STAT170 Workshop Notes prepared by Nan Carter for Numeracy ...

STAT170 Workshop Notes prepared by Nan Carter 

for Numeracy Centre, Macquarie University, 2010. 

Week 3 H/Y. Questions prepared for Stat170 workshop held at the Numeracy Centre, 

Macquarie University. Prepared by Nan Carter with examples from various sources 

and adapted for Stat170. 

The role of a statistician is to obtain information about a ‘population’ using a sample. 

Question 1. For each of the following cases, give the 

• Population of interest (also called ‘the target population’) 

• Sample to be selected (is it a random/representative/convenient sample?) 

• Variable(s) of interest 

• Type of variable – categorical: binary, nominal, ordinal 

Continuous (interval) 

a) We are interested in the average annual income of all registered medical 

practitioners in the Sydney Metropolitan area. From the register we randomly 

select 50 names and hence collect information regarding their annual incomes. 

b) We are interested in determining, for the residents of NSW, which (if any) of the 

four major blood groupings (A, B, AB, O) is the most common and which may be 

considered rare. A sample of 200 people is randomly selected from blood donors in 

all major cities and the blood group of each sample is recorded. 

c) A quality control supervisor in a factory is interested in the proportions of defective 

items produced per day. At various times during each day he randomly selects 

finished items so that by the end of the day he has 300 items. He then records the 

number of defective items found. Data is collected over 21 days. 

d) In a sociological study, a researcher is interested in the number of children in the 

family of couples who have been married for ten years. She selects a random 

sample of 25 couples married for that length of time, and records the number of 

children in the family. 

e) You are interested in the relationship between sex and income of all adults who 

reside in Sydney. You select a random sample of 75 adult males and 75 adult 

females from the electoral roll, and ask each person to give you their income. 

Question 2. For each part of question 1, state what you think would be an appropriate 

graphical summary to use to display the data collected. 

1 

Nan Carter: workshop notes prepared for Numeracy Centre Macquarie University.

Question 3. A certain bank has many branches throughout Australia. The age of the 

branch managers of a random sample of 156 of these branches is given as follows: 

Age Frequency 

30-34 8 

35-39 21 

40-44 24 

45-49 32 

50-54 40 

55-59 23 

60-64 8 

Give a graphical summary for these data. 

CHECK ANSWERS: 1.a) 

• Popln: All registered medical practitioners in the Sydney Metropolitan Area 

• Sample: the 50 randomly selected names 

• Variable: annual incomes 

• Type of data: continuous (numerical) 

1.b) 

• Popln: all residents in NSW 

• Sample: 200 blood donors from all major cities 

• Variable: major blood grouping 

• Type of data: categorical, nominal (A,B,AB,O) 

1.c) 

• Popln: days of production 

• Sample: the 21 days for which data were collected 

• Variable: the number of defectives found 

• Type of data: numerical, but is it ordinal or continuous? 

1.d) 

• Popln: all couples married for 10 years 

• Sample: 25 such couples 

• Variable: number of children in the family 

• Type of data: categorical, ordinal 

1.e) 

• Popln: adults residing in Sydney 

• Sample: 75 males and 75 females registered to vote 

• Variables: income and sex 

• Type of data: continuous for income and binary for sex. 

Question 2. a): histogram, boxplot, stem and leaf b): barchart (or pie chart) c): 

barchart or histogram d): barchart e): comparative boxplot 

Question 3. Histogram. 

2 


Week 4 H/Y. Exercises for STAT170 workshop held at Numeracy Centre, Macquarie 

University; prepared by Nan Carter. Wk4_5h 

Observations on many variables are ‘normally distributed’ (the histogram is close to 

being symmetrical, and has one hump, is unimodal). We can relate the area under a 

curve between any two values of the variable to the probability of a random value 

occurring by chance. This concept of relating the probability of a value occurring by 

chance and the area under a curve, is used for other distributions e.g. the ‘tdistribution, 

the chi-square distribution. By convention, the symbol z is used for a 

variable that is normal, with mean zero and standard deviation 1. N(0,1) is the 

‘standard normal variable’. 

Suppose we have a normal distribution with mean μ = 10, and standard deviation σ = 2. 

What is the probability that a value, x, selected at random from the population, will be: 

a) between 7 and 14? b) between 12 and 15? c)less than 6.5?; c) equal to 15 or 

more extreme? 

1. a) If 37.7% of a normal population has a standard score between zero some 

positive z-value, find that z-value. 

b) what is the probability that a z-score will lie between zero and 1.16? 

c) what is the probability that a randomly chosen z-score will lie between zero 

and 1.16? 

In the following questions, use the table of single-tail areas for the standardised normal 

curve that is given in your study guides and textbook. Y is the symbol for the normal 

variable. 

3. If μ = 5, σ = 1 and z = 2.05, what is the value of y? and 

if probability, p, is 0.02, what is the value of z? 

4. Let μ = 150 and σ = 24 

a) if p = 0.0188, what is z? 

b) if z = -1.25, what is y? 

b) if z = 0.42 what is p? 

c) what is the probability of a value of z that is more extreme than z = -2.08? 

Finding the probability that a value could be more extreme than a given value is 

used in testing nhypotheses. 

5. If the probability that a box of kiwi fruit is underweight is 0.1992, how many 

underweight boxes would you expect in a consignment of 150 boxes? If the probability 

were 0.0150, how many would you expect? 

6. a) what is the probability of a value of z that is more extreme than z = -1.43? 

b)If z = 0.71, what is p? 

c)What is the area under the curve between z = -1.43 and z = 0.71? 

d) If the area under the curve in the upper tail is 0.22, what is the value of z? 

7. a) If z = 1.00 what is p? 

b) If p = 0.03, what is the value of z? 

3 


CHECKS: Q1: a) 0.9104 b) 0.1525 c) 0.0401 Q2: a) 1.16 

b) 0.3770 c) 0.3770 Q3: 7.05 and 2.05 Q4: a) 2.08 b) 120 

c) 0.3372 d) 0.0188 Q5: 29 ; 27 Q6: a) 0.0764 b) 0.2389 

c) 0.6847 d) 0.77 Q7: a) 0.1587 b) 1.88 

ALL the questions on this page need to be interpreted as finding z-values, y-values or 

probabilities (i.e. area under the standard normal curve). The calculations that you 

need have already been done above; BUT, not all the above answers are needed, AND 

the order of the questions has been scrambled. 

1. The height of adult women in Australia is approx normally distributed with mean 

μ=160cm, and standard deviation σ=7cm. 

a) What is the probability that a single randomly selected woman has height 

between 150cm and 165cm? 

b) What is the probability that a randomly selected woman has a height less than 

or equal to your own height? (eg my height is 158cm). 

c) What is the greatest height exceeded by 22% of the population of adult women? 

2. The useful life of a certain brand of batteries is normally distributed with a mean of 

150 hours and a standard deviation of 24 hours. If a battery is selected at random, 

find the probability that its useful life is: 

a) less than 100 hours 

b) more than 120 hours 

c) between 160 and 200 hours. 

3. The weights of bags of fertiliser are normally distributed with mean 210 kg and 

standard deviation of 2.3 kg. For a consignment of 150 bags, what is the expected 

number of bags that will weigh less than 208 kg? How many of the bags weighing 

less that 208 kg would be expected to weigh more than 205 kg? 

4. A soft drink machine discharges on average 250ml per cup, with a standard 

deviation of 14 ml. The ml of fill is approximately normally distributed. 

e) What is the probability of overfilling a 264 ml cup? 

f) How large a cup is required so that the probability that the cup will overflow is 

only 3%? 

5. The average life of an electric sandwich maker is 5 years with a standard deviation 

of one year. If the manufacturer replaces free any appliance that fails while under 

guarantee, how long a guarantee should be offered if the profit depends on no more 

than 2% being replaced? (Assume lifetimes are normally distributed.) 

