Course Notes - Department of Mathematics and Statistics

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Fred wanted to address the following hypotheses.1. H 0 The mean IQs of children with biological parents of high SESare the same as those with biological parents of low SES2. H 0 The mean IQs of children with adoptive parents of high SESare the same as those with adoptive parents of low SES3. H 0 The relationship between IQs and adoptive parents’ SES is notaffected by the SES of the biological parents.To accomplish this he decided to do a two factor ANOVA,in R. Using the command aov.ex2 = aov(IQ ∼ AdoptparentSES*BioparentSES,data=Nature)then summary(aov.ex2).Fred can obtain the following output.From this printout, Fred can address each of his hypothesis1. H 0 The mean IQs of children with biological parents of high SESare the same as those with biological parents of low SESLooking at the p-value associated with the biological parents, Fred sawthat it was 0.0009, which is highly signficant. Fred can reject thishypothesis, and accept the alternative that there is a difference in theIQs between children of biological parents with high SES, compared tobiological parents with low SES.200
2. H 0 The mean IQs of children with biological parents of high SESare the same as those with biological parents of low SESLooking at the p-value associated with the adoptive parents, Fred sawthat it was 0.0064, which is also highly signficant. Fred can reject thishypothesis, and accept the alternative that there is a difference in theIQs between children of adoptive parents with high SES, compared toadoptive parents with low SES.3. H 0 The mean IQs of children with biological parents of high SESare the same as those with biological parents of low SESLooking at the p-value associated with the interaction, Fred saw that itwas 0.9174, which is not signficant. Fred cannot reject this hypothesis.So from these 3 results, Fred came to the following conclusions, there isstrong evidence that the SES of parents both biological and adoptive,have a strong effect on the IQ’s of children. Fred drew up a table ofthe mean IQ’s of each of the four groups.High B SES Low B SES TotalHigh A SES 119.60 103.6 111.60Low A SES 107.50 92.40 99.1Total 114.22 98.00 105.68The table (where B = biological and A = Adoptive) suggests, thatchildren with biological parents of high SES have IQ’s roughly 7.7 pointsabove average, and children with adoptive parents of high SES have IQ’sroughly 6.6 points above average. However there is no evidence of aninteraction between the two sets of parents.201
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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
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STANDARD DEVIATIONIf P = X n : σP
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6.4.2 Sample Size CalculationEXAMPL
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Fred for MayorFred has decided to t
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Fred’s mate Bob pointed out that
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ALTERNATIVE HYPOTHESESTwo types:Stu
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• If we use the sample standard d
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Calculate the test statistic:Test s
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Finding the pooled variance:Recall
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Caluclate the estimated standard er
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7.6 Significance and Conclusiveness
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Conclusion: There is no evidence th
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Some practical pointers• Aim for
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Fred’s Parking & Extracurricular
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Fred calculated the p-value using t
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Fred told Carol he would use odds r
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Flu vaccine example• There were 1
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8.3 Attributable Risk (AR)The attri
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• Therefore w x ≈ww+x and y z
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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 162 and 163: Summary - Confidence IntervalsFacto
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 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
Page 212 and 213: Focussed look on the limits of our
Page 214 and 215: Height example• We will perform a
Page 216 and 217: Normality Assumption FAILPP PlotExp
Page 218 and 219: 10.3 Confidence Intervals and Regre
Page 220 and 221: 0.236 < β 1 < 0.496• We can conc
Page 222 and 223: 95% confidence and prediction inter
Page 224 and 225: 100m World Record Progression●●
Page 226 and 227: Negative correlation• The denomin
Page 228 and 229: Stress exampleStress(x) Blood Press
Page 230 and 231: 10.5 Multiple regression• Simple
Page 232 and 233: • Appears that total lung capacit
Page 234 and 235: Multiple linear regression predicti
Page 236 and 237: Three variable modelIntepreting Mod
Page 238 and 239: • So for men we have:y = −7.066
Page 240 and 241: • Is the variable age an importan
Page 242 and 243: Extra sum of squares principle• T
Page 244 and 245: Diagnostics• Remember our Assumpt
Page 246 and 247: TableTreatmentControlBP(Y) AGE(X) B
Page 248 and 249: • 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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