it. (These are also known as the
25 th and 75 th percentiles.)
Notes: 1. Interquartile range can be a helpful
measure of variability. It is not affected by
extreme values.
2. Computer packages also give Q L and Q U for
large data sets and the approximations for
grouped data are no longer needed.
Example: For the 56 blood pressure readings
Q L = 82.2 and Q U = 96.6 with Q U – Q L = 14.4
49
Section 1

Box-and-whisker plot
This is a second way of summarising data
graphically. Like relative frequencies it is useful
when comparing samples of unequal size.
Ex Blood pressures
Q L = 82.2; Q U = 96.6; Median = 89.3
Suppose 63 and 116 are lowest and highest
values.
X
60 70 80 90 100 110 120
X
The centre of the data, its variation, its symmetry
(or lack of symmetry) and extreme values are
displayed.
Notes: (1). Two samples can be compared
X
X
X
X
Both samples skew, the second is more variable
(larger interquartile range) with a larger median.
50
Section 1

(2) The points at the ends of the whiskers depend
on the package and are
• the extreme values or
• the 2 2 1 % and 97 2 1 % values (centiles) or
• points 1
1
2
times the interquartile range
away from the boxes
Outliers beyond these points are shown in R-
cmdr by an asterix or a small circle (as below)
where there are obvious changes in the ozone
readings recorded over summer in a New
Zealand city. An asterix will represent an
extreme outlier.
11 12 1 2 3
51
Section 1

Example: Thirty-two traps were placed in each
of three habitats: pasture, replanted forest and
tussock on Stephens Island. The data are the
counts of skinks per trap totalled over a ten-day
period in each habitat. The boxplots are below.
Summarize conclusions about skink density.
Pasture 4 3 0 2 2 1 4 1 2 5 0 1 5 6 5 6
11 3 1 1 4 8 5 14 6 8 10 7 4 8 13 6
Replant 15 24 31 8 4 18 14 33 11 16 20 1 17 12 27 26
forest 18 6 12 16 11 8 13 12 11 8 10 17 29 3 12 5
Tussock 14 23 15 14 5 16 10 16 14 10 7 10 8 12 19 17
7 12 29 10 11 11 10 10 6 13 7 10 8 12 6 12
Greater skink density in replanted forest and
tussock. Greater variation in replanted forests.
Some outliers in the three habitats:
Means: 4.88; 14.63; 12.00
Medians: 4.50; 12.50; 11.00
Std Deviations: 3.64; 8.18; 5.07
52
Section 1

Example: Thirty-four adult hoki caught off the
Kapiti coast. Individual lengths as follows:
Males: 18.7 19.0 18.8 18.4 19.3 19.6 20.3 19.9 19.3 18.9
18.9 19.0 19.7 20.4 18.6 19.5 20.3 19.9 19.2 18.7
Females: 18.6 19.6 18.3 17.5 18.3 19.0 18.5 18.7 19.3 18.5
19.1 18.7 19.1 18.8
Boxplots indicate male hoki longer than female
hoki. Slightly greater variation in the males but no
outliers. Distributions almost symmetric
Mean: 19.32; 18.71
Median: 19.25; 18.70
Std Deviation: 0.61; 0.51
53
Section 1

Interpreting Box whisker plots (Ref: Professor
Chris Wild, Auckland University)
Observed data:
A
B
A
B
The call is
B values bigger
B values bigger
The above two hold for all sample sizes. Larger
random samples have more information about the
populations from which they come. With large
random samples we can make the “B values
bigger” call from smaller shifts. Avoid using the
box whisker plots for samples smaller than about
20.
54
Section 1

Observed data:
A
B
A
B
A
B
A
B
A
B
The call is
B values bigger if
both sample sizes >20
What is my call
What is my call
Cannot tell unless
both samples are huge
Cannot tell for
all sample sizes
55
Section 1

How to make the call.
This is based on a confidence interval idea. (See
later). But the result is easy to calculate. In the
following IQR is the interquartile range and n is a
sample size.
Med
Med – 1.5 IQR n
Med + 1.5 IQR n
In the following we can claim the values of B tend
to be bigger than the values of A back in the
populations from which the samples have been
taken if these horizontal lines (intervals) do not
overlap.
A
B
56
Section 1

SECTION 2
This covers the measures of disease frequency and disease association with several examples looking
at prevalence, incidence, relative risks, attributable risk and odds ratios.
Prevalence and Incidence
Cumulative Incidence
Incidence Rate
Disease Association
Relative Risk
Attributable Risk
Odds Ratio
57
Section 2

[C] Measures of Disease Frequency
All measures of disease frequency are ratios of the
form numerator/denominator.
There are two types of ratio:
1. Proportion: everyone in numerator must be
included in the denominator.
2. Rate: a measure of time is included in the
denominator.
The measures of disease frequency are:
1. Prevalence
• gives frequency of existing cases of disease
• is useful for measuring the disease burden in a
community
• often measured in a cross-sectional survey
e.g. proportion of Otago students at 3pm Tuesday
who have swine flu’.
58
Section 2

2. Incidence:
• measures frequency of new cases of disease
• is useful for looking at causes of disease
e.g. number of new cases of cold that develop in a
week.
Example: Frequency of hepatitis in two regions.
New cases Reporting
Location of hepatitis Period Population
Region A 58 1985 25,000
Region B 35 1984-1985 7000
Region A:
58/25000/year
= 232 per 100,000 per year
= 23.2 per 10,000 per year
= 2.32 per 1000 per year
Region B:
35/7000/2years = 17.5/7000/year
= 250 per 100,000 per year
= 2.50 per 1000 per year
Note: The time period must be specified for the
results and comparisons to be meaningful.
59
Section 2

Example: In a survey of eye disease among 2477
people aged 52-85 in Framingham, Massachusetts,
there were 310 with cataracts and 22 blind.
Prevalence of cataracts
= 310 = 0.125 = 125 per 1000 (or 12.5%)
2477
Prevalence of blindness
22
= = 0.009 = 9 per 1000 (or 0.9%)
2477
60
Section 2

Example: In the following diagram the time a
person has the disease is shaded.
Subject
Number
5
4
3
2
1
Prevalence →
1 / 5
t
2 / 5
3 / 5
2 / 5
Time
Note on Prevalence:
Prevalence is the proportion of people in a
population who have the disease at a given point in
time. The time point may refer to calendar time, or
to a fixed point in the course of events.
e.g. the proportion of people free from back pain 2
months after back injury.
Note on Incidence
Incidence on the other hand quantifies the number
of new cases of disease in a given time period.
There are two measures:
• cumulative incidence
• incidence rate
61
Section 2

2.1 Cumulative incidence is the proportion of
people who become diseased during a specified
period of time
number of new cases of disease
=
total population at risk
This provides an estimate of the probability, or risk,
that an individual will develop the disease during
the specified period of time.
Example: In a study in Evans County, Georgia,
there were 609 men aged 40 – 76 who had no
detected heart disease in 1960. These men were
followed for 7 years and 71 cases of heart disease
were detected during this period.
Cumulative incidence = 71/609
= 0.117 (or 11.7%)
over the 7 year period
Notes (1) The time period over which cumulative
incidence is calculated must be specified for it to be
interpretable.
(2) Cumulative incidence assumes the entire
population at risk at the beginning of the study
period has been followed for the whole study
period. But often -
62
Section 2

• people are lost to follow-up
• people are enrolled in the study at different
times
The length of the follow-up period is not therefore
the same for everyone in the study. It is the
incidence rate that takes account of varying amounts
of follow-up time.
2.2 Incidence rate:
=
the number of new cases of disease
total person-time at risk
Same amount of person time if follow:
16 people for one year
4 people for four years
All have 16 person-years of observation.
Example: Calculation of person-years for
incidence rate
Total
Jan Jan Jan Jan Jan Jan time
Subject 1997 1998 1999 2000 2001 2002 at risk
A • (lost to follow-up) 2.0
B • ×
3.0
C • 5.0
D • 4.0
E • ×
2.5
Total years at risk 16.5
63
Section 2

• = Initiation of follow-up
× = Development of disease
Number of new cases = 2
Number of person-years at risk = 16.5
Incidence rate = 2/16.5 = 0.121
That is, 12.1 cases per 100 person years of
observation
Example:
A study in the United States measured the incidence
rate of stroke in a group of 118,539 women aged
30-55 years of age. The women were free from
stroke in 1986, and were followed for 8 years.
Person-years
of
observation
(over 8 years)
Stroke
incidence
rate
(per 100,000
person years)
Smoking
category
No. of cases
of stroke
Never smoked 70 395,594 17.7
Ex-smoker 65 232,712 27.9
Smoker 139 280,141 49.6
Total 274 908,447 30.2
64
Section 2

274
Incidence rate = × 100,000 = 30.2 cases of
908,447
stroke per 100,000 person-years of observation.
Average follow-up per woman
= 908,447
118,539
= 7.7 years
Note: The denominator for measures of incidence
should include only those who are at risk of
developing the disease. It should exclude
• those who already have the disease
• those who cannot develop the disease
Failure to do this will lead to an underestimate of
the true incidence since fewer will develop the
condition.
For example when studying the incidence of
endometrial cancer we should exclude women
who have had a hysterectomy.
65
Section 2

Example In (a)-(c) calculate a relevant measure
of disease frequency and give its name.
(a) You survey 346 travellers returning from overseas
travel and find that 95 of them experienced a
diarrhoeal illness on their trip.
(1 mark)
(b) A tour of 143 people is travelling through Central
America for 2 weeks. During this trip 28 of the
people experience a diarrhoeal illness. (1 mark)
(c) A group of 18 Peace Corps volunteers in Guatemala
kept daily records of their exposure to various risk
factors (such as untreated water) and whether or not
they had diarrhoea. The following values are the
numbers of new episodes of diarrhoea with the
number of weeks of records (in brackets) for each
of the 18 individuals:
12(88) 12(46) 19(77) 7(102) 8(73) 15(110) 7(101) 9(94) 2(62)
8(25) 1(90) 1(17) 15(28) 9(30) 5(101) 7(21) 14(109) 17(93)
NOTE: You should assume that the reported number
of weeks does not include weeks in which the
individual had diarrhoea when the week started (i.e.,
each person was disease free at the start of each
week).
(1 mark)
66
Section 2

Solution
(a) 95/346 = 0.275. Prevalence = 27.5 per 100
overseas travellers report experiencing
diarrhoea during their trip.
(b) 28/143 = 0.196. Cumulative incidence = 19.6
cases per 100 exposed per 2 weeks.
(c) In this problem you are calculating an
incidence rate. You generally calculate the
incidence rate as the total number of
episodes divided by the total exposure
time:
12+12+19+7+8+15+7+9+2+8+1+1+15+9+5+7+14+17
88+46+77+102+73+110+101+94+62+25+90+17+28+30+101+21+109+93
= 169/1269 = 0.133
Thus, incidence rate = 13.3 cases per 100
person-weeks of observation.
67
Section 2

Relationship between prevalence and incidence
Example: Disease A
Subject
1
2
3
4
5
L t
Cumulative Incidence = 5 / 5 in t-years
Prevalence at time L = 2 / 5
Disease B
Subject
1
2
3
4
5
Cumulative Incidence = 5 / 5 in t-years
Prevalence at time L = 5 / 5
L
t
Time
Time
68
Section 2

Note: Prevalence depends on
• incidence rate
• duration of disease
Diabetes (adult onset)
• annual incidence rate is low
• duration is long as disease is neither curable
or total
so prevalence is high relative to incidence
Cold
• incidence is high
• duration is short
So prevalence is low relative to incidence
69
Section 2

HIV/AIDS
Many with HIV will live for a long time.
Prevalence of HIV in the community will be high.
There is also an issue related to the fact that a
person may not know they are HIV positive.
Hence likely to underestimate the prevalence.
70
Section 2

If diagnosed with AIDS death is quick, ie few
living with AIDS. Hence AIDS prevalence
relatively low.
There are obvious issues related to health care
provision and planning.
71
Section 2

[D] Measures of disease association
The comparisons of disease frequency in different
groups of people are made. In the simplest (and
very common) setting there are two groups, one
exposed and the other unexposed.
Example: Data from cohort study of oral
contraceptive use (OC) and bacteria in the urine
among women aged 16-49 years over 3 years.
Bacteria present
Yes No Total
OC use Yes 27 455 482
No 77 1831 1908
Total 104 2286 2390
Data from D.A. Evans et al. NEJM (1978)
Bacteria is the Disease Category. (Outcome
measure.)
OC use is the Exposure Category.
72
Section 2

Cumulative Incidence
OC users: 27/482 = 0.056
56 cases per 1000 in 3 years
Non users: 77/1908 = 0.040
40 cases per 1000 in 3 years
Measures of Association:
Difference (Absolute effect)
56-40 = 16 cases per 1000 in 3 years
Ratio (Relative effect)
56/40 = 1.4
The number of OC users with bacteria is 1.4
times the number for non users.
[Note that the ratio does not include the time
interval]
73
Section 2

1. Relative effect = Relative Risk (RR)
• ratio of incidence in exposed group (I e ) to
incidence in unexposed group (I 0 )
⎧> 1 (exposure → disease)
Ie
⎪
RR = ⎨=1 if Ie
= I0
I0
⎪
⎩ < 1 (exposure is protective)
• indicates how much more likely disease is to
develop in the exposed group than in the
unexposed group
• no association between exposure and disease:
RR = 1 (I e = I 0 )
• good measure of strength of an association
• the usual measure in studies of causation of
disease
• can also calculate ratios of prevalences, but the
interpretation is different
74
Section 2

2. Absolute effect = Attributable Risk (AR)
• difference in incidence between exposed and
unexposed groups
AR = I e – I 0
• indicates how many more people with disease
there are in the exposed than the unexposed
group
• no association between exposure and disease:
AR = 0 (I e = I 0 )
• assuming a cause-effect relationship between
exposure and disease, we say:
if AR>0, AR is the number of cases of the disease
among the exposed that can be attributed to their
exposure;
if AR

Example: A randomised trial of the effectiveness
of infra-red stimulation compared with placebo on
pain caused by cervical osteoarthritis (degenerative
joint disease in the neck) carried out over two
months.
(Placebo or Control: mock stimulation)
Treatment Control
Improvement in pain 18 8
No improvement in pain 7 17
Total 25 25
Exposure is Treatment/Control
Disease is Improvement/No improvement in pain
[The outcome classification]
Cumulative incidence of improvement (in 2
months)
Treatment group:18/25
Control group: 8/25
18/ 25
Rel. Risk =
8/25 = 2.3
The chance of improvement in the treatment group
is 2.3 times the chance in the control group.
76
Section 2

Example: Prevalence of coronary heart disease
(CHD) at initial examination among 4469 persons
age 30-62 years of age in the Framingham Study
Number Number Prevalence
examined with CHD per 1,000
Males 2024 48 23.7
Females 2445 28 11.5
Note that 23.7 = (48/2024) x 1,000 hence
called prevalence per 1,000
Similarly, 11.5 = (28/ 2445) x 1,000
Relative risk = (23.7/11.5) = 2.1
[Heart disease is twice as common in males as in
females]
Attributable risk = 23.7-11.5 = 12.2 per 1000
[There are 12.2 more cases of heart disease in 1000
men than in 1000 women]
77
Section 2

Example: Data from a cohort study of
postmenopausal hormone use and coronary heart
disease among female nurses
Coronary heart
disease
Yes No Person-years
Postmenopausal
hormone use
Yes 30 - 54,308.7
No 60 - 51,477.5
Data from Stamfer et al, NEJM (1985)
Incidence rate:
Users: 30/54308.7 = 55 per 100,000 person-years
Non-users: 60/51477.5 = 117 per 100,000 person
years
Attributable Risk:
55-117 =-62 cases of CHD per 100,000 person
years
Hormone use prevents 62 cases per 100,000 person
years
Relative Risk: 55/117 = 0.47
The risk of CHD among users is 0.47 times the risk
in non-users (ie a 53% reduction in risk)
78
Section 2

Example: Relative and attributable risks of
mortality from lung cancer and coronary heart
disease among cigarette smokers in a cohort study
in British male physicians
Annual mortality rate per 100,000
Lung cancer Heart disease
Cigarette smokers 140 669
Non-smokers 10 413
Relative risk 14.0 1.6
Attributable risk 130 256
(per 100,000 per year)
Data from Doll and Peto, Br Med J (1976)
RR: 140/10 = 14.0 669/413 = 1.6
AR: 140 – 10 = 130 669 – 413 = 256
Heart disease is more common therefore a smaller
relative increase in risk produces more people with
disease.
79
Section 2

Note
Relative risks
• provide information on the strength of an
association
• can be used to assist in assessment of the
likelihood of a causal association
Attributable risks
• measure the impact of an exposure, (assuming
that it is causal)
If a disease is common a small relative risk will
translate to a large attributable risk.
[see previous example]
80
Section 2

3. Odds Ratio: A third measure of association
This can be used in case-control studies, where
measures of disease frequency in the study
population are not available
Odds of disease =
Chance (or Probability) of disease
Chance (or Probability) of no disease
See later
81
Section 2

SECTION 3
This section covers a brief introduction to probability definitions, notation, rules and random
variables with examples, several involving tree diagram use.
Definitions including mutually exclusive and independent events
The Addition Rule for combining probabilities
The Multiplication Rule for probabilities
Tree diagrams with examples
Screening test terminology
Probability Distributions and Random Variables
Rules for combining Random Variables
83
Section 3

Introduction To Probability
To define what we mean by probability we need
to talk about experiments and events
• An experiment is the process by which
observations or measurements are obtained.
• The outcome of an experiment is referred to as
an event and may also represent a group of
possible outcomes.
• The set of all possible individual outcomes is
the sample space.
Example: Toss a coin once. Observe event A –
the coin comes up a head (H) or B – the coin
comes up a tail (T). The sample space is {H, T}.
An experiment results in outcomes that cannot be
predicted in advance. This uncertainty about an
outcome is measured by the probability of the
event. Different events have different
probabilities. We define the probability of an
n
event A as Pr(A) =
A
N where n A is the number of
experiments resulting in event A in a very large
number (N) of repetitions of the experiment.
84
Section 3

A probability is therefore like a relative
frequency. It is a measure on a scale from 0
representing absolute impossibility to 1
representing absolute certainty. Subjective
estimates of probability are “unlikely”,
“possibly”, “almost never”, etc which all convey
an idea of likelihood of occurrence of an event.
But different people attach different values to
these (and this is a problem). For example, what
is the probability that God exists (0 or 1).
Probability calculations began with games of
chance over 3000 years ago. The games involve
coins, dice, cards, roulette etc. With such objects
we can develop exact probabilities of possible
outcomes or events by making sensible
assumptions:
• a die (plural dice) is fair ( 1 6
is the probability of
any outcome)
• a coin is fair ( 1 2
is probability of a head)
• a card is drawn (
52 1 is probability)
• a birth date (
365 1 is prob of particular day)
Probabilities associated with these objects can be
calculated using our knowledge of the properties
of these objects.
85
Section 3

Example: An experiment involves throwing a
fair die. Event is “obtaining” an even number.
The answer is 3 6 or 1 (easy). This probability
2
could also be found by experiment involving
tossing the die many times.
In practice, experiments are much more complex
than this in situations of interest to researchers.
Events result from such experiments and event
probabilities are needed if we are to draw
conclusions from the sample data collected.
Further Examples
1. An experiment treats 20 patients in a clinical
investigation involving a new drug.
An event is “at least 12 patients are cured”
What is the probability of the event
2. An experiment selects 500 voters in a survey.
An event is “at least 300 support windmill
farms in Central Otago”.
3. Experiment treats two “equal” samples of
cancer patients, one by surgery and one by
chemotherapy.
86
Section 3

An event is “more chemotherapy patients are
cured”. The probability will give insight into
the better treatment.
Theoretical probabilities are unknown in such
situations, hence these probabilities must be
estimated from experimental data by observing
outcomes or noting historical information.
Combining Probabilities for Multiple Events
Example: Consider the probability of being in
each of the four blood groups. The probabilities
from the Dunedin blood donor centre are:
Blood Type Pr(Blood Type)
A 0.38
B 0.11
AB 0.04
O 0.47
(These numbers can also be estimated by
“experiment” and will take these values if
many people are sampled)
1. What is the probability that a person is either
A or B
87
Section 3

2. What is the probability that 3 unconnected (or
independent) people are all in blood group O
Solution:
1. For any two independent outcomes the
probability of either occurring is (in this case)
the sum of the individual probabilities.
The probability of being either A or B is
Pr(A) + Pr(B) = 0.38 + 0.11 = 0.49
Note: Pr(A) + Pr(B) + Pr(AB) + Pr(O) = 1
Here we have assumed that the outcomes are
mutually exclusive: that is, a person cannot be
in blood groups A and B.
2. For any two independent outcomes, the
probability that both are observed is the
product of the individual probabilities. This
can be extended to three people in the obvious
way.
Therefore, probability three people have blood
group O can be shown to be (see later)
88
Section 3

Pr(O) × Pr(O) × Pr(O)
= 0.47 × 0.47 × 0.47
= 0.104
Note: Independent events arise if the outcome of
one event tells us nothing about the other
event. We obviously must exclude the
possibility that the three people are in the
same family.
Note: This example illustrates the two laws for
combining probabilities:
• the addition rule in part 1.
• the multiplication rule in part 2.
89
Section 3

Properties of Probabilities and Probability
Laws.
Notation: There is a convenient notation for
representing event probabilities. Suppose S
represents all possible outcomes of an experiment,
A is the collection of these outcomes representing
an event and A is the collection of outcomes
which are not in A.
• A is the event called the complement of A
• A and A are said to be mutually exclusive (no
overlap)
• Also Pr(A) + Pr( A) = 1 since A and A must
represent every possible outcome.
Now suppose two events A and B may overlap.
• Event A or B denoted by A∪ B occurs if at
least one of A or B occurs. Called the union of
A and B.
90
Section 3

• Event A and B denoted by A∩ B occurs if both
A and B occur. Called the intersection of A and
B.
Example: A fair die is thrown. A is the event “a
number greater than 3 is thrown” and B is the event
“an even number is thrown”.
Then S = {1, 2, 3, 4, 5, 6}
A = {4, 5, 6} Pr(A) = 3 6
B = {2, 4, 6} Pr(B) = 3 6
A ∩ B = {4, 6} and A ∪ B = {2, 4, 5, 6}
Pr(A ∩ B) = 2 Pr(A ∪ B) =
6
4 6
Set of all outcomes
A
A
B
B
A ∩ B
Fig (i) A∩ B not empty Fig (ii) A∩ B empty
(mutual exclusiveness)
The addition rule for combining probabilities
Pr(A or B) = Pr(
Set of all outcomes
A∪ B) = Pr(A) + Pr(B) – Pr ( A∩ B)
91
Section 3

since values in the intersection A∩ B are counted
twice. The special case when A and B are
mutually exclusive is
Pr( A∪ B) = Pr(A) + Pr(B)
This was illustrated in blood group example, part (1)
Example: The dice again:
Pr(A ∪ B) = 3 6 + 3 6 – 2 6 = 4 6
using addition rule.
The Multiplication Rule
The intersection of two events A and B is the
event that both occur. The probability of this is
Pr(A and B) = Pr(A ∩ B) = Pr(A) Pr(B|A)
In words this says that for both of the two events
to occur, first one must occur [Pr(A)] and then
given that the first has occurred, the second must
occur [Pr(B|A)].
If both Pr(A) and Pr(A and B) are given, this rule
can be used to define conditional probability as
Pr( B| A)
=
Pr( A∩
B)
Pr( A)
92
Section 3

Independence
The idea behind the term Pr(B|A) is that the
occurrence of event A may cause a reassignment
of probability to event B that makes it differ from
the original value Pr(B). When the occurrence of
A gives no additional information about B, A and
B are independent.
That is Pr(B|A) = Pr(B)
In this situation the multiplication rule is
Pr( A∩
B) = Pr(A)Pr(B)
Otherwise it is the original
Pr( A∩
B) = Pr(A) Pr(B|A)
This first rule was illustrated in the blood group
example where the probability of 3 independent
people all having blood group O was
Pr( A∩
B ∩C) = Pr(A) Pr(B)Pr(C)
= 0.47 × 0.47 × 0.47 = 0.104
93
Section 3

Example: A survey of hospital patients shows
that the probability a patient has high blood
pressure given he/she is diabetic is 0.85. If 10%
of patients are diabetic and 25% have high blood
pressure:
(a) Find prob. a patient has both diabetes and
high blood pressure.
(b) Are the conditions of diabetes and high
blood pressure independent
Solution (a) A is event “patient has high blood
pressure”
B is event “patient is diabetic”
Pr( A| B ) = 0.85, Pr( B ) = 0.10 and Pr( A ) = 0.25
∴Pr( A∩ B)
= Pr( A| B ) Pr( B ) by multiplication rule
= 0.85 x 0.10
= 0.085
(b) Pr( A ) = 0.25 ≠ Pr( A| B)
. Hence not independent
94
Section 3

A tree diagram is useful for helping calculate the
probability of a combined event. The stages of
the combined event can be dependent or
independent.
Example: Independent Stages.
Stephens Island is an uninhabited island in Cook
Strait where tuatara are being re-established. For
some years three locations have been visited on
the island and tuatara have been found at a
location with probability 0.4. At any visit X
represents the number of locations out of three at
which tuatara are observed. X can take values 0,
1, 2 or 3. Find the probabilities that 0, 1, 2, or 3
locations have tuatara on a visit.
T is the event “location has tuatara’’ and N is the
complementary event “location has no tuatara”.
LOC 1
LOC 2
LOC 3
95
Section 3

Location
1
Location
2
Location
3
Outcome Pr(Outcome) No
0.40
T
TTT
0.064
3
0.40
T
0.40
0.60
T
N
N
T
N
TTN
TNT
TNN
0.096
0.096
0.144
2
2
1
0.60
N
0.40
T
0.60
N
T
N
T
NTT
NTN
NNT
0.096
0.144
0.144
2
1
1
N
NNN 0.216
0
Then Pr(T) = 0.40 (known historically)
The second location is independent of the first
Pr(both T) = Pr(T ∩ T) = Pr(T)Pr(T)
= (0.40)(0.40) = 0.160
using the multiplication rule and
Pr(TTT) = (0.4) (0.4) (0.4) = 0.064
96
Section 3

The tree diagram shows all possible outcomes.
Branch probabilities are multiplied to give the
probabilities of the 8 possible outcomes.
The addition rule tells us that the probability of
seeing tuatara at two of the three sites, Pr(X = 2),
adds the probabilities of the three possible
outcomes, TTN, TNT and NTT.
That is, Pr(X = 2) = 0.096 + 0.096 + 0.096
= 0.288
Similarly, Pr(X = 0) = 0.216, Pr(X = 1) = 0.432
and Pr(X = 3) = 0.064.
In the next examples, the probability at each
branch of the tree is conditional on earlier
outcomes. i.e. no longer are the events
independent, but branch probabilities are still
multiplied according to the multiplication law for
probabilities.
97
Section 3

Example: Dependent stages. Andrew, John,
and Mark play a game. There are six similar cars,
two of which have had the brake cylinders
removed. The player chooses a car at random,
drives at high speed towards a cliff, and brakes in
time to stop. The boys decide to proceed in
alphabetical order. Find Pr(each will lose) and
Pr(no loser), assuming that the game stops when
the first boy drives over the cliff.
2
6
4
6
Andrew loses
2
5
3
5
John loses
2
4
2
4
Mark loses
No loser
Pr(Andrew loses) =Pr(Andrew picks a faulty car) = 2 6
Pr(John loses)
Pr(Mark loses)
= Pr(Andrew picks a good car and John
picks a faulty car)
⎛4⎞⎛2⎞ 4
= ⎜ ⎟⎜ ⎟ =
⎝6⎠⎝5⎠
15
= Pr Andrew and John pick good cars,
and Mark picks a faulty car)
⎛4⎞⎛3⎞⎛2⎞
3
= ⎜ ⎟⎜ ⎟⎜ ⎟=
⎝6⎠⎝5⎠⎝4⎠
15
98
Section 3

In probability notation we get:
A is event Andrew loses
A is event Andrew does not lose
Pr( A ) = 2/6 Pr( A ) = 4/6
J is event John loses
J is event John does not lose.
It is not Pr( J ) = 2/6
Instead, Pr( J ) is revised using extra information:
Pr( J ) = Pr( JA | )Pr( A )
⎛
=
5
2 ⎞ ⎛ ⎞
⎜ ⎟ ⎜4
⎟
⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜6
⎟
⎝
⎠ ⎝
= 4/15
and so on.
⎠
99
Section 3

Example: Screening Programmes
A patient with certain symptoms consulted her
doctor to be checked for a cancer. The patient
undergoes a biopsy. With this test there is a
probability of 0.90 that a woman with the cancer
shows a positive biopsy, and a probability of only
0.001 that a healthy woman incorrectly shows a
positive biopsy.
Historical information also suggests that 1 in
10,000 women have the cancer. [This is the
prevalence of the cancer in the population.]
Find the probability that a woman has the cancer
given the biopsy says she does.
(Essentially the problem is to decide the ability of
the biopsy to diagnose true patient status. The
principle applies to breast and cervical cancer in
New Zealand.)
Solution: A is event “woman has the cancer”
B is event “biopsy is positive” (indicating cancer)
100
Section 3

Pr(A) = 0.0001 (disease prevalence)
Pr(B|A) = 0.90 (a conditional prob.)
Pr(B|A) = 0.001 (A is complement of A)
The problem is to find Pr(A|B)
Pr(A)
= 0.0001
A
Pr(B|A)
= 0.90
Pr(B|A)
= 0.10
B
B
Biopsy + ve
(true positive)
Biopsy – ve
(false negative)
Pr(A)
= 0.9999
(the
complement)
A
Pr(B|A)
= 0.001
Pr(B|A )
= 0.999
By the multiplication rule for dependent events,
Pr(True positive) = Pr(A ∩ B)
= Pr(B|A)Pr(A)
= 0.90 × 0.0001
= 0.00009 (nine out of 100
000 show true positive)
Pr(False negative) = Pr(B|A)Pr(A)
= 0.10 × 0.0001
= 0.00001
B
B
Biopsy + ve
(false positive)
Biopsy – ve
(true negative)
101
Section 3

Pr(False positive) = 0.001 × 0.9999
= 0.00100 (100 out of 100
000 show false positive)
Pr(True negative) = 0.999 × 0.9999
= 0.99890
Pr(Test positive) = Pr (B)
= 0.00009 + 0.00100
= 0.00109 (109 out of
100 000 show positive
test)
Therefore,
0.00009
Pr(A ∩ B)
Pr(A|B) =
=
0.00009 + 0.00100 Pr(B)
0.00009
=
0.00109
= 0.083 (nine of the 109 with the
positive biopsy have the cancer)
Conclusion: Only 8.3% of those women
identified as having the disease actually do.
(This is not at all what we would expect and is
rather unsatisfactory.)
102
Section 3

1. Pr(B|A) is called the sensitivity of the test (the
probability a person with the disease returns a
positive result or the proportion of positives
that are correctly identified).
2. Pr( B | A) is called the specificity of the test
(the proportion of negatives that are correctly
identified by the test).
3. Sensitivity and specificity are from a practical
point of view not helpful as the point of
diagnostic testing is to make a diagnosis i.e.
we need to know the probability of the test
giving the correct diagnosis, whether it is
positive or negative. That is pr(A|B), not
pr(B|A).
4. Pr(A|B) is the positive predictive value (the
proportion of patients with positive test
results who are correctly diagnosed).
5. The negative predictive value is the
proportion of patients with negative test
results who are correctly diagnosed i.e.
Pr( A | B).
103
Section 3

Example: A patient consulted his GP because
he had intermittent chest pain. The description
of such pain is known to suggest a patient has
heart disease with a probability of 0.48. The
patient took an ECG test which has a sensitivity
of 0.90 and a specificity of 0.84. The patient
returns a positive ECG. Now find the
probability he has heart disease in light of this
additional information. Also find the positive
and negative predictive values.
Solution:
Pr(H) = 0.48
H
Sensitivity
= 0.90
0.10
T
T
(0.90)(0.48) = 0.4320
(0.10)(0.48) = 0.0480
Pr(H)
= 0.52
H
0.16
Specificity
= 0.84
T
T
(0.16)(0.52) = 0.0832
(0.84)(0.52) = 0.4368
H is event “patient has heart disease”
T is event “ECG test is positive”
Pr(T) = 0.4320 + 0.0832 = 0.5152
Pr(H|T) = 0.4320/0.5152 = 0.839
104
Section 3

Notice how the probability of heart disease has
been revised up from 0.48 to 0.839 as a result of
the test.
Positive predictive value = 0.839
Pr(Test negative) = 0.0480 + 0.4368 = 0.4848
Negative predictive value = 0.4368/0.4848 = 0.901
105
Section 3

Example
Like swine flu’ today, about six years ago SARS was a threat to world health. In the early days
of the SARS epidemic emergency measures were put in place by the World Health
Organisation in an attempt to control the spread of SARS and to identify the condition. But no
adequate screening tests existed to identify the condition when it first appeared in Hong Kong.
A study was carried out in the early days to evaluate a WHO criteria for identifying patients
with SARS in the SARS screening clinic in Hong Kong. Of 556 consecutive clinic attendees,
97 were confirmed with SARS. Of these 97 patients with confirmed SARS, 25 met the WHO
criteria for suspected SARS. Of the 459 patients in whom SARS was not confirmed, 438 were
negative according to the WHO criteria.
(a)
Find the prevalence of confirmed SARS at the clinic (i.e. the proportion
with SARS).
(1 mark)
(b) Estimate the sensitivity and specificity of the WHO test from the numbers above. (2
marks)
(c) Estimate the probability that the WHO test produces a positive result. (1 mark)
(d) Estimate the positive predictive value of the test. (1 mark)
(e) Estimate the negative predictive value of the test. (1 mark)
(f)
How would the positive predictive value of the test be affected if the prevalence of
SARS among clinic attendees were to decrease
(1 mark)
106
Section 3

WHO SARS Confirmed
Result Yes No Total
Positive 25 [21] 46
Negative [72] 438 510
Total 97 459 556
(a) Prevalence = 97/556 = 0.174
(b) Sensitivity = 25/97 = 0.258; specificity = 438/459 = 0.954
• T +
0.174
S
•
0.258
0.742
• T –
0.826
•
S
0.046
0.954
• T +
• T –
(c) Pr (T + ) = (0.174)(0.258) + (0.826)(0.046)
= 0.0449 + 0.0380
= 0.083
(d) Positive predictive value = 0.045/0.083 = 0.542
(e) Pr(T – ) = (0.174)(0.742) + (0.826)(0.954)
= 0.917
Negative predictive value = 0.788/0.917 = 0.859
(f) The positive predictive value will decrease.
107
Section 3

Example: Sensitive Survey Questions.
This is an important way of gaining information
on sensitive or controversial issues.
The question is: do you have or have you ever
had a sexually transmitted disease (STD)
It is unlikely a truthful response or any response
will be given.
In a mail survey of 268 young people five said
they had a STD.
Probability = 5/268 = 0.019 (or 19 per 1000)
Instead, proceed as follows:
1. Roll a die, allowing no one to see the
outcome.
2. Toss a fair coin.
3. If the die shows “1” answer truthfully the
question: “Have you thrown a head”
4. If the die shows 2, 3, 4, 5 or 6 answer
truthfully to the question:
108
Section 3

“Have you ever had a sexually transmitted
disease”
A tree diagram summarises this procedure where
θ is the proportion of response “YES” to the STD
question.
Roll
Die
1/6
5/6
1
2 to 6
Pr (Yes) = 1 5 θ
+
12 6
1/2
1/2
Head Yes 1/12
Head No 1/12
STD Yes 5θ/6
STD No 5(1 - θ)/6
There were 54 “Yes” and 214 “No” for 268
people.
Estimate Pr(Yes) = 54/268 = 0.2015
∴ 0.2015 = 1 5 θ
+
12 6
∴ 12(0.2015) = 1 + 10θ
θ
1 – θ
109
Section 3

∴ 2.418 – 1 = 10θ
∴ 1.418 = 10θ
∴ θ = 0.1418
or 142 per 1000 have STD
(compare 19 per 1000 previously)
110
Section 3

Probability Distribution and Random Variables
A random variable has values which depend on
the outcome of a random experiment. Random
variables are labelled with a capital letter (X
say). They can be discrete or continuous. The
number of locations with tuatara on Stephens
Island is discrete (possible values 0, 1, 2, 3)
while cholesterol levels are continuous.
Example: (Tuatara again) Three locations are
visited on 50 occasions in the tuatara study and
the number of locations with tuatara found are
recorded each time. Results follow along with
values calculated previously in the fourth column.
X = x j f j f j /n f j /n = Pr(X = x j )
(Tuatara at
locations)
(frequency)
(rel.freq) (as n becomes large)
0 8 0.16 0.216
1 22 0.44 0.432
2 15 0.30 0.288
3 5 0.10 0.064
Total n = 50 1.00 1.000
111
Section 3

