Course Notes - Department of Mathematics and Statistics

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6.3.2 Transforming DataTRANSFORMING DATAIf data are continuous but not normally distributed, we need totransform each value to create new values which are normally distributed- we use logs, square roots or reciprocals. We do this because:1) statistical procedures require normally distributed data2) to compare two samples, the standard deviations need to be similar3) to reduce the effect of outliers (log transformation).TRANSFORMING DATA ExampleFrom Assignment 3, we saw that daily river flow is typically skewedand is bounded below by zero (negative values are impossible in thiscontext).One set of 128 observations of Rangitata river flow during springand summer had mean = 105.29 and standard deviation 44.44. Theminimum and maximum values are 40.35 and 233.14 respectively.The data are not normally distributed.When the data are transformed by taking logs to base e the meanis 4.51 and the standard deviation is 0.43. The normal fit to thetransformed data is much better.106
Confidence Interval for raw dataTo find the range of values containing the central 95% of river levelobservations we want all values within 1.96 standard deviations of themean i.e. mean ± 1.96 × standard deviationIf we use the raw data the interval is: 105.29 ± 1.96(44.44) =(18.19, 192.39) This interval cannot be correct as it contains valuesbelow the minimum.Confidence Interval for transformed dataFor the transformed data the corresponding interval is:4.51 ± 1.96(0.43) =(3.67, 5.35)Back-transformingWe can back-transform these to the original scale using e:= (e 3.67 , e 5.35 ) = (39.14, 211.20)Hence 95% of river level observations would have values between39.14 and 211.20.107
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Contents1 Course Administration 42
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10 Regression 20210.1 Introduction
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Resource PageDepartment of Mathemat
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3 Data and Study Designs3.1 Basic d
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Parameter• Parameter: Fixed numbe
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TypesDiscrete - can put in one-to-o
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- How do geologists know?- Sediment
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Source: skyscrapercity.comSampling
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- 1 in 3.8 million - 3.8 million di
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Important designs• Completely ran
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B. Randomised control - of all chil
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- Reason why observational studies
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Things that go wrong• Even the be
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4 ProbabilityFred’s DayFred awoke
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• The event that the mouse we tra
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Blood donor example - Multiplicatio
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Fair Die Example• A fair die is t
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Tree diagram rules• Add Verticall
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Tree Diagrams - Independent Stages0
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Dependent Stages• Andrew, John an
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0.90BBiopsy +ve (true positive)0.00
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Calculating Probabilities• Estima
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4.3 Random VariablesRandom Variable
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Calculating the Variance• The sam
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Examples• Consider a data set XX
Page 55 and 56: Combining 2 Random Variables• If
Page 57 and 58: Fred’s DayShould Fred be worried
Page 59 and 60: Now using the complementary event r
Page 61 and 62: Mean of Binary DistributionThe mean
Page 63 and 64: • Mean number of successes• Var
Page 65 and 66: Using the Formulan = 3, x = 2, π =
Page 67 and 68: • A probability less than 0.05 is
Page 69 and 70: Endangered bird egg example solutio
Page 71 and 72: RELATIVE FREQUENCY HISTOGRAMNormal
Page 73 and 74: Finding Areas Under the Curve• In
Page 75 and 76: • Find P r(−1 < Z < 1.64)pnorm(
Page 77 and 78: INVERSE PROBLEMS USING R-COMMANDER
Page 79 and 80: CALCULATING PROBABILITIESAssume tha
Page 81 and 82: CONTINUITY CORRECTION• Normal pro
Page 83 and 84: Sample size = 10Consider the situat
Page 85 and 86: ConclusionWhat conclusion would you
Page 87 and 88: 10%5.5 6.0 6.5 7.0 7.5 8.0 8.5XWe k
Page 89 and 90: 6 Sampling Distributions and Estima
Page 91 and 92: DERIVATION IWe can think of the sam
Page 93 and 94: Solution IIPr(X > 172) = 1 - pnorm(
Page 95 and 96: qnorm(0.975) as 2.5% of the area is
Page 97 and 98: 6.2.1 Sample Size CalculationEXAMPL
Page 99 and 100: The 99% Confidence IntervalThe 99%
Page 101 and 102: 6.3 Comparing Two SamplesCOMPARING
Page 103 and 104: THE POOLED VARIANCEIf the variances
Page 105: SOLUTION - Calculating the confiden
Page 109 and 110: = 6.6 ± 24.6That is, -18.0 < µ in
Page 111 and 112: STANDARD DEVIATIONIf P = X n : σP
Page 113 and 114: 6.4.2 Sample Size CalculationEXAMPL
Page 115 and 116: Fred for MayorFred has decided to t
Page 118 and 119: Fred’s mate Bob pointed out that
Page 120 and 121: ALTERNATIVE HYPOTHESESTwo types:Stu
Page 122 and 123: • If we use the sample standard d
Page 124 and 125: Calculate the test statistic:Test s
Page 126 and 127: Finding the pooled variance:Recall
Page 128 and 129: Caluclate the estimated standard er
Page 130 and 131: 7.6 Significance and Conclusiveness
Page 132 and 133: Conclusion: There is no evidence th