CHECKS: Q1. a) 0.6847 b) for my height of 158cm: 0.3859 c) 165.4 

Q2. a) 0.0188 b) 0.8944 c) 0.3184 

Q3. 29 ; 27 Q4. a) 0.1587 b) 276 Q5. 7 years 

4 


Week 5 H/Y. Questions on SAMPLING DISTRIBUTION of MEANS and 

PROPORTIONS prepared for Stat170 workshop held at the Numeracy 

Centre, Macquarie University, by Nan Carter, with examples 

from various sources and adapted for Stat170. 

Here, we need to keep in mind that we are working with two 

distributions: 

• the distribution of the individuals in the population from 

which we take a random sample; this distribution has mean μ 

and standard deviation σ. 

• the distribution of means of samples that are all the same 

size (n); this is the sampling distribution of means of 

samples of size n; this distribution has mean μ and standard 

deviation σ/√n (also referred to as the ‘standard error’). 

Notice that 

• The two distributions have the same mean μ 

• the means of samples are less variable than the individuals 

in the ‘parent’ population (the means have s.d. σ/√n where 

the individuals have s.d. σ) 

• if you take larger samples (i.e. n is larger) the means of 

these samples are even less variable (s.d. σ/√n gets smaller 

as n increases). 

EXERCISE 1. From past experience it is known that IQs are 

normally distributed with mean μ = 100 and standard deviation 

σ = 14. 

• what are the mean and the standard deviation of the sampling 

distribution of means of samples of size 16? 

• what would be the mean and standard deviation of the 

sampling distribution if the samples were of size 49? 

EXERCISE 2. Recall that if our variable of interest in the 

parent population is normally distributed then the sampling 

distribution of means is also a normal distribution. 

• If we take a sample of size 16 from our population and find 

their IQs, what is the probability that the mean IQ for the 

sample will lie between 95 and 105? 

• If the sample were of size 49, what would the probability 

be? 

• If the sample were of size 100, what would the probability 

be? 

5 


EXERCISE 3. The heights of the adult males in a particular 

country are known to be normally distributed with a mean of 

1.7 metres and a standard deviation of 0.09 metres. Find the 

probability that 

• an individual man, randomly selected, has a height of less 

than 1.5 metres. 

• a random sample of 25 men has average height of more than 

1.72 metres. 

**** In all the above exercises the distribution of the variable of interest for 

individuals in the parent population was ‘normal’. What happens when it is not 

normally distributed (or when its distribution is unknown)? There is a theorem that 

proves that if the samples are large enough then the sampling distribution of means is 

approximately normal (Central Limit Theorem). At Macquarie we can apply the CLT if 

n>25. This is a most important and useful result: we can always get a normal variable by 

taking a large-enough sample and working with the sample mean. 

EXERCISE 4. A certain brand of rope is known to have a 

breaking strength on average of 25 kg with a standard 

deviation of 0.5 kg. A random sample of 50 pieces of rope is 

tested for breaking strength. What is the probability that 

the mean strength is found to be 

• between 24.9 and 25.1 kg? 

• less than 24.9 kg? 

EXERCISE 5. A wire company that manufactures wires for circus 

acts has taken a sample of 100 pieces of wire and wishes to 

see if the thickness of a batch of wires meets minimum 

specifications. Assume that the population data have a mean 

of μ = 0.45 cm with a standard deviation of σ= 0.03 cm. 

• Calculate the mean and standard error of the sampling 

distribution of means of such samples. 

• What may be said about the shape of this sampling 

distribution? 

• What is the probability that the sample mean is at least 

0.448 cm? 

EXERCISE 6. A certain trucking company wants to estimate the 

average tonnage of freight handled per month, and it has taken 

a sample of 36 months. Assume the true average tonnage per 

month is 225 tonnes, with SD of 30 tonnes. What is the 

probability that the sample mean will have a value within 7 

tonnes of the true mean? 

6 


EXAMPLES ON WORKING WITH PROPORTIONS. 

EXERCISE 7. A manufacturer claims that 95% of the components 

supplied for a new jet transport meet a specified rigid 

standard of performance (ie 5% do not). 

If a sample of 400 of the components is tested, what is the 

probability that 30 do NOT meet the performance standard? 

EXERCISE 8. A report claims that the percentage of 

miscarriages among LSD users is 12%. From the files in the 

maternity ward of a Sydney hospital, out of 100 pregnant women 

who used LSD, 24 had miscarriages. Comment on the claim made 

in the report 

EXERCISE 9.The president of a certain company believes that 

30% of the company’s orders come from new or first-time 

customers. What would be the probability that, in a random 

sample of 100 orders, 40 are from new customers? 

CHECKS (not solutions): Q1: 100 and 3.5; 100 and 2.0 Q2: 

0.8472 ; 0.9876 ; 0.99964 Q3: 0.0132 ; 0.1335 Q4: The 

distribution of breaking strength is not known, but the sample 

is large (50) so, by the CLT, the sample mean is normally 

distributed; 0.8414 ; 0.0793. 

Q5: 0.45 and 0.003; it is reasonably symmetric about 0.45 cm; 

0.7486 Q6: 0.8384 Q7: z=2.294 Q8: z = 3.693; 

Q9: z = 2.182. 

7 


Week 6 H/Y. Questions on CONFIDENCE INTERVALS & HYPOTHESIS 

TESTING prepared for Stat170 workshop held at the Numeracy 

Centre, Macquarie University. 

NOTE. FOR THIS TOPIC we still need to keep in mind that we 

are working with two distributions: 

• the distribution of the individuals in the population from 

which we take a random sample; this distribution has mean μ 

and standard deviation σ. 

• the distribution of means of samples that are all the same 

size (n); this is the sampling distribution of means of 

samples of size n; this distribution has mean μ and standard 


• If we need to use the sample standard deviation, s, in a 

test of hypothesis about a mean μ, the variable is no longer 

a z variable but a t-variable with n-1 degrees of freedom. 

Similarly, in a 95% confidence interval the z-value 1.96 is 

replaced by the appropriate t-value with (n-1) degrees of 

freedom. 

EXERCISE 1. The daily consumption of electric power in a 

certain city is known to be normally distributed, and it is 

claimed to have a mean of 6.5 units with a standard deviation 

of σ = 1.5. In order to estimate the true mean daily power 

consumption, the consumption was sampled on 18 randomly 

selected days and the mean consumption was 6.973. (a) Test 

the hypothesis that that daily consumption has not, on 

average, changed. (b)Using the sample, find a 95% confidence 

interval for the true mean daily power consumption, μ. How 

confident are you that the claim that mean daily consumption 

is 6.5 units is true? 

EXERCISE 2. Washers are sold as having an inside diameter of 1.75cm. A sample of 

200 washers produced by the machine has mean inside diameter of 1.77 cm. It is known 

from past experience that the standard deviation of the diameter of washers produced 

by this machine is 0.18 cm. (a) Test the hypothesis that the machine is still producing 

washers with an inside diameter of 1.75 cm on average. (b) Using the sample, find a 

95% confidence interval for the mean inside diameter of all washers currently being 

produced by the machine. Are you at least 95% confident that the washers do on 

average meet the claimed diameter size. 

EXERCISE 3. Telecom wants to estimate the average length of telephone calls made 

between two cities. Manager A claims it is close to 2 minutes, and Manager B disputes 

this saying it is shorter, more likely 1.5 minutes. From a sample of 36 randomly selected 

calls, it finds that the average length is 1.90 minutes. Historically, the standard deviation 

of lengths of calls has been found to be 0.53 minutes. 

