Course Notes - Department of Mathematics and Statistics

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Summary - Confidence IntervalsFactor 2Factor 1 Level 1 Level 2 TotalLevel 1 w x r 1 = w + xLevel 2 y z r 2 = y + zc 1 = w + y c 2 = x + z n = w + x + y + z• s.e.(ln(RR)) =√1w− 1w+x + 1 y − 1y+z• Confidence interval for ln(RR): ln(RR) ± 1.96 × s.e.(ln(RR))√p1 (1−p• p 1 −p 2 ±1.96×1 )n 1+ p 2(1−p 2 )n 2where p 1 =w+x w and p 2 =yy+z• s.e.(ln(OR)) =√1w+ 1 x + 1 y + 1 z• Confidence interval for ln(OR): ln(OR) ± 1.96 × s.e.(ln(OR))162
Fred’s GardenSo Fred recorded the survival of cuttings collected in both winter andsummer, and wants to know if there is any difference in the survival ofcuttings from winter or summer. Fred needs to calculate the odds ratiofor this 2 × 2 table.OutcomeTime of cutting Cutting Alive Cutting Died TotalWinter 263 217 480Summer 115 365 480378 582 960The odds of a cutting planted in winter not dying is 263217= 1.2120 andthe odds of a cutting planted in the summer not dying is 115365 = 0.3151.So to calculate the odds ratio, Fred just divided the odds of not dying,i.e. OR =0.3151 1.2120 = 1.5271. This indicates that there is evidence thatplants are more likey to survive if the cutting was taken in winter.To be sure though Fred needs to generate a 95% confidence interval forthis confidence interval. Recalling the formula for s.e. (log e OR)s.e.(log e OR) =√1a + 1 b + 1 c + 1 d√1s.e.(log e OR) =263 + 1217 + 1115 + 1365s.e.(log e OR) = 0.1409To calculate the 95% confidence interval Fred needs to transform hisOR onto a log e scale add the s.e × 1.96, and then back transform theconfidence interval.NOW BACKTRANSFORMlog(OR) ± 1.96 × s.e.(log e OR)0.4234 ± 1.96 × 0.1409(0.1472 , 0.6996)(1.1586 , 2.0129)163
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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
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Combining 2 Random Variables• If
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Fred’s DayShould Fred be worried
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Now using the complementary event r
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Mean of Binary DistributionThe mean
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• Mean number of successes• Var
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Using the Formulan = 3, x = 2, π =
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• A probability less than 0.05 is
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Endangered bird egg example solutio
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RELATIVE FREQUENCY HISTOGRAMNormal
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Finding Areas Under the Curve• In
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• Find P r(−1 < Z < 1.64)pnorm(
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INVERSE PROBLEMS USING R-COMMANDER
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CALCULATING PROBABILITIESAssume tha
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CONTINUITY CORRECTION• Normal pro
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Sample size = 10Consider the situat
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ConclusionWhat conclusion would you
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10%5.5 6.0 6.5 7.0 7.5 8.0 8.5XWe k
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6 Sampling Distributions and Estima
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DERIVATION IWe can think of the sam
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Solution IIPr(X > 172) = 1 - pnorm(
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qnorm(0.975) as 2.5% of the area is
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6.2.1 Sample Size CalculationEXAMPL
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The 99% Confidence IntervalThe 99%
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6.3 Comparing Two SamplesCOMPARING
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THE POOLED VARIANCEIf the variances
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SOLUTION - Calculating the confiden
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Confidence Interval for raw dataTo
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= 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
Page 156 and 157: • We can calculate the p value us
Page 158 and 159: χ 2 =(100 − 108.99)2 (107 − 11
Page 160 and 161: • The reason for the discrepancy
Page 164 and 165: Since 1 is excluded from this 95% i
Page 166 and 167: 9 ANOVAFred’s Public ImageSince F
Page 168 and 169: • Previously used two sample t-te
Page 170 and 171: • This tells us if the observed d
Page 172 and 173: RESIDUAL MEAN SQUARE (s 2 e)Group A
Page 174 and 175: 9.2 Post ANOVA AnalysisPOST ANOVA A
Page 176 and 177: • Hence no need to additionally c
Page 178 and 179: Since zero is excluded, and the int
Page 180 and 181: SUM OF SQUARESTotal SS = 93 2 + 100
Page 182 and 183: General Level SS = 12 x ( 9812 )2=
Page 184 and 185: ANOVA TABLEA one factor analysis of
Page 186 and 187: Data > Manage variables in active d
Page 188 and 189: DATA COMPONENTSEach data value can
Page 190 and 191: ANOVA TABLE - Two Factor FactorialS
Page 192 and 193: A B C D EChristchurch 120.5 119 122
Page 194 and 195: DataFERTILISER LEVELSEED TYPE Low M
Page 196 and 197: Interaction PlotsHypothesis test fo
Page 198 and 199: Fred’s WeightUsing the informatio
Page 200 and 201: Fred wanted to address the followin
Page 202 and 203: 10 RegressionFred’s Dog/BeardFred
Page 204 and 205: 2. studies which have measured bina
Page 206 and 207: • What if we colour code the poin
Page 208 and 209: Statistically• We use slightly di
Page 210 and 211: 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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