X is the random variable. X is discrete here
because all possible outcomes x j can be counted.
The 50 results in the study are summarised by the
relative frequencies.
If many trials (n large) are carried out, the relative
frequencies of each x j stabilise to give
probabilities
Pr(X = x j )
for each outcome. Together these probabilities
form the probability distribution rather than a
relative frequency distribution.
NB (1)
4
∑ Pr( X = xj
) = 1 as for relative
j=
1
frequencies
(2) All probabilities are between 0 and 1.
112
Section 3

Describing Probability Distributions
Let X be a symbol for a probability distribution
and let μ X be the mean of X. (Assume X is
discrete for the moment.)
For a sample of n values from the distribution
suppose each possible x j occurs f j times and
there are k possible values of j. Then the sample
mean is
x = 1 k k f
n ∑ x j f j = x j
∑
⎛ ⎞
j ⎝ n ⎠
j=1
j=1
As the sample size becomes large, the relative
frequencies become probabilities and the mean of
the probability distribution X is μ X where
k
∑
μ X = x j Pr(X = x j )
j =1
A similar argument shows that the variance σ X 2 of
the probability distribution X is
k
σ 2 X = ∑ (x j −μ X ) 2 Pr(X = x j )
j=1
113
Section 3

Take the square root to get the standard deviation
of the probability distribution σ X .
Note: The sample mean x and variance s 2 are
estimates for population mean μ X and variance
σ X 2 .
Ex: Find the mean and standard deviation of the
distribution of the number of locations at which
tuatara are found.
X=x j Pr(X=x j ) x j Pr(X=x j ) (x j - μ X ) 2 (x j - μ X ) 2 Pr(X=x j )
0 0.216 0.000 (0 – 1.2) 2 = 1.44 0.311
1 0.432 0.043 0.04 0.017
2 0.288 0.576 0.64 0.184
3 0.064 0.192 3.24 0.207
Total 1.00 1.200 5.36 0.720
4
μ = ∑ = =
x Pr( X x ) 1.20
X j j
j=
1
On average just over one location per visit will
have tuatara present.
σ
4
2 2
X
xj X
X xj
j=
1
= ∑ ( − μ ) Pr( = ) = 0.72
and σ
X
= 0.85
114
Section 3

Example: A person infected with a disease can
pass it on to others. Let the random variable, X,
be the number of others infected by this person.
X is found to have the following probability
distribution.
2
Find μ
X
and σ
X
X = x
j
Pr(X = x j
)
0 0.10
1 0.25
2 0.40
3 0.20
4 0.05
μ
X
= 0(0.10)+1(0.25)+2(0.40)+3(0.20)+4(0.05)
= 1.85
2
σ
X
= (0 – 1.85) 2 0.10+(1 - 1.85) 2 0.25+(2 - 1.85) 2 0.40
+(3 - 1.85) 2 0.20+(4 - 1.85) 2 0.05
= 1.0275
Also, σ
X
= 1.0275 = 1.0137
115
Section 3

Rules for combining random variables
Often we are interested in the mean and
variance of a rescaled random variable, or in the
mean and variance of sums (or differences) of
random variables. The following properties are
true of all numerical random variables, discrete
or continuous.
If X and Y are independent random variables
and a and b are constants, then:
1. The mean of the new random variable
a + bX is
μ a+bX = a + bμ X
2. The variance of a + bX is:
σ 2 a+bX = b 2 σ 2 X
3. The mean of the new random variable
aX + bY is
μ aX+bY = aμ X + bμ Y
4. The variance of aX + bY is
σ 2 aX+bY = a 2 σ 2 X + b 2 σ 2 Y
116
Section 3

Note: Properties 3 and 4 tell us that
μ
X+ Y= μX+ μY
and
σ
+
= σ + σ .
2 2 2
X Y X Y
Also, μ
X− Y= μX− μYand
σ
−
= σ + σ .
2 2 2
X Y X Y
Example: Temperatures used to be recorded in
degrees Fahrenheit. Suppose a random variable F
measures January temperature (in Fahrenheit) in
Dunedin and daily maximum summer temperatures
have a mean of 70°F with a standard deviation of
5°F.
Use the conversion formula C = 5 ( F − 32) to find
9
the mean and standard deviation for the temperatures
in degrees Celsius.
Solution:
We will let the random variable C represent the
temperature in Celsius. The equation
C = 5 ( F − 32) may be rearranged by expanding the
9
brackets to become
C = 5 F − 5 × 32 or C = 5 F −
160
9 9
9 9
117
Section 3

We have μa+ bX
= a + bμ
X
160
Therefore a = − and b = 5 9 9
μ = a + bμ
C
160 5
= − + ×
9 9
= 21.1° C
2 2 2
We also have σ b σ
σ
a+ bX
=
X
2
2 ⎛5
⎞ 2
C
= × 5
⎜ ⎟
⎝9
⎠
25
= × 25
81
= 7.716
F
70
Therefore σ = 7.716 = 2.78° C
C
Example: What is the difference
between T = X + X + X
and T = 3X
118
Section 3

Note: These results can be extended to several
random variables.
Example: (Infected person continued)
Three people living in separate areas have the
disease. Random variables X 1 , X 2 , X 3 are
numbers of other people infected by them. Find
mean and variance of total number infected by
the original three.
Total T = X 1 +X 2 + X 3 (X 1 , X 2 , X 3 assumed
independent as people in different areas)
μ = μ + μ + μ =1.85+1.85+1.85 = 5.55
T X X X
1 2 3
σ = σ + σ + σ =1.0275+1.0275
2 2 2 2
T X X X
1 2 3
+ 1.0275 = 3.0825
Note: Do not say T = 3 X 1 . Although
μ
T
= 3μ
= 5.55,
X
1
σ
= 9σ
≠ 3.0825
2 2
T X
1
This is a very common source of error.
119
Section 3

120

SECTION 4
This section introduces both the Binomial and Normal Distributions which model many
phenomena arising in the real world. Consequently the distributions allow us to answer some
important and relevant questions.
The Binomial Distribution: Definition, mean and variance
The Binomial Table: Examples
The Normal Distribution: Definition
Standard Normal Distribution and Table
General Normal Distribution
Normal Approximation to the Binomial
Transforming Data to Normal
121
Section 4

The Binomial Distribution
The binomial distribution arises when
investigating proportions. e.g. the proportion of
adult population with diabetes. Each individual
has or does not have diabetes.
Let Y be the random variable for an individual
outcome of a person in the population. Two
outcomes occur, namely Y = 1 (e.g. diabetes
present or success) and Y = 0 (e.g. diabetes not
present or failure). The parameter π represents
the unknown proportion of 1’s occurring.
The probability distribution of Y is
Y =
y Pr(Y =
j
y j
)
1 π “success”
0 1 – π “failure”
Then μ Y = 1(π) + 0(1 – π) = π
σ = (1 – π) 2 π + (0 – π) 2 (1 – π)
2
Y
= (1 – π) [π(1 – π) + π 2 ]
= π(1 – π)
122
Section 4

Now suppose that we take a sample of size n
from the underlying population. What is the
distribution of the number of successes
The total number of successes is X where
X = Y 1 + Y 2 + Y 3 + … + Y n
with all the Y j independent of each other.
∴ μ X = π + π + π + … + π = n π
2 2 2
σ = σ
Y 1
+ σY 2
+ … + σY n
2
X
= π (1 – π) + π (1 – π) + … + π (1 – π)
= nπ(1 – π)
X is called a binomial distribution and
μ
X
= nπ
2
σ = nπ(1 − π)
X
where π is the parameter giving Pr(“success”) or
Pr(diabetes present).
123
Section 4

The mean number of successes is nπ and the
variance of the number of successes is nπ(1 – π)
The binomial distribution results from n trials
involving independent binary outcomes.
e.g. melanoma (Yes/No)
Smoking (smokes/does not smoke)
Diabetes (present/absent)
Tuatara (present/absent)
Example: X = number of locations in group of
n that have tuatara present.
It is known that Pr(success) = π = 0.40 and
Pr(failure) = 1 – π = 0.60.
Each location is assumed independent of other
locations.
Also assume the probability of tuatara being
present remains constant at each location.
124
Section 4

Notes 1. If these conditions are met, if n (the
number of trials) and π (the probability of
success) are known, all probabilities in the
distribution are known exactly.
2. n and π are said to be the parameters of the
distribution.
3. The binomial distribution requires
independent trials and a probability of
success which remains constant for each
trial.
4. We use binomial tables to approximate
these binomial probabilities for values of n
up to 20. (See table section of these notes.)
125
Section 4

For example suppose n = 8 and π = 0.40 are the
two defining parameters.
π
n x 0.05 0.10 0.15 … 0.40 0.50
8 0 0.6634 -- -- 0.0160 0.0039
1 0.2793 -- -- 0.0896 0.0312
2 0.0515 -- -- 0.2090 0.1094
3 0.0054 -- -- 0.2787 0.2187
4 0.0004 -- -- … 0.2322 0.2734
5 0.0000 -- -- 0.1239 0.2188
6 0.0000 -- -- 0.0413 0.1094
7 0.0000 -- -- 0.0079 0.0313
8 0.0000 -- -- 0.0007 0.0039
9 0 -- --
1 -- … --
2 -- --
3 -- --
etc
Notice that Pr(X = 3) = 0.2787 for π = 0.40 and n = 8
Example: Records show that twenty percent of
violin pupils are known to develop OOS during
the course of their training. Define X to be the
number of violin pupils out of 9 who develop
OOS during their training.
126
Section 4

(a) Find the probability distribution of X.
(b) What is the probability that none of the 9
pupils develop OOS
(c) What is the probability that more than 4 out
of the 9 pupils develop OOS
(d) In 2005 a certain violin teacher had 9 new
pupils and 5 developed OOS during training.
What conclusion would you draw about the
training methods of this teacher
Solution
(a) Here X is binomial with n = 9; π = 0.20
(and assume the pupils are all independent
of each other). The binomial table gives
n x π = 0.20
9 0 0.1342 = Pr(X = 0)
1 0.3020 = Pr(X = 1)
2 0.3020 etc
3 0.1762
4 0.0661
5 0.0165
6 0.0028
7 0.0003
8 0.0000
9 0.0000
127
Section 4

(b) Pr(X = 0) = 0.1342
(c) Pr(X > 4) = Pr(X = 5) + Pr(X = 6)
+ Pr(X = 7) + Pr(X = 8) + Pr(X = 9)
= 0.0196
(d) It would be rare or unusual (probability = 0.0196)
for more than four violin pupils to develop OOS
if 20% is the overall percentage known to develop
OOS historically. We conclude the training
methods of this teacher are likely to result in a
greater occurrence of OOS among pupils.
If the violin teacher has no effect on OOS,
π will remain on 0.20 and the probability
that more than four of the pupils will
develop OOS is 0.0196.
This is viewed (by convention) to be a small
probability indicating a rare or unusual event
has arisen if the value of π = 0.20 still holds
for the pupils of this teacher.
Either π = 0.20 is unchanged for this teacher
and a rare event has been observed
128
Section 4

or the teacher is at fault and more pupils
develop OOS. This second alternative is
usually taken and therefore we conclude the
teacher has a higher incidence of OOS.
Notes
1. It is the size of the probability of this
observed “event” or a more extreme and
convincing event which results in the
conclusion (more than 4).
2. 0.0196 is a chance of just under 2 per 100
(2%).
3. A probability less than 0.05 is (by
convention) taken to imply an event is rare or
unlikely to occur.
4. A probability above 0.05 often means an
event is not unusual. If the violin teacher had
produced such a probability then the teaching
would not be at all unusual in relation to
incidence of OOS.
129
Section 4

Binomial Examples and Normal Distribution
Example: (artificial data) A sociological
report suggests that 75% of Maori children
under 18 live with both parents. A random
sample of 20 Maori children is selected, and X
is the binomial random variable for the number
of these 20 who live with both parents.
(a) Define the parameters of the distribution of X.
(b) Find Pr(X = 15).
(c) Find the probability that 11 or fewer live
with both parents (i.e. Pr(X ≤ 11)).
(d) A random sample of 20 New Zealand
Caucasian children had only 11 living with
both parents. Does this result provide any
evidence to support the claim that 75% of NZ
Caucasian children live with both parents
130
Section 4

Solution
(a) X is binomial with n = 20, π = 0.75.
(b) The problem is that 0.75 does not occur in
the binomial table directly.
Whenever π > 0.50, we replace the event
“success” by its complement “failure”. This
is because the binomial table does not have
values greater than 0.50. In this case,
“failure” is the event “child does not live
with both parents”. For easy analysis, define
new random variable
Y = number not living with both parents.
Y is binomial, n = 20 and new π′ = 0.25
[here y = n – x and π′ = 1 – π]
∴ Pr(X = 15 given π = 0.75)
= Pr(Y = 5 given π′ = 0.25)
= 0.2023 from table
(c) Pr(X ≤ 11) = Pr(Y ≥ 9)
= Pr(Y = 9) + Pr(Y = 10)
+ … + Pr(Y = 20)
= 0.0271 + 0.0099
+ … + 0.0000
= 0.0410
131
Section 4

(d) No. In fact there is evidence it is less than 75%
for NZ Caucasian children.
If π = 0.75 is assumed for Caucasian families,
then the probability of observing 11 or fewer
living with both parents is, by our convention,
small (less than 0.05) providing evidence
against 75%. Hence reject claim that π = 0.75
for Caucasian families and conclude fewer live
with both parents (because 11 is the direction of
fewer rather than more).
Note: Suppose instead 12 out of 20 of the NZ
Caucasian children were living with both
parents.
Pr(X ≤ 12) = Pr(Y ≥ 8) = 0.1019 if
π = 0.75 meaning π′ = 0.25.
This probability is not small, and now there is
no evidence from our data to suppose the
situation is any different among Caucasian
families.
132
Section 4

Example (Revision)
The standard drug for treating a cancer is claimed
to halve the tumor size in 30% of all patients
treated. Suppose X is the binomial random
variable for the number of patients in a sample of
seven who have their tumor size halved.
(a) List the conditions which must be met if X is
binomial.
Patients independent. Two outcomes only.
Constant probability tumor size halved over
all the patients.
(b) Using the appropriate table, write down the
distribution of probabilities for the number
(X) who have their tumor size halved.
X = x j Pr(X = x j )
0 0.0824
1 0.2471
2 0.3177
3 0.2269
4 0.0972
5 0.0250
6 0.0036
7 0.0002
133
Section 4

(c) Write down the probability that three of the
patients have their tumor size halved.
Probability = 0.2269
(d) Find the probability that three or more of the
patients have their tumor size halved.
Probability = 0.3529
(e) In a pilot study in Auckland, three out of seven
patients given a new drug had their tumor size
halved. What conclusion if any can be drawn
about the new drug Explain how you reach
your conclusion.
Conclusion: There is no reason to suppose the
new drug is any different to the standard.
Explanation: Prob. of three or more is 0.3529
which is large meaning the result with the new
drug is consistent with the 30% before.
Note: This study involves a very small number of
patients and will be reconsidered later with a larger
sample.
134
Section 4

The Normal Distribution
This distribution will allow us to calculate
probabilities associated with observed sample
results when we are dealing with continuous
outcome measures and sample means. First we
develop properties of the normal distribution.
A relative frequency histogram tends to a
probability distribution as sample size n becomes
large.
HISTOGRAM
DISTRIBUTION
f(X)
a
b
X
n increases
and class
width decreases
Shaded area
Shaded area
= proportion of = probability of
observations
value between
between a and b
a and b
(This represents a
(This represents a
sample with a
population with a
small number of
very large number
individuals.)
of individuals.)
135
a
b
X
Section 4

The resulting curve is known as a probability
function (or probability density function) and is
described by a curve y = f(X).
The area under this curve, say between two points
X = a and X = b, is the probability Pr(a < X < b)
X is a random variable taking values on a
continuous scale.
We have seen several sets of sample data which
produce symmetrical histograms, bell shaped
with a concentration of values at the centre and
few values at extremes. (e.g. cholesterol levels in
the pravastatin study) Such data are said to be
collected from a normal distribution or from a
population of values which are normally
distributed.
[Gauss, 1777-1855, first developed the equation
of such a normal curve while observing pattern in
errors made while making measurements in
astronomy]
136
Section 4

μ
Y
Y = f(X)
X
The equation of such a normal curve is
f( X)
1
= e
σ 2π
−
1
2
X −μ
( ) 2
σ
where parameter μ is the mean and parameter σ is
the standard deviation of the distribution (in
practice, μ and σ will be estimated from sample
data by the values x and s).
Notes 1. The graph is symmetrical about centre
point denoted by μ.
2. The two parameters μ and σ completely define
a normal distribution (recall that parameters n
and π define a binomial distribution).
Notation: X ∼ N(μ,σ 2 )
3. Increasing μ moves the curve but does not
alter its shape
Section 4
137

μ 2 > μ 1
σ unchanged
μ 1 μ 2
X
4. Increasing σ spreads the curve more widely
about X = μ, but does not alter the centre of the
distribution.
σ 1
σ 2
μ
σ 2 > σ 1
μ unchanged
X
Both the above could be normal distributions.
5. Areas under these curves can be found from
tables. The table is based on what is known as
the standard normal distribution which has μ =
0 and σ = 1.
138
Section 4

Normal distribution calculations.
The Standard Normal Distribution (Z)
Z ∼ N(0, 1) i.e. Z distributed with μ Z = 0, σ Z
2 = 1
∴ f(Z) = 1 2π e−1 2 Z2 Shaded area
= Pr(0 < Z < z)
(see tables)
O z Z
z .00 .01 .02 .03 .04 .05 …… .09
.0 .0000
.1
.2
.3
1.5
1.6 0.4484 0.4495
1.7
3 0.4990
139
Section 4

Some calculations:
1. Find Pr(0 < Z < 1.63)
From table choose z = 1.63
∴ Pr(0 < Z < 1.63) = 0.4484
O 1.63 Z
Also, Pr(0 < Z < 1.64) = 0.4495
3
10
∴ Pr(0 < Z < 1.633) ≈ 0.4484 + (0.0011)
= 0.4487
[final calculation need not be this accurate
+ 0.4484 would be accepted for our purposes
using this table.]
2. Find Pr(Z > 1.64)
Pr(Z > 1.64)
= 0.5 – Pr(0 < Z < 1.64)
= 0.5 – 0.4495
= 0.0505
3. Pr(1 < Z < 1.64)
= Pr(0 < Z < 1.64) - Pr(0 < Z < 1)
= 0.4495 – 0.3413
= 0.1082
140
O 1.64 Z
Section 4

4. Pr(-1 < Z < 1.64) = Pr(0 < Z < 1.64)
+ Pr(-1 < Z < 0)
= Pr(0 < Z < 1.64)
+ Pr(0 < Z < 1) by symmetry
= 0.4495 + 0.3413
= 0.7908
–1
O 1.64 Z
5. Pr(-1 < Z < 1) = 2Pr(0 < Z < 1)
= 2(0.3413)
= 0.6826
Pr(-2 < Z < 2) = 2Pr(0 < Z < 2) = 0.9546
Since σ Z = 1, a value z of Z is a count of the number
of standard deviations to this point. Notice that
approx 68% of the area is within one and 95%
within two standard deviations of the centre.
6. Find the value z above which 25% of the area lies.
Here, find a value close to 0.25 in the centre of
normal table, then read back to margins.
0.25
0.50
O
0.25
z Z
Pr(0 < Z < 0.67) = 0.2486
Pr(0 < Z < 0.68) = 0.2517
Hence, z = 0.675 approx.
141
Section 4

The General Normal Distribution (X)
2
X ~ N( μ X , σ X ) say.
Areas under this curve cannot be found directly
from the normal table but X is related to the
standard normal Z ~ N(0, 1 2 ) by
Z
=
X − μ
σ
X
X
Notes 1. The distribution X is said to be
standardised when μ X subtracted and the
result divided by σ .
X
2. Z is essentially the number of standard
deviations ( σ X ) from μ X to a value x of X.
142
Section 4

Some calculations
1. Pr( μ X - σ X < X < μ X + σ X )
= Pr(-σ X < X - μ X < + σ X )
X − μ
= Pr(-1 <
X
< + 1)
σ X
= Pr(-1 < Z < + 1)
= 2 Pr(0 < Z < 1) = 0.6826
[68.26% of distribution within one standard
deviation of the centre]
2. In general, Pr(a < X < b)
= Pr(a - μ X < X - μ X < b - μ X )
a − μ
= Pr(
X X − μ
<
X b − μ
<
σ X σ X σ X
a − μ
= Pr(
X b − μ
< Z <
X
)
σ
σ
X
X
X
)
μ X a b X
143
Section 4

Example: Assume that diastolic blood pressures
for men aged 35-44 have a normal distribution with
mean μ X = 80 and standard deviation σ X = 12
(a) Find Pr(90 < X < 100)
(b) The percentage of men in this age range who
are hypertensive (a level over 100).
Solution
(a) Pr(90 < X < 100) =
⎛ 90 −80
100 −80
Pr
⎞
⎜ < Z < ⎟
⎝ 12 12 ⎠
= Pr(0.833 < Z < 1.667)
= Pr(0 < Z < 1.667)
– Pr(0 < Z < 0.833)
= 0.4525 – 0.2967
= 0.1558
(b) X ~ N(80, 144). Find Pr(X > 100)
⎛ 100 −80⎞
Pr(X > 100) = Pr⎜
Z > ⎟
⎝ 12 ⎠
= Pr(Z > 1.67)
= 0.5 – Pr(0 < Z < 1.67)
= 0.5 – 0.4525
= 0.0475
We expect 4.8% of men in this age group to be
hypertensive.
144
Section 4

(c) Find the diastolic blood pressure which is
exceeded by 10% of men aged 35-44.
X ~ N(80, 144)
80
O
0.40
x
z
0.10
X (original scale)
Z (standard scale)
(It is helpful, initially, to sketch the standard
scale as well as the original scale).
From the standard normal table, find the
value, z, which cuts off area 0.40 as shown.
Reading to the margins from the value 0.40 in
centre of table gives z = 1.282 (part way
between 1.28 and 1.29).
x − μ
Use z =
X
to get 1.282 =
σ X
∴ x = 80 +12(1.282)
= 95.38
x −80
12
145
Section 4

The Normal Approximation to the Binomial
(n, π)
If there is a large sample selected from a
population of binary values (e.g. people with or
without diabetes) probabilities of observed
outcomes are found from the normal N( μ X , σ )
distribution where μ X = nπ and
σ = nπ ( 1−π
)
X
2
X
σ
X
μ X = nπ
= nπ ( 1−π
)
x–1 x x+1
1
x – 2
1 x + 2
X
Area of shaded block (if x integer) is the binomial
probability of obtaining x successes.
This is approximately the area under the normal
curve between x − 1 and x + 1 .
2
2
146
Section 4

∴ Pr( X = x)
⎛ 1 1
( x −
2) − nπ
( x+ 2)
−nπ
⎞
≈ Pr
< Z <
⎜
nπ (1 −π) nπ(1 −π)
⎟
⎝
⎠
Notes: 1. This approximation is good provided n
is large and π is not too close to 0 or 1. (Under
these conditions the binomial distribution is
reasonably close to symmetrical and hence the
normal curve is seen to be a good
approximation.)
2. The normal approximation is good if
nπ± 3 nπ(1 −π )
gives two values between 0 and n (the min.
and max values of the binomial counts) since
95% of the possible values should lie within
these limits indicating a near symmetrical
distribution.
147
Section 4

Probability
We know Pr(blood group B) = 0.11
n = 2 nπ = 0.22
π = 0.11 n π ( 1−π
)
= 0.44
Hence 0.22 ± 3(0.44)
Figure 1 Binomial distribution of number of people out of two in blood
group B.
Probability
1.0
0.8
0.6
0.4
0.2
0.0
0.4
0.3
0.2
0.1
0.0
0 1 2
Number in blood group B
0 1 2 3 4 5 6 7
Number subjects
n = 10 nπ = 1.10
π = 0.11 n π ( 1−π
)
= 0.99
Hence 1.10 ± 3(0.99)
Figure 2 Binomial distribution showing the number of subjects out of ten
in blood group B based on the probability of being in blood group B.
Probability
0.15
0.10
0.05
0.0
n = 100 nπ = 11
π = 0.11 n π ( 1−π
)
= 3.13
Hence 11 ± 3(3.13)
0 5 10 15 20
Number subjects
Figure 3 Binomial distribution showing the number of subjects out of 100 in blood group B based
on the probability of being in blood group B.
148
Section 4

More on the normal and Statistical Inference
Example: One in 40 adults on average develops
a respiratory condition. A random sample of 400
workers in a certain occupation has 16 with the
condition. Find the probability that 16 or more
suffer from this condition in general. What
conclusion would you draw about the possible
effect of this occupation on the occurrence of the
condition Justify your answer.
Solution: Let X be the distribution of the number
in a sample of 400 with the condition.
Then X ~ Binomial (n =400; π = 1/40)
μ X = nπ = 10; σ
X
= n π ( 1−π
) = 3.123
Since n π ± 2 nπ
(1 −π
) is 10 ± 6. 2, the normal
approximation can be used.
Pr(X ≥ 16)
≈ Pr( Z
15.5 −10
> )
3.123
15 1
2
16
149
16 1
2
X
= Pr(Z > 1.761)
= 0.0391
Section 4

This is the p-value associated with a study result
of 16. There is evidence of a higher incidence of
the respiratory condition than expected in this
occupation. (The probability 0.0391 is small
indicating that the event X = 16 or more is rare if
π = 1/40 were to hold in this occupation.)
Therefore, π is likely to be greater than 1/40 for
workers in this occupation. (If this is the case,
the event observed would not be unusual.)
150
Section 4

Example:
It is claimed cancer tumor size is halved in 30%
of all patients using current treatment. A new
drug was used on 70 patients with the cancer.
(Last week we looked at a case where the drug
was tried on 7 patients with 3 successes.)
(a) Suppose Y is the binomial random variable
for the number of patients who have their
tumor size halved. Write down the values for
the mean and standard deviation of Y.
μ Y = nπ = 70(0.3) = 21
σ = nπ(1 − π) = 21(0.7) = 3.83
Y
151
Section 4

(b) In a study, thirty out of seventy patients
(previously 3 out of 7) administered the
standard drug experience a halving of their
tumors. Find the probability that 30 or more
out of 70 have their tumors halved.
⎛ 29.5 − 21⎞
Pr(Y ≥ 30) = Pr⎜
Z > ⎟
⎝ 3.83 ⎠
= Pr(Z > 2.22)
= 0.5 – 0.4868
= 0.0132
(c) In a study 30 out of 70 patients in Auckland
administered this new drug had their tumor
size halved. What conclusion can be drawn
about the new drug
There is evidence that the new drug is more
effective than the standard because the
probability of 30 or more successes is less
than 0.05 indicating the observed 30 (or
more) is not likely to occur unless the new
drug has a beneficial effect.
152
Section 4

Transforming Data
If data being analysed are continuous but not
normally distributed, it may be necessary to
modify the data by transforming each value in
order to create new values which are normal.
Then work with the transformed values. Typical
transformations involve logs, square roots or
reciprocals.
There are three reasons for transforming data.
1. Statistical procedures which we develop may
only be valid if the data are approximately
normal, and non-normal data can be converted
to normal by transforming.
2. When comparing for example two samples of
data (e.g. cholesterol levels after treatment with
pravastatin or a control) the two groups should
have similar standard deviations for some
testing procedures to be valid. Transforming
such data can produce two sets of values with
similar standard deviations.
3. Transforming can also reduce the influence of
outlying values on the results of an analysis.
153
Section 4

(e.g. suppose most values are around 10 in a
data set with one value of 100.
Then ln10 = 2.30 and ln100 = 4.61)
EXAMPLE: A sample of 216 values of a serum
has mean = 60.7 and standard deviation 77.9
Frequency
90
80
70
60
50
40
30
20
10
0
Histogram of the serum
values in 216 patients with
fitted normal distribution is
shown. (The normal fit is
terrible!)
The data are transformed by using the ln function.
Mean = 3.547 and standard dev. = 1.03
25
20
15
10
5
0
0 100 200 300 400 500
Serum bilirubin (μmol/l)
1 2 3 4 5 6
154
Histogram of log serum
values with fitted
Normal distribution
(ln values) looks
reasonably normal.
Section 4

Now suppose we want the range of values
containing the central 95% of all patients. If data
are normal, 95% of the population lie in
mean ± 1.96 (standard deviations)
0.475 0.475
–1.96 1.96
(from standard
normal table)
For the raw data, mean = 60.7 and s.d. = 77.9.
Hence, interval could be 60.7 ± 1.96(77.9) which
cannot be correct with the negative values.
But the transformed data have approximately a
normal distribution. For transformed data,
mean = 3.547 and standard deviation = 1.030.
Hence, 95% of the patients will have ln (serum)
levels in the range
3.547 ± 1.96(1.030)
155
Section 4

That is, 95% of distribution (or values) between
3.547 – 2.019 and 3.547 + 2.019
or 1.528 and 5.566
Transforming back to original scale,
e 1.528 = 4.61 and e 5.566 = 261.4
Hence, 95% of patients would have serum levels
between 4.61 and 261.4 μmol/l
156
Section 4

REVIEW EXERCISES
4. For the standard normal distribution find the following:
(a) The area below –1.58.
(b) The two points between which the central 85% of the area lies. (2 marks)
5. In the Framingham Study, serum cholesterol levels were measured for a large number of healthy
males. The population was then followed for 16 years. At the end of this time, the men were
divided into two groups: those who had developed coronary heart disease and those who had not.
The distributions of the initial serum cholesterol levels for each group were found to be
approximately normal. Among individuals who eventually developed coronary heart disease, the
mean serum cholesterol level was μ d = 244 mg/100 ml and the standard deviation was σ d = 51
mg/100ml; for those who did not develop the disease, the mean serum cholesterol level was μ nd =
219 mg/100 ml and the standard deviation was σ nd = 41 mg/100ml.
(a) Suppose that an initial serum cholesterol level of 260 mg/100ml or higher is used to predict
coronary heart disease. What is the probability of correctly predicting heart disease for a man
who will develop it
(b)
(c)
What is the probability of predicting heart disease for a man who will not develop it
What is the probability of failing to predict heart disease for a man who will develop it
(3 marks)
6. The length of human pregnancies from conception to birth varies according to a distribution that is
approximately normal with mean 266 days and standard deviation 16 days.
(a) What percent of pregnancies last less than 240 days (that’s about 8 months)
(b) What percent of pregnancies last between 240 and 270 days (roughly between 8 months and 9
months)
(c) How long do the longest 20% of pregnancies last (3 marks)
1. The probability of recovery for patients who are administered an established treatment for a
stomach complaint is 0.8. A random sample of 100 patients with the complaint is monitored.
Suppose X is the binomial random variable for the number of patients in this sample who recover
when the established treatment is used.
(a) Specify the parameters of X.
(b) Find the mean and standard deviation of X.
(c) Find the probability that at least 90 of the patients administered the treatment recover. Here you
should first verify that the normal approximation to the binomial distribution can be used.
(d) In a trial involving a new drug for the treatment of this stomach complaint, 90 out of 100
patients who are administered the new drug recover. What conclusion can you draw about the
new drug State your reason.
(7 marks)
157
Section 4

SOLUTIONS
4. [Note to markers: Since students only have access to a table with z values to two decimal places, be
prepared to accept calculations based on the nearest values in the table. Many students will, of course,
interpolate between table values.]
(a)
Shaded area = 0.5 – Pr(0 < Z < 1.58)
= 0.5 – 0.4429
= 0.0571
–1.58 0 Z
(b)
Pr(0 < Z < 1.44) = 0.425
The central 85% lies
between –1.44 and + 1.44
5. (a)
244 260
260 − 244
For men who develop chd, Pr(X > 260) = Pr(Z > )
51
= Pr(Z > 0.314)
= 0.5 – 0.1217
X
= 0.3783
260 − 219
(b) For men who do not develop chd, Pr(X > 260) = Pr(Z >
41
= Pr(Z > 1)
= 0.5 – 0.3413
219 260 = X0.1587
(c) The probability of failing to predict chd for a man who will develop it is 1 – 0.3783 = 0.6217
6. X ~ N(266, 16 2 ) or X is normal with μ X = 266 and σ
2
X
= 256
(a) Pr(X < 240)
240 − 266
= Pr( Z < )
16
= Pr(Z < – 1.625)
= 0.5 – Pr(0 < Z < 1.625)
240 266 X
= 0.5 – 0.4479
= 0.0521
That is, 5.2% less than 8 months.
270 − 266
)
16
= Pr(–1.625 < Z < 0.25)
= 0.4479 + 0.0987
= 0.5466 i.e. 54.7%
(b) Pr(240 < X < 270) = Pr(–1.625 < Z <
(c)
0.425 0.425
–z 0 z Z
240
266 270 X
z = 0.842 approx, from table
(30% of the standard normal
lies between 0 and z = 0.842)
0.30
x − 266
0.20 ∴ = 0.842
266 x X 16
∴ x = 266 + 16(0.842)
0 z Z
= 279.47 days
That is, approximately 280 days or more.
1. (a) n = 100; π = 0.8
(b) μ X = nπ = 80; σ
X
= 80(0.2)
= 4.
(c) μ X ± 2σ X gives 80 + 2(4) or 72 to 88. Both values lie in the range of possible values 0 to 100 hence
normal approximation can be used. (1.96 instead of 2 also acceptable)
89.5
− 80
Pr(X > 89.5) = Pr(Z > )
4
= Pr(Z > 2.375)
= 0.5 – 0.4912
= 0.0088
(d) There is evidence that the new drug produces a greater number who recover from the stomach complaint
than expected from the established treatment; the probability 0.0088 is very small for a recovery rate of
80%.
Section 4
158

SECTION 5
This section defines sampling distributions, establishes the standard deviations of these distributions
called standard errors, and set up confidence intervals for population means, differences between the
means of two populations, proportions and difference between proportions based on random samples
drawn from the populations.
An outline of the Research Process
The Distribution of Sample Means
The Standard Error of the Mean
Confidence Interval for a Mean
The t-distribution and Its Use
Comparison of Two Independent Groups
The Standard Error of the Difference Between Two means
Pooled Estimate for the Common Variance
Comparison of Two Dependent Groups (Paired Data)
Confidence Interval for a Proportion
Confidence Interval for Difference Between Two Proportions
Summary of Distributions and Confidence Intervals
159
Section 5

The Research Process in Two Situations
Binomial
Underlying Population
Bernoulli Outcomes
Success or failure
Inference Y = 1 or 0
Use probability
of outcome or
estimate of the
success proportion
Sample
(n)
Statistics
Result of study
Number of successes, X
Binomial
e.g. Prevalence (π) of asthma in women aged 20 to
40.
This can be estimated as the proportion (p) in a
sample chosen from the population.
160
Section 5

Normal
Underlying Population
Continuous Outcomes
X ~ N(μ, σ 2 ) say
Inference
Use probability
of outcome or an
estimate based
on the sample mean
Sample
(n)
Statistics
Result of study
How does the sample
mean behave
What is the sample
mean distribution X
e.g. What is the mean resting pulse rate (μ) in beats
per minute for men in age range 20 to 25 years
The mean x from the sample is an estimate for the
mean μ in population of all men in this age range.
161
Section 5

Sampling Distributions
Statistical inference is process of using
information from a sample to infer something
about the population from which sample was
drawn, thus completing the research loops just
described.
How reliable are these estimates for π and μ
To answer these questions focus first on the
sample mean x for a sample of size n say.
Proportions will be discussed later. The argument
proceeds as follows:
Successive samples of size n can be drawn from
the population. These produce means x 1 , x 2 , x 3 ,
x 4 , … etc and these form what is called a
distribution of sample means, X , which is quite
different to the original distribution, X, of values
in the population.
The problem is now to find μ and σ . [Here,
X X
σ is the standard deviation of the distribution of
X
means and hence is the “typical” variation in these
means. i.e. the “typical” error.]
162
Section 5

The Distribution of Sample Means
Suppose a population with distribution X has known
mean μ X and standard deviation σ X
Ex: Female adult heights. Suppose μ X = 169 cm
and σ X = 3.20 cm
A sample of size n = 4 drawn randomly from the
population has values 163, 172, 166, 166 say with
mean x 1 = 667/4 = 166.8 cm
x
1
Distribution of
individual heights (X)
σ X = 3.20 cm
160 163 166 169 172 175 178 X
=μ
X
The four sample values and their mean are plotted.
The average x 1 is not as extreme as the individual
values in the sample. x 1 is an estimate of μ X
(usually unknown in the real situation).
A second sample of n = 4 gives x 2 = 170.5 cm
A third sample of n = 4 gives x 3 = 169.5 cm
163
Section 5

If this process is continued we can obtain a
distribution of sample means. What are the
properties of this distribution These will allow us
to decide how well a sample mean estimates μ X .
Distributions of means are for samples of size
n = 10, n = 25 and n = 100. The population from
which the samples are taken is Normal.
164
Section 5