• Using the data from the random sample, find a 95% confidence 

interval for the true average length of calls. Which 

Manager’s claim was closer? (A or B) 

8 


EXERCISE 4. Suppose we wanted to place an order for wire that 

had a breaking strength of 8kg on average with a standard 

deviation of 1.6kg. We first bought a random sample of 50 

lengths, measured the breaking strength of each and found the 

mean was 7.8kg. Could we be 95% confident that a large order 

would meet our specification on average? 

EXERCISE 5. A quality control engineer wishes to check the 

mean weight of potato chip bags produced on his machines. He 

takes a random sample of 36 bags and finds their mean weight 

is 199.0 grams. Assuming that the standard deviation of all 

weights is σ = 3.6 grams, Can he be 95% sure that the machines 

could be producing bags that are on average 200 grams? Check 

your answer using an appropriate hypothesis test. 

EXERCISE 6. A metallurgist claims that the mean hardness of a 

die-cast aluminium product is 13.7 with a standard deviation 

of 0.8. He gives us a random sample of 50 pieces and we find 

the mean hardness is 14.3. We claim these are likely to be 

too hard for our purposes. Are we making the correct 

decision? 

ANSWERS: Q1: (6.280,7.666) Q2: (1.75,1.79) Q3: (1.73,2.07) ; (1.67,2.13) 

Q4: Yes, the 95% CI is (7.36,8.24) Q5: Yes, the 95% CI is (196.8,199.2) Q6: Yes, it is 

possible because we are 95% confident that the true mean is between 14.08 and 14.52. 

EXAMPLES ON WORKING WITH PROPORTIONS. EXERCISE 7. A 

manufacturer claims that 95% of the components supplied for a 

new jet transport meet a specified rigid standard of 

performance (ie 5% do not). If a sample of 400 of the 

components is tested, what is the probability that 30 do NOT 

meet the performance standard? 

EXERCISE 8. A report claims that the percentage of 

miscarriages among LSD users is 12%. From the files in the 

maternity ward of a Sydney hospital, out of 100 pregnant women 

who used LSD, 24 had miscarriages. Comment on the claim made 

in the report. 

EXERCISE 9.The president of a certain company believes that 

30% of the company’s orders come from new or first-time 

customers. What would be the probability that, in a random 

sample of 100 orders, 40 are from new customers? 

CHECKS (not solutions): Q1: 100 and 3.5; 100 and 2.0 Q2: 

0.8472 ; 0.9876 ; 0.99964 Q3: 0.0132 ; 0.1335 Q4: The 

distribution of breaking strength is not known, but the sample 

is large (50) so, by the CLT, the sample mean is normally 

distributed; 0.8414 ; 0.0793. Q5: 0.45 and 0.003; it is 

reasonably symmetric about 0.45 cm; 0.7486 Q6: 0.8384 Q7: 

z=2.294 Q8: z = 3.693; Q9: z = 2.182. 

9 


Week 7 H/Y TESTING HYPOTHESES ABOUT POPULATION MEANS 

i. Hypotheses about a population mean, µ, using one random sample of 

observations, without value for σ . 

Without a known value for the population standard deviation, σ, the 

hypothesis is tested using the standard deviation, s, of the random sample. 

For a sample size n, the sample estimate s has (n-1) degrees of freedom. The 

test statistic is no longer a z-statistic, but a t-statistic. 

Go back to the examples in part (i) on pages 10-11, and in the description 

suppose that the given value of σ is the sample estimate, s. Now answer the 

questions Q1 to Q12 using these values of s, and the appropriate tdistribution. 

ii. One random sample of items, two variables observed on each 

item, and we are interested in the difference between the means of the 

variables. The two variables are said to be ‘paired’ or ‘matched’ because they 

are observed on the same item. 

PAIRED t-TEST: 

QUESTION 1. (from Stat171 lecture overheads, prepared by Stephen Brown) 

In an experiment using 15 hypertensive patients, the effect of the drug Captopril on their 

blood pressure is assessed by recording their blood pressure, giving them Captopril and then 

measuring their blood pressure again two hours later. Does the drug have any effect on blood 

pressure? 

10 

Patient Blood pressure Blood pressure DIFFERENCE 

BEFORE drug AFTER drug BEFORE - AFTER 

1 130 125 5 

2 122 121 1 

3 124 121 3 

4 104 106 -2 

5 112 101 11 

6 101 85 16 

7 121 98 23 

8 124 105 19 

9 115 103 12 

10 102 98 4 

11 98 90 8 

12 119 98 21 

13 106 110 -4 

14 107 103 4 

15 100 82 18 

Notice that 

• there is only one random sample of patients 

• there are two observations on each patient and we are interested in the difference between 

the two 

• the change is expressed as the drop in blood pressure. 


We now have a single sample of differences, and the researcher is expecting that there will be 

a drop in blood pressure after taking the Captopril. 

The Statistical hypothesis is that for population of treated patients the mean drop is zero. 

So, H0 : μd = 0 

Assumptions: we only assume that the differences are Normally distributed, and that the 

differences are independent. 

COMMENTS 

• To do a paired t-test, the experimenter must plan ahead to collect the data in the 

appropriate way. 

• As students in Stat170, you are expected to recognise paired data from the information 

given in the question. 

• Remember that you will be interested in the difference between the two observations, that 

is, the null hypothesis will be Ho: μd = 0. 

• The pairing is not always as obvious as the ‘before’ and ‘after’ example above; see for 

example, question 2, where we have observations on pairs of dogs, each pair taken from 

the same litter (dogs from the same litter have the same parents, and hence should be 

more alike genetically than dogs from different litters). Here, the random sample is of 

litters of dogs arriving at the pound. 

• See also question 2 in the 1998 mid-year examination, where you have a random sample 

of women and information on their incomes and the incomes of their partners: the 

question relates to the difference between a woman’s and her partner’s income. 

• ***** ASSUMPTIONS: We need make absolutely no assumptions about the original 

observations. If they are not normally distributed, it does not matter, as long as the 

differences are. Plot the differences to check that the assumption is not seriously violated. 

QUESTION 2. (from Stat171 tutorial exercises) 

In order to investigate whether the time of neutering of male 

dogs had any effect on their rate of growth, Crenshaw and 

Carter (1995) neutered dogs at 2 months and 7 months of age 

and measured their weight at 18 months of age. As litters of 

any breed became available at a pound, two male dogs were 

randomly selected from each litter. One was neuterd at 2 

months, the other was neutered at 7 months. All the dogs were 

kept under identical conditions during the 18 months of the 

trial. The weights (in kg) of the dogs at 18 months were: 

Litter 1 2 3 4 5 6 7 8 9 10 

2 months 19.0 23.2 16.1 30.3 27.1 22.3 18.1 27.8 34.2 17.3 

7 months 18.9 23.5 16.7 30.3 27.8 22.4 18.2 28.1 34.1 17.7 

a) Is there any evidence that early neutering affects the rate of growth of the dogs? 

b) Comment on why you think the experimenter chose this particular design in setting up 

the experiment. Do you think the right choice of design was made? 

11 


QUESTION 3. A researcher wants to know the average sucrose content in a certain 

concentration of sugar-beet juice. There are two methods of measurement and the researcher 

wants to find out whether they both give the same mean dry weight. Ten sample containers 

of beet juice were randomly selected from ten different batches. Half the liquid in each 

container is measured using method A and the remaining half using method B. The 

measuring process yields the following data: 

METH 

OD 

12 

BATCH 

1 2 3 4 5 6 7 8 9 10 

A 11.0 5.0 9.8 5.7 6.5 8.2 5.9 6.0 7.5 5.4 

B 11.1 5.6 9.7 5.3 6.7 8.5 5.6 5.8 7.1 5.5 

Do these data support the hypothesis that both methods of 

measuring yield the same mean concentration? 

