Course Notes - Department of Mathematics and Statistics

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ALTERNATIVE HYPOTHESESTwo types:Study based - implies we do not know at the beginning of the studyabout the effect of a new treatment. Leads to a two sided test(two parameter values are not equal to each other).Data based - suggested by the collected data (usually suggests treatmentbenefit). Leads to a one sided test (one parameter value isgreater than or less than the other).TESTING STEPS1. Set up the null hypothesis (H 0 ) about the population parameter ofinterest e.g. parameter = current (null) value.2. Propose the alternative hypothesis (H A ) e.g. parameter ≠ currentvalue.3. Calculate the test statistic.4. Calculate the p-value (probability of observing the test statisticfrom 3).The Test StatisticThis is the standardised value of the sample parameter i.e. a z-scoreor t-score:observed sample value - null valueTest statistic =estimated standard errori.e. the number of standard deviations from the null value to thesample value.The p-value• This is the probability of observing the value of the test statistic,or a value more extreme, calculated under the assumption thatH 0 is true.• We draw appropriate conclusions if the pvalue is less than 0.05 -if the p-value is less than 0.05 we have significance at the 5% leveland if less than 0.01 we have significance at the 1% level.120
• If the standard deviation is unknown the p-value is found usingthe t distribution.7.1 Hypothesis Test for MeanEXAMPLE ONE - HYPOTHESIS TEST FOR THE MEANSuppose the resting pulse rates for young women are normally distributedwith mean µ = 66 and standard deviation σ = 9.2 beats perminute. A drug for the treatment of a medical condition is administeredto 100 young women and their average pulse rate is found tobe 68 beats per minute. Because the drug had for a long time beenobserved to increase pulse rates, test the claim that the drug does infact increase the pulse rates (H A is data based).Set up the hypotheses:STEP ONE: H 0 : µ = 66 (the null hypothesis)STEP TWO: H A : µ > 66 (the research hypothesis)Calculate the test statistic:Test statistic =observed sample value - null value( )estimated standard error σ√n68 − 66=√9.2100= 2.174Find the p-value:Pr(Z > 2.174)=0.0150• Can get p-value directly from Rcmdr.• For this example use 1 - pnorm(2.174).• If p-value < 0.05 we have significance at the 5% level (and ifp-value < 0.01 we have significance at the 1% level).121
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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
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 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 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 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
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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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