Distributions of means are for samples of size
n = 10, n = 25 and n = 100. The population from
which the samples are taken is not Normal. But the
sampling distributions are normal.
165
Section 5

Derivation:
Suppose a random sample of size n is taken from a
population with distribution X. The sample can be
viewed as values from n random variables X 1 , X 2 ,
…, X n each with mean μ X and variance σ . X 1 , X 2 ,
…, X n are independent (if the population is large),
and are identical.
A value, x , from one sample is one value of X , the
distribution of sample means for sample of size n.
Then
1
X = 1 2 … +
n
( X + X + )
1
∴ μ X = X X +
n
1 2
∴ μ X = μ X
X n
( μ + μ + … μ )
X n
1
= ( nμ
X ) (X 1 , X 2 etc identical)
n
2
X
166
Section 5

The addition rule for the variance of independent
random variables gives
σ
2
2
2 1 2 1 2 1
X =
⎛
σ
⎛
X σ
⎛
⎜
⎞ ⎟ + ⎜
⎞ ⎟ X + … + ⎜
⎝ n⎠
1
=
⎛ ⎞
⎜ ⎟
⎝ n⎠
2
nσ
⎝ n⎠
⎞
⎟
⎝ n ⎠
2
σ
2
X
1 2
n
2
X
=
σ X 2
n
[i.e. if T = aX + bY, then
2
T
2
2
X
2
2
Y
σ = a σ + b σ ]
Therefore, the standard deviation of the distribution
of sample means is
σ
σ
X
X =
n
σ
X
The derivations of μ = μ
X X
and σ = need
X
n
not be known. These two formula are in fact very
important and you must know how to use them.
167
Section 5

Note: 1. σ X is called the standard error of the
mean. (It is the “typical” deviation in the mean
– i.e. a measure of precision of the error in the
mean).
2. If μ X = 169; σ X = 3.20 for heights of women,
for sample of size n = 4, μ X = 169 and
= σ 3.20
σ X
X
4 = = 1.60.
2
3. If sample size (n) is greater than 4, σ X is
smaller meaning the distribution X is more
compact about μ = μ X .
X
4. If X is normal, it can be shown that X is
normal no matter what the sample size.
5. If X is not normal, but n large, then X is
approximately normal. (This result is a
consequence of the Central Limit Theorem in
note 6.)
6. For random sample of size n, the sample means
x i fluctuate around the population mean μ X
with standard error σ X = σ X / n . As n
increases, the distribution fluctuates less and
less, getting closer to a normal distribution.
168
Section 5

Example: Suppose a population has values which
are normally distributed (distribution is X) and
μ X = 7.9 with σ X = 0.60.
Find (i) Pr(X > 7.7)
(ii) Find Pr( X > 7.7) where X is the distribution
of means for samples of size n = 9.
Solution:
⎛ 7.7 − 7.9 ⎞
(i) Pr(X > 7.7) = Pr⎜
Z > ⎟
⎝ 0.60 ⎠
= Pr(Z > - 0.333) = 0.6304
(ii) Since
σ
X
0.60
μ = 7.9 and σ = = = 0.2,
X
X
n 9
⎛ 7.7 − 7.9 ⎞
Pr( X > 7.7) = Pr⎜
Z > ⎟
⎝ 0.2 ⎠
= Pr(Z > -1) = 0.8413
169
Section 5

Example: Serum values for a sample n = 216
give x = 34.46 and s = 5.84. What is the
standard error of x
σ
Standard error = where σ is the (unknown)
n
population standard deviation.
In practice, we estimate σ by s.
s
∴ estimated standard error =
n
=
5.84
216
= 0.397.
Suppose the sample had been twice the size,
n = 432, and sample had same mean and standard
5.84
deviation. Estimated s.e. = = 0.281
432
(compare 0.397 for n = 216)
170
Section 5

A Confidence Interval for the Mean
The problem here is to use sample data to find an
estimate for an unknown population mean. This
estimate reflects the random variation in the data
collected by establishing an interval in which we
are fairly certain that the mean μ lies.
As can be seen, this will complete the research
loop concerning the unknown population.
X
To motivate the procedure we work with the
distribution of sample means, X , which is
(
2
)
⎛
2
N μ , σ or alternatively N⎜
X X
μ
X
X , σ
⎝ n
First consider the standard Normal:
⎞
⎟
⎠
Area = 0.025 Area = 0.025
0.95
of area
–1.96 +1.96 Z
171
Section 5

0.95 = Pr(-1.96 < Z < +1.96)
⎛ X − μ ⎞
= Pr⎜−1.96
<
X
< + 1. 96⎟ ⎝ σ X / n ⎠
= Pr
⎛ σ
⎞
⎜−
X
σ
1 .96 < X − μ < +
X
X 1.96 ⎟
⎝ n
n ⎠
⎛ σ
⎞
= Pr⎜
−
X
σ
μ < X < +
X
X 1 .96 μ X 1. 96 ⎟
⎝
n
n ⎠
This result is used to construct a 95% confidence
interval as follows:
For a sample x 1 , x 2 , …, x n of n values from a
population we are said to be 95% confident that
the sample mean satisfies
μ
X
σ
X
σ
X
− 1.96 < x < μ
X
+ 1.96
n
n
But
x
while
σ X
σ
< μ X +1. 96 implies x −1
. 96
X < μ X
n
n
σ X
σ
μ X −1 . 96 < x implies μ x
X
X < +1. 96 .
n
n
172
Section 5

Therefore, we are 95% confident that the unknown
population mean μ satisfies
X
x
σ X
σ
− 1 .96 < μ x
X
X < + 1.96
n
n
Alternatively, we are 95% confident that the true
population mean lies in the interval
x
σ
± 1.96
X
n
173
Section 5

Notes: 1. The sample has produced an interval
estimate for the unknown population mean.
2. A 99% confidence interval replaces the value
1.96 by 2.58 since the tail areas beyond +2.58
and –2.58 are both 0.005
0.005 0.005
0.99
area
–2.58 +2.58 Z
3. The 99% confidence interval
x
±
σ
2.58
X
n
is wider, hence less precise, but we are now
99% certain μ X is in this interval.
σ
4. As n increases,
X
n
decreases and the
confidence interval is narrower meaning a
more precise estimate.
i.e. a large sample leads to greater accuracy.
174
Section 5

Example: A pharmacologist is investigating the
length of time that a sedative is effective. Eight
patients are selected at random for a study and the
eight times for which the sedative is effective have
mean x = 8.4 (It is also known that the standard
deviation for such measures is σ X = 1.5 hours).
Find 95% and 99% confidence intervals for the true
mean number of hours μ .
X
Solution: Here, n = 8; x = 8.4;
(assuming that X is normal).
1.5
σ = = 0.53
X
8
The 95% confidence interval is
8.4 ± 1.96 (0.53)
or 8.4 ± 1.04
That is, 7.36 < μ X < 9.44 or (7.36, 9.44).
The 99% confidence interval is
8.4 ± 2.58(0.53)
or 8.4 ± 1.37
That is, 7.03 < μ X < 9.77 or (7.03, 9.77)
The second interval is much wider.
175
Section 5

Example: The pharmacologist is required to find
the value of μ X to within 15 minutes with 95%
confidence. Assuming that the standard deviation
is σ X = 1.5 hours, find the size of the sample which
must be taken in order to achieve this accuracy.
Solution: Since 15 minutes is ¼ hour, for a sample
size n we need
x ± 1
4
to be an interval which is wider than
σ
x ± 1.96
X
n
1.5
or x ± 1.96
n
1.5
∴1.96
≤ 1
n 4
Rearranging, 1.96 (1.5) 4 ≤ n
or 11.76 ≤ n
Squaring, n ≥ 138.3
Hence, 139 patients must be selected.
176
Section 5

Use of t-table when σ X is unknown
In all practical contexts, σ X is not known. In this
case it is estimated in the best possible way by the
sample standard deviation s X . In this situation, the
t-table provides alternative larger values in place of
1.96 and 2.58.
The confidence intervals are wider and hence there
is less precision.
The 95% confidence interval is
s
x
x − tν
< μ
X
< x + tν
n
s
x
n
where ν = n – 1 is the “number of degrees of
freedom” and t ν is found in the appropriate column
in the t-table for 95% confidence (see table at end
of notes)
Note: (n – 1) = ν is the divisor in the estimate
for the variance.)
2
X s
177
Section 5

Exercise: Now suppose that the pharmacologist
did not know the value of σ X and was forced to
take the sample standard deviation from the sample
of size n = 8 as the best estimate of σ X , namely
s X = 1.5 hours. Find 95% and 99% confidence
intervals for μ .
X
Solution: x = 8.4 and
s
estimated standard error = X 1.5
= = 0. 53
n 8
The 95% confidence interval for the mean sedative
time μ X for all such patients is
8.4 ± t 7 (0.53) where t 7 = 2.365
That is, 8.4 ± 1.25
or 7.15 < μ X < 9.65
The 99% interval is
8.4 ± t 7 (0.53) where t 7 = 3.500
That is, 8.4 ± 1.86
or 6.54 < μ X < 10.26
Both are wider than before.
178
Section 5

Student’s t distribution
ν
-t ν 0 t ν T
2p 0.100 0.050 0.020 0.010
p 0.050 0.025 0.010 0.005
1
2
3
4
5
6
7 1.895 2.365 2.998 3.500
8
9
10
120
∝ 1.645 1.960 2.326 2.576
Area (p)
or probability
p refers to the area of one tail 2p gives the
combined area of both tails (View the t-distribution
above as a slight modification to the normal
distribution Z.
179
Section 5

Notes: 1. The interval is wide when samples
small. That is, less precision in estimates.
2. This last example is the most common
situation where: the population is assumed to
be normal; μ X and σ X are both unknown;
σ X is estimated by s X from a random sample
of size n.
3. Even for large n the t-table is used. Last row
of table has 1.96 and 2.58 row of t.
4. From the point of view of exams we shall
accept the normal distribution value for
degrees of freedom greater than 30.
180
Section 5

Example: Tablets must be produced which weigh
200milligram. Choose sample of n = 20 from
production line. x = 201.7mg and s X = 5.13mg.
Does this sample confirm that μ = 200mg
Solution: 19 degrees of freedom and
t 19 = 2.093 for a 95% confidence interval.
Therefore,
X
5.13
201.7 – 2.093 < μ X <
20
or 199.3 < μ X < 204.1
201.7
+
5.13
2.093
20
The weight of 200milligram lies in this interval.
Hence, 200milligram is an acceptable value of the
mean μ with 95% confidence.
X
181
Section 5

The Meaning of a confidence Interval
199.3 μ X = 200 204.1
Sample 100
↑
Sample 6
Sample 5
Sample 4
Sample 3
Sample 2
Sample 1
Sample 5 does not include
μ X = 200mg.
In general, if 100 different samples construct 100
intervals, then five of the 100 will miss μ X if we
are working at 95% confidence levels.
(This is the possible error which must be accepted.
With 99% confidence intervals which are wider
only one will miss μ X .)
182
Section 5

100 Confidence Intervals (95%)
Sample 100
↑
Sample 6
Sample 5
Sample 4
Sample 3
Sample 2
Sample 1
199.3 204.1
In the above, the position of the true mean μ X is
unknown. Also, in practice we only have one of the
above intervals. We say we are 95% confident the
true mean lies in this interval.
183
Section 5

184
Section 5

185
Section 5

186
Section 5

187
Section 5

188
Section 5

Example:
It is claimed that males committed for trial for
minor offences are spending more time in prison on
remand than females committed for trial for similar
offences. A sample of 40 females and 49 males
awaiting trial gave the following information. The
outcome measure is time on remand (X days).
Female Male
Sample mean ( x i
)
16.3 29.5
Sample standard deviation (s i ) 14.6 17.2
Sample size (n i ) 40 49
The difference between the sample means is
x M – x F = 29.5 – 16.3 = 13.2 days
Is this an important difference
189
Section 5

If μ M and μ F are the population mean times for
males and females, a 95% confidence interval for
μ M – μ F is
x M – x F ±
2 2
sM sF
1.96 +
n n
M
(17.2) (14.6)
= 13.2 ± 1.96 +
49 40
= 13.2 ± 6.61
= (6.59, 19.81)
or 6.59 < μ M – μ F < 19.81
F
2 2
The population male remand time is likely to be
within 6.59 and 19.81 days longer than that for
females (alternatively, the true mean difference is
between 6.59 and 19.81 days).
190
Section 5

Case 2: Comparing means when samples small
In this situation the CLT no longer holds for the
difference between the sample means.
Instead we need to assume that the population
from which the difference is drawn is normally
distributed.
This should be the case if the populations for any
small samples are normal.
In addition to assuming normality, we assume the
two populations have equal variances.
191
Section 5

If
2
σ 1 and
2
σ 2 are similar and equal to
2
σ say.
Then the 95% confidence interval μ 1 – μ 2 is
1 1
( x 1 − x2)
± 1.96σ
+
n n
1 1
1
( x 1 − x2)
−1.96σ
+ < μ1
− μ2
< ( x1
− x2)
± 1.96σ
+
n n
n
1
2
1
2
1
1
n
2
The common variance σ 2 needs to be estimated
from sample data. If both populations have the
same variance, the best estimate for σ 2 is found
when the variation in both samples is averaged to
give the pooled estimate s 2 pwhere
∴
with
s
2 p
( n
=
1
2
1
1
−1)
s
n
+ ( n
+ n
2
2
2 ∑ ( x1
− x1
)
1
=
n1
−1
2
− 2
−1)
s
s i
and
2
2
2
2 ( x2
− x2)
s
2
= i
n2
−1
When sample estimates for the variances are used,
replace 1.96 by t-value to get
1 1
( x1
− x2)
± tν
s p +
n1
n2
with degrees of freedom ν = n 1 + n 2 – 2.
192
Section 5

Example 3: Following data are 24 hour total energy
expenditures (MJ/day) in groups of lean and obese
patients (1986 study)
Lean (n = 13) Obese (n = 9)
6.13 8.79
7.05 9.19
7.48 9.21 Question: Is
7.48 9.68 there a
7.53 9.69 difference in
7.58 9.97 energy
7.90 11.51 expenditure
8.08 11.85 between lean
8.09 12.79 and obese
8.11 patients
8.40
10.15
10.88
Mean: 8.066 10.298
S.D.: 1.238 1.398
Possible explanations for the difference between
samples in above situations:
1. bias (need to randomise)
2. confounding (e.g. gender, age)
3. chance (random variation)
4. true difference
The methods we discuss in next few lectures assume
that bias and confounding are not the explanation.
193
Section 5

n 1 = 13; x 1 = 8.066; s 1 = 1.238 (lean)
n 2 = 9; x 2 = 10.298; s 2 = 1.398 (obese)
Solution: x 2 – x 1 = 2.232 (obese – lean)
2
2
2 (13−1)1.238
+ (9 −1)1.398
s p =
13 + 9 − 2
12 (1.533) + 8(1.954)
=
20
= 1.7014
∴s
p = 1 .7014 = 1.304
ν = 20 giving t 20 = 2.086 for 95% interval
∴ 95% confidence interval is
1 1
2 .232 ± 2.086(1.304) +
13 9
or 2.232 ± 1.180
That is, 1.05 <
μ − μ < 3.41 MJ/day
obese
lean
194
Section 5

Note: This confidence interval tells us that we
can be 95% sure that the true difference in energy
expenditure between obese and lean patients is
between 1.05 and 3.41 MJ/day.
Since this interval is entirely positive, it means
that we can conclude that lean patients have lower
energy expenditure than obese patients.
Notes: 1. ν = n 1 + n 2 – 2 is the divisor in the
2
formula for s p , the variance estimate (the
degrees of freedom are always the divisor in
the variance estimate e.g. n – 1 in the single
sample case)
2. Both populations should have values which
are normally distributed if the samples are
small.
2 2
3. The two population variances, σ
1
and σ
2
should be equal approximately. (Otherwise
may need to transform data or use another
test.) R-cmdr has an option which confirms
this.
195
Section 5

4. The two samples from the two populations are
random and independent of each other.
5. Testing whether μ 1 = μ2
can be achieved by
seeing if μ1 − μ2
= 0 i.e. confirming if 0 lies in
the confidence interval for the difference.
6. It is possible to obtain the probability value
associated with the study outcome value of
2.232. (see later)
Example: A nutrition scientist is assessing a
weight-loss programme to evaluate its
effectiveness. Ten people randomly selected.
Initial weight recorded and followup weight 20
weeks later.
Subject Initial Weight (x Ii ) Weight at followup (x Fi )
1 180 165
2 142 138
3 126 128
4 138 136
5 175 170
6 205 197
7 116 115
8 142 128
9 157 144
10 136 130
196
Section 5

Find a 95% confidence interval for the reduction
in weight (Assuming the two sets of values
independent).
x
I
= 151.7 x
F
= 145.1
s 2 I = 750.76 s 2 F = 620.01
2 9(750.76) + 9(620.01)
s P =
18
= 685.17
Since ν = 18 giving t 18 = 2.101 we get
1 1
(151.7 − 145.1) ± 2.101 685.17 + 10 10
or 6.6 ± 24.6
That is –18.0 < μ I – μ F < 31.2
Note 1: Since the confidence interval includes 0,
conclude there is no evidence to indicate that the
weight loss programme has altered weights.
Note 2: In this study the two sets of data are not
independent. One person produces two values
here.
197
Section 5

Case 3: Comparing means if matched data.
It is natural to consider the differences d i in the
weights for each person rather than considering
the two samples separately. The d i are the data
now and a confidence interval is constructed for
the average difference μ d based on the single
sample of differences. The 95% confidence
interval is
d
± t
ν
sd
n
where d is the average of the d i , n is the number
of data pairs, ν = n – 1, and s d is the standard
deviation of the differences. We have
s
d
=
∑ ( di
− d
n −1
)
2
with n – 1 degrees of freedom.
198
Section 5

Example: Weight loss programme again
Subject x Ii x Fi d i = x Ii – x Fi
2
( di
− d)
1 180 165 15 70.56
2 142 138 4 6.76
3 126 128 –2 73.96
4 138 136 2 21.16
5 175 170 5 2.56
6 205 197 8 1.96
7 116 115 1 31.36
8 142 128 14 54.76
9 157 144 13 40.96
10 136 130 6 0.36
Total 66 304.40
d = 66/10 = 6.6
2
∑
2
( di
− d) 304.4
sd
= = = 33.82
n −1 9
ν = n – 1 = 9 giving t ν = 2.262 for a 95% interval.
The 95% confidence interval for the average
difference is
33.82
6.6 ± 2.262
or 6.6 ± 4.2
That is, 2.4 < μ d < 10.8
10
199
Section 5

There is evidence that the weight loss programme
has reduced weights since the difference of 0 is
not in this interval (we are 95% sure).
Notes: (1) The “profile” of each person is
constant in this study because the same
person has produced the two values.
(2) A test involving paired data based on d is
called a paired t-test. The earlier test on
μ − is called an unpaired t-test.
1 μ 2
(3) Negative differences are possible in this
analysis when subtracting. Be consistent with
subtraction process.
200
Section 5

Confidence Intervals for a Proportion
Suppose X is a binomial distribution with
parameters n and π (i.e. the number of “successes”
lies between 0 and n).
Then
μ X = nπ
σ = nπ ( 1−π
)
X
Suppose one sample produces a proportion of
successes p = in n trials.
n x
Many such samples can be taken to get different
values of p. The resulting distribution (P) of these
proportions is normal (by the Central Limit
theorem.) It follows that
P =
X
n
where X is binomial. The mean and standard
deviation of P are then
201
Section 5

1 1
μ P = μ = nπ
= π
n
X n
2 1 2 1
and, since σ =
⎛
⎜
⎞ ⎟ σ X = nπ
(1 −π
)
2
⎝ n⎠
n
2
P ,
σ
P
=
π ( 1−π
)
n
The sample proportion (p) estimates the unknown
true population proportion (π) (e.g. prevalence of
asthma in women not known.). Thus the
estimated standard error is
p ( 1−
p)
n
and the 95% confidence interval for π is
p
± 1.96
p(1
−
n
p)
Note: 1.96 (or 99% equivalent 2.58) are always
used for confidence intervals for proportions. (If
202
Section 5

small sample, the normal distribution is not a good
approx.)
Example: A random sample of 500 Aucklanders
taken in 1996 had 173 supporting aerial spraying
to eradicate tussock moth. Estimate the
proportion (π) of Aucklanders who support this.
Solution:
and
x 173
p = = = 0.346
n 500
p( 1−
p)
0.346(1 − 0.346)
=
n
500
= 0.021
The 95% confidence interval is
0.346 ± 1.96(0.021)
or 0.346 ± 0.041
Therefore, 0.305 < π < 0.387
We are 95% sure that between 30.5% and 38.7%
of the Auckland population support the spraying.
203
Section 5

Note: Alternatively, we could say 34.6% of the
population support spraying with a margin of error
of 4.1%. But ‘margin of error’ concept must be
used with caution. It is reasonable if the value of
p lies between 0.3 and 0.7 but the margin of error
should be adjusted if p outside this range. (We
omit this adjustment.)
Example: Epidemiologist estimates proportion of
women with asthma. Find the sample size (n)
needed to give an estimate for this proportion with
an error no more than 0.03 with 95% confidence.
Solution: The largest possible value of p(1 – p)
occurs when p =
1
2
(verify this by choosing several
p values).
The most conservative (or safest) sample size is
obtained using this value p =
1
2
. The requested
accuracy requires confidence interval p ± 0.03
to be the largest interval. But the actual interval is
0.5(1 − 0.5)
p ± 1.96
for sample size n.
n
Therefore
204
Section 5

0.5(1 − 0.5)
1.96
< 0.03
n
2
(1.96) (0.5)(0.5)
∴ < (0.03) 2
n
2
(1.96) (0.5)(0.5)
∴ n > = 1067.11
2
(0.03)
It follows that 1068 women must be tested.
Now consider the Confidence Interval for
Difference Between Two Proportions
(Derivation not examined)
The difference π1 −π
2is estimated by p 1 – p 2
where p 1 = x1
n1
and p 2 = x2
n2
for the two
samples.
The distribution P 1 – P 2 of sample proportion
differences is a normal distribution with
μ
P −P
1
2
= π
1
−π
2
and standard deviation (standard error)
σ
P − P
=
π
−π
1 (1 1)
2(1
2)
1 2
n n
1
205
π
+
−π
2
Section 5

using the addition rule for the mean and variance
of two independent random variables, P 1 and P 2 .
If π 1 and π 2 are estimated from sample data, the
95% confidence interval is
p1(1 − p1) p2(1 − p2)
( p1 − p2) ± 1.96
+
n
n
1 2
Exercise: To study the effectiveness of a drug for
arthritis, two samples of patients were randomly
selected. One sample of 100 was injected with the
drug, the other sample of 60 receiving a placebo
injection. After a period of time the patients were
asked if their arthritic condition had improved.
Results were
EXPOSURE
DRUG(1) PLACEBO(2)
IMPROVED 59 22
NOT IMPROVED 41 38
TOTAL 100 60
206
Section 5

Solution: The proportions improved are
59
22
p 1 = = 0.59 and p 2 = = 0. 37
100
60
p 1 – p 2 = 0.22 and the estimated standard error of
difference between the proportions is
0.59(1 − 0.59)
100
+
0.37(1 − 0.37)
60
=
0.0794
The 95% confidence interval is
0.22 ± 1.96 (0.0794)
or 0.22 ± 0.156
or 0.064 < π1 − π 2 < 0.376
Since 0 excluded from interval and the interval is
positive, there is evidence π1 − π 2 > 0. That is, we
conclude the proportion improved is higher when
the drug is used.
207
Section 5

REVIEW EXERCISES
2. A population is known to be normally distributed with a mean µx = 60 and standard deviation σx =
15. Let X be the distribution of means of samples of size 25 drawn from the population.
(a) Define completely the probability distribution X.
(b) What is the probability that a value in the population will lie between 55 and 65
(c) What is the probability that the mean of a sample of size 25 will lie between 55 and 65 (4 marks)
3. Large studies indicate that the mean cholesterol level in children aged 2 – 14 is 175 mg%/mL and
the standard deviation is 30 mg%/mL.
The problem here is to see if there is a familial aggregation of cholesterol levels. A group of fathers
who have had a heart attack and have elevated cholesterol levels (≥ 250 mg%/mL) are identified.
The cholesterol levels of their offspring within the 2-14 age range are measured. The mean
cholesterol level in a group of 100 such children is 207.3 mg%/mL. The problem is to decide if this
value is sufficiently far from 175 mg%/mL for us to believe that the underlying mean cholesterol
level μ in the population of all children selected in this way is greater than 175 mg%/mL.
(a) Construct a 95% confidence interval for μ on the basis of the sample data. State your conclusion
about familial aggregation of cholesterol levels.
(2 marks)
(b) Find the probability of obtaining the sample mean of 207.3 mg%/mL or a value which is greater
under the assumption that there is no familial aggregation. State your conclusion from this
probability.
(2 marks)
4. Patients with chronic kidney failure may be treated by dialysis, using a machine that removes toxic
wastes from the blood, a function normally performed by the kidneys. Kidney failure and dialysis
can cause other changes, such as retention of phosphorus, that must be corrected by changes in diet.
A study of the nutrition of dialysis patients measured the level of phosphorus in the blood on six
occasions. Here are the data for one patient (milligrams of phosphorous per decilitre of blood):
5.5 6.1 4.8 5.8 6.2 4.6
The measurements are separated in time and can be considered a random sample of the patient’s
blood phosphorus level.
(a)
(b)
(c)
If the level varies normally with σ = 0.8 mg/dl, find a 95% confidence interval for the mean
blood phosphorus level of this patient.
(1 mark)
If the value of σ is unknown but estimated by the sample standard deviation s = 0.669, find a
95% confidence interval for the mean blood phosphorus level of this patient. (1 mark)
The normal range of phosphorus in the blood is considered to be 2.6 to 4.8 mg/dl. Is there
evidence that the patient has a mean phosphorus level that exceeds 4.8 Explain. (1 mark)
5. A salmon fishing company is monitoring the weight of salmon in its ponds prior to harvest. A pilot
sample of ten fish, randomly selected, shows a mean weight of 2.31 kilograms with a standard
deviation of 0.17 kilogram.
(a)
(b)
Obtain a 95% confidence interval for the mean weight of all salmon in the ponds.(2 marks)
Using the standard deviation from the pilot survey as an estimate of the true variation of
weights of salmon in the ponds, establish how many fish should be sampled to obtain an
estimate of the mean weight of all the salmon in the ponds to within 0.03 kilogram with
95% confidence. (Take 2 as an approximation to the value of t.) (3 marks
208
Section 5

SOLUTIONS
2. (a) X is a normal distribution with μ X
= 60 and
i.e. X ~ N(60, 9)
55 − 60 65 − 60
(b) Pr(55 < X < 65) = Pr( < Z < )
15 15
= Pr(–0.33 < Z < 0.33)
= 2(0.1293)
= 0.2586 approx
55 − 60 65 − 60
(c) Pr(55 < X < 65) = Pr( < Z < )
3
3
= Pr(-1.67 < Z

REVIEW EXERCISES
2. The extent to which X-rays can penetrate tooth enamel has been suggested as a suitable
mechanism for differentiating between males and females in forensic medicine. Listed
below in appropriate units are the ‘spectropenetration gradients’ for eight female teeth and
eight male teeth:
Male (x 1
) 4.9 5.4 5.0 5.5 5.4 6.6 6.3 4.3
Female (x 2
) 4.8 5.3 3.7 4.1 5.6 4.0 3.6 5.0
The data give sample means:
x = 5.4250,
1
x = 4.5125
2
2
2
and the sample variances: s = 0.5536, s = 0.5784:
1
2
(a) Calculate the pooled estimate for the variance common to the male and female
populations.
(1 mark)
(b) Estimate the standard error of the difference between the population means. (1 mark)
(c) Construct a 95% confidence interval for the difference between the two population
means.
(1 mark)
(d) What conclusion do you now draw about the procedure for differentiating between males
and females
(1 mark)
SOLUTIONS
( n
−1)
s + ( n −1)
s
n + n − 2
7(0.5536) + 7(0.5784)
8 + 8 − 2
2 1 1 2 2
2. (a) s p
=
=
= (1.132)
= 0.566
1
2
2
(b) Estimated standard error of difference =
2
1
2
1 1
0 .566 + = 0.376
8 8
(c) The 95% confidence interval is x − x ± t (0.376)
1 2 14
That is, (5.4250 – 4.5125) ± 2.145(0.376)
or 0.9125 ± 0.8065
giving 0.106 < μ 1 – μ 2 < 1.719
(d) We are 95% sure that there is a difference in the mean tooth penetrations for males and
females since 0 does not lie in the confidence interval in (c). (Because the confidence
interval is positive the male tooth penetration will be greater.)
210
Section 5

[A] DISTRIBUTION SUMMARY
1. Binomial (X): n trials; π is the probability of
success (discrete)
μ = nπ
X
σ = nπ(1 − π)
X
2. Normal (X): (continuous)
Parameters are μ
X
and σ
X
⎛ X − μ ⎞
X
3. Standard Normal ⎜Z
= ⎟
⎝ σ
X ⎠
Parameters are μ
Z
= 0 and σ
Z
= 1
4. Normal Approximation to Binomial
Original binomial has parameters n and π.
The normal approx has parameters μ
X
= nπ
,
σ = nπ(1 − π)
X
5. Distribution of Sample Means ( X )
σ
X
Normal with μ = μ
X X
and σ = . The
X
n
standard deviation σ is also called the
X
standard error of the mean.
211
Section 5

6. Distribution of Differences between Sample
Means ( X1−
X2)
μ = μX
−μX
(or μ1−
μ2)
X − X
1 2 1 2
2 2
σ1 σ2 1 1
2 2
σ = + = σ +
X X
1 2
1 2
if σ = σ
−
n n n n
1 2 1 2
7. Distribution of Sample Proportions (P)
μP
= π
π (1 −π
)
σ
P
=
n
8. Distribution of Differences between Sample
Proportions (P 1 – P 2 )
μP
1− P= π
2 1−
π2
π1(1 −π1) π2(1 −π2)
σ
P1 − P= +
2
n n
1 2
Estimates for π, μ, σ are found from sample data
and given by p, x and s.
212
Section 5

[B] SUMMARY: CONFIDENCE INTERVALS
s
1. Mean x ± tν
with ν = n –1 D.F.
n
2. Difference Between Means (small samples and
independent, normal populations with equal
variances.)
1 1
( x1− x2) ± tν
sp
+ with ν = n 1 + n 2 – 2.
n1 n2
2 2
2
( n1− 1) s1 + ( n2−1)
s2
Here, sp
=
n + n −2
1 2
2 2
1 2
Note: If samples ≥ 30, x1− x2
± 1.96 s +
s
n1 n2
3. Difference Between Means (paired populations)
d
s d
± tν
with ν = n – 1
n
4. Proportion:
p
(1 )
1.96 p −
±
p
n
5. Difference Between Two Proportions
1(1 1) 2(1 2)
(
1 2) 1.96 p −
p p
p p −
− ± +
p
n n
1 2
213
Section 5

214

SECTION 6
This section reviews hypothesis testing, type 1 and type 2 errors, conclusive and inconclusive
results and the power of a study.
Null and Alternative Hypotheses
Study Based and Data Driven Hypotheses
One and Two Sided Tests
Four Steps in the Hypothesis Testing Procedure
Examples
Pooled proportion estimate
Clinical and Ecological Importance
Conclusive and Inconclusive Results
Errors in Hypothesis Testing
Power of a Study
Examples
215
Section 6

Hypothesis Testing
In most scientific studies we set up hypotheses
before about treatments (or populations) which
are the focus of the study. A null hypothesis (H 0 )
which is a claim about a treatment which is
assumed to be true unless the data collected in
our study show substantial evidence against H 0 .
At the same time we propose a research or
alternative hypothesis (H A ) which will be adopted
if there is sufficient evidence against the null
hypothesis.
There are two types of alternative hypotheses:
(i)
(ii)
a study based hypothesis which will imply
that we do not know at the outset whether a
new treatment is beneficial or possibly
harmful and
a data based hypothesis which is suggested
by the very nature of the collected data and
which will usually suggest treatment
benefit.
216
Section 6

If the data suggest harm we are likely to terminate
the study quickly but if the data suggest benefit we
need to know if the benefit is clinically important.
The study based alternative will usually lead to a
two sided test while the data based alternative
will lead to a one sided test. In the literature, the
two sided test is by far the most common.
There are FOUR STEPS in the standard
hypothesis testing procedure.
Step (1) A null hypothesis (H 0 ) is assumed about
a population parameter.
Step (2) An alternative (research) hypothesis is
proposed. This is accepted if H 0 is rejected.
217
Section 6

Step (3) A test statistic is computed from data.
It is the standardised value of a sample
mean, sample proportion or sample
difference obtained from the data. It is either
a z-score (large sample) or a t-score (for
small samples) given by
test statistic
=
observed sample value - null value
estimated standard error
That is, the number of standard deviations
from null value to the sample value. It is this
test statistic which allows calculation of the
p-value associated with the outcome of a
particular study.
Step (4) The probability of observing the value of
the test statistic in step (3), or a value which is
even more extreme, is calculated under the
assumption that the null hypothesis is true.
This probability is the p-value for the test
statistic. The test statistic has of course
summarized the data in the study. We draw
appropriate conclusions if the p-value is less
than 0.05.
218
Section 6

Examples Hypothesis Testing
Exercise: Suppose the resting pulse rates for
young women are normally distributed with mean
μ = 66 and standard deviation σ = 9.2 beats per
minute. A drug for the treatment of a medical
condition is administered to 100 young women
and their average pulse rate is found to be x = 68
beats per minute. Because the drug had for a long
time been observed to increase pulse rates, test
the claim that the drug does in fact increase the
pulse rates. (i.e. H A is data based.)
Solution:
Step (1) H 0 : μ = 66 (the null hypothesis)
Step (2) H A : μ > 66 (the research hypothesis)
Step(3) x = 68 from sample data. Assuming H 0
is true, and noting that population standard
deviation is known, standardising x leads to
z
=
observed sample mean - null mean
standard error of mean
219
Section 6

x − μ
=
σ / n
68 − 66
=
9.2 / 100
= 2.174
Step(4): Calculate p-value assuming μ = 66
0.50
66 68 X
0 2.174 Z
p-value = Pr( X > 68 given μ = 66)
68 − 66
= Pr( Z > )
9.2 / 100
= Pr( Z > 2.174) = 0. 015
This means that if H 0 is true, there is only a
probability of 0.015 of observing a sample mean
as large or larger than 68. Hence there is little
support for H 0 . Reject H 0 and conclude the mean
pulse rate has been increased by the treatment.
220
Section 6

Notes: 1. R-cmdr and other statistical
packages give a p-value directly beside the
study result or the test statistic. If the p-value
is less than 0.05 we have significance at the
5% level and if p-value is less than 0.01 we
have significance at the 1% level.
2. If σ is unknown but estimated from the sample,
the standardised statistic is t and the p-value is
found from the t-table with appropriate degrees
of freedom. (The exact p-value is not possible
since only a few values are given at top of
columns in t-table.)
e.g. Suppose s = 9.2 rather than σ = 9.2 and
sample size is n = 100.
Then,
68 − 66
p-value = Pr( t > )
9.2 / 100
= Pr( t > 2.174)
with 99 DF
t = 2.174 lies between values in columns
headed p = 0.025 and p = 0.010.
Hence p-value lies between these two
numbers (R-cmdr gives exact value)
221
Section 6

Exercise: In a large overseas city it was
estimated that 15% of girls between the ages of
14 and 18 became pregnant. Concerned parents
and health workers introduced an educational
programme in an effort to lower this percentage.
After four years of the programme, a random
sample of n = 293 18-year-old girls revealed that
27 had become pregnant.
(a)
(b)
(c)
Define null and alternative hypotheses for
investigating whether the proportion
becoming pregnant after the educational
programme has decreased. (Suppose the
alternative hypothesis is one sided.)
Calculate the probability value.
State your conclusion.
Step(1): H 0 : π = 0.15 (15% still become
pregnant)
Step(2): H A : π < 0.15 (less than 15% become
pregnant)
222
Section 6

Step(3): Sample gives p = 27/293 = 0.092
observed proportion - null proportion
z =
standard error of proportion
=
p −π
π ( 1−π
)
n
=
0.092 − 0.15
0.15(1 − 0.15)
under H 0 : π = 0.15
293
= –2.78
use 0.15 and
not 0.092 here
Step (4): p-value = pr(Z < –2.78)
= 0.5000 – 0.4973
= 0.0027
Z
–2.78 0
There is evidence that after the education
campaign the proportion becoming pregnant has
reduced.
223
Section 6