CHECKS: 

Q1: mean diffce = 9.267; s = 8.614; t=4.166; df=14; 0.0005

Week 8 H/Y. STAT170 workshop on the comparison of the means of two 

samples. (Prepared by Nan Carter for workshops at the Numeracy Centre; material 

taken from various sources and adapted for Stat170.) 

TWO INDEPENDENT RANDOM SAMPLES and the SAME VARIABLE OBSERVED 

ON BOTH SAMPLES 

We are interested in whether the two samples could come from two populations with the 

same mean: 

H0: μ1 = μ2 

OR: to test if two samples come from the same population. 

THIS TEST IS ONLY VALID if both populations are NORMAL with the SAME SD. 

13 

POOL the two estimates of the unknown SD σ to find sp, 

then use the two-independent sample t-test. 

The SE(ybar1 - ybar2) = sp√(1/n1 + 1/n2), and 

t =(ybar1 ybar2)/SE(ybar1 ybar2) with (n1+n2–2) df 

QUESTION 1: The Chapin Social Insight Test is a psychological test designed to measure 

how accurately the subject appraises other people. The possible scores on the test range from 

0 to 41. During the development of the Chapin Test, it was given to several groups of people. 

Here are the results for male and female students majoring in the liberal arts: 

Group Sex N X-bar S.D. 

1 Male 133 25.34 5.05 

2 Female 162 24.94 5.44 

Do these data support the contention that female and male students differ in average social 

insight? 

QUESTION 2: Two groups of children were given visual acuity tests. Group 1 was 

composed of 11 children who receive their health care from private physicians. The mean 

score for this group was 26 with a standard deviation of 5. The second group, consisting of 

14 children who receive their health care from the health department, had an average score of 

21 with a standard deviation of 6. Assuming normally distributed scores for the populations 

of private and health department patients, find if the two groups of children have different 

scores on average. Find 90% and 95% confidence intervals for the difference between the 

mean scores of the two populations. 

QUESTION 3: An educator believes that new, directed reading activities in the classroom 

will help elementary school pupils improve some aspects of their reading ability. She 

arranges for a third grade class of 26 students to take part in these activities for an 8-week 

period. A control class room of 28 third graders follows the same curriculum without the 

activities. At the end of 8 weeks, all students are given a Degree of Reading Power (DRP) 

test, which measures the aspects of reading ability that the treatment is designed to improve. 


The summary statistics are: 

Group number x-bar SD 

Activity group 26 51.48 11.01 

Control group 28 41.52 17.15 

Is there a difference, on average, between the two groups of students? State any assumptions 

that you have made. 

ANSWERS: Q1: pooled SD = 5.2679 SE(y1bar – y2bar) = 0.6164 

t= 0.649 df = 293 p-value < 0.05 

Q2: pooled sd = 5.5873 SE(y1bar – y2bar) = 2.2512 

t= 2.221 0.02

Week 9 H/Y. Stat170 workshop on Hypothesis Testing, prepared by Nan Carter for 

Numeracy Centre, Macquarie University. Examples taken from various sources and 

adapted for Stat170. 

This document is made up from workshops that cover hypothesis testing about means and 

proportions: the hypotheses are about 

i. population mean, µ, using one random sample of observations, with value for σ 

ii. population proportion, π, using one random sample of observations 

iii. population mean, µ, using one random sample of observations, without value for σ 

iv. two population means and interested in the difference, µd = µ1- µ2, using one 

random sample of items and the two variables observed on each item 

v. two population means using two random samples of items, without values for σ1 , σ2 

i. SAMPLING DISTRIBUTION of MEANS We start by working with 

distributions for which the standard deviation σ is known, AND we are 

interested in testing an hypothesis about the mean of the population using data. 

from a single random sample of items. 

NOTE 1. Recall that we need to keep in mind that we are working with two distributions: 

• the distribution of the individuals in the population from which we take a random sample; 

this distribution has mean μ and standard deviation σ. 

• the distribution of means of samples that are all the same size (n); this is the sampling 

distribution of means of samples of size n; this distribution has mean μ and standard 


Notice that 

• The two distributions have the same mean μ 

• the means of samples are less variable than the individuals in the ‘parent’ population (the 

means have s.d. σ/√n where the individuals have s.d. σ) 

• if you take larger samples (i.e. n is larger) the means of these samples are even less 

variable (s.d. σ/√n gets smaller as n increases). 

{RECALL: the Central Limit Theorem, which proves that, if the parent distribution is 

NOT NORMAL (or NOT KNOWN) BUT the samples are large enough, then the sampling 

distribution of means is approximately normal. At Macquarie we can apply the CLT if n>25. 

THIS MEANS THAT, ONE WAY OR ANOTHER, WE CAN HAVE A SAMPLING 

DISTRIBUTION OF MEANS THAT IS NORMAL (APPROX at least).} 

NOTE 2. If we have a normal distribution with mean μ and standard deviation σ/√n, then 

95% of the values in the distribution lie in the interval from (μ - 1.96 x σ/√n to μ + 1.96 x 

σ/√n). This means that, if we took lots and lots of samples and found the mean of each, we 

would expect that 95% of those means would lie in that interval. 

NOTE 3. IT ALSO MEANS THAT (think this through carefully): 

• if we do not know μ, but have a lot of estimates ybar from samples (all of the same size, 

n) and for each ybar we calculate the interval (ybar - 1.96 x σ/√n to ybar + 1.96 x σ/√n), 

then we expect that 95% of these intervals will include the true value of μ. IN OTHER 

WORDS: we can expect that the true value of μ will be included in 95% of the intervals. 

15 


• SO: if we have only one sample and find its mean, ybar, we can say that we are 95% 

confident that the true mean μ lies within the interval (ybar - 1.96 x σ/√n to ybar + 1.96 x 

σ/√n). This is what is called a 95% confidence interval for the population mean μ. 

EXERCISE 1. The daily consumption of electric power in a certain city is known to be 

normally distributed with a standard deviation of σ = 1.5. In order to estimate the true 

mean daily power consumption, the consumption was sampled on 18 randomly selected days 

and the mean consumption was 6.973. Find a 95% confidence interval for the true mean 

power consumption, μ. 

EXERCISE 2. The mean inside diameter of a sample of 200 

washers produced by a machine is 1.77 cm. It is known from 

past experience that the standard deviation of the diameter of 

washers produced by this machine is 0.18 cm. Find a 95% 

confidence interval for the mean inside diameter of all 

washers produced by the machine. 

EXERCISE 3. Telecom wants to estimate the average length of 

telephone calls made between two cities. From a sample of 36 

randomly selected calls, it finds that the average length is 

1.90 minutes. Historically, the standard deviation of lengths 

of calls has been found to be 0.53 minutes. 

• Find a 95% confidence interval for the true mean length of calls. 

• (optional) What would be a 99% confidence interval? 

EXERCISE 4. Suppose we wanted to place an order for wire that had a breaking strength of 

8kg on average with a standard deviation of 1.6kg. We first bought a random sample of 50 

lengths, measured the breaking strength of each and found the mean was 7.8kg. Could we be 

95% confident that a large order would meet our specification on average? 

EXERCISE 5. A quality control engineer wishes to check the mean weight of potato chip 

bags produced on his machines. He takes a random sample of 36 bags and find their mean 

weight is 198 grams. Assuming that the standard deviation of all weights is σ = 3.6 grams, 

Can he be 95% sure that the machines are producing bags that are on average 198.8 grams? 

EXERCISE 6. A metallurgist claims that the mean hardness of die-cast aluminium is 13.7 

with a standard deviation of 0.8. He gives us a random sample of 50 pieces and we find the 

mean hardness is 14.3. We claim these are likely to be too hard for our purposes. Are we 

making the correct decision? 

EXERCISE 7. Go back to Q1, and use a test of hypothesis to answer the research question: is 

the consumption of power equal to 6.5? 

