Course Notes - Department of Mathematics and Statistics

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• Standard error of the difference (standard √deviation of the distributionof differences in sample means):2 n 1σ1+ σ2 2n 2• 95% confidence interval for the difference in means of two populations√ (LARGE SAMPLES where n 1 and n 2 ≥ 30): (x 1 − x 2 ) ±1.96σ 2 1n 1+ σ2 2n 2EXAMPLEIt is generally accepted that males are taller than females. Fromthe STAT110 questionnaire the sample of 176 females and 133 malesgave the following information. The outcome measure is height (cm).Is the observed difference between the sample means statisticallysignificant?Female MaleSample mean (x i ) 166.6 180.6Sample standard deviation (s i ) 8.04 7.97Sample size (n i ) 176 133Observed difference = 180.6 - 166.6 = 14 cmConfidence IntervalThe confidence interval for µ male − µ female is:That is,(x male − x female ) ± 1.96√s2male√= (180.6 − 166.6) ± 1.96x= 14 ± 1.80n male+ s2 femalen female7.97 2133 + 8.04217612.20 < µ male − µ female < 15.80Interpreting the Confidence IntervalThe height of males is likely to be within 12.20 and 15.80 cm tallerthan that for females. This confidence interval does not include zero,hence the difference is significant.102
THE POOLED VARIANCEIf the variances of the two distributions (σ1 2 and σ2 2 ) are similar thenwe can simplify the 95% confidence interval to:√1(x 1 − x 2 ) + 1.96σ + 1 n 1 n 2The common variance σ 2 is estimated from sample data by the pooledvariance s 2 p where:s 2 p = (n 1 − 1)s 2 1 + (n 2 − 1)s 2 2n 1 + n 2 − 2EXAMPLE - POOLED VARIANCEFor the height data in the previous example the variances of the twodistributions (σmale 2 and σ2 female) are similar. The common varianceσ 2 is:s 2 p = (n 1 − 1)s 2 1 + (n 2 − 1)s 2 2n 1 + n 2 − 2= (133 − 1)7.972 + (176 − 1)8.04 2133 + 176 − 2= 132 × 7.972 + 175 × 8.04 2307= 19697.0388307= 64.1597355Example - Calculating the Pooled VarianceThe pooled standard deviation s p is found as the square root of thepooled variance:s p = √ 64.1597355= 8.01(2dp)103
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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
Page 51 and 52: Calculating the Variance• The sam
Page 53 and 54: 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: 6.3 Comparing Two SamplesCOMPARING
Page 105 and 106: SOLUTION - Calculating the confiden
Page 107 and 108: Confidence Interval for raw dataTo
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
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OR = 0.81 with CI (0.39,1.70)• p
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• 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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