Exercise: The birthweight of a baby is thought to
be associated with the smoking habits of the
mother during pregnancy. The means and
variances of the INDIVIDUAL values in the two
samples of birthweights, one for non-smoking
and the other for smoking mothers, are in the
following table.
Mother
non-smoker
Mother
smoker
Sample Size (n i ) 100 50
Sample Mean ( x i
) 3.45 3.30
Sample Variance ( s 2 i
) 0.36 0.32
Investigate the claim that the mean birthweights
are different in the two groups. In this case we
shall suppose the alternative is study driven rather
than data driven.
Step(1): H 0 : μ NS – μ S = 0 (no difference in the
mean birth weight)
Step(2): H A : μ NS – μ S ≠ 0 (there is a difference
in the mean birth weight)
224
Section 6

Step(3): Sample gives x − x = 3.45 − 3.30 = 0.15
Standardising gives the test statistic
NS
S
observed difference of means - null difference
t =
estimated standard error of difference
0.15 − 0
=
1 1
s p
+
100 50
2
2
2 ( n 1) ( 1)
where
1 − s1
+ n2
− s
s =
2
p
n1
+ n2
− 2
99(0.36) + 49(0.32)
=
148
= 0.3468
0.15
=
1 1
0.3468 + 100 50
= 0.15
0.102
= 1.47
(use of pooling optional)
Since the sample is large we can use the standard
normal z in place of t with 148 degrees of
freedom.
225
Section 6

Step(4): In this case (two sided H A )
p-value = Pr(|z| > 1.47)
= Pr(z > 1.47 or z < –1.47)
–1.47 0 1.47
p-value = 2(0.5 – 0.4292) = 2(0.0708) = 0.1416
There is no evidence that the mean birthweights
for the smoking and non-smoking groups are
different.
Note: If the test had been one-sided
[H A : μ NS – μ S > 0] p-value = Pr(z > 1.47)
= 0.0708
z
z
0 1.47
There is again no evidence the non-smoking
group has a greater mean birthweight than the
smoking group.
226
Section 6

227
Section 6

228
Section 6

Notes on Hypothesis Testing
1. There is some terminology for reporting the
result of a test.
(a) If the p-value < 0.05 the result is
“significant at α = 0.05 level” (5% level)
or “There is some evidence that …”
(b) If the p-value < 0.01 the result is
“significant at α = 0.01 level” (1% level)
or “There is strong evidence that …”
(c) If p-value > 0.05 the result is “not
significant” or “There is no evidence that
…”
In the above α is generally a pre-selected cutoff
value.
229
Section 6

2. Choosing a smaller level of significance
requires the test statistic to be more
extreme before H 0 rejected.
3. Whether the test is one or two sided
depends on whether the alternative
hypothesis is data based or study based.
4. If H A is one sided, the p-value is the area
in one tail of the distribution of the
standardised test statistic.
5. If H A is two-sided, the p-value is the area
in the two tails of the distribution of the
standardised test statistic.
6. If using t-table choose column heading 2p
for a two sided alternative hypothesis and
p for a one sided alternative hypothesis.
7. When a test statistic leads to rejection of
H 0 , there are two possible explanations
(a) H 0 is true but random variation has given
an improbable test statistic.
230
Section 6

(b) H 0 is not true, and the observed statistic is
consistent with H A .
The second alternative (b) is taken but
there is possible error. This error is the
value α, the level of significance, which is
usually 0.05 or 0.01. α is called the type
one error (it is the chance of a false
conviction in a court of law – i.e. must
operate beyond reasonable doubt hence
choose small α).
8. In published work a p-value is quoted
beside the study result (indicating whether a
new treatment, say, has an effect) and a
confidence interval is reported (giving some
idea of the magnitude of an effect).
But one problem still remains when reporting
conclusions from a scientific study. It is possible
to obtain a result which is statistically
significant (with a small p-value) yet from a
clinical point of view the result is unimportant.
That is, it is not clinically important.
(Ecological importance is an equivalent concept.)
231
Section 6

Example: There are two treatments for raising
iron levels in infants, a standard treatment A and
a new treatment B.
A mean for treatment B that is 20 units greater
than the mean for treatment A is recognised as a
clinically important improvement which would
lead to widespread introduction of treatment B.
An experiment produces the following mean
differences, xB
− xA, with a 95% confidence
interval. Decide in each case whether the p-value
is less than or greater than 0.05. Report whether
the scientific result is conclusive or inconclusive
by considering clinical importance.
(a) Mean Diff = 40. Confidence Interval is
(33, 47)
The confidence interval does not include the
null hypothesis value so the p-value is less
than 0.05 (a statistically significant result).
The point estimate of 40 is in the direction
indicating treatment benefit. The result is
conclusive and there is evidence the benefit
is enough to be important.
232
Section 6

(b)
Mean Diff = 36. Confidence interval is
(18, 54)
p-value < 0.05. The result is conclusive.
There is treatment benefit but it may not be
as large as hoped.
(c)
Mean Diff = 27. Confidence interval is
(–4, 58)
p-value > 0.05 and inconclusive result. The
confidence interval includes H 0 . The new
treatment is probably better than treatment
A but we cannot completely rule out the
possibility that it is worse.
(d)
Mean Diff = –7. Confidence interval is
(–55, 41)
p-value > 0.05and result inconclusive. The
new treatment is likely to be harmful but we
cannot rule out the possibility that there is a
clinically important benefit.
(e) Mean Diff = –12. Confidence Interval =
(–34, 10)
233
Section 6

p-value > 0.05 and result is conclusive. Any
benefit is not clinically important and it is
more likely there will be treatment harm.
Treatment B should not be pursued as a
potential treatment.
(f) Mean Diff = –13. Confidence interval =
(–19, –7)
p-value < 0.05 and result very conclusive.
The new treatment is harmful.
(g) Mean Diff = 11. Confidence interval =
(4, 18)
p-value < 0.05. The result is conclusive.
There is treatment benefit but not enough to
lead to the introduction of treatment B.
Note: In practice you decide what is clinically
important. This is difficult but as you gain
experience with your own area of research it
becomes easier and you are able to critique any
published research.
234
Section 6

Summary of previous results
0 = null value
20 = clinically important improvement.
p-value < 0.05 implies confidence interval
excludes the null value of zero.
p-value > 0.05 implies null value included
The result can be conclusive or inconclusive.
(a) ( )
(b) ( )
(c) ( )
(d) ( )
-55 -7 0 20 41
(e) ( )
-34 -12 0 10 20
(f) ( )
-19 -13
-4
0 20 27
-7 0 20
(g) ( )
0 20 33 40 47
0 1820
36 54
0 4 111820
58
235
Section 6

(a) Conclusive p-value < 0.05
(b) Conclusive p-value < 0.05
(c) Inconclusive p-value > 0.05
(d) Inconclusive p-value > 0.05
(e) Conclusive p-value > 0.05
(f) Conclusive p-value < 0.05
(g) Conclusive p-value < 0.05
Clearly, if the confidence interval is too large,
there is greater chance for an inconclusive result.
236
Section 6

Example
A clinical trial is set up to compare two drugs
(pravastatin, A, and a control, B) for lowering
cholesterol. The mean cholesterol reductions in
the two groups are compared. The probability
that such a study will correctly detect a clinically
important difference between the effects of the
drugs is called the power of the study. Power
depends on the size of the difference, the
variability of estimates, sample size, and the level
of significance.
Figure 12.4: 95% confidence intervals for different sample sizes
Mean reduction
greater in A
target treatment
difference
(clinically
important)
0
no difference
Mean reduction
greater in B
n = 10 n = 20 n = 50 n = 200
237
Section 6

If the two samples are of size 5 (giving total
n = 10), the three 95% confidence intervals
include zero difference and the important
difference. As n increases, the confidence
intervals become smaller and it is possible to
detect the difference.
NB 1. It is helpful to aim for a confidence
interval which has diameter (or range) no
greater than the clinically important treatment
difference as in this case the result obtained
must be conclusive (rather than inconclusive).
2. If the clinically important effect size is large,
the confidence interval can be wider and hence
a smaller sample taken.
3. A larger sample gives smaller confidence interval.
4. Less random variation in the data gives
smaller confidence interval. (That is, the value
of σ is smaller.)
5. A smaller level of significance (α), say 0.01,
gives a wider confidence interval and hence
smaller power as there is less chance of
detecting a clinically important effect in a
conclusive way.
238
Section 6

Errors in Hypothesis Testing
The level of significance (α) is chosen by the
researcher, usually 0.05, and is the chance that the
null hypothesis (H 0 ) will be rejected when in
actual fact it is true. It would seem sensible for α
to be made as small as possible. Then the
probability of correctly not rejecting H 0 when it is
true will be large. But this is not the real issue in
a scientific study involving hypothesis testing.
The real issue is to have high probability of
rejecting H 0 when in fact H 0 is false or needing to
be rejected. That is, a high probability that a test
will correctly detect a real treatment effect of a
given magnitude. This is known as the power of
the test, and involves detecting clinically
worthwhile improvements as defined by
researchers. Power is related to the level of
significance. A smaller value for the level of
significance results in a smaller power. A power
between 80% and 90% is desirable.
239
Section 6

These ideas have a parallel in the courts of law in
this country. To illustrate, suppose we are
interested in testing a new treatment to see if it
has an effect.
1. The treatment is “arrested”.
2. The treatment is charged with having an
effect (H A ).
3. It is assumed treatment is “innocent” (has no
effect, H 0 ) until the evidence (data) shows
otherwise. The evidence is summarized in the
test statistic.
4. The level of significance (α) is the probability
that an innocent treatment will be convicted.
This error must be made small. That is, the
probability of a false conviction.
5. The power is the probability that a guilty
treatment will be convicted. This is the best
outcome for a court case as it is a correct
conviction. This probability should be large
since then we correctly convict the treatment
concluding there is an important treatment
effect. Power should be at least 0.80 or 0.90.
240
Section 6

Some Computer packages (Minitab is one) have
an excellent routine for analysing the power of a
study and showing how power, data variability,
sample size, level of significance and clinically
important effects are related.
EXAMPLE: The problem is to design a milk
feeding trial in 5 year old children to see if a daily
supplement of milk for a year leads to an
increased gain in height compared with a control
group (such a study would be both expensive and
difficult for practical and ethical reasons). It is
known that at this age children grow 6cm in a
year with a standard deviation of 2cm (σ). The
effect of milk on height gain is important if it
results in a gain of at least 0.5cm. We want a
high probability of detecting such a difference so
we set the power to be 0.9 (90%) and choose a
1% (α = 0.01) significance level.
Known:
Find
σ = 2 (data variability)
α = 0.01 (chosen level of sig.)
Clinically important diff = 0.5 cm
Target power is 0.90 (90%)
Sample Size.
241
Section 6

(a)
Find the sample size required to meet these
conditions. (i.e. σ = 2.0cm; clinically
important difference = 0.5cm; power = 0.9;
α = 0.01)
Step 1. STAT > POWER AND SAMPLE
SIZE > 2-SAMPLE t (i.e. choose an
unpaired t-test)
Step 2. Specify power value of 0.9, a clinically
important difference of 0.5 and sigma 2.0
Step 3. Choose Not equal for a study based two
sided alternative hypothesis and
significance level alpha of 0.01
242
Section 6

A printout is as follows:
Power and Sample Size
2-Sample t Test
Testing mean 1 = mean 2 (versus not =)
Calculating power for mean 1 = mean 2 + difference
Alpha = 0.01 Sigma = 2
Sample Target Actual
Difference Size Power Power
0.5 478 0.9000 0.9001
There need to be 478 children in each sample
meaning 956 children in total.
(i.e. the size of one sample is given)
[Note: the actual power will be different as a
result of rounding to the sample size.]
243
Section 6

(b)
Now consider clinically important
differences of 0.5, 0.6, 0.7, 0.8, 0.9, 1.0
A printout gives
Power and Sample Size
2-Sample t Test
Testing mean 1 = mean 2 (versus not =)
Calculating power for mean 1 = mean 2 + difference
Alpha = 0.01 Sigma = 2
Sample Target Actual
Difference Size Power Power
0.5 478 0.9000 0.9001
0.6 333 0.9000 0.9007
0.7 245 0.9000 0.9006
0.8 188 0.9000 0.9006
0.9 149 0.9000 0.9009
1.0 121 0.9000 0.9008
Notice that smaller samples will detect the larger
clinically important differences. Necessary
sample size reduces from the 956 to 242 [similar
to moving from a high resolution microscope to
pocket magnifying glass which is all that is
needed to detect 1.0]
244
Section 6

(c)
Halve the value of sigma to 1.0 and repeat
the analysis in (b)
Power and Sample Size
2-Sample t Test
Testing mean 1 = mean 2 (versus not =)
Calculating power for mean 1 = mean 2 + difference
Alpha = 0.01 Sigma = 1
Sample Target Actual
Difference Size Power Power
0.5 121 0.9000 0.9008
0.6 85 0.9000 0.9027
0.7 63 0.9000 0.9032
0.8 49 0.9000 0.9058
0.9 39 0.9000 0.9051
1.0 32 0.9000 0.9060
Notice how greater precision (decreased standard
deviation) in the data results in smaller sample
sizes required to achieve the desired power which
is now only 64 for a difference of 1.0.
245
Section 6

(d) A doctor set up a study involving 100
children (50 in each group) and monitored
the children for one year. The doctor
wanted to detect a clinically important
difference of 0.5, knew from historical
information that sigma = 2.0, and set up a
study based (two sided) test at α = 0.05(5%)
level of significance. The printout obtained
for the doctor after the study was carried out
follows.
Power and Sample Size
2-Sample t Test
Testing mean 1 = mean 2 (versus not =)
Calculating power for mean 1 = mean 2 + difference
Alpha = 0.05 Sigma = 2
Sample
Difference Size Power
0.5 50 0.2358
The power for this study is only 0.2358. The
probability of detecting the clinically important
difference of 0.5 is too small. The study was a
waste of effort in the sense that it is unlikely to
detect a difference as small as 0.5 when this size
difference is important.
If α = 0.01, power = 0.0891
246
Section 6

Revision Examples
1. Exam 2006:
In a study to assess the impact of an industrial
development on a nearby river, water temperature
was measured. It has been suggested the mean
water temperature is higher in this river than in a
similar river 30 km away that is not affected by
the development. Daily temperature in degrees
Celsius were taken at mid day for a fortnight in
February from both rivers. Two readings from
the “unaffected” river were spoiled. The data are
summarised below:
Unaffected Affected
river river
Sample Size (n i ) 12 14
Sample Mean ( x i
) 15.41 16.49
Sample Variance ( s 2 i
) 1.963 2.132
(a)
(4 marks) Assuming that temperature has a
common variability in both rivers and the
values are approximately normal, calculate
the pooled estimate for the common
variance and an estimate for the standard
error of the difference between the two
means.
247
Section 6

2
11(1.963) + 13(2.132)
s p
= =
24
Pooled variance = 2.055
2.055
standard error =
1 1
2.055 + = 0.564
12 14
Estimated standard error = 0.564
(b)
(2 marks) Using the appropriate value from
the t-table construct the 95% confidence
interval for the difference in mean
temperature in the affected and unaffected
rivers.
1.08 ± t 24 (0.564) where t 24 = 2.064
or 1.08 ± 1.164
Confidence interval:
–0.084 < μ A – μ u < 2.244
248
Section 6

(c)
(d)
(e)
(2 marks) A mean temperature increase of
0.6 degrees Celsius is ecologically
important. State your conclusion about the
true mean temperature from the confidence
interval in (b).
Conclusion:
Result inconclusive. There is no evidence of a
temperature mean difference but an important
increase cannot be ruled out.
(1 mark) State one way in which you might
increase the power of this study.
Statement:
Increase sample size.
(5 marks) A more powerful study is to be
set up which has a 95% confidence interval
for the difference between the mean river
temperatures no greater than 0.6 degrees
Celsius. Assuming the same number of
measurements is taken from each river and
the pooled estimate for the common
variance from (a) is the best estimate for the
variability, approximately how many
readings should be taken from each river
249
Section 6

Taking 1.96 as multiplier, the 95% C.I. is
1 1
( μ2− μ1) ± 1.96 2.054 + n n
But required precision needs (μ 2 – μ 1 ) ± 0.3
2
Therefore, 1.96 2.054 0.3
n ≤
2
(1.96) (2.054)2
∴
≤ n
2
(0.3)
∴ 175.3 ≤ n
Number of readings from each river: 176
(f)
(2 marks) The 95% confidence interval from
the study in (e) is (0.49, 1.12). What
conclusion would you now reach about the true
mean temperature difference
Conclusion:
Result conclusive. There is evidence of
increased temperatures but the increase may
not be ecologically important.
250
Section 6

2. Exam 2005
An ecologist must determine whether a cleanup
project at a lake has been effective. This is to be
done by recording dissolved oxygen content (in
parts per million, ppm) in the lake, with higher
values indicating less pollution. Prior to the
cleanup project a random sample of 50 dissolved
oxygen readings was recorded around the lake. Six
months after the initiation of the cleanup a second
random sample of 70 readings was recorded.
Results are summarised in the following table.
Before Cleanup After Cleanup
Sample Size (n i ) 50 70
Sample Mean ( x i ) 10.30 10.46
2
s 0.32 0.36
Sample Variance ( )
i
(a) (1 mark) State null and alternative hypotheses
for testing the data driven hypothesis that the
cleanup has resulted in an increase in the
dissolved oxygen content.
Null hypothesis, H 0 : μ BC = μ AC
Alternative hypothesis, H A : μ BC < μ AC
251
Section 6

(b) (6 marks) Calculate the pooled estimate for
the common variance of the two samples, an
estimate for the standard error of the difference
between the two means, and a standardised
normal z statistic for testing the hypotheses.
2
49(0.32) + 69(0.36)
= = 0.3434
118
s p
Pooled variance = 0.3434
estimated standard error
=
1 1
0.3434 + = 0.1085
50 70
Standard error = 0.1085
z
10.46 −10.30 = = 1.475
0.1085
Standardised z statistic = 1.475
252
Section 6

(c) (2 marks) Find the probability value (p-value)
for the z statistic in (b) and state your
conclusion from the p-value (using a 5% level
of significance).
p-value = 0.5 – 0.4306 = 0.0694
Conclusion: There is no evidence that the
clean-up has raised the mean dissolved
oxygen reading.
(d) (2 marks) Construct the 95% confidence
interval for the difference in the dissolved oxygen
means for the readings before cleanup and the
readings after cleanup.
(10.46 – 10.30) ± t 118 (0.1085) where
t 118 = 1.98 (accept 1.96)
i.e. 0.160 ± 0.215
Confidence interval:
–0.055 < μ AC – μ BC < 0.375
253
Section 6

(e) (1 mark) The power of this study is small.
Suggest one way in which you might increase the
power of this study.
Answer: Select a larger sample
(f) (3 marks) A more powerful study produced
the 95% confidence interval (0.04, 0.27).
What conclusions would you reach about the
p-value of this study result and the effect of
the cleanup project if an increase of 0.25 in the
dissolved oxygen mean is ecologically
important
Conclusion: p-value < 0.05
There is evidence the oxygen mean has
increased after the cleanup but it may not be
an important increase (or may not be as great
as hoped)
[Question 4 : 15 marks]
254
Section 6

SECTION 7
One factor analysis of variance, post analysis of
variance tests on means, and multiple comparison
procedures.
255
Section 7

ONE FACTOR ANALYSIS OF VARIANCE
This section of the course returns to the
continuous outcome theme.
In the studies of this type considered so far there
have been two treatments when usually a new
treatment is compared with a control or placebo.
In the first half of the semester we answered the
question about the effect of the new treatment by
using the two sample t-test to find p-values and
confidence intervals for the comparison of means.
These studies involved an outcome measured on a
continuous scale and the scores in the two
treatments were compared.
Regression procedures were developed which
allowed us to introduce potential confounding
variables and hence obtain adjusted or modified
confidence intervals and different p-values.
We are now going to investigate how to analyse
continuous data when there are more than two
treatments of interest.
256
Section 7

Example A general surgeon believes that
providing pain relief immediately following
surgery improves the level of comfort postsurgery.
Three pain killing drugs and a placebo
are randomly administered to patients
immediately following tonsillectomies. The
times in hours until onset of pain are as follows.
The study is double blind.
Placebo Drug A Drug B Drug C
1.6 2.6 1.2 3.6
0.3 12.6 1.7 3.2
1.1 2.8 0.9 3.4
0.4 4.5 2.1 3.9
1.4 5.3 1.3 4.9
2.4 4.4
3.9
Which drugs, if any, may be better than placebo
Notice that there are now three comparisons with
placebo. We can do better than just make the
three comparisons using three unpaired t-tests.
257
Section 7

Example: A comparison was made of protein
intake among three groups of post-menopausal
women: (1) women eating a standard American
diet (STD), (2) women eating a lacto-ovovegetarian
diet (LAC), and (3) women eating a
strict vegetarian diet (VEG). It was hypothesized
that protein intake was affected by diet. The
protein intakes (mg) for 30 women are:
STD LAC VEG
76 62 47
63 76 75
84 71 32
72 61 40
66 35 52
83 56 37
77 44 56
79 58 35
72 55 27
69 49 66
What are the effects of diet on protein intake
Notice that there are three comparisons which
could be of interest.
258
Section 7

We now investigate the problem of how to deal
with multiple comparisons. The unpaired t test
for comparing two sample means will be
extended to situations involving more than two
samples. As with simple linear regression the
idea is again to partition the total variability of a
response or outcome measure into components
due to different sources of variation.
Example: The effect of five drug treatments (A
to E) on reduction of fever is investigated. Four
children are assigned each treatment and
temperature reductions measured in appropriate
units with high values showing greater reduction.
Responses as follows:
A B C D E
9 7 2 4 4
8 4 3 8 9
6 9 4 1 6
9 6 3 3 3
Total 32 26 12 16 22 108
Mean 8.0 6.5 3.0 4.0 5.5 5.4
259
Section 7

One source of variation is due to differences
between the effects of the drugs, the other source
of variation is the random variation between the
individual children within each drug treatment.
But which of these is most responsible for
explaining the variation in the responses
The Method
Each response can be divided into three
components as follows:
Response = overall effect present in each value
+ a drug treatment (factor) effect
+ random error (or residual effect)
From the estimates for these components we find
a number measuring treatment variation and a
number measuring residual (including error)
variation. These values are compared using an F
statistic as in regression.
260
Section 7

Estimation of Components (for reference)
1. Overall mean = 5.4 (this is the estimate for
the overall effect with one degree of
freedom)
2. The five treatment effects estimated as
follows:
A: 8.0 – 5.4 = 2.6
B: 6.5 – 5.4 = 1.1
C: 3.0 – 5.4 = – 2.4
D: 4.0 – 5.4 = – 1.4
E: 5.5 – 5.4 = 0.1
These add to zero (as they are deviations
from their mean).
There are 5 – 1 = 4 degrees of freedom.
Note: The responses for A are, on average,
2.6 units above the overall mean, while
responses for D are, on average 1.4 units
below overall mean.
261
Section 7

3. The residuals (including random error) are
estimated by subtracting the overall mean
and the treatment effect from each response
to get:
A: 9 = 5.4 + 2.6 + 1.0
8 = 5.4 + 2.6 + 0.0
6 = 5.4 + 2.6 – 2.0
9 = 5.4 + 2.6 + 1.0
B: 7 = 5.4 + 1.1 + 0.5
4 = 5.4 + 1.1 – 2.5
9 = 5.4 + 1.1 + 2.5
6 = 5.4 + 1.1 – 0.5
C: 2 = 5.4 + (– 2.4) – 1.0
3 = 5.4 + (– 2.4) + 0.0
4 = 5.4 + (– 2.4) + 1.0
3 = 5.4 + (– 2.4) + 0.0
D: 4 = 5.4 + (– 1.4) + 0.0
8 = 5.4 + (– 1.4) + 4.0
1 = 5.4 + (– 1.4) – 3.0
3 = 5.4 + (– 1.4) – 1.0
E: 4 = 5.4 + 0.1 – 1.5
9 = 5.4 + 0.1 + 3.5
6 = 5.4 + 0.1 + 0.5
3 = 5.4 + 0.1 – 2.5
The residuals are the third values on right.
262
Section 7

There are 20 data values altogether and hence 20
degrees of freedom but 5 degrees of freedom
have been used up leaving 15 for the residual
effect.
Sums of Squares Computation
2
∑ (responses)
= 9 2 + 8 2 + 6 2 + 9 2 + … +
6 2 + 3 2
= 714 (with 20 DF)
2
∑ (overall means) = 5.4 2 +… + 5.4 2
= 20 (5.4) 2
= 583.2 (with 1 DF)
2
∑ (treatment effects) = 2.6 2 + … + 2.6 2 + … +
0.1 2 + … + 0.1 2
= 4[(2.6) 2 + (1.1) 2 + (–2.4) 2
+ (–1.4) 2 + (0.1) 2 ]
= 62.8 (with 5 – 1 = 4 DF)
2
∑ (residuals)
= (1.0) 2 + (0.0) 2 + (–2.0) 2 +
… + (–2.5) 2
= 68.0 (with 15 DF)
From these 714 = 583.2 + 62.8 + 68.0
In general Total response Sum of Squares
= overall mean SS + treatments SS
+ residuals (error) SS
263
Section 7

Notes: 1. If there are no treatment differences,
treatment effects will all be close to zero,
hence treatments SS will be small. But how
does this compare with the random SS
measured by the residuals.
2. We find the mean (or average) squares (MS)
for treatment and residual effects and
compare these with an F statistic. Sums of
squares are divided by degrees of freedom.
The Analysis of Variance (ANOVA) Table
The calculations are summarised in a table similar
to those arising with a regression analysis.
Source of Sum of DF Mean F
Variation Squares Squares
Overall mean 583.2 1
Treatment effects 62.8 4 15.70 3.47
Residual (error) 68.0 (15) 4.53
Total 714.0 20
F = 15.70/4.53 = 3.47 (giving effect of the
treatments on the responses compared with the
chance (residual) effect on responses.
Is this value large enough to be significant
The critical value is found from F table (5%)
264
Section 7

υ 2 υ 1 1 2 3 4 … 30
1
15 3.056
120
If υ 1 = 4 and υ 2 = 15 then critical F = 3.056
meaning Pr(F 4,15 > 3.056) = 0.05. Since 3.47 >
3.056 we have significance at the 5% level. This
means that treatment effects outweigh the chance
(residual) effect.
Conclusion: There is evidence of a difference
between the mean temperature reductions
resulting from the five treatments.
Note:
Because the overall mean appears in each data
value, it makes no impact on variability between
data values and the ANOVA table becomes.
Source SS DF MS F
Treatment effects 62.8 4 15.70 3.47
Residual (error) 68.0 (15) 4.53
Total (mean deleted) 130.8 19
265
Section 7

SYSTEMATIC CALCULATIONS
The calculations for a one factor analysis of
variance can be carried out easily using statistical
software or by the following computation method
which is quicker than the previous partitioning
approach.
A B C D E
9 7 2 4 4
8 4 3 8 9
6 9 4 1 6
9 6 3 3 3
Col Total (C j ) 32 26 12 16 22 108
2
j
C 1024 676 144 256 484 2584
The between treatments (or samples) sum of
squares is
C
n
2 2 2
1 C C
+
2
+ +
k
−
1
n
2
…
n
k
(overall mean SS)
where n 1 , n 2 etc are sample sizes and k = 5 here.
266
Section 7

If n 1 = n 2 = … = n k = n (say) this becomes
[
2 2
] C + C + … + C − (overall mean SS)
1 2
1 2 k
n
Total SS = 9 2 + … + 3 2 = 714.0 as before.
Overall mean SS = 20(108/20) 2 = 583.2 as before
Treatment effects SS
1
= [ 1024 + 676 + 144 + 256 + 484] − 583. 2
4
= 62.8 as n 1 = n 2 = … = 4
SOURCE SS DF MS F
Overall mean 583.2 1
Treatment effects 62.8 4 15.70 3.47*
Residual (error) (68.0) (15) 4.53
Total 714.0 20
Brackets indicate numbers found by subtraction.
If the effect of the overall mean is deleted again,
the reduced table is produced.
SOURCE SS DF MS F
Treatment effects 62.8 4 15.70 3.47*
Residual (error) (68.0) (15) 4.53
Total 130.8 19
267
Section 7

A Note on the Residual Mean Square
s or s )
( 2 p
2
e
The four treatment A residuals were 1.0, 0.0, –2.0,
1.0. These are values 9, 8, 6, 9 with A mean of 8
subtracted. i.e. they are of form x − x . An
Ai A
estimate of the variance for treatment A is therefore
s
2
A
2
Ai A A
1 2 2 2 2
= ∑ ( x − x ) /( n −1)
= (1.0 + 0.0 + [ −2.0]
+ 1.0 )
3
For the other four treatments the variance estimates
are
2
2
s = ∑ ( x − x ) /( n −1)
B Bi B B
2
2
s = ∑ ( x − x ) /( n −1)
E
Ei
E
where in this case n A = n B = n C = n D = n E = 4
If it is assumed that the variance is the same at all
five treatments, then the common or pooled
variance estimate is
E
268
Section 7

[
2 2 2 2 2
]
s + s + s + s s
2 1
s =
+
p 5 A B C D E
1 ⎡1
2 1
= ∑ ( x − x ) + … + ∑ ( x − x
5 ⎢⎣ 3
Ai A
3
Ei
1
2
= ∑ ( x − x ) + … +∑ ( x − x )
15 Ai A
Ei E
1
[
2 2 2 2
]
= 1.0 + 0.0 + ( −2.0)
+ 1.0 +…
15
= Residual SS/Residual DF
= Residual Mean Square ( s 2 e
)
[
2]
The residual mean square is just the pooled
variance estimate for all five samples. (It is a
direct extension of the pooled variance estimate
in an unpaired t test.)
Notes (1) For the F test to be valid, the
variances in all samples compared (here 5)
should be approximately equal.
(2) The square root of the residual mean square
s is the standard deviation of the residuals.
2
e
E
)
2
⎤
⎥⎦
269
Section 7

(3) In the R-cmdr printout for such an analysis
the overall mean effect is deleted from the
ANOVA table (as in the equivalent
regression printout). The important section
of the table remains.
SOURCE SS DF MS F
Treatment effects 62.8 4 15.70 3.47*
Residual (error) effect 68.0 15 4.53
Total (less mean) 130.8 19
Example: 20 children allocated randomly to four
equal groups subjected to different treatments.
After 3 months, progress measured by a test, with
responses below (one child in group 3 died). Test
for treatment mean differences.
TREATMENT
1 2 3 4
4 31 30 19
12 49 41 66
44 22 13 65
9 56 26 46
17 19 89
C 86 177 110 285 658
j
2
C 7396 31329 12100 81225
j
270
Section 7

Total SS = 4 2 + 12 2 + … + 89 2 = 32214
Overall mean SS = 19(658/19) 2 = 22787.58
Total SS (less mean SS) = 9426.42
Treatment effect SS
7396 31329 12100 81225
= + + + − 22787. 58
5 5 4 5
= 4227.43
The ANOVA table becomes
SOURCE SS DF MS F
Treatment effect 4227.43 3 1409.14 4.066
Error (residual) (5198.99) (15) 346.60
Total (less mean) 9426.42 18
Critical value at 5% level of significance is 3.287
< 4.066 (Using 3 and 15 DF)
Conclusion: There is some evidence that the
mean outcomes in the four treatments differ.
271
Section 7

272
Section 7

POST ANALYSIS OF VARIANCE RESULTS
It is important for further interpretation to set up
confidence intervals for individual sample means
or for differences between pairs of sample means.
The useful new development here is that the
residual mean square is an excellent estimate for
the data variance meaning there is no need to
additionally calculate the usual pooled variance
estimate for each pair of samples. The advantage
of using the residual mean square is that it
involves all the data, not just data in individual
samples.
Example: Set up a 95% confidence interval for
the mean of treatment 2.
Solution: Here,
x = 177/5 = 35.4
2
s s 346.60
Estimated standard error = 2
e
= =
n n 5
= 8.33
which has 15 degrees of freedom (same as
residual)
273
Section 7

The 95% C.I. is 35.4 ± t 15 (8.33)
where t 15 = 2.132
That is, 35.4 ± 17.76
or 17.64 < μ 2 < 53.16
N.B. (1) Use 15 DF rather than 5 – 1 = 4 DF for
the single second sample. Hence greater
precision as t 15 < t 4 (note t 4 = 2.776)
(2) R-cmdr gives confidence intervals for these
treatment means automatically.
(3) As we have seen, use of the residual mean
square requires the variances to be equal in
each sample.
Example: Compare the mean scores for
treatments 3 and 4 by setting up a 95% C.I. for
the difference.
274
Section 7

Solution: x = 110/4 = 27.5 x = 285/5 = 57.0
3
4
estimated standard error of difference
=
1
s p
+
n
2 1 1
= s
e
+
4 5
=
3
1
n
4
1 1
346.60 + 4 5
= 12.49
with 15 DF again rather than n 3 + n 4 – 2 = 7 DF
as for the usual unpaired t-test.
The 95% C.I. for μ 4 – μ 3 is
(57.0 – 27.5) ± t 15 (12.49)
where t 15 = 2.132
That is 29.5 ± 26.63
or 2.87 < μ 4 – μ 3 < 56.13
Since zero excluded, there is evidence treatment 4
has a higher average score than treatment 3.
275
Section 7

A NOTE ON ASSUMPTIONS IN ANOVA
Residuals and residual plots can be used to check
the required assumptions. As in a regression
analysis, the residuals should be
(i) normally distributed,
(ii) randomly distributed about 0,
(iii) have similar variation within each of the
samples chosen
The following graph shows the variability in each
of the drugs in the temperature reduction fever
data. There could be some concern about unequal
variation within each of the five treatments (but
the samples are very small in this case so this is
not too surprising).
276
Section 7

The next two residual plots confirm that the
variation is similar for each drug treatment and
the residuals are close to being normally
distributed.
277
Section 7

278

SECTION 8
This section covers the analysis of count data including the Chi-square test for contingency, the chisquare
test for trend as well as relative risks, attributable risks and odds ratios along with their
confidence intervals. The analysis of a three way table and Simpson’s paradox are investigated as a
way of introducing the concept of a confounding variable in the lead up to regression analyses.
Categorical Data Examples
Relative Risk and its Confidence Interval
Attributable Risk and its Confidence Interval
Odds Ratio and its Confidence Interval
Chi-square Test for Contingency
Chi-square Test for Trend
Interpretation of Confidence Intervals
Simpson’s Paradox and Confounder Control
279
Section 8

Analysis of categorical data
Categorical Data arise when individuals or
experimental units are classified into one of two
or more mutually exclusive groups. For example,
• binary e.g. sex (M/F); dead/alive;
diseased/disease free;
treatment/placebo; smoker (yes/no)
Tuatara present/absent
herpes present/absent
melanoma present/absent
• nominal e.g. ethnicity
• ordinal e.g. disease severity; socio economic
status; smoking (never/ex/current)
In a sample of units, the number falling into a
particular group is the frequency. The analysis of
such data is sometimes referred to as the analysis
of frequencies or counts.
280
Section 8

Examples of research questions that we shall
look at.
Estimation of one proportion:
Ex 1. What is the prevalence of asthma in a
population
Associations between two factors:
Ex 2. Is a vaccine effective in reducing the risk
of catching influenza
Ex 3. Is there an association between exposure
to chlorinated water and dental enamel
erosion
Ex 4. Does infra-red stimulation (IRS) provide
effective pain relief in patients with
cervical osteoarthritis
Ex 5. Is there an association between income
level and severity of cardiovascular
disease in a group of people presenting for
treatment
281
Section 8

What tools do we need to answer these types of
questions Recall the research loop
Underlying Population
Selection
bias
Inference
Sample
Confounding
Statistical
analysis
Information
bias
Possible explanations for an association include
• bias (controlled with study design when
selecting the people for a study or
systematic error arising from the way
information was collected from study
participants)
• confounding (must be allowed for)
• chance (or random error)
• true association
We shall use proportions, relative and attributable
risks, odds ratios, confidence intervals and
probability values.
282
Section 8

Example 1: What is the prevalence of asthma in
a population
Population: adult males on a general practice
register.
Study
• random sample from population, n = 215
• 39 have history of asthma
Sample proportion p = 39/215 = 0.18
0.18(1 − 0.18)
Standard error of proportion =
215
= 0.026
95% confidence interval for the true proportion
(0.13, 0.24)
Conclusion
We can be 95% sure that the true prevalence of
asthma among men attending this general practice
is between 13% and 24%.
Confidence intervals for very small proportions
• If the number of events is small the
distribution of sample proportions is not
normal and values would be negative.
• an ‘exact’ method based on the binomial
distribution must be used.
283
Section 8

Evaluating associations in 2 × 2 tables
Example 2: is a vaccine effective in reducing the
risk of catching influenza
Study
169 people were randomly allocated to receive a
flu vaccine or a placebo. At end of winter they
were asked if they had contracted flu’.
Flu’ No Flu’ Total
Vaccine 9 75 84
Placebo 22 63 85
Total 31 138 169
This is what is called as a prospective cohort
study. In a cohort study the cohort of people is
followed into the future. Such studies can be
expensive as they may be of long duration. Also
if a disease is rare (say a cancer) many
participants will be needed. The Dunedin
Multidisciplinary Study is one of these. Recall
the example on circumcision and sexually
transmitted disease.
284
Section 8