EXERCISE 8: In Q2 the research question is whether the mean inside diameter is on average 

1.80cm. 

EXERCISE 9: In Q3 test the hypothesis that the mean length of call is 2 minutes. 

EXERCISE 10: In Q4 use a hypothesis test to find if a large order would meet our 

specification on average. 

16 


EXERCISE 11: In Q5 use a hypothesis test to check the average weight of potato chip bags. 

EXERCISE 12: Use a hypothesis test to answer our question 6. 

CHECK ANSWERS: Q1: (6.280,7.666) Q2: (1.75,1.79) Q3: (1.73,2.07) ; (1.67,2.13) 

Q4: Yes, the 95% CI is (7.36,8.24) Q5: Yes, the 95% CI is (196.8,199.2) Q6: Yes, it is 

possible because we are 95% confident that the true mean is between 14.08 and 14.52. 

Q7: H 0: μ = 6.5; z = 1.33; do not reject H 0 and conclude that the mean daily 

consumption could be 6.5. 

ii. Examples on working with hypotheses about population proportions, 

prepared for STAT170 workshop at Numeracy Centre, by Nan Carter from 

various sources. 

17 

A population proportion, π,is estimated by a random 

sample proportion, p. The population proportion has SE = 

√{ π(1- π) / (n-1)} and, if π is not known, SE is estimated 

by substituting the sample proportion p for π. The 

proportion in a random sample of size n can be assumed 

normally distributed if n π and n(1- π) are BOTH at least 5. 

1. A manufacturer claims that 95% of the components supplied 

for a new jet transport meet a specified rigid standard of 

performance (ie 5% do not). A random sample of 400 of the 

components is tested, and 30 do not meet the performance 

standard. Comment on the manufacturer’s claim. 

2. A report claims that the percentage of miscarriages among LSD users is 12%. From the 

files in the maternity ward of a Sydney hospital, out of 100 pregnant women who used 

LSD, 24 had miscarriages. Comment on the claim made in the report 

3. The president of a certain company believes that 30% of the 

company’s orders come from new or first-time customers. In 

a random sample of 100 orders, 40 are from new customers. 

Comment on the president’s claim. 

CHECKS: Q1 z=2.294 Q2: z = 3.693; Q3: z = 2.182; 

iii. Hypotheses about a population mean, µ, using one random sample of 

observations, without value for σ . 

Without a known value for the population standard deviation, σ, the 

hypothesis is tested using the random sample standard deviation, s. For a 

sample size n, the sample estimate s has (n-1) degrees of freedom. The test 

statistic is no longer z, but the t-statistic. 

Go back to the examples in part (i) on pages 2-3, and in the description 

suppose that the given value of σ is the sample estimate, s. Now answer the 


questions Q1 to Q12 using these values of s, and the appropriate tdistribution. 

iv. One random sample of items, two variables observed on each 

item, and we are interested in the difference between the means of the 

variables. The two variables are said to be ‘paired’ or ‘matched’ because they 

are observed on the same item. 

18 


PAIRED t-TEST: 

QUESTION 1. (from Stat171 lecture overheads, prepared by Stephen Brown) 

In an experiment using 15 hypertensive patients, the effect of the drug Captopril on their 

blood pressure is assessed by recording their blood pressure, giving them Captopril and then 

measuring their blood pressure again two hours later. Does the drug have any effect on blood 

pressure? 

19 

Patient Blood pressure Blood pressure DIFFERENCE 

BEFORE drug AFTER drug BEFORE - AFTER 

1 130 125 5 

2 122 121 1 

3 124 121 3 

4 104 106 -2 

5 112 101 11 

6 101 85 16 

7 121 98 23 

8 124 105 19 

9 115 103 12 

10 102 98 4 

11 98 90 8 

12 119 98 21 

13 106 110 -4 

14 107 103 4 

15 100 82 18 

Notice that 

• there is only one random sample of patients 

• there are two observations on each patient and we are interested in the difference between 

the two 

• the change is expressed as the drop in blood pressure. 

We now have a single sample of differences, and the researcher is expecting that there will be 

a drop in blood pressure after taking the Captopril. 

The Statistical hypothesis is that for population of treated patients the mean drop is zero. 

So, H0 : μd = 0 

Assumptions: we only assume that the differences are Normally distributed, and that the 

differences are independent. 

COMMENTS 

• To do a paired t-test, the experimenter must plan ahead to collect the data in the 

appropriate way. 

• As students in Stat170, you are expected to recognise paired data from the information 

given in the question. 

• Remember that you will be interested in the difference between the two observations, that 

is, the null hypothesis will be Ho: μd = 0. 


• The pairing is not always as obvious as the ‘before’ and ‘after’ example above; see for 

example, question 2, where we have observations on pairs of dogs, each pair taken from 

the same litter (dogs from the same litter have the same parents, and hence should be 

more alike genetically than dogs from different litters). Here, the random sample is of 

litters of dogs arriving at the pound. 

• See also question 2 in the 1998 mid-year examination, where you have a random sample 

of women and information on their incomes and the incomes of their partners: the 

question relates to the difference between a woman’s and her partner’s income. 

• ***** ASSUMPTIONS: We need make absolutely no assumptions about the original 

observations. If they are not normally distributed, it does not matter, as long as the 

differences are. Plot the differences to check that the assumption is not seriously violated. 

QUESTION 2. (from Stat171 tutorial exercises) 

In order to investigate whether the time of neutering of male 

dogs had any effect on their rate of growth, Crenshaw and 

Carter (1995) neutered dogs at 2 months and 7 months of age 

and measured their weight at 18 months of age. As litters of 

any breed became available at a pound, two male dogs were 

randomly selected from each litter. One was neuterd at 2 

months, the other was neutered at 7 months. All the dogs were 

kept under identical conditions during the 18 months of the 

trial. The weights (in kg) of the dogs at 18 months were: 

Litter 1 2 3 4 5 6 7 8 9 10 

2 months 19.0 23.2 16.1 30.3 27.1 22.3 18.1 27.8 34.2 17.3 

7 months 18.9 23.5 16.7 30.3 27.8 22.4 18.2 28.1 34.1 17.7 

c) Is there any evidence that early neutering affects the rate of growth of the dogs? 

d) Comment on why you think the experimenter chose this particular design in setting up 

the experiment. Do you think the right choice of design was made? 

QUESTION 3. A researcher wants to know the average sucrose content in a certain 

concentration of sugar-beet juice. There are two methods of measurement and the researcher 

wants to find out whether they both give the same mean dry weight. Ten sample containers 

of beet juice were randomly selected from ten different batches. Half the liquid in each 

container is measured using method A and the remaining half using method B. The 

measuring process yields the following data: 

METH 

OD 

20 

BATCH 

1 2 3 4 5 6 7 8 9 10 

A 11.0 5.0 9.8 5.7 6.5 8.2 5.9 6.0 7.5 5.4 

B 11.1 5.6 9.7 5.3 6.7 8.5 5.6 5.8 7.1 5.5 

Do these data support the hypothesis that both methods of 

measuring yield the same mean concentration? 


QUESTION 4. “You can taste the luscious grapes in Lindeman’s 

Montillo Sherry” - advertisement in ‘The SUN’. 

On the presumption that this claim was meant to imply that this product is more appealing 

than its competitors, ten allegedly experienced tasters were asked to rate samples of this 

product and also of another sherry on a scale from 0 to 100. 

Taster A B C D E F G H I J 

Montillo 83 80 94 81 90 68 74 81 86 80 

Competitor 64 82 88 86 91 72 66 71 82 80 

Is there significant evidence of a substantial preference for Montillo over the competing 

brand. 

