Course Notes - Department of Mathematics and Statistics

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DataFERTILISER LEVELSEED TYPE Low Medium High14 18 18R i R 2 i1 18 18 12 148 2190417 19 1413 16 112 13 18 11 116 134568 17 9C j 75 83 106 264 35360C 2 j5625 6889 11236 23750Computational FormulaeGeneral Level SS = nȳ 2Group SS = C2 1n 1+ C2 2n 2= 1 n g∑ C2i− nȳ 2+ . . . + C2 kn k− nȳ 2n 1 = n 2 = . . . = n k = #replicates × #levels <strong>of</strong> other factor = n gResidual SS = ∑ (residuals) 2Total SS = ∑ (data value) 2 = 4076Interaction SS= Total SS - General Level SS - Group SS - Resid SS194
General Level SS = 18 x ( 26418 )2= 3872Seed type SS = 219049+ 134569- 3872= 353609- 3872= 56.89Fertiliser Level SS = 56256+ 68896+ 11236= 237506- 3872= 86.336- 3872Residual SS = (14 − 16.33) 2 + . . . + (9 − 10.33) 2= 49.33ANOVA Table (reduced) - Seed YieldSource <strong>of</strong> Variation SS DF MS F pSeed type effect 56.89 1 56.89 13.84 0.003Fertiliser level effect 86.33 2 43.17 10.50
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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
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 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 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
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
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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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