Example 3: Is there an association between
exposure to chlorinated water and dental enamel
erosion
Study
Of 49 swimmers with enamel erosion (the cases)
32 reported swimming 6 or more hours per week
compared with 118 to 245 swimmers without
enamel erosion (the controls).
Swim time Erosion of enamel Total
per week Yes No
(Cases) (Controls)
≥ 6 hrs 32 118 150
< 6 hrs 17 127 144
Total 49 245 294
This is what is called a retrospective case control
study. Advantage is that such a study is relatively
quick and smaller than a cohort study particularly
for rare diseases. But greater potential for bias as
there may be inaccurate recall.
The analysis of this 2 × 2 table is not the same as
the analysis in the 2 × 2 table in the previous
cohort study. (We shall see that odds ratio rather
than relative risk must be used.)
285
Section 8

Both these data summaries are in the form of a
2 × 2 table. Usually there is an exposure (or
predictor) category and an outcome (response
category).
Outcome (disease)
Exposed Present Absent Total
Yes a b a + b
No c d c + d
Total a + c b + d n
We know how to summarize data from tables like
these
• the choice of measure depends on the study
design
• options include relative risk, attributable risk
(difference in proportions), odds ratio
The tools needed for statistical inference are
• confidence intervals for relative risks
attributable risks and odds ratios
• hypothesis tests (p-values) for these
associations
286
Section 8

Prospective Studies
• groups are followed up to see if an outcome
of interest occurs
• the proportions in each group who develop
the outcome are found (these are often called
the incidence which defines numbers of new
cases of a disease)
• the ratio of these proportions is the relative
risk
• the difference in these proportions is the
attributable risk
General form of 2 × 2 table:
Outcome (disease)
Exposed Present Absent Total
Yes a b a + b
No c d c + d
Total a + c b + d n
Relative risk, RR =
a/( a+
b)
c/( c+
d)
Attributable risk, AR = a/( a+ b) − c/( c+
d)
Section 8
287

Example 2: Is a vaccine effective in reducing the
risk of catching influenza
Flu’ No Flu’ Total
Vaccine 9 75 84
Placebo 22 63 85
Total 31 138 169
Risk in vaccine group = 9/84
Risk in placebo group = 22/85
Relative risk, RR = 9/84
22/85 = 0.4
Those who were vaccinated were 0.4 times as
likely to develop the flu as those who were not
vaccinated. So flu vaccine was associated with a
60% reduction in risk of flu.
Notes:
• if a RR = 1.00, then rates are equal and there
is no association between flu’ and vaccine
• the convention is to calculate the relative risk
this way round so that a ‘protective’ exposure
gives a relative risk less than 1.
288
Section 8

Confidence interval for relative risk
One method for finding confidence intervals for
RR is as follows:
The sampling distribution for ln(RR) is
approximately normal with standard deviation (or
standard error) given by
[ ]
s.e. ln(RR)
1 1 1 1
= − + −
a a+ b c c+
d
Then the 95% confidence interval for ln(RR) is
ln(RR) ± 1.96 s.e.[ln(RR)]
For example,
1 1 1 1
s.e. [ ln(RR) ] = − + − = 0.364
9 84 22 85
Now RR = 0.414, giving ln(RR) = –0.882
289
Section 8

The confidence interval (95%) becomes
–0.882 ± 1.96 (0.364)
i.e. –0.882 ± 0.714
Therefore, –1.596 < ln(RR) < –0.168
Taking exponentials, 0.20 < RR < 0.85
So the 95% confidence interval for the true
relative risk is (0.20, 0.85)
Since 1 is not contained in this confidence
interval we conclude that there is evidence of
association between vaccine use and a reduced
risk of contracting flu’
Note:
• this method will give a correct CI only if the
numbers in each cell are not too small
• in order to complete our evaluation of the
effectiveness of the vaccine we need to also
consider possible sources of bias and
confounding
• regression procedures allow us to take
account of confounding effects (see later).
290
Section 8

Confidence interval for attributable risk
Once we have determined treatment is effective
we may also wish to consider how many cases of
flu’ vaccine is likely to prevent:
Attributable risk:
22/85–9/84 = 0.26 – 0.11 = 0.15
Use the normal approximation to get a confidence
interval for this difference in proportions. The
estimated standard error for the difference
between the proportions is
p1(1 − p1) p2(1 − p2) 0.26(0.74) 0.11(0.89)
+ = +
n1 n2
85 84
= 0.059
and the 95% confidence interval for the
attributable risk (risk difference) is
giving
0.15 ± 1.96(0.059)
(0.04, 0.27)
291
Section 8

So, assuming the treatment is effective, in every
100 people vaccinated there will be between 4
and 27 fewer cases of flu than if they had not
been vaccinated (i.e. vaccination prevents
between 4 and 27 cases of flu in every 100
people)
292
Section 8

Case control studies
• a group of individuals with a disease (called
the cases) is compared to a control group who
do not have the disease. In these cases we
choose the number of people with the disease
and the number without.
General form of 2 × 2 table
Outcome (disease)
Exposed Present Absent Total
Yes a b a + b
No c d c + d
Total a + c b + d n
The measure of association used in case-control
studies is the odds ratio, not the relative risk
• In terms of probabilities, the odds of an event
Pr( A) Pr( A)
A is defined as = . With the
Pr( A) 1 − Pr( A)
notation in the table above, in the exposed
group the odds of disease present equals
293
Section 8

⎛ a ⎞ ⎛ b ⎞
⎜ ⎟ ⎜ ⎟ which simplifies to a/b.
⎝a+ b⎠ ⎝a+
b⎠ For unexposed group, odds = c/d.
Example: Is there an association between
exposure to chlorinated water and dental enamel
erosion
Study
Of 49 swimmers with enamel erosion (the cases)
32 reported swimming 6 or more hours per week
compared with 118 of 245 swimmers without
enamel erosion (the controls).
Swim time Erosion of enamel Total
per week Yes No
(Cases) (Controls)
≥ 6 hrs 32 118 150
< 6 hrs 17 127 144
Total 49 245 294
For ≥ 6 hrs, odds = a/b = 32/118
For < 6 hrs, odds = c/d =17/127
294
Section 8

a/
b
The odds ratio, OR =
c/
d
= 32/118
17 /127
= 2.026 (= 2.0)
Note 1: why we use the odds ratio
Compare the numbers in the previous table to a
study which is identical except that we chose to
have only 49 controls:
Swim time Erosion of enamel Total
per week Yes No
(Cases) (Controls)
≥ 6 hrs 32 24 56
< 6 hrs 17 25 42
Total 49 49 98
The values 24 and 25 give the same proportions
with slight rounding
32/ 24
Odds ratio = = 2.0 with rounding which is
17 / 25
the same as the previous result.
295
Section 8

But now suppose we were to try and calculate the
relative risk in both cases:
‘Risk’ ‘RR’
Study 1 ≥ 6 hrs 32/150
< 6 hrs 17/144 1.75
Study 2 ≥ 6 hrs 32/56
< 6 hrs 17/42 1.43
Notice that there is disagreement. The
consequence is that the relative risk can be made
to take any value by choice of numbers of cases
and controls. This is unacceptable.
Note 2: When are the odds ratio and relative risk
close
Consider a retrospective case-control study:
If disease (the outcome of interest) is rare,
then a and c will be small in the table.
Disease No Disease
Exposed (Case) (Control) Total
Yes a b a + b
No c d c + d
296
Section 8

so
a a c c
≈ and ≈
b a+ b d c+
d
Then relative risk =
⎛ a ⎞ ⎛ c ⎞ a/
b
⎜ ⎟ ⎜ ⎟ ≈
⎝a+ b⎠ ⎝c+
d ⎠ c/
d
Thus, in a case-control study investigating a rare
disease the odds ratio gives a good estimate of the
true unestimable relative risk.
Confidence interval for odds ratio
In repeated sampling, ln(OR) are normal with
standard deviation (or standard error) given by
[ ]
s.e. ln(OR)
1 1 1 1
= + + +
a b c d
The 95% confidence interval for ln(OR) is
For the example
ln(OR) ± 1.96 s.e. [ln(OR)]
297
Section 8

1 1 1 1
s.e. [ ln(OR) ] = + + + = 0.326
32 118 17 127
and ln(OR) = ln (2.026) = 0.706
The confidence interval becomes
0.706 ± 1.96 (0.326)
i.e. 0.706 ± 0.639
Therefore, 0.067 < ln(OR) < 1.345
∴ e 0.067 < OR < e 1.345
∴ 1.069 < OR < 3.838
We conclude the odds of erosion in dental enamel
are raised among those swimming more than 6 hours
per week. We would reject the null hypothesis as p-
value < 0.05.
Note: An odds ratio simply measures if an
association is present between outcome and
exposure. With a relative risk we are interested if
treatment improves outcome status. A protective
exposure gives a relative risk less than 1.
298
Section 8

Chi Square Test for Contingency Tables
The above examples (2 × 2 tables) are very
common in health research and other areas.
However, we may want:
• p-values to formally test for an association
• to answer questions relating to larger
contingency tables.
Note:
• as long as one of the variables is binary we
can think of comparing proportions and
calculate RRs or ORs
• if both variables have more than 2 categories
the analysis is more complex
299
Section 8

Example 4
Does infra-red stimulation (IRS) provide effective
pain relief in patients with cervical osteoarthritis
A randomised controlled trial was carried out
with 100 patients: 20 were randomly allocated to
a double dose and 40 each to a single dose and
control (placebo) treatment. The patients were
classified according to improvement levels over a
period of one week as follows:
(hypothetical data)
Pain score
IRS Improve No Worse Total
change
Double dose 10 5 5 20 = r 1
Single Dose 15 20 5 40 = r 2
Control 5 20 15 40 = r 3
Total 30 = c 1 45 = c 2 25 = c 3 100 = n
• we can look at the percentage improved, no
better and worse for each treatment category
300
Section 8

We wish to know whether the data indicate that
either
or
IRS does provide effective pain relief (and in
what dose)
it is no better than the control.
Calculating a p-value for the following
hypotheses will tell us whether there is evidence
that IRS is effective, or whether the differences
we have observed between the treatment groups
are consistent with random variation.
Hypotheses:
H 0 : The response and the type of treatment are
independent (i.e. no association)
H A : response and type of treatment are not
independent (i.e. are associated in some way
or one of the responses may occur more often
with one of the treatments)
301
Section 8

If there were no association between treatment and
outcome (H 0 ), I would expect to have the same
fraction of improved responses using the three
treatments and this fraction should be
c 1 /n = 30/100 (i.e. 30 of the 100 patients show
improvement).
Suppose E 11 , E 21 and E 31 are the numbers of
improvements expected if RESPONSE and
TREATMENT are independent. Then
30 E11
E
= =
21
E = 31
100 20 40 40
20(30)
∴E 11 = = 6 100
40(30)
E 21 = 100
40(30)
E 31 = 100
= 12
= 12
In general,
E
ij =
r c
i
n
j
for each “cell” or “class” in the contingency table.
302
Section 8

Using this formula, expected numbers can be
calculated for each cell:
RESPONSE
TREATMENT Improve No change Worse Total
Double dose 6 9 [5] 20
Single Dose 12 18 [10] 40
Control [12] [18] [10] 40
Total 30 45 25 100
Each row and column total has to be met by the
entries in the table and for this reason the numbers
in brackets can be found by subtraction.
The observed frequencies (the data counts) are now
compared with the expected counts calculated
under H 0 .
If H 0 is true, then the expected counts will agree
closely with those observed. [But how closely
must they agree]
This is answered by calculating the chi-square (χ 2 )
statistic
303
Section 8

χ
2
=
∑
over
all cells
(Observed - Expected)
Expected
2
i.e.
χ
2
=
∑
over
all cells( i,
j)
( O
ij
−
E
E
ij
ij
)
2
Observed Counts (O ij )
Treatment Response
1 2 3
1 10 5 5
2 15 20 5
3 5 20 15
Expected Counts (E ij ) [Under H 0 : independent]
Treatment
Response
1 2 3
(Improved) (No change) (Worse)
Double 1 6 9 5
Single 2 12 18 10
Control 3 12 18 10
χ 2 is large if O ij and E ij seriously disagree – hence
χ 2 being large will result in H 0 rejection.
304
Section 8

Example: For the drug responses,
χ 2 =
(10 − 6)
6
2
+
(5 − 9)
9
2
+
(5 − 5)
5
2
+
(15 −12)
12
2
+
(20 −18)
18
2
+
(5 −10)
10
2
+
(5 −12)
12
2
+
(20 −18)
18
2
+
(15 −10)
10
2
= 14.72 (χ 2 will always be positive)
In repeated sampling these χ 2 values are distributed
as a chi-square distribution which has
υ = (number of rows – 1) × (number of columns – 1)
degrees of freedom.
Here, υ = (3 – 1) × (3 – 1) = 4
which is just the number of values that can be
freely inserted in the table!! (the remaining values
are fixed if the row and column totals are to be
met.)
The critical χ 2 value is found from the table at the
end of the notes.
305
Section 8

0
Critical value
α (area, or the
level of
significance)
α
υ 0.1 0.05 0.025 0.01 0.005 0.001
1
2
3
4 9.488 14.86
5
100
Since 14.72 > 9.488, the null hypothesis of no
association is rejected.
Note: when we do this on the computer we get the
exact p-value, p = 0.005
χ
2
υ
306
Section 8

• the p-value gives the probability of observing a
difference this large or larger between what we
observed and what is expected under H 0 , if H 0
is true.
• since the p-value is small, it is unlikely we
would observe a difference this big just by
chance, it is more likely that the null hypothesis
is false.
• there is evidence that the pain levels depend on
the treatment administered.
• closer inspection of the observed frequencies
indicates
• more patients improved on double dose
than expected
• few patients experiencerd improved
response on the control
• fewer than expected being worse on single
dose.
307
Section 8

Notes
1. Check the observed counts in order to interpret
a significant association.
2. Maximum power is achieved if there are equal
numbers in each ‘exposure’ group. This is
often not possible to achieve in observational
studies.
3. This chi-square procedure is unreliable if
counts are small, in particular less than 5.
• For larger contingency tables it is possible
to combine classes in order to raise
frequencies.
• For 2 × 2 tables if expected frequencies
are between 5 and 10, a correction called
Yates correction will modify the χ 2
statistic.
• For 2 × 2 tables, if expected frequencies
are less than 5, there is a test called
Fisher’s Exact Test which can be used.
308
Section 8

Example 5
Is there an association between income level and
severity of cardiovascular disease in a group of
people presenting for treatment
Study
A group of people presenting to a hospital with
acute myocardial infarction or unstable angina are
enrolled in a study. Cross-sectional data are
collected at baseline.
Income level (Exposure)
Disease
level 1 2 3 4 Total
(Outcome)
0 100 107 111 122 440
≥1 (Severe) 115 112 104 97 428
Total 215 219 215 219 868
% ≥1 52.0 51.1 48.4 44.3
RR 1.00 0.96 0.90 0.84
115/ 215
115/ 215
112/ 219
115/ 215
104/ 215
115/ 215
97 / 219
115/ 215
309
Section 8

To test whether or not there is an association
between disease severity and income level:
H 0 = there is no association between disease
severity and income (i.e. the proportion
with severe disease is the same for all
income levels)
H A =
there is some association (i.e. the
percentage with severe disease varies by
income)
Expected frequencies:
Income level
Disease 1 2 3 4 Total
level
0 108.99 111.01 108.99 111.01 440
≥1 106.01 107.99 106.01 107.99 428
Total 215 219 215 219 868
440
215 108.99
868
E
11
= × =
12
440
E
13
= 215× = 108.99
868
440
E = 219× = 111.01
868
310
Section 8

χ 2
2
(100−108.99)
(107−111.01)
(111−108.99)
=
+
+
108.99 111.01 108.99
2
2
2
(122−111.01)
+
(115−106.01)
+
(112−107.99)
111.01 106.01 107.99
2
2
(104−106.01)
(97−107.99)
+
+
106.01 107.99
= 4.1
2
2
+
The appropriate sampling distribution is a χ 2 with
3 d.f.
From the χ 2 table
Pr(χ 2 (3 d.f.) > 6.251) = 0.1
so p-value > 0.1
From the computer, p-value = 0.25
Hence the observed differences in proportions we
have seen are of the order we might expect to see
by chance. There is no evidence supporting
rejection of the null hypothesis.
We conclude that there is no evidence of an
association between disease severity and income.
311
Section 8

Contingency Tables (Continued)
Tests for trend
Example 5 (continued): Do people with lower
incomes tend to present with more severe
disease
The chi-squared test of association may not
provide the best answer to this question. It does
not take account of the ordering in the income
variable. Specifically, our prior hypothesis is
that the percentage with severe disease decreases
as income increases.
We can test this hypothesis directly using a χ 2
test for trend. The main difference is that this
test has only one degree of freedom rather than
the three for the test of association.
Note: You will NOT be asked to calculate a test
for trend in this course. You may be asked to
2
interpret the p-value or a χ
trend
value with one
degree of freedom.
312
Section 8

This page for reference only
Income level (x i )
Disease 1 2 3 4 Total
level
0 100 107 111 122 440
≥ 1 (r i ) 115 112 104 97 R = 428
Total (n i ) 215 219 215 219 N = 868
r i x i 115 224 312 388
n i x i 215 438 645 876
n i x i
2
215 876 1935 3504
p
= R N = 428 = 0.49 ∑ rx
i i
2174
868 x = = =
N 868
χ
2
trend
[ ∑ rx − Rx] 2
i i
=
2 2
p(1 − p)
⎡⎣∑
nx
i i−Nx
⎤⎦
1039 − 428×
2.50
=
0.49(1 −0.49) ⎡⎣
6530 − 868×
2.50
= 4.06
[ ] 2 2
⎤⎦
2.505
313
Section 8

The trend statistic has only 1 degree of freedom.
From χ 2 table, Pr(χ 2 (1 d.f.) > 3.841) = 0.05
Since 4.1 > 3.841 the p-value < 0.05, so we
conclude there is evidence that the proportion
with severe disease decreases as income
increases.
Overview
• interpretation of confidence intervals for RR
and OR
• relationship between confidence intervals, p-
values and sample size
Example: (Hypothetical Data)
The following confidence intervals are from a
study into the erosion of tooth enamel as a result
of exposure to chlorinated water.
They are the ratio of odds for those exposed
(swim ≥ 6 hours per week) to those not exposed
(swim < 6 hours per week).
Suppose an odds ratio greater than 1.5 is
considered clinically important.
314
Section 8

(a) OR = 1.90 with CI (1.23, 2.92)
• p < 0.05 and conclusive.
• 1 is not contained in the CI, so there is
evidence of an association between
exposure and outcome.
• the CI is above 1 indicating harm.
(Swimming bad for teeth.)
• note we have not ruled out a non-clinically
important association
(b) OR = 1.69 with CI (0.83, 3.45)
• p > 0.05 and inconclusive.
• point estimate indicates possible clinically
important association but “protection”!! of
tooth enamel (rather than “harm”) is also
plausible.
(c) OR = 0.81 with CI (0.39, 1.70)
• p > 0.05, inconclusive.
• conclude no evidence of an association
even though CI includes clinically
important effects.
• the point estimate is in the “protection”
range (harm is above 1).
315
Section 8

(d) OR = 0.85 with CI (0.53, 1.37)
• p > 0.05, conclusive.
• point estimate in protection range and CI
excludes any clinically important harm.
(e) OR = 0.81 with CI (0.67, 0.97)
• p < 0.05 and conclusive
• CI excludes 1
• CI entirely less than 1, indicating benefit
from swimming
(f) OR = 1.23 with CI (1.03, 1.48)
• p < 0.05 and conclusive
• CI excludes 1
• CI entirely above 1, but excludes the
clinically important difference
• there is evidence of an association between
exposure to chlorinated water for more than
6 hours per week but the increased odds are
not clinically important.
(g) OR = 1.15 with CI (0.73, 1.80)
p > 0.05 and inconclusive. A clinically
important association is not ruled out.
Advice: Probably continue swimming.
316
Section 8

0 1 1.5 2 3 3.5
a
x
b
x
c
d
e
x
x
x
f
x
g
x
0 1 1.5 2 3 3.5
Notice that these confidence intervals are not
symmetric.
317
Section 8

A Problem when Contingency Tables are
combined
Example: A University has a Law School and a
Medical Sciences School with men and women
being admitted or declined admission as follows:
Admit Decline Total
Male 490 210 700
Female 280 220 500
Total 770 430 1200
Is there gender bias concerning admission (i.e.
is there an association between gender and
admission decision)
Expected frequencies under H 0 : no association are
Admit Decline Total
Male 700(770)
[250.8] 700
= 449.2
1200
Female [320.8] [179.2] 500
Total 770 430 1200
χ 2 =
(490 − 449.2)
449.2
2
+ … + … + … = 24.82
318
Section 8

with υ = 1 degree of freedom. Since critical
value at α = 0.01 level of significance is 6.635,
there is strong evidence of an association.
Inspection of the observed frequencies shows a
tendency to admit a higher number of men than
expected i.e. O 11 = 490 but E 11 = 449.2. This
means fewer women are admitted than expected
under equal opportunity. The admission patterns
for the two schools are also known as follows:
LAW SCHOOL MEDICAL SCIENCES
Admit Decline Total Admit Decline Total
M 480 120 600 M 10 90 100
F 180 20 200 F 100 200 300
Total 660 140 800 Total 110 290 400
The expected frequencies under H 0 are:
Admit Decline Admit Decline
M 495 105 M 27.5 72.5
F 165 35 F 82.5 217.5
For Law School χ 2 = 10.38**
For Medical Sciences School, χ 2 = 20.45**
319
Section 8

There is strong evidence of an association in both
schools.
HOWEVER, inspection of the observed counts
indicates a higher number of women than
expected are admitted to both schools.
For LAW, O 21 = 180 with E 21 = 165
For MEDICAL SCIENCES, O 21 = 100 with
E 21 = 82.5
This is the opposite conclusion to that when the
schools are combined. Is there discrimination
against men or women!!
This is known as Simpson’s Paradox.
The reason for this discrepancy is that more
women applied to the Medical Sciences school to
which it was more difficult to be admitted. The
final conclusion is therefore unclear.
Notice that there are essentially three factors of
classification here, and we have summed over
one of these factors, namely the “TYPE OF
SCHOOL”
320
Section 8

COMBINED
ADMIT DECLINE
Male 490 (449.2) 210 ( )
Female 280 (320.8) 220 ( )
LAW
MEDICAL
Admit Decline Admit Decline
M 480 (495) 120 ( ) M 10 (27.5) 90 ( )
F 180 (165) 20 ( ) F 100 (82.5) 200 ( )
(Expected numbers are in bold)
“Variable” 1 = GENDER
“Variable” 2 = ADMISSION DECISION
“Variable” 3 = SCHOOL TYPE
Note how careful we must be with such an
observational study which fails to recognise an
important “variable” (here school type).
This phenomenon can occur whenever we sum
over a classification in categorical data.
321
Section 8

REVIEW EXERCISES
1. A randomized double blind study (prospective) was set up to test for an association between
the use of aspirin and the incidence of fatal or nonfatal strokes in a five year period from the
start of the study. The results (Journal of the American Medical Association, 243: 661-669)
are summarised in the following contingency table:
Stroke No stroke
Placebo 45 2257
Aspirin 29 2238
(b)
(c)
Calculate and interpret the risk of stroke for people in the placebo group relative to the
aspirin group. Set up a 95% confidence interval for the relative risk. (3 marks)
The use of aspirin was felt to increase the occurrence of gastrointestinal irritation. In
the study, 229 of 2267 patients in the aspirin treatment suffered irritation as opposed to
22 of the 2302 in the placebo treatment. Calculate the relative risk of gastrointestinal
irritation for people in the aspirin group compared with those in the control. Set up a
95% confidence interval for the relative risk and interpret the result. (3 marks)
(d) Calculate the attributable risk for aspirin compared with control Set up a 95%
confidence interval for the attributable risk and interpret the result.
(3 marks)
3. Long-term Mobile Phone Use and Brain Tumour Risk.
Lonn et al (2005), American Journal of Epidemiology, 161: 526-535
Human exposure to radiofrequency has increased dramatically during recent years from
widespread use of mobile phones. If radiofrequency radiation has a carcinogenic effect, the
exposure poses an important public health problem, and intracranial tumours would be of
primary interest. Handheld mobile phones were introduced in Sweden during the late
1980’s. This case-control study was carried out to test the hypothesis that long-term mobile
phone use increases the risk of brain tumours.
(a)
(b)
This was a case-control study. Describe one advantage and one disadvantage of using a
case-control study instead of a cohort study to investigate the association between longterm
use of mobile phones and the risk of brain tumour.
The information is summarised below.
Brain Tumour (Outcome)
Mobile phone use Yes No Total
Never/rarely 155 275 430
Regularly 118 399 517
Total 273 674 947
(i) Calculate the odds ratio for the association between long-term mobile phone use
and the risk of brain tumour.
(ii) Interpret the odds ratio.
(iii) Calculate the 95% confidence interval for the odds ratio.
(iv) Interpret the confidence interval.
322
Section 8

SOLUTIONS
29
1. (b) Risk (aspirin group) =
2267
45
and risk (placebo group) = 2302
45 / 2302
Relative risk, RR = = 1.53
29 / 2267
The risk is 1.53 times greater for those in the placebo.
Also, s.e. (ln RR) =
1
45
−
1
2302
+
1
29
−
1
2267
= 0.236
and since ln(RR) = 0.424 the 95% confidence interval is
0.424 ± 1.96 (0.236)
or 0.424 ± 0.463
or -0.039 < ln(RR) < 0.887
Therefore 0.96 < RR < 2.43, taking exponentials
(notice that the null value for the relative risk is 1 hence no evidence against the null hypothesis)
(c) Irritation No irritation Total
Placebo 22 2280 2302
Aspirin 229 2038 2267
229 / 2267
RR =
22 / 2302
= 10.57
ln(RR) = 2.358
s.e. ln(RR) =
1 1 1 1
− + − = 0.221
229 2267 22 2302
The 95% C.I. for ln(RR) is 2.358 ± 1.96 (0.221)
That is, 2.358 ± 0.433
Giving 1.925 < ln RR < 2.791
Taking exponentials,
6.86 < RR < 16.30
The null value of equal risk is rejected.
The true relative risk of irritation if aspirin used is between 6.86 and 16.30
(d)
229 22
Attributable risk = − = 0.10101 – 0.00956 = 0.09145
2267 2302
Estimated standard error =
0.10101(0.89899) 0.00956(0.99044)
+ = 0.00665
2267 2302
The 95% C.I. for attributable risk is 0.09145 ± 1.96(0.00665)
or 0.091 ± 0.013
or 0.078 < AR < 0.104
Between 78 and 104 in every 1000 people have increased occurrence of gastrointestinal irritation
as a result of using aspirin.
3. (a) Advantage: A case control study is quick and cheaper since information on exposure and disease
status are obtained at same time. Brain tumours also are rare so number of participants for cohort
study would be large.
Disadvantage: Information collected likely to be affected by recall bias since events have already
occurred.
(b) (i) OR = (118/399)/(155/275) = 0.52
(ii) Those who use mobile phones have 0.52 times the odds of a brain tumour compared with those
who do not. [Protective effect from using mobile phones – the odds are 48% less for mobile
phone users compared with those who do not use mobile phones.]
(iii) ln(0.52) = –0.654
The 95% C.I. for ln(OR) is
1 1 1 1
− 0.65 ± 1.96 + + +
155 275 118 399
or –0.654 ± 0.284
or –0.938 < ln(OR) < –0.370
Therefore, 0.39 < OR < 0.69
(iv) 95% confident true OR between 0.39 and 0.69. The value (1) is excluded hence
chance is an unlikely explanation.
323
Section 8

324

SECTION 9
This section introduces the topic of Simple Linear Regression which sets out to fit a straight line
through what is called a scatter diagram. One purpose of this analysis is to establish whether one
predictor variable is influencing the outcomes of a response variable and also measuring the
magnitude of the effect of this predictor variable on the outcome. It is possible to use the fitted
straight line to make predictions.
Simple linear regression is also the first step in controlling for a confounder variable. This occurs
with the extension to multiple regression which will be considered in the next section.
Scatter Diagrams and Examples
Equation of Fitted Straight Line
Analysis of Variance for Regression Model
Confidence Interval for Slope
Confidence Interval for Prediction
Correlation as Measure of Linear Association
Review Exercises
325
Section 9

Regression Procedures Introduction
During the semester we have analysed data from
1. studies which have measured outcomes on
continuous scales [e.g. blood pressure; lung
capacity; cholesterol] resulting from different
treatments
2. studies which have measured binary
outcomes, establishing odds ratios and
relative risks as a result of exposure to
certain conditions. [e.g. effect of chlorine on
tooth enamel; effect of sun exposure on
melanoma]
In both cases there are potentially other variables
which have an effect and/or possible confounding
factors other than the treatments or exposures
which influence the outcomes.
We must allow for these confounders otherwise
invalid conclusions will be drawn about the real
effects of the treatments or exposures.
326
Section 9

Regression methods are used to introduce these
controls. We now develop:
1. Simple linear Regression (now)
• to describe the relationship between two
variables and test whether changes in an
outcome measure may be linked to
changes in the other variable.
• to enable the prediction of the value of
the outcome measure from the other
variable.
2. Multiple Regression (later)
• to identify the main factors influencing a
continuous outcome
• to adjust the means of outcomes for
confounders or other factors.
3. Logistic Regression (later)
• to identify the main factors influencing
binary outcomes and hence odds ratios
and relative risks
• to adjust odds ratios for confounding or
other factors.
Show Hans Rosling’s website gapminder.
327
Section 9

Example: Blood Alcohol Concentration in
mg/100mL and Body Mass in kg for 8 adults after
drinking 12 glasses of regular beer.
0.04
0.02
0.00
MASS (kg) BAC (mg/100mL)
55 0.140
85 0.102
69 0.120
65 0.126
80 0.106
90 0.092
67 0.128
73 0.120
BAC (mg/100mL)
0.14 X
X X
0.12
X
0.10
0.08
0.06
X
X X
MASS (kg)
50 60 70 80 90 100
Does BAC drop as Body Mass increases
Other variables which could be important are:
gender amount eaten alcohol level of the beer
Eventually we shall see how to determine which
of these may be important.
328
X
Section 9

BAC
X
X X
•
•
• •
X
X
X X
X
• •
•
•
MASS
• Women consistently above men
• Lines could be parallel
BAC
X
X X
X
X
X X
X
• •
• •
• •
• •
MASS
• Lines not parallel. (If low body mass, large
difference, if high body mass there is no
difference.)
329
Section 9

Example: Lung function in children as measured
by a lung capacity variable called FEV.
FEV
+
+ +
+
+
+
+
+
+
+ +
+ + + + + + +
+ + +
+
+ + + + +
+
+
+
+
+ +
+ + + +
+
+ + + +
+ +
+
+
+
+ +
+ + + + +
3 5 7 9 11 13 15 17 19 Age
FEV values are increasing as the children grow.
But now see the next two graphs.
330
Section 9

FEV
+
+ +
+
+
+
+
+
+
+ +
+ + + + + + +
+ + +
+
+ + + + +
+
+
+
+
+ +
+ + +
+
+ + + +
+ + +
+
+
+ +
+ + + + +
3 5 7 9 11 13 15 17 19 Age
• Once start smoking FEV is reduced for the smokers.
FEV
+
+ +
+
+
+
+
+
+
+ +
+ + + + + + +
+ + +
+
+ + + + +
+
+
+
+
+
+
+ + +
+
+ + + +
+ +
+
+
+
+ +
+ + + + +
Non-smoker
Smoker
Non-smoker
Smoker
3 5 7 9 11 13 15 17 19 Age
• This is more accurate as children may only begin
smoking at age 9 and the rate of increase is much
smaller with the non-parallel lower line.
• Multiple regression needed for this analysis.
331
Section 9

With a simple linear regression take one variable
as response and one variable as a predictor.
The response is plotted on the vertical Y axis.
The predictor is plotted on the horizontal X axis.
Equivalent terms for response and predictor:
⎧outcome
⎪
response = ⎨dependent variable
⎪
⎩(
Y - variable)
⎧explanatory variable
⎪
covariate
predictor = ⎨
⎪independent variable
⎪⎩
( X - variable)
Simple regression deals with the case where the
relationship is approximately a straight line.
Example: The values of a response variable (Y)
and the values of a predictor variable (X) are as
follows
X Y
100 39.7
200 51.1
300 49.9 The scatter diagram
400 69.8 below shows the
500 65.2 relationship between
600 65.1 Y and X.
700 80.7
332
Section 9

80
Y
X
70
60
X
X
X
50
X
X
40
X
100 200 300 400 500 600
700
Y increases as X increases. The question is
whether this apparent increase in Y is caused by
changing X, or has it been caused by some other
factor, or has it arisen by chance alone.
The values of X, the independent variable, are
known exactly (i.e. no error) whereas the values
of Y, the dependent variable, have some random
error associated with them.
The relationship between Y and X could be linear
so we attempt to “fit” a straight line through the
data. This line gives the predicted yields ŷ for
i
each value x i of X.
X
333
Section 9

80
Y
X
70
60
50
40
X
d X X
4
{
⎫
X ⎪
X
⎬ŷ
4
X
⎪
⎪⎭
100 200 300 400 500 600
⎫
⎪
⎪
⎬
⎪
⎪
⎪⎭
700
y
4
X
An attempt is made to minimise the differences d i
= y i – ŷ between the observed values (y
i
i ) and the
predicted values ( ŷ i
). The d i are positive for
points above the fitted line and negative for
points below the line. The expression ∑ n
i = 1
d
i
where there are n data points (i.e. the sample is of
size n) does not measure “fit” due to cancellation
of negative and positive values.
Therefore, minimise ∑ i=
1d 2 = ∑i=
y − y
i 1( ˆ ) .
i i
Suppose the straight line which does this has
slope “β 1 ” and intercept “β 0 ”. That is,
n
n
2
y
= β + β x
0 1
334
Section 9

The method of least squares finds the values of β 0
and β 1 which minimise
n
2
n
2
( y ˆ ) ( [
1 i− yi = y
1 i− β0+
β1xi])
i= i=
∑ ∑
The estimates of β 0 and β 1 are
0
ˆβ and ˆβ 1
which
turn out to be
ˆ
∑
n
i=
1
β
1
=
∑
( x − x)( y − y)
n
i
i=
1
i
( x − x)
ˆ β = y − ˆ β x
0 1
i
2
The line which best “fits” the data is
ŷ = ( y − ˆ β ˆ
1x)
+ β1x
= y + ˆβ 1(x – x )
⎡∑( x −x)( y −y) ⎤
y+ ⎥ ( x−x
)
⎢⎣
∑ ⎥⎦
i i
= ⎢
2
( xi
−x)
335
Section 9

Example:
x i y i (x i – x ) (x i – x ) 2 (y i – y )(x i – x )(y i – y )
100 39.7 –300 90000 –20.51 6153
200 51.1 –200 40000 –9.11 1822
300 49.9 –100 10000 –10.31 1031
400 69.8 0 0 9.59 0
500 65.2 100 10000 4.99 499
600 65.1 200 40000 4.89 978
700 80.7 300 90000 20.49 6147
2800 421.5 280000 16630
x = 400 y = 60.21
Therefore, ˆβ 1
= 16630/280000 = 0.059
0
ˆβ = 60.21 – 0.059 (400) = 36.61
giving ŷ = 36.61 + 0.059 x
To draw this line on the scatter diagram two
points are needed:
e.g. if x = 400, = 36.61 + 0.059 (400) = 60.21
if x = 100, ŷ = 42.51
336
Section 9

N.B. 1. In this situation we have regressed Y on
X.
This implies the X values are known without
error but the Y values are influenced by
random variation.
2. Numerically, we could regress X on Y. But
the “slope” of this regression is not the same
as that for Y on X. The reason is that now the
Y values are known exactly with the X values
influenced by random variation.
3. ŷ = y + ˆβ 1(x – x )
When x = x, yˆ
= y+ ˆ β1(0)
= y
This means that the point ( x,
y)
always lies
on the least squares straight line. i.e. the
regression line always passes through the
centre of the scatter diagram.
337
Section 9