CHECKS: 

Q1: mean diffce = 9.267; s = 8.614; t=4.166; df=14; 0.0005

QUESTION 1: The Chapin Social Insight Test is a psychological test designed 

to measure how accurately the subject appraises other people. The possible 

scores on the test range from 0 to 41. During the development of the Chapin 

Test, it was given to several groups of people. Here are the results for male and 

female students majoring in the liberal arts: 

Group Sex N X-bar S.D. 

1 Male 133 25.34 5.05 

2 Female 162 24.94 5.44 

Do these data support the contention that female and male students differ in 

average social insight? 

The variable is a score on a psychological test. It is reasonable to assume the 

variable, score, is normally distributed. 

Do these data support the contention that female and male students differ in 

average score? 

CHECKS: sp = 5.2679; SE(y1bar – y2bar) = 0.6164; t = 0.649; df = 293; pvalue 

> 0.05; no. 

METHOD Two independent random samples: do the samples come from 

populations with the same mean? 

Population 1 male 2 female 

Mean μ1 μ2 

Standard deviation σ1 σ2 

Sample size n1 133 n 2 162 

Sample mean y1bar 25.34 y2bar 24.94 

Sample SD s1 5.05 s 2 5.44 

Null hypothesis H0: μ1 = μ2 

Assumptions: 

• the populations are both normal 

• the standard deviations are equal, or in symbols: σ1 = σ 2; 

• to check this second assumption use: 

smax / s min < about 2. 

The value of the t-statistic is given by 

y1bar – y2bar 

t = __________________ where 

SE(y1bar – y2bar) 

SE(y1bar – y2bar) = sp √( 1/n1 + 1/n2) and 

22 


sp =√[{(n1-1)s1 2 + (n2-1)s2 2 } /{n1 + n2 – 2}] 

the t-statistic has (n1 + n2 - 2) degrees of freedom. 

QUESTION 2: Two groups of children were given visual acuity tests. Group 1 

was composed of 11 children who receive their health care from private 

physicians. The mean score for this group was 26 with a standard deviation of 

5. The second group, consisting of 14 children who receive their health care 

from the health department, had an average score of 21 with a standard 

deviation of 6. Assuming normally distributed scores for the populations of 

private and health department patients, find if the two groups of children have 

different scores on average. Find 95% confidence intervals for the difference 

between the mean scores of the two populations. 

QUESTION 3: An educator believes that new, directed reading activities in the 

classroom will help elementary school pupils improve some aspects of their 

reading ability. She arranges for a third grade class of 26 students to take part 

in these activities for an 8-week period. A control class room of 28 third 

graders follows the same curriculum without the activities. At the end of 8 

weeks, all students are given a Degree of Reading Power (DRP) test, which 

measures the aspects of reading ability that the treatment is designed to 

improve. 

The summary statistics are: 

Group number x-bar SD 

Activity group 26 51.48 11.01 

Control group 28 41.52 17.15 

Is there a difference, on average, between the two groups of 

students? 

ANSWERS: Q1: pooled SD = 5.2679 SE(y1bar – y2bar) = 0.6164 

t= 0.649 df = 293 p-value < 0.05 

Q2: pooled sd = 5.5873 SE(y1bar – y2bar) = 2.2512 

t= 2.221 0.02

Week 9 H/Y. STAT170 workshop, REGRESSION ANALYSIS, 

held at Numeracy Centre, Macquarie University. 

Examples taken from various sources and adapted for 

Stat170 by Nan Carter. 

EXAMPLE: Prior to opening a new shop, management 

requires an estimate of yearly sales revenue. Such 

an estimate is used in planning the appropriate shop 

capacity, making initial staffing decisions, and 

deciding whether or not the potential revenues 

justify the cost of the operation. 

Suppose management believes that the size of the 

student population on a nearby campus is related to 

annual sales revenues. On an intuitive basis, 

management believes that shops located near larger 

campuses generate more revenue than those near small 

campuses. 

To evaluate the relationship between student 

population (X) 

and annual sales (Y), management collected data from 

a random sample of 10 of its shops located near 

university campuses. These data are summarised 

below: 

Shop Student 

population 

(thousands) 

(X) 

24 

Annual sales 

($thousands) 

(Y) 

1 2 58 

2 6 105 

3 8 88 

4 8 118 

5 12 117 

6 16 137 

7 20 157 

8 20 169 

9 22 149 

10 26 202 

MEAN 14.0 130.0 


a) Draw the scatterplot of the data, and comment. 

(Look first for an overall pattern; its direction, 

form, and strength of relationship.) 

b) Use your calculator to confirm the values of the 

means, and to find 

SX = 7.94425. 

25 

Then find S = (n-1)( sx 2 ) 

XX 

= 568 

c) Given that SX = 2840, find the value of the slope 

Y 

b = SXY / SXX = 5 

d) Given that a = y-bar – b(x-bar), find the value of 

the intercept a = 130 – 5(14) = 60 

e) state the regression equation 

f) interpret the regression equation 

g) Use two values of X to find two corresponding 

predicted values of Y, and then draw the line on 

your scatter plot. 

h) What are the assumptions of the linear model for 

the relationship between annual sales and student 

population? 

i) Given that the standard error of the estimated 

slope is 0.5803, find the 95% confidence interval 

for the true value of the slope β. 

sales 

relayionship between annual sales and 

nearby student population 

250 

200 

150 

100 

50 

0 

0 10 20 30 

student population 

sales 


Relation: Survey of stores 

outcome: sales 

df: 8 

predictor 

coeff SE t p-value 95% C.I. 

constant 

60.0000 9.226 6.5033 0.000 38.725 81.275 

no. of students 5.0000 0.580 8.6167 0.000 3.662 6.338 

r-sq: 0.903 Resid SS: 1530.000 s: 13.829 

Fitted line: sales = 60 + 5 no. of students 

sales 

220 

200 

180 

160 

140 

120 

100 

80 

60 

40 

Residual 

20 

15 

10 

5 

0 

-5 

-10 

-15 

-20 

26 

Survey of stores 

0 10 20 

no. of students 

Normal Scores Plot 

-25 

-30 

p = 00.65 

-1.5 -1 -0.5 0 0.5 1 1.5 

normal score 

Nan Carter: workshop notes prepared for Numeracy Centre Macquarie University. 

30

Residual 

20 

15 

10 

5 

0 

-5 

-10 

-15 

-20 

-25 

EXERCISE 1. The relationship between production 

level and price of bananas. 

The following sample data have been collected over a 

period of twenty years for Queensland bananas: 

Production 

000s cases 

(X) 

27 

Residuals vs Fitted 

-30 

60 110 160 

Fitted value 

Price $ per 

case (Y) 

Production 

000s cases 

(X) 

32.0 1.27 38.3 0.57 

29.0 0.63 37.4 0.89 

33.0 0.26 31.1 0.98 

30.2 0.67 35.2 1.04 

24.2 1.79 30.5 1.05 

33.2 0.94 31.6 0.96 

36.0 0.52 35.0 0.67 

32.5 0.76 30.0 1.24 

42.0 0.49 26.3 2.24 

34.8 0.63 31.9 1.47 


Price $ per 

case (Y) 

a) plot the data on a scatterplot and comment on the 

display. 

b) Given that X-bar = 32.71, sX = 4.08681, 

Y-bar = 0.9535, 

SS XY = -24.5867, 

find the equation of the regression line. 

c) Draw the regression line on the scatterplot.

d) Interpret the regression line. 

e) Check the assumptions of the model. 

f) What average price would be expected if 

• production were 30,000 cases? 

• Production rose to 45,000 cases? 

• Production were 40,000 cases? 

g) A grower needs a price of $2 per case to break 

even, what production should she aim for? 