Page 134 and 135: Some practical pointers• Aim for
Page 136 and 137: Fred’s Parking & Extracurricular
Page 138 and 139: Fred calculated the p-value using t
Page 140 and 141: Fred told Carol he would use odds r
Page 142 and 143: Flu vaccine example• There were 1
Page 144 and 145: 8.3 Attributable Risk (AR)The attri
Page 146 and 147: • Therefore w x ≈ww+x and y z
Page 148 and 149: Aspirin study example - gastrointes
Page 150 and 151: Mobile phone example• Human expos
Page 152 and 153: OR = 0.81 with CI (0.39,1.70)• p
Page 154 and 155: • We calculate these expected cou
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• We can calculate the p value us
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χ 2 =(100 − 108.99)2 (107 − 11
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• The reason for the discrepancy
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Summary - Confidence IntervalsFacto
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Since 1 is excluded from this 95% i
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9 ANOVAFred’s Public ImageSince F
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• Previously used two sample t-te
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• This tells us if the observed d
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RESIDUAL MEAN SQUARE (s 2 e)Group A
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9.2 Post ANOVA AnalysisPOST ANOVA A
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• Hence no need to additionally c
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Since zero is excluded, and the int
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SUM OF SQUARESTotal SS = 93 2 + 100
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General Level SS = 12 x ( 9812 )2=
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ANOVA TABLEA one factor analysis of
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Data > Manage variables in active d
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DATA COMPONENTSEach data value can
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ANOVA TABLE - Two Factor FactorialS
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A B C D EChristchurch 120.5 119 122
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DataFERTILISER LEVELSEED TYPE Low M
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Interaction PlotsHypothesis test fo
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Fred’s WeightUsing the informatio
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Fred wanted to address the followin
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10 RegressionFred’s Dog/BeardFred
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2. studies which have measured bina
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• What if we colour code the poin
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Statistically• We use slightly di
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Blood pressure exampleStress(x) Blo
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Focussed look on the limits of our
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Height example• We will perform a
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Normality Assumption FAILPP PlotExp
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10.3 Confidence Intervals and Regre
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0.236 < β 1 < 0.496• We can conc
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95% confidence and prediction inter
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100m World Record Progression●●
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Negative correlation• The denomin
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Stress exampleStress(x) Blood Press
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10.5 Multiple regression• Simple
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• Appears that total lung capacit
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Multiple linear regression predicti
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Three variable modelIntepreting Mod
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• So for men we have:y = −7.066
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• Is the variable age an importan
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Extra sum of squares principle• T
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Diagnostics• Remember our Assumpt
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TableTreatmentControlBP(Y) AGE(X) B
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• So for people in the control gr
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Discussion• The value of d is the
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Multiple Linear RegressionTreatment
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• Logistic regression is sometime
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• Given the s.e.(OR) = 0.751. Thi
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• We get the same as when we calc
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Female(0)Male(1)Born Left(1) Right(
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Fred’s Dog/BeardTo help solve his
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This is a very high r 2 value and i
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Attractiveness = 4.85 − 0.03 × (
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• Because the calculator does div
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• Install R Commander from the R
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- In large random samples (n 1 and
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Course Notes - Department of Mathematics and Statistics

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