4. We say the least squares line “fits” or
“models” the relationship between Y and X.
5. A straight line may give poor fit e.g.
Y
X
X
X X
X
X
X
X
X
X
X
X
X
X X
X
X
Here, it is not appropriate to use the line to
predict values of Y for given values of X.
The next step in our regression analysis is to
establish how well this fitted line is able to model
or explain the effect X has on Y; and also, if the
fitted line is used to make forecasts of the values
of Y, how accurate these forecasts turn out to be.
(We set up confidence intervals for these
forecasts.)
Definition: The value d i = y i – ŷ is called the
i
residual at the value x i of X. These residuals are
338
Section 9

important as they represent the error made when
using the line to make a forecast.
Analysis of Variance for a Regression Model
Y
y
The diagram shows that any numerical value y i
can be partitioned into three components as
follows:
That is,
any value,
d
i
⎧
= ( y − yˆ
)
i i ⎨
⎩
x
} 1
ˆ β ( x − x )
⎫
⎪
⎬y
⎪
⎭
y = y+ ˆ β ( x − x) + ( y − yˆ
)
i 1 i i i
x
x i
( x
y i = an overall average
+ an amount explained by a
predictor variable X
+ a residual (or random error)
i
i
,
y
i
)
Regression
Line
X
339
Section 9

The amount explained by the independent
variable X is called the regression effect. This is
also known as the explained component of the
outcomes y i .
The magnitude of the regression effect is related
to the slope of the line and the distance x i is away
from the overall mean x of the values x i .
The mean y is the overall average effect.
The term ( y − yˆ
) is the residual effect. This is
i i
also known as the unexplained component of the
outcomes.
Therefore,
data value = overall average effect
+ regression effect + residual (error)
effect.
= overall average effect
+ explained amount + unexplained
amount
340
Section 9

To illustrate, the example has x = 400,
y = 60.21 and ˆβ 1
= 0.059
x i y i = y + 0.059(x i – 400) + residual
100 39.7 = 60.21 + (–17.82) + (–2.69)
200 51.1 = 60.21 + (–11.88) + 2.77
300 49.9 = 60.21 + (–5.94) + (–4.37)
400 69.8 = 60.21 + 0.00 + 9.59
500 65.2 = 60.21 + 5.95 + (–0.95)
600 65.1 = 60.21 + 11.88 + (–6.99)
700 80.7 = 60.21 + 17.82 + 2.67
overall mean explained unexplained effect
common to effect. chosen to give
each data
equality.
value.
It is important to establish if the explained effect
has a much greater impact on the values y i than
the unexplained residual effect i.e. does the
regression effect explain more of the variation in
the y i values. It turns out that the total variation
in the y i values can be partitioned into an overall
mean component, a regression component and a
residual component as follows:
341
Section 9

[This page just for reference]
Total sum of Squares (SS) of y i values
= (39.7) 2 + (51.1) 2 + (49.9) 2 + (69.8) 2
+ (65.2) 2 + (65.1) 2 + (80.7) 2
= 26550.89
The overall mean SS
= (60.21) 2 + … + (60.21) 2 (7 times)
= 7(60.21) 2
= 25380.32
The regression effect SS
= (–17.82) 2 + (–11.88) 2 + … + (17.82) 2
= 987.70
The residual effect SS
= (–2.69) 2 + (2.77) 2 + … + (2.67) 2
= 182.87
Now notice that
26550.89 = 25380.32 + 987.70 + 182.87
i.e. Total SS = overall mean SS + regression SS
+ residual SS
342
Section 9

That is, the total variation is partitioned into these
components which should now be compared. But
the three component values cannot be compared
directly. Note that:
(i) There are seven data values y i hence seven
degrees of freedom.
(ii) One overall mean has one DF.
(iii) The seven regression values depend on the
one slope estimate 1
ˆβ , hence one DF.
(iv) The seven residuals have the remaining
7 – 2 = 5 DF.
The average or mean squares (MS) are then found
by dividing the sums of squares by the degrees of
freedom. These mean squares can be compared.
The procedure is summarised in the following
analysis of variance table:
SOURCE OF VARIATION SS DF MS
Overall mean 25380.32 1
Regression effect 987.70 1 987.70
Residual effect 182.87 (5) 36.57
Total 26550.89 7
343
Section 9

The average regression effect (or the average
effect of X on the Y values) far exceeds the
average residual effect (unexplained) since
987.70 far exceeds 36.57. But is this difference
large enough to be important. The question of
whether the average regression effect is large
enough is answered by defining F = 987.70/36.57
= 27.01 and testing this F-statistic for
significance by reference to the F-table as
follows: (note that the DF here are 1 and 5
respectively for numerator and denominator).
Since 27.01 > 6.608 there is evidence that the
regression (or explained) effect dominates the
residual (or unexplained) effect. Since the key
part of the regression effect is the slope 1
ˆβ , this
effectively means ˆβ 1
≠ 0 (or alternatively that
there is evidence that changes in the values x i of X
explain the variation in the values y i of Y (and this
dominates any left over residual or unexplained
effects).
344
Section 9

The F-distribution (Table in Appendix)
υ 1 = num DF
υ 2 = denom DF
α = 0.05 (say)
0
F
υ 1 ,υ 2
F
υ 1 1 2 3 4 … 60
υ 2
1 … … … … …
2
3
4
5 6.608 5.786 5.409 … … …
6
120 3.920 3.072 2.680 … … …
345
Section 9

Note: 1. The residual effect includes any
random error plus the effects of other
variables which may be affecting the
outcome Y values.
2. Computer software produces the analysis of
variance table directly.
3. It is a slightly modified form because the
overall mean effect is never used. Therefore,
this is subtracted (with appropriate changes
to the total SS and the degrees of freedom)
SOURCE OF VARIATION SS DF MS F
Regression effect 987.70 1 987.70 27.01*
Residual effect 182.87 (5) 36.57
Total (overall mean
removed)
1170.87 6
4. The “fitted” straight line should pass through
the middle of the scatter diagram, and hence
the residuals should take positive and
negative values as X increases. (This can be
checked by studying plots of the residuals
produced by the program.)
346
Section 9

5. For the validity of the F-test, residuals should
be approximately normally distributed. This
can also be checked by obtaining the normal
probability plot using the program.
Analyse > Regression > Linear with Y in
the Dependent Variable box and X in the
Independent Variable box produces the
following printout.
347
Section 9

A Confidence Interval for Slope of line.
Our sample of n = 7 produced an estimate
ˆβ = 0.059
1
Repeated samples of size n = 7 give values ˆβ 1
which follow a normal distribution (just the
Central Limit Theorem again).
If β
1
is the true slope of the regression line then
the standard error of β
1
is
σ
β
1
=
∑
n
i=
σ
e
( x )
1 i
− x
2
where σ 2 e is estimated from the data by the
formula.
s
2
e
n
∑i
1 ( − ˆ
= = y y
i
n−2
Notes
1. ( y ˆ
i− yi)
is the residual (or error) at the value
x i of X.
i
)
2
348
Section 9

2. The divisor is (n – 2) rather than (n – 1) used
in the calculation of an ordinary variance
because here two values “ β 0
” and “ β 1
” are
estimated from the data and used to find the
ŷ from which the deviations are measured.
i
[For an ordinary variance,
only x is estimated.]
2
2 ∑ ( x − x)
s =
i
,
n −1
The estimated standard error of the slope of the
regression line is
s
β
1
=
∑
n
i=
s
e
( x )
1 i
− x
2
Therefore, the 95% confidence interval for β
1
is
ˆ β ± t
1 n−2
∑
n
i=
1
s
e
( x − x)
i
2
Notes.
(1) There are υ = n – 2 degrees of freedom for
use with the t-table.
349
Section 9

(2) If σ e was known exactly (which it never is)
the 95% confidence interval would be
ˆ β ± 1.96
1
∑
σ
e
( x − x)
i
2
.
(3) In practice, σ e is always estimated by
s
e
=
∑ ( y
i
− yˆ
n − 2
i
)
2
(4)
2
s is just the residual mean square and this
e
can be read directly from the analysis of
variance.
Example
Refer to the earlier data which gave
2
∑ ( x i
− x) = 280000, 1
= 0.059 and
yˆ = 36.6 + 0. 059x
i
i
350
Section 9

x i y i ( y − yˆ
)
i
i
( y −
100 39.7 -2.69 7.24
200 51.1 2.77 7.67
300 49.9 -4.37 19.10
400 69.8 9.59 91.97
500 65.2 -0.95 0.90
600 65.1 -6.99 48.86
700 80.7 2.67 7.13
182.87
i
yˆ
i
2
)
Therefore,
Residuals (see earlier)
Residual
sum of squares
2 182.87
s = = 36.58 (the residual mean square)
e 7 − 2
with
n – 2 = 7 – 2 = 5 D.F. giving
t 5 = 2.571 for 95% confidence.
The standard error of the slope is estimated to be
s
e 36.58
= = 0.0114
2
∑ ( x − x)
280000
i
351
Section 9

The 95% confidence interval is 0.059 ±
2.571(0.0114) or 0.059 ± 0.029
Hence 0.030 < β 1 0
0
X
As X changes, the values
of Y tend to show an
increasing trend with
random variation about
the trend line.
Example
A test has been designed to measure patient stress
level (X). Blood pressure (Y) is recorded for
different stress levels.
352
Section 9

Stress (X) 55 94 64 73 96 86
Blood Pr. (Y) 72 91 76 78 94 81
These data give x = 78; y = 82;
2
∑ ( x i
− x) = 1394 and ∑ ( x − i
x)( yi
− y)
= 686.
Find the least squares line, 95% confidence
interval for slope and test the research proposal
that higher stress results in higher blood pressure
levels.
353
Section 9

Solution:
ˆ ∑( xi
− x)( yi
− y) 686
β1 = = = 0.492
∑
2
( xi
− x) 1394
yˆ = y+ ˆ β ( x− x) = 82 + 0.492( x−
78)
∴
1
Suppose a computer analysis gives the analysis of
variance as follows:
SOURCE OF VARIATION SS DF MS F
Regression effect 337.59 1 337.59 33.41
Residual effect 40.41 4 10.10
2
2 ∑ ( y − yˆ
) 40.41
Then s =
i i
= = 10. 10
e n − 2 4
giving s e = 3.178 as residual standard deviation.
For 95% confidence, t 4 = 2.776 and standard error
3.178
of slope = = 0.085.
1394
The 95% confidence interval is
0.492 ± 2.776(0.085)
It follows that 0.256 < ˆβ 1
< 0.728
The test has the p-value less than 0.05.
354
Section 9

Confidence Interval for Prediction using a
Regression Line
The prediction value at value x i of X is found by
substituting the value x i in the regression equation
e.g. For our data, ŷ = 36.6 + 0.059x
When x i = 750, ŷ = 36.6 + 0.059(750)
i
= 80.85
But what error is associated with this prediction
At value X = x k say the estimated standard error
of the prediction is
s
1 ( x − x)
2
k
yˆ = se
1+ +
2
n ( xi
− x)
∑
where s e is the residual standard deviation.
But s e = 36 . 58 = 6.05 (see ANOVA table)
∴
s y ˆ
2
1 (750 − 400)
= 6.05 1+ + = 7.604
7 280000
355
Section 9

The 95% confidence interval is
yˆ ± t s where t 5 = 2.571
That is 80.85 ± 2.571(7.604)
5
yˆ
Therefore, 61.30 < ŷ < 100.40
750
where ŷ is the prediction at x k = 750.
750
Notes
(1) R-cmdr (and other packages) give this
interval when requested.
(2) A graph showing the confidence bands
around the regression line can also be
produced as follows
(3) Essentially the confidence interval for the
prediction involves line error and natural
variation to predict a data point.
356
Section 9

Y
Prediction Interval
X
EXAMPLE: 2003 EXAM
The data for this question are a sample of 100 low
birth weight infants. Measurements of systolic
blood pressure (sbp) and values of gestational age
(gestage) are recorded. The following table
shows the layout of the data along with the results
of some calculations using the 100 data values.
357
Section 9

sbp
(Y mm Hg)
gestage
(X weeks)
43 29 y= 47.31
51 31 x = 28.89
42 33
2
∑ ( i
− x)
= 635.69
39 31
2
∑ ( i
− y)
= 15222.24
40 33 ∑ ( x i
x)( yi
− y)
= 806.31
50 28
(a) (4 marks) Using systolic blood pressure as
the response and gestational age as the
predictor variable, compute the least squares
regression line. Interpret the slope of this
regression line.
(b) (5 marks) The standard deviation of the
sample points about the regression line in (a)
is s e = 3.47. Obtain an estimate for the
standard error of the slope of the regression
and hence set up a 95% confidence interval
for the slope of the regression line. State
whether you would reject the null hypothesis
that the true slope is equal to 0.
358
Section 9

(c) (3 marks) What is the predicted systolic
blood pressure for a low birth weight infant
whose gestational age is 31 weeks
Construct a 95% confidence interval for the
prediction.
(d) (1 mark) The value of the coefficient of
determination is R – Sq = 67%. Interpret this
value. (discussed next lecture)
(e) (3 marks) What conclusions would you draw
from the two residual plots below arising
from the fitted regression in (a)
359
Section 9

SOLUTION
(a) ˆβ 1
= 806.31/635.69 = 1.27
0
ˆβ = 47.31 – 1.27(28.89) = 10.62
ŷ = 10.62 + 1.27x
For infants with gestation age one week
higher, the model predicts sbp increases by
1.27 mmHg.
360
Section 9

(b) Estimated standard error
= 3.47 / 635.69 = 0.138
95% C.I. is 1.27 ± 1.98(0.138)
giving 1.27 ± 0.273
or 1.00 < ˆβ 1
< 1.54
The confidence interval excludes zero (pvalue
< 0.05) hence reject null hypothesis.
(c) Prediction = 10.62 ± 1.27(31) = 49.99
95% C.I. is
49.99 ± 1.98(3.47)
giving 49.99 ± 6.92
or 43.07 < ŷ
31
< 56.91
1 (31−
28.89)
1+ + 100 635.69
(d) 67% of the total sum of squares of the sbp
values is explained by changes in the number
of weeks of gestation. (Alternatively, 67% of
the variation in the sbp values is explained.)
(Discussed next lecture.)
(e) Variation about the fitted line is constant for
different gestation times. The residuals
appear close to pa normal distribution except
for a possible outlier at x = 29.
2
361
Section 9

Correlation
The correlation coefficient is a measure of linear
association. The Pearson correlation coefficient r
is defined
r
=
∑
∑
n
( x −x)( y − y)
i=
1 i i
n
2
n
2
( x ) ( )
i 1 i−x ∑ y
i 1 i−
y
= =
This measures the ‘strength” of linear association
between X and Y (as we shall now see). Recall
that the regression line passes through the point,
x,
y .
( )
y
+
2
+ + +
+ + +
+ + +
+ + + +
+ + + +
3
x
+
+
+
+
+
+
+
+
4
1
X
Section 9
362

The denominator in formula for r is always
positive. In quadrant 1 , x i – x > 0 and
y i – y > 0 meaning ( xi− x)( yi− y) > 0. In
quadrant 3 , ( xi
− x)
< 0 and y i – y < 0 giving
( xi−x)( yi− y) > 0. In quadrant 2 and 4 ,
( x −x)( y − y) < 0.
i
i
Therefore, r is large and positive if points mainly
in quadrants 1 and 3 ; it is large and negative if
points in quadrants 2 and 4.
Y
Y
+ +
+ + + +
+
+
+ + +
+ +
+
+ +
+
+
+ +
+ +
+
X
+ +
+
+ + +
+
+
+
+
+
+ +
+ +
X
(i)
(ii)
In case (i) the contribution is equal from each
quadrant, the contributions cancel, and therefore r
= 0. i.e. there is no relationship between Y and X.
In case (ii) there is again cancellation and r = 0,
but here there is a strong relationship between Y
and X but it is non-linear.
363
Section 9

therefore measures the strength of the linear
association between X and Y. But we must be
careful as r = 0 in the following case (iii) where
β 1 = 0. In fact r is directly related to β 1 and zero
if β 1 is zero.
Y
+ +
+ + + + + + +
+ + + +
+ + + +
(iii)
X
Example: A researcher investigates the
relationship between reading and spelling tests
administered to nine students
Student 1 2 3 4 5 6 7 8 9
X (spelling) 52 90 63 81 93 51 48 99 85
Y (reading) 56 81 75 72 50 45 39 87 59
364
Section 9

2
( y i
− y) ( x − x)(
y − y)
x i y i ( x )
2
i
− x
i i
52 56 … … …
90 81 … … …
63 75 … … …
81 72 … … …
93 50 … … …
51 45 … … …
48 39 … … …
99 87 … … …
85 59 … … …
3220.2225 2258.0001 1718.6665
x = 73.55 y = 62. 67
1718.6665
r = =+ 0.6374
3220.2225(2258.0001)
But what does this mean
Very strong correlation:
x
x x
x x
x
x x x
x xx
Near +1
x
x
x
x
Near –1
x
x
x x x x
x x
Section 9
365

Smaller
x
x
x x x
x x x
x
x
x
x
x
x
x
x
x
x
x
+ 0.7
x
x
x
x
x x
x
– 0.7
x
x
x x
x Ho x
lla
nd
x x
x x
Very small
x x
x
x x
x x x
x
x
x
x x
x
x x x x x
x x x
x x
x + 0.2
x
– 0.2
x
x
x x x x
x x x x
x x
x x x x x
x x x x
x
Notes:
1. The largest value of r turns out to be +1. In
this case all points lie on a straight line in
quadrants 1 and 3 . This implies perfect
positive linear association. i.e. as X increases,
Y increases in the same ratio (if the increase
of X is doubled, the increase in Y would also
be doubled).
366
Section 9

2. r = –1 is smallest value which implies perfect
negative linear association when all points lie
in quadrants 2 and 4. i.e. as X increases, Y
decreases in same ratio.
3. |r| > 0.7 implies strong linear relationship.
|r| < 0.3 implies negligible linear relationship.
4. The correlation coefficient is an index. It
does not depend on the units of either X or Y.
(numerator and denominator in same units)
5. r is called the Pearson Correlation
Coefficient.
6. An important correlation does not imply a
causal link between the two variables. (The
correlation is often caused by the effect of a
third variable influencing both X and Y).
e.g. smoking and lung cancer incidence
correlated – not smoking causing lung
cancer.
7. If r is large, a regression line will fit the data
well.
367
Section 9

8. If r 2 gives the fraction of variability in the Y
values associated with the predictor variable
X.
e.g. In the example, r = 0.6374 so 0.406
40.6% of the variability in Y is explained
by changes in X.
That is,
2 SS(Regression)
r =
SSTotal (Reg+Resid)
for a simple linear regression.
368
Section 9

Some examples on correlation and association discussed in lectures.
Correlation measures association but association is not the same as causation.
Example: For school children, shoe size is strongly correlated with reading skills.
Learning new words does not make the feet get bigger.
Instead, there is a third factor, age. As children get older, they learn to read better and they outgrow
their shoes.
Age is a confounder. Here, this confounder is easy to spot. Often this is not so easy. The
arithmetic of the correlation coefficient does not give protection against third factors.
Example: Education level and unemployment.
In the Great Depression (1929 – 1933), better educated people had shorter spells of unemployment.
(Education level and days unemployed were very highly correlated: negatively as more education
associated with less days unemployed). Does education protect you against unemployment.
Discussion:
Perhaps, but the data were observational. Age is a confounding variable. Younger people were
better educated as education level had been increasing over time. (It still is!!)
Employers seemed to prefer younger job seekers.
Controlling for age made the effect of education on unemployment much weaker.
Example:
In countries where people eat lots of fat, rates of breast and colon cancer are high. This correlation
is often used to argue that fat in the diet causes cancer. How good is this evidence
Death
Rate
(per 100000)
25
20
15
10
+
+
+
+
+
+
+ +
+
+ Finland
+
+ Spain
+ Holland
+ UK
+ Denmark
+ + NZ
+
5
+
+ Japan
+ Sri Lanka
Thailand
25 50 75 100 125 150 175
Fat intake per capita per day (grams)
369
Section 9

Discussion: There is a very high correlation as shown by the scatter diagram which is very
elongated. If fat in diet causes cancer, then the points should slope up as shown. So the diagram is
some evidence for the theory. But the evidence is weak.
For example, countries with lots of fat in diet also have lots of sugar, and a similar plot for sugar
would be found.
As it turns out, fat and sugar are relatively expensive. In rich countries people can afford to eat fat
and sugar rather than starchier grain products.
Some aspects of diet in these countries or these life-style factors probably do cause certain kinds of
cancer. Epidemiologists can identify only a few of these factors with confidence. Fat is not among
them.
Example: Ultrasound and low birthweight.
Babies can be examined in the womb using ultrasound. Several experiments on lab animals have
shown ultrasound exams can cause low birthweight. If true for humans, there are grounds for
concern. Scientists at Johns Hopkins Hospital in Baltimore ran an observational study to find out.
Babies exposed to ultrasound differ from unexposed babies in many ways beside exposure; this
investigation was only an observational study.
The scientists found a number of confounding variables and adjusted for them. There was still an
association. Babies exposed to ultrasound in the womb had lower birthweight, on average.
Is this evidence that ultrasound causes lower birthweight
Discussion: Obstetricians suggest ultrasound examination when something seems wrong. The
investigators concluded that the ultrasound exams and low birthweights had a common cause –
problem pregnancies.
Later, a randomized controlled experiment was carried out to get more definite evidence. If
anything, ultrasound was protective.
Journal of Obstetrics and Gynaecology. Volume 71 (1988) pp 513-517
Also Lancet (1988) pp 585-588
370
Section 9

REVIEW EXERCISES
1. Physical fitness testing is an important aspect of athletic training. A common measure of the magnitude of
cardiovascular fitness is the maximum volume of oxygen uptake during a strenuous exercise. A study was
conducted on 18 middle-aged men to study the influence of the time that it takes to complete a 2-mile run.
The oxygen uptake measure was accomplished with standard laboratory methods as the subjects performed
on a motor driven treadmill. The data (Ribisl et al. Journal of Sports Medicine, 9: 17-22) are below:
Maximum Volume of O 2 (Y) Time in Seconds (X)
42.33
53.10
918
805
Data summary
x = 831.40
42.08 892 y = 47.67
42.45 968
42.46 907 2
∑ ( i
− x)
= 160613.28
49.92 743
36.23
49.66
1045
810
∑ ( x i
x)( yi
− y)
= -8698.33
41.49 927 2
∑ ( y ˆ
i
− yi)
= 55.25
46.16 813
48.18 858
51.81 760
53.28
53.29
747
743
47.18 803
56.91
47.80
683
844
53.69 700
(a) Use the data summary to find an estimate for the equation of the least squares regression line of Y on X.
(2 marks)
(b)
(c)
(d)
Find an estimate for the standard error of the slope of the regression line and set up a 95% confidence
interval for the slope of the regression line.
(4 marks)
What does the confidence interval in (b) tell you about the effect of time (X) on maximum volume of
oxygen uptake (Y).
(1 mark)
If a man in this age group takes 50 seconds longer to run the 2-mile distance what is the change in his
maximum volume of oxygen update Write down the 95% confidence interval for this change using the
result from (c).
(2 marks)
(e) Set up a 95% confidence interval for the maximum volume of oxygen uptake for a man who takes 11
minutes (660 seconds) to complete a two mile run.
(3 marks)
371
Section 9

SOLUTIONS
−8698.33
1. (a) b YX = =− 0.054
160613.28
ŷ = 47.67 – 0.054(x – 831.4)
= 92.566 – 0.054x
(b) Estimated standard error =
∑
se
( x − x)
i
2
where
55.25/16
That is, standard error = = 0.004637
160613.28
A 95% confidence interval for the true slope is
–0.054 ± t 16 (0.004637) where t 16 = 2.120
That is, –0.054 ± 0.0098
giving – 0.064 < β YX < –0.044
y ˆ
2 i
− yi
e
=
s
∑
( )
n − 2
(c) The maximum volume of oxygen uptake is smaller for men who take longer to run 2
miles.
2
(d)
Oxygen uptake reduces by 50(0.054) = 2.7 units.
The 95% confidence interval extends from 50(0.064) to 50(0.044) or a reduction from
2.2 to 3.2 units.
(e) When x = 660 seconds, ŷ = 92.566 – 0.054(660) = 56.93
The 95% confidence interval is
1
56.93 ± t s 1+ +
( x − x)
2
k
16 e
2
n ( xi
− x)
∑
That is, 56.93 ± 2.120
or 56.93 ± 4.38
52.55 < ŷ
660
< 61.31
1 (660 − 831.4)
55.25/16 1+ + 18 160613.28
2
372
Section 9

SECTION 10
Multiple regression models and logistic regression models are introduced in this section. In the case
of ordinary multiple regression the response or outcome variable is on a continuous scale whereas
in the case of a logistic regression the outcome measure is binary taking therefore only two possible
values interpreted as success versus failure.
The models allow us to identify those variables which have an effect on the outcomes and those
variables which do not.
Adding additional variables leads to adjusted values for estimated parameters and it is this that
allows us to control for confounding.
The Multiple Regression Model
R-cmdr Printout for Multiple Regression
Dummy Variables
Checking Model Fit
Parallel Regression Lines and Analysis of Covariance
Binary Outcomes and Logistic Regression
373
Section 10

Multiple regression
• Simple linear regression (SLR) allowed us to
assess the effect of a single independent
variable (X) on a response variable (Y).
• But what do we do if we think that the
response may change according to more
than one independent variable
• SLR regression can be extended.
• Multiple regression allows us to assess the
effects of several independent variables on
the outcome variable and it allows the
prediction of a response from the values of
several independent variables.
• In multiple regression, there is a single
dependent (outcome) variable and two or
more independent (explanatory, predictor)
variables or covariates.
• The predictor variables can be:
Continuous (e.g. blood pressure, height)
Categorical – binary (e.g. sex)
374
Section 10

• The type of multiple regression that is
performed depends on the data type of the
outcome variable.
• If the outcome variable is continuous, we use
multiple linear regression.
• If the outcome variable is binary, we use
multiple logistic regression.
The possible applications of multiple
regression include:
1. Adjusting for the effect of confounding
variables.
2. Establishing which variables are important in
explaining the values of the outcome
(response) variable.
3. Predicting values of the outcome variable.
375
Section 10

4. Describing the strength of the association
between the outcome variable and explanatory
variables and reducing residual variation by
introducing further effects as predictor
variables.
Multiple regression investigates and tests the joint
effect of all predictors on the outcome variable as
well as the measurement of individual effects of
each predictor.
Example: Predict lung capacity from age, sex
and height of patient.
Lung capacity itself is difficult to measure. For
heart lung transplants to have best chance of
success it is desirable to have donor and recipient
lungs of similar size.
376
Section 10

The multiple linear regression model:
y = β + β x + β x + β x + … +
0 1 1 2 2 33
error
For simple linear regression the model is:
y
= β + β x+
ε
0 1
The fitted straight line then becomes
ŷ
= ˆ β + ˆ β x
0 1
where
0
ˆβ and ˆβ 1
are chosen to minimise the sum
of the squared errors (residuals).
In the case of two explanatory variables, the
multiple linear regression model can be written in
the following form:
y = β + β x + β x + ε
0 1 1 2 2
where ε is the residual (including random error)
with mean of zero (for all data values i) and
constant variance.
377
Section 10

The fitted regression equation is
ŷ = ˆ β + ˆ β x + ˆ β x
0 1 1 2 2
The estimates ˆ β0, ˆ β ˆ
1
and β
2
are found from the
data in such a way that the sum of the squared
residuals (errors), that is
y − ( β + β x + β x ) , is minimised.
∑
[ ] 2
i
0 1 1 2 2
The results are complicated and statistical
software is always used for calculations.
378
Section 10

Example
For lung transplantation it is desirable for the
donor’s lungs to be of a similar size as those of
the recipient. Total lung capacity (TLC) is
difficult to measure, so it is useful to be able to
predict TLC from other information. The
following table shows the pre-transplant TLC of
32 recipients of heart-lung transplants, and their
age, sex and height
Age Sex Height(cm) TLC(1) Age Sex Height(cm) TLC(1)
1 35 F 149 3.40 17 30 F 172 6.30
2 11 F 138 3.41 18 21 F 163 6.55
3 12 M 148 3.80 19 21 F 164 6.60
4 16 F 156 3.90 20 20 M 189 6.62
5 32 F 152 4.00 21 34 M 182 6.89
6 16 F 157 4.10 22 43 M 184 6.90
7 14 F 165 4.46 23 35 M 174 7.00
8 16 M 152 4.55 24 39 M 177 7.20
9 35 F 177 4.83 25 43 M 183 7.30
10 33 F 158 5.10 26 37 M 175 7.65
11 40 F 166 5.44 27 32 M 173 7.80
12 28 F 165 5.50 28 24 M 173 7.90
13 23 F 160 5.73 29 20 F 162 8.05
14 52 M 178 5.77 30 25 M 180 8.10
15 46 F 169 5.80 31 22 M 173 8.70
16 29 M 173 6.00 32 25 M 171 9.45
379
Section 10

Step 1: First look at some plots in order to gain an
understanding of the data
1. Plot each predictor variable against the
outcome.
Relationship between total lung capacity and age
Total lung capacity(l)
2 4 6 8 10
10 20 30 40 50
age(yrs)
It appears that total lung capacity is not affected
by age.
380
Section 10

It appears total lung capacity increases as height
increases.
The effect of sex is not clear.
381
Section 10

Step 2: Fit (in R-cmdr) Simple Linear
Regression models for each predictor variable.
1. Age alone:
TLC= 5.07 + 0.036 age
If age increases by one year, TLC increases
by 0.036 litre (which is not significant if
tested).
2. Height alone:
TLC= -9.74 + 0.095 x height
If height increases by 1 cm, TLC increases
by 0.095 (which is significant if tested).
382
Section 10

Step 3: Fit (in R-cmdr) Multiple Linear
Regression Model.
3. Age and height
383
Section 10

From regression equation for the model including
age and height, the predicted TLC for someone
aged 25 and with a height of 160 cm is:
TLC = -11.218 – 0.030 × 25 + 0.108 × 160
= 5.322 litres
Regressions which include binary (e.g. sex)
predictor variables
The predictor variable, SEX, has two categories
only, female and male. We need a technique for
including such binary variables in the regression
models.
Define a dummy variable (D) as follows:
D
=
⎧0 if
⎨
⎩1if
female
male
If there are two other predictors X 1 and X 2 then
we fit the model
y = β
0
+ β
1
x
1
+ β
2
x
2
+ β
3
d + ε
384
Section 10

The fitted equation is therefore
ŷ = ˆ β + ˆ β x + ˆ β x + ˆ β d
0 1 1 2 2 3
We find estimates ˆ β0, ˆ β ˆ ˆ
1, β2 and β
3
by minimising
the squared residuals as before (using the
computer).
4. Model with age, height and sex.
385
Section 10

Model interpretation:
* TLC decreases with increasing age.
For a person 10 years older, the predicted TLC will
be 0.25 litres lower.
* TLC increases with increasing height.
For a person 10 cm higher, the predicted TLC will
be 0.9 litres higher.
* Males have higher TLC than women:
For males, the predicted TLC is 0.697 litres higher
than for women with same age and height.
females, sex = 0
so TLC = –8.54 – 0.025age + 0.0895height + 0.697 × 0
males sex = 1
so TLC = 8.54 – 0.025age + 0.0895height + 0.697 × 1
Therefore, the difference in average TLC between
males and females is 0.697.
Note: compare this to the crude difference in mean
TLC between males and females
386
Section 10

It is 6.98 – 5.20 = 1.78 litres
where 6.98 and 5.20 are male and female
averages
Some of this difference between males and
females can be explained by differences in age
and height.
Overall, how well does the model fit
The analysis of variance is
1. The regression effect has 3 degrees of
freedom since there are 3 predictor variables
in the model.
2. The ANOVA table shows the ‘usefulness’ of
the linear regression model – we want the p-
value to be < 0.05.
Here, p-value = 0.000, implying that at least
one of the explanatory variables has a
significant linear relationship with the
outcome variable.
387
Section 10

3. The strength of the relationship between
TLC and the three predictors can be
expressed as the proportion of the total SS
explained by the regression equation.
The coefficient of determination is:
R 2 = 44.305/81.712 = 54.2%
Thus, 54.2% of the total sum of squares
(variation) is explained by age, height and sex
together.
Notice how the value of R 2 has increased from
0.510 or 51.0% to the value of 0.542 or 54.2%
when all three predictor variables are included.
388
Section 10

Are all three variables needed in the model
There are 3 ways of evaluating the importance of
a variable in the model:
1. Construct a test of the null hypothesis that
the regression coefficient = 0.
2. Calculate a 95% confidence interval for the
regression coefficient.
Note: Regardless of whether an additional
variable is significant or not the real point
at issue is that the other regression
parameters are adjusted for the influence
of these new confounding variables to
produce adjusted test or confidence
intervals.
Model is
TLC = β
0
+ β
1
age + β
2
height + β
3
sex + ε
giving R-cmdr printout as follows:
389
Section 10

Std Error is the standard error of the
corresponding regression coefficient. (See how
the coefficients of age and height change when
allowance made for sex).
1. Test of the hypothesis H 0 : β 3 = 0
Is the variable sex an important predictor in the
model
T
ˆ β
3
− 0 0.697 − 0
= =
s.e.( ˆ β ) 0.499
3
= 1.396
p – value = 0.174. There is no evidence sex is
important in predicting TLC – the coefficient is
not significantly different from 0.
(Note: the t-test has 28 degrees of freedom, the
DF of the residual (error) effect).
390
Section 10

Test of the hypothesis H 0 : β 1 = 0
Age: t = –0.025/0.024 = –1.063
with 28 degrees of freedom (Residual DF)
p-value = 0.297
No evidence age affects TLC
Test of the hypothesis H 0 : β 2 = 0
Height: t = 3.647 (p-value = 0.001)
Strong evidence height is important
in predicting TLC.
2. Calculating a confidence interval for a
regression parameter
A true parameter β
i
is estimated by ˆi β .
For sex, the parameter estimates the difference in
average TLC between males and females after
taking into account age and height.
The C.I. for ˆi β is: ˆ β ˆ
i± t28se
..( β
i)
391
Section 10

For sex, this becomes
0.697 ± t 28 (0.499)
where t 28 = 2.048 for 95% confidence interval.
That is 0.697 ± 1.022
That is (–0.326, 1.720)
This includes zero so there is no evidence of
difference in average TLC between men and
women.
Note:
The above interval is called an adjusted confidence
interval. Recall unadjusted difference in means was
-1.78. The unadjusted 95% confidence interval for
the true difference in mean TLC between males and
females is (-2.77, -0.79).
Adjusting for age and height has removed the
statistically significant association between sex and
TLC.
392
Section 10

95% confidence interval for coefficient of age:
–0.0250 ± t 28 (0.024)
or ( –0.073, 0.023)
95% confidence interval for coefficient of height
0.0895 ± t 28 (0.025)
or (0.039, 0.14)
Note the correspondence between the 95%
confidence interval and the t-test carried out at
the 0.05 (2 sided) significance level.
393
Section 10

Note:
(i) The effect of sex was contained in the
residual when TLC was expressed in terms of
age and height only. The effect of the
residual was therefore greater.
(ii) The real effect of interest can be hidden by
residual variability - reducing this residual
variability by including more predictors in
the model can improve the analysis (and
therefore the study). The p-values associated
with hypothesis tests for the parameters of
interest will generally be smaller.
(iii) Confounders can affect the parameter
estimates of the predictor variables of interest
as well as the residual variability. Therefore
including confounders in the model is
important for obtaining valid estimates of the
coefficients of interest regardless of the
reduction in residual variability.
394
Section 10

Checking the fit of the model
We do not expect our model to be correct. We
want it to capture the important aspects of the
process under investigation, but also to simplify
things enough to aid understanding. Choosing an
appropriate model is a complex art which is
covered more fully in higher level courses on
regression. Here we consider some basic
principles.
Rule of thumb:
We should not perform a multiple linear
regression analysis if the number of variables in
the model is greater than the number of
individuals divided by 10.
Residual plots
1. The residuals associated with each data value
should be normally distributed with mean = 0
and constant variance. (In R-cmdr we can
save the residuals for subsequent plotting e.g.
normal probability plot).
395
Section 10

2. The printouts also identify any unusual data
point which has a very large residual. The
residuals can be standardised to have mean
zero and standard deviation one. Hence we
can see clearly the unusual cases. (One of
the options in R-cmdr is to save the
standardised residuals).
(1) Checking the normality assumptions of
the residuals.
The matching histogram will present the usual
bell-shaped pattern for the 32 residuals.
396
Section 10

The points in the normal P-P plot lie along a
straight line, confirming the distribution of the
residuals is close to normal.
Two extreme points correspond to:
i) female, aged 20, height 162 cm. Predicted
value from model is 5.46 and actual TLC is
8.05.
ii) male, aged 25, height 171. Predicted TLC
from model is 6.84, actual TLC is 9.45
(2) Plot of residuals vs independent variables
Residuals versus age plot
397
Section 10

This plot identifies the negative residuals for the
people under 20 years and also shows the two
large outliers. Otherwise the plot is reasonably
random about zero.
Residuals versus height plot.
Again the plot has negative residuals for the
shorter people and identifies the two large
outliers. These plots indicate special thought
should be given to whether the young people
should be retained in the model.
398
Section 10