SCATTERPLOT 

of PRICE against PRODUCTION 

2.5 

2.0 

1.5 

1.0 

0.5 

0.0 

22 27 32 37 42 47 

28 

price per case $ 

No. of cases (ooos) 

Relation: 

Banana Production 

outcome: 

price per case $ 

df: 18 

predictor 


constant 

3.4878 0.670 5.2085 0.000 2.081 4.895 

No. of cases (ooos) -0.0775 0.020 -3.8126 0.001 -0.120 -0.035 

r-sq: 0.447 Resid SS: 2.359 s: 0.362 

Fitted line: price per case $ = 3.4878 - 0.0775 No. of cases (ooos) 


Residual 

Residual 

1 

0.8 

0.6 

0.4 

0.2 

0 

-0.2 

-0.4 

29 


-0.6 

-0.8 

p = 00.46 

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 

price per 

case $ 

2.5 

2 

1.5 

1 

0.5 

0 

1 

0.8 

0.6 

0.4 

0.2 

0 

-0.2 

-0.4 

-0.6 

-0.8 

normal score 

Banana Production 

22 27 32 37 42 

No. of cases (ooos) 

Residuals vs Fitted 

0 0.5 1 1.5 

Fitted value 


Question 2. 

The following table gives a sample of advertising 

expenditures and associated sales volumes for a 

company during 10 randomly selected months. We are 

interested in being able to predict gross monthly 

sales volume from the amount spent on advertising. 

(from J. Gosling, Introductory Statistics) 

Month 1 2 3 4 5 6 7 8 9 10 

Advertising 

expenses 

(x $10 000) 

1.2 0.8 1.0 1.3 0.7 0.8 1.0 0.6 0.9 1.1 

Sales Volume 101 

(x $10 000) 

92 110 120 90 82 93 75 91 105 

a) Draw, and comment on, a scatter plot of the data. 

b) State the slope of the regression line 

c) State the intercept of the regression line. 

d) State the equation of the regression line. 

e) Draw the line on your scatter plot. 

f) Interpret the regression equation; would it be 

useful for prediction? 

g) Check the assumptions of the model. 

Scatterplot of sales against advertising expenses 

1.4 

1.3 

1.2 

1.1 

1.0 

0.9 

0.8 

0.7 

0.6 

0.5 

Advertising expenses (x $10000) 

70 80 90 100 110 120 130 

30 

Sales Volume (x $10 000) 


Relation: 

Monthly sales volume 

outcome: 

Sales Volume (x $10 000) df: 8 

predictor 


constant 

46.4865 9.885 4.7029 0.002 23.693 69.280 

Advert. Exp (x $10000) 52.5676 10.261 5.1231 0.001 28.906 76.229 

r-sq: 0.766 Resid SS: 373.973 s: 6.837 

PLOT of sales against expenses 

PLOT of residuals against fitted 

NORMAL PROBABILITY PLOT 

Residual 

130 

response 

120 

15 

10 

110 

100 

5 

0 

-5 

90 

80 

70 

31 

Fitted line: Sales (x $10 000) = 46.4865 + 52.5676 Advert. Exp.(x $10000) 

0.5 0.7 0.9 1.1 1.3 1.5 


-10 

p = 00.88 

-1.5 -1 -0.5 0 0.5 1 1.5 

normal score 

determ. 1 

15 

residuals 

10 


5 

0 

-5 

-10 

predicted 

75 85 95 105 115 125

EXERCISE 3. In a large random sample of 91 homesales, 

two of the variables reported were selling 

price (thousands of $) and size of the home 

(thousands of square feet). The scatter plot is shown 

below, and an extract from the computer analysis of 

the relationship between price and size is 

reproduced. 

PRICE 

a) comment on the display. 

b) Given that the estimate of the intercept is –10.3, 

and the estimate of the slope is 65.7, state the 

equation of the regression line. 

b) Draw the regression line on the scatterplot. 

c) Interpret the regression line. 

d) Given that the standard error of the slope, b, is 

4.007, find the 95% confidence interval for 

the population slope, β. 

e) Check the assumptions of the model using the plot 

below: 

32 

200 

100 

0 

0.0 

SIZE 

.5 

1.0 

1.5 


2.0 

2.5 

3.0 

3.5

Unstandardized Residual 

33 

40 

20 

0 

-20 

-40 

-60 

0.0 

SIZE 

.5 

1.0 

1.5 

2.0 


2.5 

3.0 

3.5

Week 11 STAT170 workshop on CATEGORICAL variables. Prepared for 

Numeracy Centre Macquarie University, by Nan Carter. 

Examples taken from various sources and adapted for STAT170. 

CATEGORICAL VARIABLES: WE CONSIDER ONLY ONE RANDOM SAMPLE IN 

STAT170 

SUMMARY: 

1. ONE CATEGORICAL VARIABLE OBSERVED ON EACH ITEM 

• TWO CATEGORIES: use EITHER 

34 

z-TEST OF PROPORTIONS FOR H0: π = π0 where π0 has a 

particular value, and 

provided that nπ > 5 and n(1-π) > 5 

z = (p - π0 )/SE(p) where p is the sample estimate of π 

and SE(p)= √[(π)(1-π)/n] 

The 95% confidence interval for π uses the 

estimated SE(p) = √[(p)(1-p)/n] 

and is given by (p – 1.96 SE(p), p + 1.96 SE(p)) 

OR χ 2 GOODNESS OF FIT TEST: 

The same null hypothesis as for the z-test, and it 

is used to find expected values, E, for both 

categories. The observed COUNTS are denoted by ‘O’. 

The χ 2 -test is valid if all expected values, E, are 

>5; 

χ 2 = Σ[(O-E) 2 /E] with 1 degree of freedom 

• MORE THAN TWO CATEGORIES: use only the χ 2 

goodness of fit test with degrees of freedom = 

(no. of categories–1) 

Use info from the question to form the null 

hypothesis, and thence the expected values. The 

observed counts are given in the data. 

WARNING: SMALL EXPECTED FREQUENCIES: The chi-square 

test is not valid when any expected frequency is less 

than 5. To counter this problem, we group classes 

together to create larger observed counts and larger 

expected counts until the problem goes away! There 

is a good example of the need to do this in the 1998 


mid-year exam paper (See below: the question on no. 

of children and level of education of working women.) 

2. TWO CATEGORICAL VARIABLES OBSERVED ON EACH ITEM 

and we are INTERESTED IN WHETHER THE 

VARIABLES ARE INDEPENDENT. THE DATA ARE A TABLE OF 

OBSERVED COUNTS, O. 

• χ 2 TEST OF INDEPENDENCE: 

H0: the variables are independent 

For ‘expected values’, E, use the totals of 

observed counts in the table: 

E = (row total)(column total) / (grand total) 

WE STILL REQUIRE THAT ALL Es BE >5. 

Then, χ 2 = Σ[(O-E) 2 /E] where the sum is over all 

cells; 

and degrees of freedom are (r-1)(c-1) where there 

are r rows and c columns. If H0 is rejected then 

the variables are not independent, they are 

associated (or related). 

35 


Examples on working with proportions, prepared for 

STAT170 workshop at Numeracy Centre, Macquarie 

University, by Nan Carter from various sources. 

CATEGORICAL VARIABLE with ONLY TWO CATEGORIES. 

4. A manufacturer claims that 95% of the components 

supplied for a new jet transport meet a specified 

rigid standard of performance (ie 5% do not). 

A sample of 400 of the components is tested, and 30 

do NOT meet the performance standard. Comment on 

the manufacturer’s claim. 

5. A report claims that the percentage of miscarriages 

among LSD users is 12%. From the files in the 

maternity ward of a Sydney hospital, out of 100 

pregnant women who used LSD, 24 had miscarriages. 

Comment on the claim made in the report 

6. The president of a certain company believes that 

30% of the company’s orders come from new or firsttime 

customers. In a random sample of 100 orders, 

40 are from new customers; does this meet the 

company’s claim? 

