Course Notes - Department of Mathematics and Statistics

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Use R commander or R excel4. In a pilot study in Auckland, three out of seven patients given anew drug had their tumour size halved. What conclusion if anycan be drawn about the new drug? Explain how you reach yourconclusion• There is no reason to suppose the new drug is any different to thestandard.• Probability of three or more is 0.3529 which is large meaning theresult with the new drug is consistent with the 30% before.Cancer drug example solutions note• The reason that 0.3529 is means that there is no difference is NOTbecause 0.3529 is close to 0.3.• The reason is this number is greater than 0.05.Endangered bird egg exampleA scientist has established over a long period of time that only 30%of the eggs laid by an endangered bird species result in the successfulrearing of a chick1. A sample of 10 of these eggs is monitored. Find the probabilitythat at least half of the 10 eggs result in the successful rearing ofa chick2. A second sample of 20 eggs is monitored. Find the probabilitythat at least half of the 20 eggs result in the successful rearing ofa chick3. Two breeding programmes for this endangered bird on two seperateoff-shore islands were investigated. On Island A, 5 out of the10 eggs, and on Island B, 10 out of 20 eggs resulted in the successfulrearing of chicks. Comment on the success or otherwise ofthe two breeding programmes in light of your answers in (1) and(2)68
Endangered bird egg example solutions1. A sample of 10 of these eggs is monitored. Find the probabilitythat at least half of the 10 eggs result in the successful rearing ofa chick2. A second sample of 20 eggs is monitored. Find the probabilitythat at least half of the 20 eggs result in the successful rearing ofa chickUse R commander3. Two breeding programmes for this endangered bird on two seperateoff-shore islands were investigated. On Island A, 5 out of the10 eggs, and on Island B, 10 out of 20 eggs resulted in the successfulrearing of chicks. Comment on the success or otherwise ofthe two breeding programmes in light of your answers in (1) and(2)Island B has a higher rate of success than 30% there is no evidencethat Island A has a higher success rate.69
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Contents1 Course Administration 42
Page 3 and 4:
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
Page 17 and 18: Source: skyscrapercity.comSampling
Page 19: - 1 in 3.8 million - 3.8 million di
Page 23 and 24: Important designs• Completely ran
Page 25 and 26: B. Randomised control - of all chil
Page 27 and 28: - Reason why observational studies
Page 29 and 30: Things that go wrong• Even the be
Page 31 and 32: 4 ProbabilityFred’s DayFred awoke
Page 33 and 34: • The event that the mouse we tra
Page 35 and 36: Blood donor example - Multiplicatio
Page 37 and 38: Fair Die Example• A fair die is t
Page 39 and 40: Tree diagram rules• Add Verticall
Page 41 and 42: Tree Diagrams - Independent Stages0
Page 43 and 44: Dependent Stages• Andrew, John an
Page 45 and 46: 0.90BBiopsy +ve (true positive)0.00
Page 47 and 48: Calculating Probabilities• Estima
Page 49 and 50: 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: • A probability less than 0.05 is
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
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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
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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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