Analysis of Covariance
This analysis uses a multiple regression to
compare simple regressions coinciding with the
categories of a qualitative explanatory variable.
Example: A study investigates the effect of a
treatment for hypertension on systolic blood
pressure (BP) compared with a control treatment.
Age for all patients also known and it was
thought that age might confound the differences
in BP between the groups.
TREATMENT CONTROL
BP(Y) AGE(X) BP(Y) AGE(X)
120 26 109 33
114 37 145 62
132 31 131 54
130 48 129 44
146 55 101 31
122 35 115 39
136 40 133 60
118 29 105 38
Control mean = 121.00 mm (of mercury)
Treatment mean = 127.25 mm (of mercury)
399
Section 10

But note:
average age of control group = 45.13 years
average age of treated group = 37.63 years
[A] First, an ordinary unpaired t-test will be
performed on the BP(Y) values using the
pooled variance of the Y values.
Analyze > compare Means > Independent-
Samples t-test
y is the test variable and d is the grouping
variable. d is 0 for control and 1 for
treatment.
There is no evidence of a difference between
the two means as t = -0.932. The 95%
confidence interval for μ − μ is (-8.1,
T C
20.6) which includes 0 confirming no
evidence of a difference between the means.
Also p-value = 0.367
400
Section 10

At this stage the ages have been ignored.
Age could be increasing the residual
variation hiding the true treatment difference.
i.e. Age could be a confounder.
[B] Second, a regression analysis of Y on d is
performed where d = 0 for the control and d
= 1 for the treatment.
Analyze > Regression > Linear
(Again, the 16 Y values are in one column
and the values of d in a second column)
The estimated regression equation is
ŷ = 121 + 6.25d
401
Section 10

The estimated coefficient for d is 6.25 with a
standard error of 6.708. Note that when d = 0,
ŷ = 121.00 and when d = 1, ŷ = 127.25 so
coefficient of d is the difference between the
two means. The 95% confidence interval for
treatment difference is
6.25 ± t 14 (6.708) where t 14 = 2.145
giving 6.25 ± 14.39 or (-8.14, 20.64)
as before. This regression is equivalent to the
unpaired t-test. The age variable effect remains
hidden in the residual.
Note: The Confidence Interval can also be obtained
on the printout if requested.
402
Section 10

[C] Third, a regression analysis of Y on X and D
together is performed where D = 0 for control,
otherwise 1.
Analyze > Regression > Linear
(Values of X are now in a third column)
The estimated regression equation is
ŷ = 73.9 + 1.04x + 14.1d
The estimated coefficient of d is now 14.082
with a standard error of 3.818. The coefficient
of d represents the difference between patients
of the same age, one in the control and one in
the treated group.
403
Section 10

e.g. Let X = x k be age of two such patients.
Then ŷ T – ŷ C = (73.9 + 1.04 x k + 14.1)
– (73.9 + 1.04 x k + 0)
= 14.1
The 95% confidence interval for the difference
is
14.082 ± t 13 (3.818) where t 13 = 2.160
giving 14.082 ± 8.247
or (5.84, 22.33)
Now there is evidence that the treatment raises
blood pressure as 0 excluded from the
confidence interval. The 13 DF are n – 3,
namely those of the residual.
Also note that the t test value associated with d is
3.69 with a p-value of 0.003.
Also note how the effect of age has effectively been
removed from the residual which is substantially
reduced from 2519.5 to 669.4.
The value of R 2 has risen from 0.058 or 5.8% to
0.750 or 75% when X is added to the model
involving d only.
404
Section 10

The confidence interval here is the ADJUSTED
CONFIDENCE INTERVAL after allowing for
the effect of age.
Unadjusted interval: (– 8.14, 20.64)
Adjusted interval: (5.84, 22.33)
It is helpful to put a geometrical interpretation on
this analysis. The scatter diagram of Y(BP)
against X(age) follows for all 16 patients.
150
140
130
120
110
100
Y(BP)
x
x
x
x
x
x
x
x
25 30 35 40 45 50 55 60 65
X(Age)
Notice the difference between the treated group
(dots) and the control group (crosses)
Suppose we fit the equation (by least squares)
ŷ =
0
ˆβ + ˆβ 1x +
2
ˆβ d
where d = 0 for control, d = 1 for treatment.
Section 10
405

If d = 0, ŷ =
0
ˆβ + ˆβ 1x
If d = 1, ŷ =
0
ˆβ + ˆβ 1x +
2
ˆβ
= (
0
ˆβ +
2
ˆβ ) + ˆβ 1x
These two lines are PARALLEL (same slope ˆβ 1) but
the intercepts are
0
ˆβ and (
0
ˆβ +
2
ˆβ ). Thus,
2
ˆβ is the
vertical distance between the two parallel straight
lines.
150
140
130
120
110
100
Y(BP)
x
x
x
x
x
X(Age)
25 30 35 40 45 50 55 60 65
2
ˆβ is the effect of the treated group relative to the
control. If
2
ˆβ is significant, then there is evidence of
different blood pressure values in the two groups.
We see how to test
2
ˆβ for significance shortly. The
next printout gives Y regressed on X only, and Y
regressed on X and d together.
x
x
x
⎫
⎬
⎭
ˆβ 2
406
Section 10

Notes:
1.
2
ˆβ (the coefficient of d) = 14.082 is the
increase in blood pressure level due to
administering the treatment (regardless of the
age of a patient since the two lines, being
parallel, have constant difference).
2. The 95% confidence interval for the
coefficient of d (namely β 2 ) is
14.082 ± t 13 (3.818) where t 13 = 2.160
It follows that 5.84 < β 2 < 22.33
3. Without taking age into account, the treatment
raised blood pressure by 6.25 mm of mercury
only. Taking age into account, the treatment
raised blood pressure by 14.082 mm.
4. The ordinary unpaired t-test originally
suggested for this problem is equivalent to
regressing Y on d alone. In this case, the
variable x (or Age) remains as part of the
residual which is therefore inflated hiding the
true treatment effect. In addition correlation
between age and treatment group distorts the
estimate of treatment effect on blood
pressure.
407
Section 10

Binary outcomes: Logistic Regression
Recall: For simple and multiple linear regression
the outcome variable was continuous.
What do we do if the outcome variable Y is binary
e.g. disease present: yes/no
e.g. tuatara: present / absent
e.g. claim to ACC goes to litigation : Yes / No
e.g. depression: yes/no in 18 yr olds if bullied at
school earlier
We use logistic regression (LR).
In a logistic regression the explanatory or predictor
X variables can be either continuous or
categorical(binary).
Like multiple regression, we can use logistic
regression to:
(1) control for confounding;
(2) investigate the effect of several variables on
the outcome variable at one time.
We can use the method of LR with data from any
study type as long as we have a binary outcome.
408
Section 10

The logistic regression model is:
⎛ p
⎜
⎞ = β + β + β + + β + ε
− p
⎟
⎝ ⎠
ln
0 1X1 2X2
…
kXk
1
where
Y is the binary outcome variable (values 0 or 1)
p is the probability that a particular event will
occur, i.e. Pr(Y = 1).
X1, X2,..., X
k
are the explanatory variables
β0is the intercept
β1, β2,..., β
k
are the regression coefficients
ε is the random error
Interpreting the model:
p
is the ‘odds’ of the event occurring
1−
p
⎛ p ⎞
ln ⎜ 1 − p
⎟ is the ‘log odds’
⎝ ⎠
The regression coefficient β
i
represents the change
in the log odds for a 1-unit change in X .
i
Fitted logistic model:
The formulae to estimate the values β
0
and β
1
etc
are computationally complex. We shall not worry
Section 10
409

about the details here and we shall instead focus on
understanding the results from a logistic regression
R-cmdr printout.
Example:
A study was conducted to investigate the
relationship between physical inactivity and
myocardial infarction (MI). It was found that
people who were physically inactive had an
increased risk of MI. Age was considered to be a
potential confounder.
Compared to younger people, older people:
• are more likely to be physically inactive.
• have a higher risk of MI.
Hence, we would expect that age can explain some
of the association between physical inactivity and
MI.
Outcome:
whether a person has a MI (Y) where Y = 0 or 1
Exposure of interest:
whether a person was physically inactive
(exposure variable, X 1 )
Possible confounder:
age (X 2 ) of the person.
410
Section 10

(1) Investigating the relationship between
physical inactivity and MI.
Option 1: Calculate the odds ratio as shown
earlier in the semester.
The 2 × 2 contingency table for outcome and
exposure is constructed from the 924 people.
Outcome - MI
Exposure (X 1 ) Yes No
Physically inactive 136 98
Physically active 343 347
Odds ratio of MI in exposed to unexposed:
136/ 98
OR = = 1.40
343/ 347
with 95% confidence interval 1.04< OR < 1.89.
Interpretation: The odds of having a MI is 40%
higher for a person that is physically inactive
compared to a physically active person. The
result is significant.
411
Section 10

Option 2: Alternatively, we can fit a logistic
regression model, using R-cmdr
Y = MI
1 = Yes 0 = No
X 1 = Physically inactive 1 = Yes 0 = No
Fitted Regression Model:
⎛ pˆ ⎞
ln ˆ ˆ
⎜ = + X
1−
pˆ
⎟ β β
⎝ ⎠
0 1 1
where ˆp is the probability that a person has a MI.
R-cmdr commands:
Analyze > Regression > Binary Logistic
Dependent: enter MI
Covariate: enter Physical Inactivity. OK.
Results from R-cmdr
⎛ pˆ
⎞
Fitted equation is ln⎜
1−
pˆ
⎟
⎝ ⎠ = -0.01+ 0.34 X
1
Odds ratio = 1.40 as before
95% confidence interval for OR is (1.04, 1.89)
412
Section 10

BUT what about the potential confounding
effect of age How can we control for that
Note: The odds ratio calculated previously is a
crude odds ratio – it (and its corresponding 95%
confidence interval) is not adjusted for the
potential confounder age.
To control for age, we include age as a second
explanatory variable in our logistic regression.
(2) Investigating the relationship between
physical inactivity and MI, adjusting
(controlling) for age.
Now add age (X 2 ) to the regression in order to
obtain the adjusted OR and its 95% confidence
interval.
Y = MI
1 = Yes 0 = No
X
1
= Physically inactive 1 = Yes 0 = No
X = age
2
Results from R-cmdr:
413
Section 10

The fitted regression is
⎛ pˆ
⎞
ln⎜
1−
pˆ
⎟
⎝ ⎠ = -0.41+0.17 X
1+0.68 X 2
This leads to the age adjusted odds ratio of 1.19
which has 95% confidence interval (0.87, 1.62)
These values are read from the printout and
compare with the crude ratio of 1.40 with
confidence interval (1.04, 1.89).
Conclusion: After adjusting for age, the OR
decreased from 1.40 to 1.19. Therefore, age was
making the association between physical inactivity
and MI more extreme than it actually was.
414
Section 10

SECTION 11
Study design principles, critical appraisal, sources
of bias and confounding.
415
Section 11

Study Design and Critical Appraisal
Research process:
1. Development of research question
2. Design of study
3. Collection of information
4. Description of data
5. Interpretation of results
Study design
• Study design refers to the methods used to select the
study participants, control any experimental
conditions, and collect the information.
• Interpretation of results depends on the study design.
• The study design should be tailored to the research
question.
• Methods of statistical analysis and information
produced will depend on the study design.
“The data from a good study can be analysed in many
ways, but no amount of clever analysis can compensate
for problems with the design of the study.” Altman.
416
Section 11

Critical appraisal
Critical appraisal is the process of reviewing a study with the
goal of identifying its strengths and weaknesses, the major
results, and its broader implications.
Why teach study design and critical appraisal
• it is not possible to sensibly interpret the results of
statistical analysis without understanding the context in,
and methods with which the data were collected
• health sciences practice and policy needs to be based on
sound evidence (as far as possible).
• poorly conducted research should not influence policy or
practice.
• because even well conducted research is not perfect, it is
necessary to understand the nature of evidence so that
you can begin to learn to interpret research findings for
yourselves.
• for this you need to gain an understanding of the
scientific method as used in the health sciences.
• this understanding is enhanced by learning to critique
research.
417
Section 11

Outline of next four lectures
1. Introduction to critical appraisal (lecture 1)
• process for critical appraisal
• structure of a research paper
2. Design and appraisal of surveys (lecture 1)
• review of surveys
• internal validity
bias
chance
• external validity
• example
3. Design and appraisal of analytic studies
(lectures 2 – 4)
• review of analytic study designs
• internal validity
bias
confounding
chance
• external validity
• causation
• examples: randomised controlled trials
cohort studies
case-control studies
418
Section 11

1. Introduction to critical appraisal
Guideline for critical appraisal
Study summary
What were the study objectives
Why was the study necessary
What type of study design was used
How were the participants selected
What information was collected
What were the key results
Internal validity
What do the findings of the study tell us about the
population studied
External validity / Generalisability
Can the findings of the study be applied to other
populations
Causation (for analytic studies only)
Implications
What are the implications of the study
419
Section 11

Structure of a scientific paper
Abstract or summary
• usually contains the key results of the study.
Introduction
• gives the background, necessity and objectives.
Methods
• summarises the study design including source of
participants and methods used to collect data.
Results
• description of the study participants including response
rates.
• summary of analyses.
Discussion
• provides the authors views of the internal and external
validity of the study, and their conclusions about the
implications of the study.
420
Section 11

2. Design and appraisal of Descriptive studies
Aim: To describe characteristics of a group or groups of
people at a given point in time.
Generally, a sample is taken from the population and the
distribution of variables within that sample is described.
Examples: A descriptive study can be used to
• describe characteristics of a group of people,
e.g. prevalence of asthma, prevalence of smoking,
average cholesterol level.
• find out peoples’ opinions and attitudes,
e.g. attitudes to alternative health care; satisfaction
with health care delivery.
• find out extent of peoples knowledge,
e.g. knowledge of risk factors for melanoma, risk factors
for coronary heart disease.
• comparisons of subgroups may well be part of a survey,
e.g. comparison of attitudes of men and women to
alternative health care; comparison of prevalence of
smoking among different ethnic groups in NZ.
A descriptive study is concerned with and designed only to
describe the existing distribution of variables, without regard
to causal or other hypotheses.
Descriptive studies can generate hypotheses.
Descriptive studies are often called surveys or cross-sectional
studies.
421
Section 11

Descriptive studies generally use a sample from a population.
Descriptive studies
underlying population
(parameters eg μ, π)
other populations
(external validity)
sample
inference
(internal validity)
statistics
(eg x, p)
422
Section 11

Recall
Suppose we want to estimate mean cholesterol in the
population:
sample
mean
=
population
mean
+ "error"
systematic error
(bias)
random
variation
random error (chance):
• due to natural biological variability.
• increasing the sample size will reduce the random
fluctuations in the sample mean.
systematic error (=bias)
• due to aspects of the design or conduct of the study which
systematically distort the results.
• occurs if a sample is not representative of the population.
• cannot be reduced by increasing the sample size.
423
Section 11

Internal validity for descriptive studies
Bias
• bias
• chance (random error)
Selection bias
• systematic error arising from the way people are selected
for the study.
• includes biases from sample selection and from nonresponse
to study.
Information bias
• systematic error arising from the way information was
collected from the study participants.
Chance
• Confidence intervals around estimates indicate the degree
of precision with which the sample value estimates the
population value.
424
Section 11

Selection bias
• systematic error arising from the way people are selected
for the study.
• includes biases from sample selection and from nonresponse
to study.
Questions to ask:
• Is the sample representative of the population
• What was the response rate
Example: A study was conducted to estimate the prevalence
of smoking among males and females in NZ.
Design:
A random sample of households was selected using random
digit dialling. If the call was not answered, the machine
automatically went on to the next number. All interviews
were conducted from 8am – 5pm (weekdays only).
63% of people agreed to participate in the study.
425
Section 11

Information bias
• systematic error arising from the way information was
collected from the study participants.
Question to ask:
Is the information gathered correct
Example: Suppose an investigator wished to estimate the
prevalence of depression in NZ. To do this, he carried out
face-to-face interviews around the country with a random
sample of adults. Can you think of how information bias
might enter into his study
426
Section 11

Example
Life in New Zealand Survey, Hillary Commission for
Recreation and Sport, 1990, David Russell and Noela
Wilson.
Objectives
• to provide a snapshot of New Zealanders from a health
perspective.
• included questions on physical activity, leisure patterns,
dietary habits and other risk factors for disease.
Necessity for the study
• study provides a benchmark for comparison in future
years.
• the information is useful for generating hypotheses and
for designing interventions to improve health.
Type of study design
• survey of New Zealanders 15 years and over.
• carried out April 1989 – May 1990.
Selection of participants
• over 18 years:
- selected from electoral rolls.
- each month 10 people were selected at random from
each of the 97 electoral rolls, plus 19 from each of
the 4 Maori rolls.
• 15 – 18 years:
- snowball sample was used.
- people already selected were asked to identify up to 5
people aged 15 – 18.
• total number selected: 12,463.
427
Section 11

Results
Physical activity
Activity level
low moderate high
Male
15 – 18 17 20 64
19 – 24 23 27 51
25 – 44 34 31 35
45 – 64 50 34 16
64+ 58 39 3
All 37 31 32
Female
15 – 18 24 22 54
19 – 24 30 30 40
25 – 44 20 53 26
45 – 64 25 64 11
64+ 34 63 3
All 25 51 24
Can you summarise these results
428
Section 11

Internal validity
Bias
Selection bias:
• random sampling was used for those 18 and over.
• bias from snowball sample (note multiple starting
points based on random sampling)
• response rate
Information bias:
• questionnaire
• accuracy of recall
• tendency to report what people think the researchers
will want to see
Chance
• the study is large so the confidence intervals for overall
proportions will be fairly narrow, but for smaller
subgroups the proportions may not be so well estimated.
e.g.: women aged 64+, n=814
proportion with low activity level
= 34%, CI= (30.8 to 37.3)
proportion with high activity level
= 3% CI= (2.0 to 4.5)
429
Section 11

External validity
Are the results applicable to other populations
• this calls for a judgement as to whether the other
populations are likely to be similar to New Zealand in
terms of their exercise patterns.
Implications
• high activity levels are the levels recommended to
maintain cardio-respiratory fitness.
• programmes to increase activity levels may be useful in
preventing cardiovascular disease.
• efforts to increase activity levels of men over the age of
45 may be particularly useful.
430
Section 11

Study design and critical appraisal sessions: 2
1. Introduction to critical appraisal (lecture 1)
• process for critical appraisal
• structure of a research paper
2. Design and appraisal of surveys (lecture 1)
• review of surveys
• internal validity
bias
chance
• external validity
• example
3. Design and appraisal of analytic studies
(lectures 2 – 4)
• review of analytic study designs
• internal validity
bias
confounding
• chance
• external validity
• causation
• examples: randomised controlled trials
cohort studies
case-control studies
431
Section 11

3. Design and appraisal of analytic studies
Review of analytic study designs
Purpose
To test hypotheses regarding
• causes of disease
• disease prevention strategies
• effectiveness of treatments
Example:
• Is a statin drug more effective than a diet high in plant
sterols, soy proteins and almonds in reducing serum
cholesterol levels
• Do people who are physically inactive have an increased
risk of developing colon cancer
When we are conducting an analytical study we are studying
associations among two or more variables. We will have
an outcome variable (eg
exposure variables (eg
confounding variables – these are variables which distort
the association of interest (eg age)
432
Section 11

Types of design:
• experimental (intervention)
• e.g. randomised controlled trials.
• observational
• e.g. case-control studies, cohort studies.
Key features of common designs
Randomised controlled trials
• people are assigned to an intervention or control group
using random allocation, then followed up over a period
of time.
Cohort studies
• participants are selected before they develop disease.
• exposure status is measured, and they are followed up
over a period of time.
Case-control studies
• two groups of people are chosen: a group with disease
(cases) and a group without disease (controls).
• information is collected from both groups about
exposures that occurred in the past.
433
Section 11

Key ideas:
• control (or comparison) groups are essential.
• experimental studies provide much stronger tests of
hypotheses than observational studies.
• experimental studies allow testing of causal relationships
• with observational studies it is much harder to isolate the
effects of the exposure of interest, so much harder to
determine whether an association is causal
434
Section 11

Example
Does smoking cause coronary heart disease
1. Estimate the association between smoking and coronary
heart disease (eg relative risk).
2. Does this relative risk represent the true association
between smoking and CHD in the population studied
(internal validity)
if yes
3. Can this result be generalised to other populations
(external validity)
4. Is the association causal
435
Section 11

Internal validity
Does the observed association represent the true association
Specifically:
What are the possible explanations for the observed results
• bias
• confounding
• chance
• true relationship
436
Section 11

Assessing internal validity:
Bias
Selection bias – systematic error arising from the way
participants are selected for inclusion in the study.
In an analytic study, selection bias occurs if the selection
processes cause a systematic difference between the groups
of people selected for the study.
It includes bias from non-response.
Information bias – systematic error arising from the way
study information is obtained, interpreted and recorded.
In an analytic study information bias is a particular problem
if there are systematic differences in the information obtained
from the different groups of people in the study.
Information bias may be introduced by the:
• Observer
• Study individual (respondent)
• Instruments used to collect the data (e.g. badly-designed
questionnaire)
437
Section 11

Example
Case-control study to examine relationship between stress
and coronary heart disease:
cases:
controls:
people with coronary heart disease
identified through opportunistic
screening by GPS
random sample from the population
Information on stress collected through a structured interview
Selection bias:
Information bias:
438
Section 11

Evaluation and control of bias
• Statistical methods cannot control for bias in the selection
of subjects or in the measurement of the variables of
interest. Control of bias can only be done during the
design and data collection phases of the study.
• General inaccuracy which is the same in both groups
generally results in an underestimate of the true
association.
• If inaccuracy is different in the two groups, the
association can be an over or under estimation.
• It is important to identify sources of bias and estimate the
magnitude and direction of their effect on the association.
439
Section 11

Confounding
A distortion of the association between exposure and disease
caused by the presence of a third factor.
• A confounder is a variable which causes this distortion
• To be a confounder a variable must be
• associated with the exposure (independent of disease).
• associated with disease (independent of exposure).
• it must not just be an intermediate link in the causal
chain.
440
Section 11

Example of confounding:
A study was conducted to investigate the relationship
between coffee consumption and oral cancer. It was found
that coffee drinkers had an increased risk of oral cancer.
Smoking is a potential confounder in this study.
Compared to non-smokers:
• Smokers are more likely to drink coffee;
• Smoking is an independent risk factor for oral cancer.
Hence, the observed association may be due to smoking
habits rather than coffee drinking.
Can you think of any other potential confounders
441
Section 11

Example of non-confounding:
diet cholesterol level coronary heart disease
In this case, the raised cholesterol levels are likely to be due
in part to diet, so are part of the causal pathway. Therefore in
studies of diet and coronary heart disease raised cholesterol
would not be considered a confounder.
Example of a confounder in a cohort study:
Results from a cohort study investigating the relationship
between myocardial infarction and exercise.
Myocardial
infarctions
Personyears
Table A: all subjects (n=8000
person-years)
Low exercise 105 4000 26.25
High exercise 25 4000 6.25
Relative risk = 26.25/6.25 = 4.2
Subgroup Analysis
Obese subjects (n=4000)
Low exercise 90 3000 30.0
High exercise 10 1000 10.0
Relative risk = 3.0
Non-obese subjects (n=4000)
Low exercise 15 1000 15.0
High exercise 15 3000 5.0
Relative risk = 3.0
Incidence/1000
442
Section 11

Positive and Negative Confounding
Positive confounder – a confounding variable which makes
an association look more extreme or create a spurious
associations.
Example: A study was conducted to investigate the
relationship between physical inactivity and MI. It was found
that people who were physically inactive had an increased
risk of MI. Age was considered to be a potential confounder.
Physical inactivity
Myocardial infarction
Age
Crude odds ratio =2.5
But compared to younger people, older people:
• are more likely to be physically inactive.
• have a higher risk of MI.
Hence, age can explain some of the association between
physical inactivity and MI.
After “adjusting” for the confounding association of age the
OR decreases to 1.4. So confounding by age is making the
association between physical inactivity and MI seem more
extreme than it should be, i.e. it is a positive confounder.
443
Section 11

Negative confounder – a counfounding variable which
makes an association look less extreme or even in the
opposite direction. It can mask a real difference.
Example: A study was conducted to investigate the
relationship between physical inactivity and MI. It was found
that people who were physically inactive had an increased
risk of MI. Sex was considered to be a potential confounder.
Physical inactivity
Myocardial infarction
Sex
Crude OR = 2.5
But compared to females, males:
• are less likely to be physically inactive.
• have a higher risk of MI.
Hence, sex masks some of the association between physical
inactivity and MI.
After “adjusting” for the confounding effect of sex, the OR
becomes 3.9.
So confounding by sex makes the association between
physical activity and MI seem less extreme than it should be,
i.e. it is a negative confounder.
444
Section 11

Some comments on confounding:
AGE and SEX are the most common confounding variables.
This is because these two variables are not only associated
with most exposures we are interested in such as diet,
smoking habits etc., but they are also independent risk factors
for most diseases.
Control of confounding
Confounders can be controlled for during the study design,
during the analysis, or both in the design and the analysis.
The aim is to make the groups being compared as similar as
possible with respect to the confounders.
(1) Identify potential confounders. A review of previous
literature in the area should give you an idea of potential
confounders.
Also: What are the known risk factors for the outcome of
interest; What factors are associated with exposure
Data should be collected on all potential confounders since if
you do not obtain the information you cannot control for it.
445
Section 11

(2) Control of confounding during the study design.
Restriction:
• Limits participation in a study to specific groups that
are similar to each other with respect to the
confounder.
e.g. Include only non-smokers in a study of exercise
and risk of CHD.
• Disadvantages
• residual confounding if restriction criteria are too
wide.
• lack of generalisability.
• smaller number of available participants.
Matching:
Particular subjects are selected in such a way that the
potential confounders are distributed in an identical
manner among each of the study groups.
Case-control study: Matching cases and controls.
Cohort study: Matching exposed and unexposed.
Matching needs to be accounted for in the analysis
Randomisation
446
Section 11

(3) Control of confounding during the analysis.
Multivariate analysis – multiple regression.
Evaluating confounding
• Check for associations between suspected confounder and
exposure and disease.
• See whether controlling for confounding affects the
association.
Chance
• Study design: ensure study has sufficient power.
• Confidence intervals and p-values for the association
indicate the role of chance in the study.
• When multiple statistical tests are carried out in a study,
there is an increased chance of “false positive” results.
447
Section 11

Study design and critical appraisal sessions: 3
Randomised controlled trials (RCTs)
Aim: To study evaluate the effects of an intervention
• considered the “Gold standard” for evaluation of
interventions
Why
• allows isolation of the effects of the intervention through
controlling the experimental condition
• experiment (“trial”)
• comparison/control group (“controlled”)
• randomisation (“randomised”)
Randomisation
• process for deciding who will get the experimental
intervention and who will be the control
448
Section 11

Basic structure of a RCT
• population to be studied
• choice of comparison group
• allocation of subjects to intervention or control group
• choice of outcome measure
Population to be studied:
Usually not a representative sample from the population
• eg in trials of treatments they will be patients coming to
see the doctors who have agreed to take part in the
study
Chosen to maximise internal validity with some cost in terms
of generalisability.
• eg we may choose participants who are likely to be able
to complete the requirements of the trial
Choice of comparison group:
• the control group should provide information on what
would have happened without the experimental
intervention
• in trials of disease treatment or prevention the control
group should in general receive the best available
“standard” treatment.
• sometimes there is no standard treatment or practice, in
which case a “placebo” control group may be used.
449
Section 11

• “placebos” are substances with no biological effect on
the disease process.
• placebos are used to isolate the particular effect of
interest from effects that may occur because of people’s
belief they are getting a particular intervention
• use of a placebo allows “blinding” of intervention and
control groups, so that the results are not biased through
knowledge of who got the new intervention
Allocation of subjects to treatment groups:
Example: Is the new treatment more effective than the
standard treatment
How would we test this
(1) We could compare the results of the new treatment
on patients with records of previous results from
other patients using the old treatment (historical
controls).
Do you think this is a good idea
(2) Ask people to volunteer for the new treatment and
give the standard treatment to those who do not
volunteer
Do you think this is a good idea
(3) Allocate patients to the new treatment or the old
treatment using an “objective” method and observe
the outcome.
450
Section 11

The way in which patients are allocated to treatments can
influence the results enormously.
We need a method of allocation to treatments in which the
characteristics of subjects will not affect their chance of
being put into any particular group – RANDOM
ALLOCATION
Volunteers are assigned to intervention groups using
randomisation, then followed up over a period of time.
Randomisation:
• best way to control for both known and unknown
confounders.
• but does not guarantee control of confounding.
• is ethical when there is genuine uncertainty about whether
the new intervention or the comparison strategy is better
(“equipoise”).
451
Section 11

Choice of outcome measure:
• needs to be sensitive to the effects of intervention
• early in the process of evaluation short term outcomes are
used to screen for promising interventions
• ultimately, need to demonstrate that the intervention has a
tangible benefits for the individual and society
Example: Zidovudine in treatment of people with
asymptomatic HIV infection.
Studies found
• statistically significant improvement in immune function
(measured by CD4 count).
but
• no difference in survival at 3 years.
452
Section 11

Randomised controlled trials : Example
Nichol et al. The effectiveness of vaccination against
influenza in healthy working adults.
New England J. Med (1995)
Objectives
• to clarify the benefits of immunisation for influenza in a
population not at high risk for complications.
Background
• most deaths from influenza occur among elderly people,
but all age groups are affected.
• influenza accounts for millions of days lost from work
each year.
• current recommendations of the US Advisory Committee
on Immunisation Practices target persons at increased risk
for complications of influenza, although all people who
wish to avoid illness are encouraged to consider
vaccination.
Type of study
Randomised controlled trial
453
Section 11

Selection of participants
• recruited in Minneapolis-St Paul through newspaper
advertisements, advertisements at work sites and
recruitment sessions at shopping malls.
• aged 18 – 64 years.
• employed full time.
• no medical conditions which would place them at high
risk for complications of influenza.
• not allergic to eggs.
• not pregnant or planned pregnancy within 3 months.
• had not had a previous vaccination for influenza.
Information collected
“Exposure” (=treatment)
• influenza group: active vaccine
• placebo group: vaccine diluent
Outcome measure:
• structured telephone interviews
Week 1:
side effects
Monthly for 4 months:
• occurrence of upper respiratory illness
• use of sick leave
• visits to the doctor
454
Section 11

Key results
849 randomised
placebo vaccine
n=425 n=424
complete follow-up
n= 416 (98%) n=409 (96%)
455
Section 11

456
Section 11

Internal validity
Chance
• 95% confidence intervals around the differences exclude
zero.
• p-values are small, indicating that differences this large
(or larger) are very unlikely to occur by chance if the
vaccine is not effective.
• several outcome measures were used, increasing the
chance of false positive results, but since the p-values are
very small this is not likely to affect the conclusions.
457
Section 11

Confounding
Randomisation + intention to treat analysis
458
Section 11

Intention-to-treat analysis
“once randomised, always analysed”
• outcome is compared in
• the group randomised to placebo
• and the group randomised to vaccine.
• this preserves the control of confounding achieved by
randomisation.
Bias
Selection bias is not a problem in randomised controlled
trials (see generalisability though)
Information bias in randomised trials arises from
• incomplete follow-up of participants
• error in measurement of outcome
Information bias in vaccine trial:
Completeness of follow-up:
• placebo: 98% (416/425)
• vaccine: 96% (409/424)
Measurement of illness:
• definition of influenza
• recall of symptoms
459
Section 11

Blinding
• means participants experience or recall of symptoms is not
affected by knowledge of whether they had the vaccine
(single blind).
• people collecting the information from the participants
cannot introduce bias through their knowledge of whether
or not they had the vaccine (double blind).
Generalisability
• broad group of working adults
• risk of influenza
• strain of influenza
Implications
• the trial demonstrates that vaccination against influenza
can be effective in reducing symptoms, sick leave and
visits to the doctor.
460
Section 11

Study design and critical appraisal sessions: 4
1. Introduction to critical appraisal (lecture 1)
• process for critical appraisal
• structure of a research paper
2. Design and appraisal of surveys (lecture 1)
• review of surveys
• internal validity
• bias
• chance
• external validity
• example
3. Design and appraisal of analytic studies
(lectures 2 – 4)
• review of analytic study designs
• internal validity
• bias
• confounding
• chance
• external validity
• causation
• examples: randomised controlled trials
• cohort studies
• case-control studies
461
Section 11

Cohort study
Ref: “Cohort studies: marching towards outcomes”, Lancet
2002; 359:341-45.
462
Section 11

Prospective cohort study (concurrent): Cohort is defined
and characterised at the start of the study and followed up
into the future.
• Assemble the cohort.
• Measure predictor variables and potential confounders.
• Follow up the cohort and measure outcomes.
Retrospective (historical) cohort: Cohort is defined and
characterised in the past, based on data already recorded,
and followed up toward the present to some cut-off time.
• Identify a suitable cohort.
• Collect data about predictor variables from past records.
• Collect data about subsequent outcomes that occurred at
a later time.
463
Section 11

Cohort studies: example
Hart C, Davey Smith G. “Coffee consumption and coronary
heart disease mortality in Scottish men: a 21 year follow-up
study.”
J Epidemiol Commun Health (1997); 51: 461-2
Objective
• to examine the effects of coffee on coronary heart
disease mortality.
Background / Necessity
• recent studies of this hypothesis have produced
conflicting results.
• data on confounding factors has often been limited in
those studies.
Type of study design
• cohort (prospective)
464
Section 11

Selection of participants
• 5,766 men aged 35-64 from work places in an area in
the west of Scotland.
• enrolled between 1970 and 1973.
Information collected
• at enrollment:
• how many cups of coffee they usually drank per
day;
• information on confounders such as smoking, social
class.
• followed up for 20 years.
• information about deaths from coronary heart disease
was obtained from the national registry.
465
Section 11

Key results
No. of cups CHD
coffee per day Deaths RR 95% CI
0 308 1.0
1 94 0.89 (0.70, 1.12)
2 104 0.98 (0.78, 1.23)
3-4 82 0.90 (0.70, 1.16)
5+ 37 0.96 (0.67, 1.37)
p value from trend test = 0.71
Chance
• all confidence intervals include the null value, 1.
• the upper limits of the confidence intervals for < 5 cups
per day are fairly close to 1.
• for 5+ cups per day we cannot exclude a true RR as big as
1.37 (a 37% increase in risk).
• the test for trend gave a p-value >> 0.05.
466
Section 11

Bias
Selection bias
• because there is only one selection process, selection
bias is minimised.
• the study sample may not be representative of the
population in west Scotland, but in analytic studies that
issue is addressed under generalisability.
Information bias
• information bias could come from:
• inaccurary in exposure information;
• loss to followup;
• inaccurary in determining death from CHD.
• crude measure of coffee consumption used, may bias
RR towards null.
• followup will be nearly complete using national
registry.
• may be some misclassification of cause of death.
467
Section 11

Confounding
• RR presented were adjusted for a number of
confounding factors including: age, diastolic blood
pressure, cholesterol, smoking, social class and body
mass index.
Generalisability
• type of coffee drunk (instant vs ground).
Implications
• found no clear evidence of an association between
instant coffee use and risk of CHD.
• cannot rule out an increase in those drinking 5+ cups
per day (small numbers).
• other types of coffee may have detrimental effects on
CHD risk.
468
Section 11

Case-control studies
Ref: “Case-Control studies: research in reverse”, Lancet
2002; 359:431-34.
• Subjects are ascertained based on whether they have
experienced the outcome of interest (cases) or not
(controls).
• Information is collected from cases and controls about
their past exposures.
469
Section 11

Case-control studies: example
Shinton R and Sagar G. “Lifelong exercise and stroke.”
BMJ (1993); 307: 231-4.
Objective
• to examine the potential of lifelong patterns of increased
physical activity to prevent stroke.
Background / Necessity
• there is growing evidence that exercise can protect
against stroke.
• the importance of exercise in early adult life in
protection from stroke has received little attention.
• previous studies had not adequately controlled for
confounding.
Type of study design
Case-control study
470
Section 11

Selection of participants
Study population: people registered with a GP in west
Birmingham, England.
Cases:
• men and women aged 35-74 who had just had their first
stroke.
• obtained by phoning GPs weekly, and by checking
admissions at the local hospital.
Controls:
• randomly selected from the general practice population.
• no history of stroke.
471
Section 11

Information collected
• structured questionnaire.
• one interviewer for all cases and controls.
• when disability prevented an adequate response the
closest friend or relative was interviewed.
• people were classified by their responses into those who
did or did not engage in vigorous exercise during:
youth (15-25)
early middle age (25-40)
late middle age(40-55)
• information on confounders (e.g. age, sex, smoking)
472
Section 11