CHECKS: Q1 z=2.294 Q2: z = 3.693; Q3: z = 2.182; 

Now we want to do the same questions using χ 2 . 

The problem is how to set up the hypothesis H0, and 

how to find values for the expected numbers E. 

Try thinking of it this way: we are often interested 

in 

• testing a theory which is used to give the 

hypothesis H0 and the expected numbers if the 

hypothesis is true. 

• testing a ‘claim’ made by others, and which is used 

to give the hypothesis H0 and the expected numbers 

36 


if the hypothesis is true. FOR EXAMPLE: In a study 

of the genetics of eye colour in fruit flies, there 

is a theory that the numbers of red-eyed and whiteeyed 

flies will be in the proportions of 1:3. So 

we have 

H0: the proportions are as 1:3 (i.e ¼ are red-eyed and 

¾ white-eyed) 

If the total number of flies is 9693, we can find the 

expected numbers in each class, assuming H0 is true. 

Expected values are 

1/4 x 9693 = 2423.25 and ¾ x 9693 = 7269.75. The data 

supplies the observed numbers (1981 and 7712 

respectively), and so we can calculate 

37 

χ 2 = Σ[(O-E) 2 /E] where the sum is over all 

cells, 

and degrees of freedom = no of classes – 1 

χ 2 = (1981-2423.5) 2 /2423.25 +(7712-7269.75) 2 /7269.75 = 

107.6 with 1 df. 

From χ 2 table, p-value < 0.05, so we reject H0 and 

conclude that the theory does not hold for the 

inheritance of red and white eyes in these fruit 

flies. In the population from which this sample was 

taken, there are fewer red-eyed and more white-eyed 

flies than the theory claims. 

4. Do the above exercise using a different 

statistical test. 

5. In tossing a die 300 times the frequency of 

occurrence of 1, 2, 3, 4, 5, 6 are, respectively, 43, 

49, 56, 45, 66 and 41. 

• Is there any basis for the claim that the die is 

biased? If there is a bias, which number appears 

more frequently than expected? 

6. In a survey taken 5 years ago, 25% of students 

watched less than 5 hours of TV per week, 10% watched 

20 or more hours per week, and the rest watched 


etween 5 and 20 hours. A recent survey of 450 

students showed that 117 students watched TV for less 

than 5 hours per week, 267 watched between 5 and 20 

hours, and the remainder watched 20 or more hours per 

week. Has the pattern of watching TV changed for 

students in the five years? If there has been a 

significant change, describe where it has occurred. 

CHECKS: Q4: z = -11.14, p

sandwich, only 4 became ill. We would like to test 

whether there is an association between the 

consumption of a sandwich and the onset of 

gastroenteritis. 

Question 9. In a sample of 110 working women in 

Sydney, all aged between 30 and 50 years, the 

following table shows the data classified by ‘highest 

level of education attained’ [coded 1: did not 

complete secondary school; 2: graduated from 

secondary school; 3: attended TAFE college or other 

institution; 4: bachelor degree from university; 5: 

higher degree from university] and by ‘number of 

children’. 

Number of Children 

Education 

39 

0 1 2 3 4 5,6 


All 

1 4 2 2 9 3 0 20 

2 1 4 7 11 4 2 29 

3 0 3 11 7 2 0 23 

4 1 6 12 10 0 0 29 

5 2 3 3 1 0 0 9 

All 8 18 35 38 9 2 110 

a) Are all the expected frequencies at least 5? 

b) If not, collapse the table until this criterion 

is met. 

Note: to identify the smallest expected frequency, 

multiply smallest row total by smallest column 

total; if

Week 12 Questions on ODDS RATIO TESTS prepared for 

Stat170 workshop held at the Numeracy Centre, Macquarie 

University. Prepared by Nan Carter with examples from 

various sources and adapted for Stat170. 

ODDS RATIO TESTS 

The χ 2 test of independence showed that wearing a helmet and 

receiving a head injury in a bicycle accident are associated, 

but does not indicate how strong, or weak, the association is. 

To get that information we need to look at odds ratios. 

DEFINITION: if p is the probability of an event occurring, 

and (1-p) the probability of it not occurring, then p/(1-p) is 

the odds of it occurring. 

We can show that the odds of a cyclist wearing a helmet will 

receive a head injury is [17/147]/[130/147], which simplifies 

to 17/130. 

Similarly, we can show that the odds that a cyclist not 

wearing a helmet will receive a head injury is 

[218/646]/[428/646], which simplifies to 218/428. 

So, the odds ratio of a cyclist receiving a head injury if 

wearing a helmet relative to a cyclist receiving a head injury 

if not wearing a helmet is [17/130]/[218/428], which can be 

written [17x428]/[218x130] = 0.26. 

Similarly, a cyclist receiving a head injury if not wearing a 

helmet relative to a cyclist receiving a head injury if 

wearing a helmet, is [218/428]/[17/130], which can be written 

[218x130]/[17x428]= 3.89. 

40 


Notice that 

a) 3.89 = 1/0.26, so if we find one ratio, we also have the 

other as the reciprocal. 

b) If the two odds were equal, then their ratio would be 1. 

As statisticians, we calculate a 95% confidence interval 

for the ratio and if it contains 1, we say that with 95% 

confidence the ratio could be 1, which is to say that the 

odds could be equal. 

c) For interpretation, we say that ‘a cyclist in an accident 

and not wearing a helmet is nearly four times as likely 

to receive a head injury than a cyclist who was wearing a 

helmet’. 

d) Or we might say that a bicyclist wearing a helmet is only 

25% as likely to receive a head injury if in an accident 

as one who is not wearing a helmet. 

This is EcStat output for this problem: read off the 95% 

confidence interval. 

Helmets and head injuries 

Observed counts 

head injury 

wearing helmet yes no total 

yes 17 130 147 

no 218 428 646 

total 235 558 793 

Expected counts 

wearing helmet 

head injury 

yes no total 

yes 43.6 103.4 147.0 

no 191.4 454.6 646.0 

total 235.0 558.0 793.0 

Chi-squared decomposition 

head injury 

wearing helmet yes no total 

yes -4.024 2.612 23.018 

no 1.920 -1.246 5.238 

total 19.882 8.373 28.255 

Test: df chi-sq p-val 

Indep 1 28.26 0.000 

wearing helmet head injury OR 95% CI 

yes / no yes / no 0.26 0.15 0.44 

41 


EXERCISE: In a Canadian experiment to examine the claim that 

the use of vitamin C helps to prevent the common cold, 818 

volunteers were randomly divided into 2 groups (one of 411 and 

the other of 407), and received a supply of either vitamin C 

or placebo. The results are shown below: 

TREATMENT 

OUTCOME 

placebo Vitamin C TOTALS 

Has Colds 335 302 637 

No Colds 76 105 181 

TOTALS 411 407 818 

Do these data suggest that getting a cold is independent of 

whether or not the subjects received vitamin C? 

Can the risk of a cold be reduced by using vitamin C? 

vitamin C 

Observed counts 

response 

determinant yes no total 

placebo 335 76 411 

vitamin C 302 105 407 

total 637 181 818 

Expected counts 

response 


placebo 320.1 90.9 411.0 

vitamin C 316.9 90.1 407.0 

total 637.0 181.0 818.0 

Chi-squared decomposition 

response 


placebo 0.698 2.455 3.153 

vitamin C 0.704 2.479 3.184 

total 1.402 4.934 6.337 

Test: 

df 

chisq 

p-val 

Indep 1 6.34 0.012 

determinant response OR 95% CI 

placebo / 

vitamin C yes / no 1.53 1.10 2.14 

42

STAT170 Workshop Notes prepared by Nan Carter for Numeracy ...

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