Key results
Response rates:
Cases:
• 125 patients were eligible for inclusion.
• no patient or relative declined to participate.
(100% response rate)
Controls:
• 220 controls were selected and contacted.
• 13 excluded.
• 198 of the remainder (207) agreed to participate.
(95.7% response rate)
Table I. Odds ratios* (95% confidence interval) of stroke
according to when exercise undertaken.
Exercise undertaken
no
yes
Age undertaken
15-25 1.0 0.33 (0.2 to 0.6)
25-40 1.0 0.43 (0.2 to 0.8)
40-55 1.0 0.63 (0.3 to 1.5)
* Odds ratios are adjusted for age and sex
473
Section 11

Now, let’s consider possible explanations for an
association: Internal validity
Chance
• confidence intervals show the range of plausible values
of the true odds ratio which are consistent with the
study results.
• if the confidence interval for an odds ratio excludes 1,
then the study provides evidence of an association in the
population studied.
• if the confidence interval for the odds ratio includes 1,
then the study results are consistent with the possibility
that there is no true association.
• to conclude definitely that there is no association, the
confidence interval must include 1 and be narrow, so that
important differences in the risk of disease can be
excluded.
474
Section 11

In this study:
• the odds ratios increase with increasing age at which the
exercise was undertaken.
• the confidence intervals for ages 15-25 and 25-40
exclude 1, so there is some evidence of an association
between exercise at those ages and reduction in risk of
stroke.
• the odds ratio for exercise undertaken at age 40–55 is
less than 1, but the confidence interval contains 1
indicating that this apparent beneficial effect could just
be due to random variation or chance.
475
Section 11

Bias
Case-control studies are particularly susceptible to bias
because at the time the study is done both exposure and
disease have already occurred.
Selection bias
cases: all non-fatal cases which arose from the GP
population were included.
controls: randomly selected from the population the cases
arose from.
Therefore, the controls are representative of the population
the cases arose from, and selection bias is minimised.
Response rates were high.
(100% for cases and 95.7% for controls)
476
Section 11

Information bias
Things done to minimise bias:
• cases and controls all interviewed by the same
interviewer.
• structured questionnaire was used.
Possible sources of information bias:
recall bias:
• cases and controls may both have trouble recalling
accurately exercise patterns when they were young.
• similar patterns of poor recall in cases and controls will
bias an odds ratio towards 1, so it could not explain the
observed association.
• cases have had a stroke so they may be less likely to
remember than the controls.
• if cases were less likely than controls to report exercise,
an apparent protective association between exercise and
stroke would be created.
bias from surrogate interviewee:
• information on exercise for cases unable to respond was
obtained from a friend or relative.
477
Section 11

Interviewer bias:
• the interviewer will have known whether or not people
were case or controls.
• If he/she prodded the controls harder for information on
exercise an apparent protective effect would be created.
Confounding
• risk factors for stroke include age, sex, and smoking.
• since all 3 of these are likely to be associated with
exercise, they may be confounding the relationship
between exercise and stroke.
• analyses were adjusted to remove confounding effects
of confounding variables including age, sex and
smoking.
478
Section 11

Generalisability
Could we apply the results of this study to the New Zealand
population
• need to think about whether or not New Zealanders
would be likely to experience the same apparent benefit
from exercise.
• depends on the nature of the exercise and the biological
mechanism by which exercise reduces risk of stroke.
Causation
• it is difficult to show causation conclusively with a
single observational study, primarily because of the
susceptibility to bias and confounding.
• an association is more likely to be causal if :
• the observed association is very strong;
• a dose-response effect can be demonstrated;
• the results from several different studies are
consistent;
• there is a known biological mechanism.
479
Section 11

480

Appendix One: The Basics
This appendix contains some background material to help you prepare for the course.
1. Basic Mathematical Rules
1. BEDMAS – how to work things out in the right order
2. Rounding
3. Dealing with Negatives
4. Fractions
5. Solving Equations
6. Powers and Logarithms
7. Sigma means Add Up
2. Basic Statistical Concepts
1. Mean
2. Median
3. Range
4. Variance and Standard Deviation
5. Quartiles and Interquartile Range
6. Scatterplot
3. Sample Exercises
MATHERCIZE
Practice examples for many of the topics covered in this booklet are available on the computer
package MATHERCIZE. This program is available at: http://mathercize.otago.ac.nz, and the
login password is line.
Appendix 1 – Basic rules and concepts

Section 1: Basic Mathematical Rules
1. BEDMAS – how to work things out in the right order
Brackets
Exponents (also known as Powers)
Division and Multiplication
Addition and Subtraction
When Division and Multiplication occur together, work from the left. Similarly when Addition and
Subtraction occur together, work from the left. Otherwise follow the guidelines suggested by the
word BEDMAS.
Note that a scientific calculator will maintain this order, provided care is taken, but other calculators
do not.
Example 1
Evaluate ( 3+ 2) × 6 + 9 2 ÷ ( 2+ 7−
6)
• First evaluate both brackets: ( 3+ 2) = 5 and ( 2 + 7 − 6)
= 3
• Then the exponent:
2
9 = 81
• Then the division and multiplication: 5× 6 = 30 and 81÷ 3 = 27
• Finally the addition: 30 + 27 = 57
Setting this out on paper:
2
3+ 2 × 6 + 9 ÷ 2+ 7− 6 = 5 × 6 +
Example 2
5 +
( ) ( )
2
Evaluate ( 9 − 5 ÷ 5×
2 ) − 9
2
9 ÷ 3
= 5 × 6 + 81 ÷ 3
= 30 + 27
= 57
• First evaluate the brackets. The exponent is evaluated first:
9 5 5 2 2
− ÷ × = 9 − 5 ÷ 5 × 4
( ) ( )
( 9 1 4)
( )
= − ×
= 9− 4 = 5
• Finally the addition 5 + 5 – 9 = 1.
Appendix 1 – Basic rules and concepts

Example 3
If Z = X − μ calculate Z if X = 15, μ = 8, and σ = 2.75
σ
• First carry out a “clean” substitution. This means that each variable whose value is known
is replaced by that value without any calculation being done:
Z = 15 − 8
2.75
• The division sign implies brackets. The expression could be rewritten as
(15 − 8)
Z =
2.75
although the brackets are seldom shown. Nevertheless the expression 15 – 8
is evaluated first.
•
7
Finally the division: = 7 ÷ 2.75 = 2.55 (to two decimal places).
2.75
• Note that using brackets on a standard calculator should let you evaluate the expression
directly. Try ( 15 – 8 ) ÷ 2.75 = (Missing out the brackets will almost
certainly lead to an incorrect answer.)
Example 4
s
If t = 2.086, s = 3.44, and n = 21, evaluate the expression t
n
• Clean substitution: 2.086 × 3.44
21
s
• Note the multiplication sign: t
n means t × sn
• A square root is an exponent, so evaluate 21 = 4.583 (to three d.p.)
• There is no addition or subtraction involved, so work from the left:
2.086 × 3.44 ÷ 4.583 = 1.57 (to two d.p.) (Rounding is discussed below.)
• Again this may be calculated directly on a calculator. Press the buttons:
2.086 × 3.44 ÷ 21 =
Example 5
Evaluate the expression x − μ
if x = 215.8, μ = 246, s = 64.5, and n = 10.
s
n
• For this example, only the calculator working is shown. Press the buttons:
( 215.8 – 246 ) ÷ ( 64.5 ÷ 10 ) =
The answer is –1.48 (to two d.p.)
• Try to obtain the same answer using the rules of BEDMAS.
Example 6
Evaluate 1.96
2 2
4.5 3.6
+
18 22
• The square root sign implies brackets around the expression
2 2
4.5 3.6
+
18 22
Appendix 1 – Basic rules and concepts

⎛ 2 2
4.5 3.6 ⎞
i.e. we have to evaluate 1.96
+
⎜ 18 22 ⎟
⎝
⎠
• All the exponents inside the brackets are calculated first, followed by the divisions:
2
2
4.5 20.25
3.6 12.96
= = 1.125 and = = 0.589
18 18
22 22
• Next the addition, followed by the remaining exponent (the square root)
1.125 + 0.589 = 1.714 and 1.714 = 1.309
• Finally the multiplication: 1.96 × 1.309 = 2.56 (to two d.p.)
• Again note that this could be calculated directly on a calculator (although a single small
mistake will make everything wrong). Try
1.96 × (4.5 x 2 ÷ 18 + 3.6 x 2 ÷ 22) =
The result should be 2.566 which rounds to 2.57, a little different to the answer above due to
rounding. Note that x 2 refers to the button on a Casio calculator. Other brands may have
different notations for squaring, although they should be similar.
2. Rounding
When you have decided how many digits you want to round to, look at the next digit. If this value
is 0, 1, 2, 3, or 4, the previous digit is rounded down. Otherwise (if the value is 5, 6, 7, 8, or 9), the
previous digit is rounded up.
Example:
By calculator, 8 30 = 1.460593487
• To three d.p. (decimal places) 8 30
= 1.461 because the next digit (5) causes the third
decimal value (0) to be rounded up.
• To four d.p. 8 30 = 1.4606
• To five d.p. 8 30 = 1.46059
• To six d.p. 8 30 = 1.460593
Appendix 1 – Basic rules and concepts

There are no hard and fast rules concerning how many digits you should round a value to, although
a few general principles should be noted:
• When you are calculating an expression do not round too soon. For example, consider the
expression 150 . To eight decimal places, 10 = 3.16227766.
10
• If you use a calculator to evaluate 150 , and round your final answer to three decimal
10
places, the result is 47.434.
• However, if you first round 10 to 3.16 and you then calculate 150 , the result is 47.468
3.16
(to three d.p.). This may not appear to be much different to the value 47.434, but it could
make a substantial difference if you have to use the value in further calculations.
• Do not round your working to fewer figures than your final answer. In the previous
example, the value 3.16 has three significant figures, while the (slightly incorrect) given
answer 47.468 has five figures. Having rounded to three figures in the working, three
figures (or fewer) should be used for the final answer.
You should not give an answer “more” accurate than the data or working.
• As a rule of thumb, round probabilities to four decimal places.
• Historically, Z-scores have been rounded to two decimal places. The reason for this is that
normal distribution tables use two decimal place Z-scores.
3. Dealing with Negatives
Adding a negative number is the same as subtracting the corresponding positive number:
• Example 5 + (-4) = 5 – 4 = 1
Subtracting a negative number is like adding a positive number:
• Example 5 – (-4) = 5 + 4 = 9
Multiplying two negative numbers give a positive number:
− 5 × − 4 =
• Example ( ) ( ) 20
Multiplying a negative number by a positive number gives a negative number:
− 5 × 4 = −
• Example ( ) 20
Appendix 1 – Basic rules and concepts

4. Fractions
Many people have difficulty with fractions. Sometimes the difficulty is in the interpretation rather
than with actual calculations.
Example
Imagine that you have attended a course and you are trying to work out your final mark. You have
been told that you scored:
• 8.5 out of 10 for the assignments
• 20 out of 40 for the test
• 32 out of 50 for the exam
If you add these three values up as if they were fractions, you would get
8.5 20 32
+ + = 1.99 (Check this using a calculator.)
10 40 50
This is clearly a silly answer because the values were not actually fractions as such, but marks from
different sections of the assessment scheme for the course.
If you just add up the marks you get 60.5. This is a more reasonable answer, because it gives a total
out of 100.
But suppose that in the course mentioned in this example, the assessment scheme states that if the
internal mark is higher than the exam mark, your final mark is the average. Otherwise the final
mark is the exam mark. For this example, the internal total is 28.5 out of 50, or 57%, while the
exam mark translates to 64%. As the exam mark is higher than the other combined marks, the final
mark in this case would be 64.
Using Calculators for Fractions
When probabilities are involved, dealing with fractions is important. This section aims to show
how to use a calculator to handle problems involving fractions.
As long as you estimate whether the final answer is sensible, practically all fraction work can be
carried out using a calculator. The key button to use is a b c
on a Casio. Other calculators should
have equivalent buttons.
Simplifying Fractions
12
Example 1:
20
Example 2:
On your calculator type 12
a b c
20 =
The answer is given as 3 5 i.e. 12
20 = 3 5
21
105 Type 21 b c
a 105 = i.e. 21
105 = 1 5
Appendix 1 – Basic rules and concepts

Converting Fractions to Decimals
The a b c
button will often do this, although not always!
Example 1: Convert 11 into decimal form
15
On the calculator type 11 a b c
15 =
The screen shows 11 15. Now press the a b c
button and the fraction is
converted to the decimal 0.733333 . . . Press a b c
again, and the fraction
version reappears.
Example 2:
Example 3:
Convert 0.6875 to a fraction.
Type .6875 = Now press the a b c
button. The screen shows
11 16 i.e. 0.6875 = 11
16
Convert 0.1234567 to a fraction.
Type .1234567 = . Now press the a b c
button.
Nothing happens. The calculator leaves the decimal alone. If you want to
convert this one to a fraction you will have to carry out the working yourself:
1234567
0.1234567 =
10000000
Adding and Subtracting Fractions
3 2
Example:
+
5 3
On your calculator type 3 a b c
5 + 2 a b c
3 =
The screen shows 1 4 15 i.e. 3 + 2 = 1
4
5 3 15
(Incidentally, if you now press the a b c
button, the decimal equivalent to this
fraction appears on screen: 1.266666. . .)
Remember that if these two fractions represent probabilities that you are adding together, and the
final answer was also meant to represent a probability, then there has to be an error somewhere
because a probability cannot be larger than 1.
Appendix 1 – Basic rules and concepts

Multiplying and Dividing Fractions
5 5
Example 1: ×
8 3
Type 5 a b c
8 x 5 a b c
3
The result is 1 1 24 i.e. 5 × 5 =
25
8 3 24
Example 2:
5 10
÷
7 11
Type 5
a b c
7 ÷ 10 a b c
11
The result is 11 14 i.e.
5 10 11
÷ =
7 11 14
More Complicated Calculations
As soon as you have a problem involving both addition and multiplication, brackets become very
useful.
3⎛1 3⎞
Example:
⎜ + ⎟
4⎝8 7⎠
Note that the fraction in front of the brackets implies multiplication.
Type 3 a b c
4 × (1 a b c
8 + 3 a b c
7) =
The answer is 93 or 0.4152 (to four d.p.)
224
Note that as an alternative approach you could use BEDMAS and work out the brackets first:
1 a b c
8 + 3 a b c
7 = gives 31
56
Now type × 3 a 4 = to reach 93 224 as before.
b c
5. Solving Equations
Solving equations involves more than evaluating expressions, which was covered earlier. To solve
an equation you should make a clean substitution, then rearrange the expression so that the required
variable is on its own.
Loosely speaking, solving equations involves “undoing BEDMAS”. For example, anything inside
brackets is dealt with last.
In STAT 115 one particular type of equation will need to be solved:
Appendix 1 – Basic rules and concepts

X − μ
Example 1: If Z = , calculate X if Z = 1.96, μ = 8.5, and σ = 1.8
σ
• First make a “clean substitution”, i.e. substitute each of the known variables into the
equation without trying to simplify at all:
X − 8.5
1.96 =
1.8
• The division sign implies brackets around X – 8.5. We are “undoing” the equation, so
this part will be left to last.
( X − 8.5)
1.96 =
1.8
• This means we “undo” the value 1.8 first. Because the right hand side of the equation reads
“(X – 8.5) divided by 1.8”, we will multiply by 1.8, since multiplication is the inverse
operation to division:
1.96 × 1.8 = (X – 8.5)
• Because we added the brackets because of the original division sign and we have dealt with
the division, the brackets are no longer needed:
3.528 = X – 8.5
• To undo subtraction we perform the opposite operation, addition:
3.528 + 8.5 = X
• We have now rearranged the equation so that X is on its own.
X = 12.0 (one decimal place)
X − μ
Example 2: If Z = calculate X if Z = 2.58, μ = -2.5, σ = 0.85,
σ
n
and n = 60.
• Clean substitution: 2.58 =
X −−2.5
0.85
60
• Simplify a little:
2.58 =
X + 2.5
0.1097
• Solve the equation: 2.58 × 0.1097 = X + 2.5
0.2830 = X + 2.5
0.2830 − 2.5 = X
X = − 2.22 (to two d.p.)
Appendix 1 – Basic rules and concepts

6. Powers and Logarithms
The following power rules may be needed occasionally, and examples will be given where
necessary.
a b a b
x x x +
a
= ( x ) b
=
ab
x
x
x
a
b
=
x
a−b
x
− a =
1
x
a
1 2 x
=
x
The following log rules may also be needed. Note that in this paper, log means log e (or natural log
i.e. ln).
log
ln
e x
= ln x
ln x= y ←⎯→ e = x (where e=
2.71828 (five d.p.))
y
( x ) yln
( x)
= ln ( x) + ln ( y) = ln ( xy)
⎛
ln ( x) ln ( y)
ln x ⎞
− = ⎜ ⎟
⎝ y ⎠
Example:
y
⎛ ˆ π ⎞
If log⎜
1 ˆ
⎟ = 3.1305 – 1.1499 – 0.027729 x 45 , find the value of the expression
⎝ − π ⎠
• First use BEDMAS to evaluate the RHS (Right Hand Side) of the expression:
ˆ π
.
1 − ˆ π
3.1305 – 1.1499 – 0.027729 × 45 = 3.1305 – 1.1499 – 1.247805
= 0.732795
⎛ ˆ π ⎞
• We now have log⎜
1 ˆ
⎟ = 0.732795.
⎝ − π ⎠
Remembering that log here means ln we are able to rewrite this in exponential form using
the formula
y
ln x = y ←⎯→ e = x
Therefore
⎛ ˆ π ⎞
⎜
1 ˆ
⎟
⎝ − π ⎠
=
0.732795
e = 2.08 (two d.p.)
Appendix 1 – Basic rules and concepts

7. Sigma means Add Up
The Greek letter Σ (capital sigma) means “add up what follows”.
Example 1:
Example 2:
3
Evaluate ∑ 3 i
i= 1
Each of the values 1, 2, and 3 are substituted into the expression one by one
in place of the variable i. Then the three values are added:
3
i 1 2 3
∑ 3 = 3 + 3 + 3
i= 1
= 3 + 9 + 27 = 39
n
Expand the expression ∑ xi
, where x 1 is the first observation, x 2 the
i=
1
second observation, etc. in a data set.
There are n observations. Write out the sum of the first two or three
observations, use three dots to indicate the other values, and add on the final
observation:
n
∑ i
i=
1
x x x x ... x
= 1 + 2 + 3 + +
n
Notation
x i is the i th term from the data set x1, x2, x3 , ..., xn
1,
x
x ij is the (i, j) th term from the data set
x11, x21, ..., xn
1,
1
x12, x22, ..., xn2
2,
...,
x1k, x2k, ..., xn k
k
− n.
Example 3:
If we select 50 female and 50 male Stat 115 students and measure their
heights, we obtain the data set
xij
i = 1, 2 j = 1, 2, . . . , 50
Here i represents sex (1 for female and 2 for male), and j the individual.
For example, x29
is the height of the 9 th male in the sample.
Example 4: Evaluate the expression x =
set {4, 7.5, 3.5, 8}
1
4
4
∑ xi
where x i is the i th
i=
1
• Substitute each of the x i values into the expression and follow BEDMAS:
observation in the
Appendix 1 – Basic rules and concepts

1
x = 4 + 7.5 + 3.5 + 8
4
1
= ( 23 )
4
= 5.75
( )
Example 5: Evaluate the expression v = ( x − x )
4
1
2
∑ i where x i is the i th observation
3
i=
1
in the set {4, 7.5, 3.5, 8} and x = 5.75 (calculated in Example 4).
• Substitute each of the x i values into the expression, along with x = 5.75:
v = 1 ( (4 − 5.75) 2 + (7.5 − 5.75) 2 + (3.5 − 5.75) 2 + (8 − 5.75) 2
)
3
• Follow BEDMAS and evaluate each one of the four inner brackets:
v = 1 ( ( − 1.75) 2 + (1.75) 2 + ( − 2.25) 2 + (2.25) 2
)
3
• The exponents (squares) are calculated and then the four terms are added:
v = 1 ( 3.0625 + 3.0625 + 5.0625 + 5.0625 )
3
1
= ( 16.25 )
3
• The multiplication by 1 is outside the brackets so it is calculated last:
3
v = 5.417 (to three d.p.)
Example 6: Evaluate the expression χ 2 (observed - expected)
= ∑
expected
allcells
for the table below where the expected values are given in brackets and the
observed values are not in brackets:
15 (26) 50 (39)
33 (22) 22 (33)
2
• Note that for this type of sigma expression, the notation means we have to add up the result
from each of the four cells.
χ 2
• Substitute each value into the expression:
χ 2 (15 − 26) (50 − 39) (33 − 22) (22 − 33)
=
+ + +
26 39 22 33
• Evaluate each bracket, and then square the result:
χ 2 =
=
2 2 2 2
2 2 2 2
( −11) ( −11) (11) ( −11)
+ + +
26 39 22 33
121 121 121 121
+ + +
26 39 22 33
a (or equivalent) button to calculate the sum:
• Use the b c
= 16.923 (to three d.p.)
Appendix 1 – Basic rules and concepts

Section 2: Basic Statistical Concepts
1. Mean
The mean x is commonly referred to as the “average”. It is used as a measure of the “centre” of a
data set. To find the mean, simply add up all your data values (observations) and divide by the
number of values (sample size):
n
x1+ x 2 + ... + xn
1
x = or ∑ xi
n n
i =
1
Example:
Calculate the mean of the data set 2, 4, 6, 8, 10, 12.
There are six values in the data set i.e. n = 6.
2 + 4 + 6 + 8 + 10 + 12 42
x = = = 7
6 6
2. Median
The median is defined as the middle observation in the data set, and is another measure of the centre
of the data. Note that the data must be in order before you calculate the median!
• In general, the median is the ( n + 1)
th observation, where n is the sample size.
2
• If there is an odd number of observations, the median will be the middle observation.
• If there is an even number of observations, the median will be the mean of the two middle
observations.
Example 1: Calculate the median of the data set 10, 1, 3, 8, 9.
• First sort the data into order: 1, 3, 8, 9, 10
• There are n = 5 observations so the median in the data set is the ( + )
i.e. 8.
Example 2: Calculate the median of the data set 32, 2, 36, 14, 6, 33.
• First sort the data into order: 2, 6, 14, 32, 33, 36
6 + 1
2
• There are n = 6 observations so the median is the ( ) = 3. 5
14 + 32
2
• Take the mean of the 3 rd and 4 th observations i.e. ( ) = 23
5 1 = 3 rd observation,
2
th observation.
.
Appendix 1 – Basic rules and concepts

3. Range
The range is the difference between the largest and smallest observations in the data set. It is a
measure of the variation in the data.
Example:
The range of the data set 2, 5, 6, 9, 16, 2, 13 is 16 – 2 = 14.
4. Variance and Standard Deviation
• The variance (s 2 ) is calculated as follows:
n
2 1
s = ∑ xi
−x
n −1
i=
1
( ) 2
The standard deviation (s) is the most commonly used measure of variation in a set of data. It is the
square root of the variance
n
1
i.e. s = ∑ ( xi
−x) 2
n −1
i=
1
Usually we calculate the variance first, then we take the square root to give the standard deviation.
(This follows the order of operation indicated by BEDMAS)
Example:
The mean for the data set 9, 5, 6, 4, 16, 2 is 7.0. Calculate the standard deviation:
• First calculate the variance. Substitute in each value, including x = 7 and n = 6:
( )
2 1 (9 7)
2 (5 7)
2 (6 7)
2 (4 7)
2 (16 7)
2 (2 7)
2
s = − + − + − + − + − + −
5
• Evaluate the expression, following BEDMAS:
( )
2 1 2 2 2 2 2 2
s = (2) + ( − 2) + ( − 1) + ( − 3) + (9) + ( − 5)
5
1
= ( 4 + 4 + 1 + 9 + 81 + 25 )
5
1
= (124) = 24.8
5
• Take the square root of the variance to give the standard deviation:
s = 24.8 = 4.98 (to two decimal places)
Appendix 1 – Basic rules and concepts

5. Quartiles and Interquartile Range
There are two quartiles: a lower quartile (Q 1 ) and an upper quartile (Q 3 ). The lower quartile has
25% of the data below it, and the upper quartile has 25% of the data above it.
To find a quartile, first find the median of the data set. Then treat the data above the median (upper
set) and the data below the median (lower set) as separate sets. The lower quartile is the median of
the lower set, while the upper quartile is the median of the upper set.
The interquartile range is the upper quartile minus the lower quartile, and contains 50% of the data.
It is a measure of the variation in the data.
Example 1:
The data set 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21 has a median of 11.
• Therefore the lower set is 1, 3, 5, 7, 9, which has a median of 5. So the lower quartile is 5.
• The upper set is 13, 15, 17, 19, 21, and has a median of 17. So the upper quartile is 17.
• The interquartile range is 17 – 5 = 12
Example 2:
The data set 1, 5, 6, 8, 12, 16, 19, 22, 29, 31, 36, 40 has a median of 17.5.
• The lower set is 1, 5, 6, 8, 12, 16 which has a median of 7, so the lower quartile is 7.
• The upper set is 19, 22, 29, 31, 36, 40, which has a median of 30, so the upper quartile is 30.
• The interquartile range is 30 – 7 = 23.
6. Scatterplot
A scatterplot shows the relationship between two variables. Each observation consists of two
measurements. Often we are interested in the “response” of one measurement to the value of the
other. We try to distinguish between the “response” variable and the “explanatory” variable. The
response variable is plotted on the y-axis (vertical axis) and the explanatory variable on the x-axis
(horizontal axis).
Example:
The weight of 13 students and the amount of time it took them to drink a particular beverage are
plotted below: the explanatory variable is the student’s weight (x-axis) and the response variable is
the time taken to drink the beverage (y-axis).
Time taken to drink beverage
9
8
7
6
5
4
3
2
1
0
50.0 60.0 70.0 80.0 90.0 100.0 110.0 120.0
Weight of Student (kg)
Appendix 1 – Basic rules and concepts

Section 3: Sample Exercise
This sample exercise contains questions based on the Basics Booklet, plus a few questions from
material taught during the first week of the course.
1. In a recent study looking at rainbow trout, researchers measured the lengths of juvenile fish.
The lengths (in cm) for five randomly selected fish were:
18.6, 15.4, 13.4, 17.0, 12.9
Calculate to one decimal place the mean for these data.
2. For a second random sample of six juvenile fish the lengths (in cm) were:
Calculate the median for these data.
15.5, 12.6, 17.5, 17.4, 13.8, 12.2
3. Calculate the range for the data in Question 2.
4. For a third random sample of five juvenile fish the lengths (in cm) were:
14.5, 14.8, 16.5, 18.4, 13.8
The mean for these data is 15.6 (cm). Calculate to one decimal place the standard deviation
for these data.
5. The mean value of 15.6 (cm) in Question 4 is a:
A. Parameter
B. Statistic
C. Distribution
D. Population value
E. Measure of Spread
6. The following list contains five values:
3.2%
0.096
0.048
0.32
0.58%
Beside each value select “True” if the value is less than 0.05 or “False” if the value is greater
than 0.05.
7. Calculate the value of the expression
5
∑ 3i
.
i = 1
8. If
Z.
Z
X − μ
= , with X = 43.6, μ = 48, σ = 8.6 and n = 50, then calculate the value of
σ
n
Appendix 1 – Basic rules and concepts

9. If
1.96
X − 2.8
= , calculate the value of X.
5
10. In a previous STAT110 class at Otago University, 64% of students sitting the paper were
known to be first year students. In a study of students sitting the paper, a random sample of
40 students was taken, and 60% of the students in this sample were found to be first year
students.
To earn the mark in this question you will have to answer correctly both questions below. For
each question select your answer from these five options:
A. 64%
B. 40 students
C. 60%
D. all first year students at Otago University
E. students sitting the paper
Question 1: The statistic in the paragraph above is: . . . . .
Question 2: What is the population . . . . .
Answers
Answers without working are provided. For the working, look through the Basic Booklet above, or
consult your Notes for the first week of the course. If you need help, go to one of the help sessions.
Details of these sessions are provided in the Course Outline at the start of this book.
1. 15.5 cm (1 d.p.)
2. 14.65 cm
3. 5.3 cm
4. 1.9 cm
5. B
6. True, false, true, false, true
7. 45
8. –3.62 (2 d.p.)
9. 12.6
10. B, E
Appendix 1 – Basic rules and concepts

Appendix Two: Some Summaries
1. Some Useful Rules of Probability
2. Random Variables
3. Binomial Distribution
4. Normal Distribution
Basic Probability Rules and Distributions
1. Some Useful Rules of Probability
• Pr(A or B) = Pr(A) + Pr(B) – Pr(A and B)
If we use set notation for this rule, it can be rewritten as
Pr(A ∪ B) = Pr(A) + Pr(B) – Pr(A ∩ B)
A
B
A
B
A ∪ B
A ∩ B
• If A and B are mutually exclusive
(disjoint) then:
Pr(A and B) = 0
or Pr(A ∩ B) = 0
A
B
• If A represents the complement of A
(every event not in A) then
Pr(A) + Pr( A ) = 1
A
A
Appendix 2 – Some summaries

• Probability of B given A: Pr( A ∩ B) = Pr( A) x Pr( B| A)
This may be rewritten as
Pr( B| A)
=
Pr( A ∩ B)
Pr( A)
Pr(B | A)
B
Pr(A ∩ B)
Pr(A)
A
Pr( B | A)
B
Pr(A ∩ B )
Pr(B)
A
Pr(B | A )
B
Pr( A ∩ B)
Pr( B | A )
B
Pr( A ∩ B )
• If A and B are independent then: (i) P(B | A) = P(B)
(ii) P(A ∩B) = P(A) × P(B)
2. Random Variables
• A random variable is one whose value is determined by a random mechanism.
• A continuous random variable can take any value in an interval.
• A discrete random variable can take one of a countable number of values.
3. Binomial Distribution
Suppose
1. We have a fixed number of trials (n)
2. Trials are independent
3. Each trial has only two outcomes (“success” or “failure”)
4. The probability of success (π) is the same for each trial
The total number of successes (X) is a discrete random variable and has a Binomial distribution,
with
⎛n⎞
x n x
Pr( X = x) = ⎜ ⎟ ( 1 )
x π − π −
⎝ ⎠
The mean and variance of the distribution are
μ nπ
σ
2
= nπ 1−
π
Example:
If n = 30 and π = 0.6 then
• μ = n π = 30 x 0.6 = 18
= and ( )
Appendix 2 – Some summaries

2 = n 1− = 30 x 0.6 x 0.4 = 7.2
• σ π( π)
• The standard deviation is σ = 7.2 = 2.68 (to two d.p.)
4. Normal Distribution
A distribution that is commonly used to describe the behaviour of continuous random variables is
the normal distribution.
2
• X ~ N( , σ )
μ means “X has a normal distribution with mean μ and variance
• X ~ N ( 0,1)
means X has a standard normal distribution
2
• If X ~ N( μ , σ )
, then the standardised random variable
For any Normal distribution, approximately:
• 68% of the observations are between μ − σ and μ + σ .
• 95% of the observations are between μ − 2σ
and μ + 2σ
.
• 99.7% of the observations are between μ − 3σ
and μ + 3σ
.
− μ
= X
σ
Z ~ N ( 0,1)
2
σ ”
Example:
If X ~ N( 45, 30 ) then
• μ = 45
• the standard deviation σ = 30 = 5.477 (to three d.p.)
• Approximately 68% of the observations are expected to be between
μ − σ = 39.5 and μ + σ = 50.5.
• Approximately 95% of the observations are expected to be between
34 and 56.
• Over 99% (i.e. almost all) of the observations are expected to be between
28.5 and 61.
• Pr(X < 40) = Pr(Z <
X − μ
)
σ
40 − 45
= Pr(Z < )
5.477
= Pr(Z < –0.913)
= 0.1806
40 45 X
–0.913 0
Z
Appendix 2 – Some summaries

Summary of Formulae
1. Normal Distribution
If X is a normal random variable with parameters µ X (mean) and σ 2 X (variance)
• Mean: µ x
• Standard deviation: σ X =
√
σ 2 X
A standard normal random variable Z has mean µ Z = 0 and σZ 2
variable X into a standard normal (and vice versa):
= 1. To transform a normal random
Z = X − µ X
σ X
and X = Zσ X + µ X .
2. Binomial Distribution
If X is a binomial random variable with n trials and probability π then
• Mean: µ x = nπ
• Standard deviation: σ X = √ nπ(1 − π)
• If nπ and n(1 − π) are both greater than 5, then X is approximately normally distributed with mean
µ X and variance σ 2 X .
3. Distributions of Statistics
• The mean ¯X of a random sample of size n has mean µ ¯X = µ X and standard deviation σ ¯X = σX √ n
.
• The sample proportion P computed from√
a binomial distribution with parameters n and π has a mean
π(1−π)
of µ P = π and standard deviation σ P =
n
. If nπ and n(1 − π) are both greater than 5, then P
will be approximately normally distributed.
• The distribution of the difference between two sample means ¯X 1 − ¯X 2 has a mean of µ ¯X1 − ¯X 2
= µ 1 − µ 2
and a standard deviation of σ ¯X1 − ¯X 2
=
√
σ 2
1
n 1
+ σ2 2
n 2
.
- In large random samples (n 1 and n 2 ≥ 30) σ ¯X1 − ¯X 2
can be estimated by ˆσ ¯X1 − ¯X 2
=
√
s 2
1
- If σ 2 1 = σ2 2 then we can estimate σ ¯X 1 − ¯X 2
by ˆσ ¯X1 − ¯X 2
=
√
(n1 −1)s 2 1 +(n 2−1)s 2 2
n 1 +n 2 −2
4. Contingency tables
√
1
n 1
+ 1 n 2
.
n 1
+ s2 2
n 2
.
Factor 2
Factor 1 Level 1 Level 2 Total
Level 1 w x r 1 = w + x
Level 2 y z r 2 = y + z
c 1 = w + y c 2 = x + z n = w + x + y + z
χ 2 =
2∑
i=1 j=1
2∑ (o ij − e ij ) 2
e ij
where e ij = r ic j
n
and o ij is the observed
value in row i column j.
Odds ratio: OR =(w/x)/(y/z) =(w × z)/(x × y)
Relative risk: RR =(w/(w + x)) / (y/(y + z))
Attributable risk: AR = w/(w + x) − y/(y + z)
Appendix 3 - Formulae

5. Confidence Intervals
All of the 100(1 − α)% confidence intervals calculated in this course are of the form:
Estimate ± multiplier × standard error.
In the following ¯x, p etc are the values calculated from the samples.
Estimate df (ν) Multiplier Standard Error
Population mean
• Random sample, σ x known ¯x NA z α/2
√
σ X
n
• Random normal sample, σ x unknown
and estimated by s
Difference between population means
• Small random samples, normal population,
σ 1 = σ 2 = σ unknown
¯x n − 1 t α/2,ν
s √ n
¯x 1 − ¯x 2 n 1 + n 2 − 2 t α/2,ν
√
(n1 −1)s 2 1 +(n 2−1)s 2 2
n 1 +n 2 −2
• Large random samples (both ≥ 30) ¯x 1 − ¯x 2 NA z α/2
√
s 2
1
• Paired difference in small random ¯d ν = n − 1 t α/2,ν
s d
samples from a normal population
After ANOVA and Regression
• Estimate, multiplier and standard errors determined from output
n 1
+ s2 2
n 2
√n
√
1
n 1
+ 1 n 2
Population proportions
√
p(1−p)
• Population proportion p NA z α/2
√ n
• Difference between 2 population proportions
p 1 − p 2 NA z
p1 (1−p 1 )
α/2 n 1
+ p 2(1−p 2 )
n 2
Odds ratio, relative risk, attributable risk (see contingency tables above for
√
w, x, y and z)
• Log (natural) odds ratio ln(OR) NA z 1
α/2 w + 1 x + 1 y + 1 z
√
• Log (natural) relative risk ln(RR) NA z 1
α/2 w − 1
w+x + 1 y − 1
y+z
• Attributable risk –as for the difference of two population proportions with p 1 = w/(w + x) and p 2 = y/(y + z)
6. Regression
ŷ = ˆβ 0 + ˆβ 1 x where ˆβ 1 =
where s e =
7. ANOVA
√ ∑(yi
−ŷ i ) 2
n−2
1. Total SS = Treatment SS + Error SS
2. Total df = Treatment df + Error df
∑ (xi −¯x)(y i −ȳ)
∑ (xi −¯x) 2 and ˆβ 0 =ȳ − ˆβ 1¯x. Standard error of the slope SE( ˆβ 1 )=
= √ MS Residual. Standard error of a forecast at x k = s e
√
1+ 1 n + (x k−¯x) 2 ∑ (xi −¯x) 2 .
3. MS Treatment = Treatment SS/Treatment df and MS Error = Error SS/Error df
4. Overall mean SS = nȳ 2 where n = n 1 + ...+ n k and ȳ = 1 n (n 1ȳ 1 + ...+ n k ȳ k ).
5. Treatment SS = C2 1
n 1
+ C2 2
n 2
+ ...+ C2 k
n k
− nȳ 2 where C j is the jth column total.
√ s e ∑(xi , −¯x) 2
Appendix 3 - Formulae