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UW-Madison STAT 371 - Chapter 9 - Comparison of Paired Samples

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Hypothesis TestsChapter 9 Comparison of Paired SamplesFall 20109.1 IntroductionOverviewIn a paired sample design, we model the data as if there is asingle bucket of balls, and each draw from the bucket resultsin a pair of numbers (that we can distinguish as first andsecond).This model applies if there are two measurments on eachindividual (often before and after), or if a pair of individualsare sampled together (such as twins, siblings, a matched pairdesign).In a paired design, the method of analysis is as follows.Take individual differences for each pair.Treat the differences as a sample from a single population.Structure of the DataThe data from a paired design can be tabulated in this form.Individual Y1Y2di= Y1− Y212...nMean ¯y1¯y2¯dSD s1s2sdThe important summary statistics are¯d =Pni =1din= ¯y1− ¯y2and sd=sPni =1(di−¯d)2n − 19.2 Paired-Sample t Test and Confidence IntervalConfidence IntervalsThe population mean difference is represented by µd.The following confidence interval formula for µdis derivedassuming that the population of differences has a normaldistribution.¯d − tα/2sd√n< µd<¯d + tα/2sd√nIf there are n pairs, there are n − 1 degrees of freedom, andthe area between −tα/2and tα/2is the confidence level 1 −α,often chosen to be 95%.Hypothesis TestsFor hypothesis tests of the null hypothesis H0: µd= 0 versuseither a one- or two-sided alternative, the test statistic istd=¯dsd/√nand p-values are found by computing areas under tdistributions with n − 1 degrees of freedom.Example — Exercise 9.3Cyclic adenosine monophosphate (cAMP) is a substance that canmediate celluar response to hormones. In a study of maturation ofegg cells in frogs, oocytes from each of four females were dividedinto two batches; one batch was exposed to progesterone and theother was not. After two minutes, each batch was assayed for itscAMP content.Example (cont.) - SummaryIn an experiment, eggs from each of four female frogs aredivided into two groups.One group of eggs from each frog is treated withprogesterone, one is not.The cAMP level is measured for each group of eggs.Frog Control Progesterone Difference1 6.01 5.23 0.782 2.28 1.21 1.073 1.51 1.40 0.114 2.12 1.38 0.74Mean 2.98 2.31 0.68SD 2.05 1.95 0.40µd= population mean decrease in cAMP due to progesteroneExample (cont.) - Question1Find a 95% confidence interval for µd.2Test the hypothesisH0: µd= 0 HA: µd6= 0 .3Interpret both results in the context of this setting.Example (cont.) - Chalkboard CalculationsDo the Calculations on the Board.Interpretation — Confidence IntervalWe are 95% confident that the mean decrease in cAMP(pmol/oocyte) due to exposure to progesterone for twominutes under the given experimental conditions for eggssampled from this population of frogs would be between0.04 and 1.32.Interpretation — Hypothesis TestingThere is evidence that exposure to progesterone underthe experimental conditions causes a change in the meancAMP levels in this population of frogs (two-sided pairedt-test, p = 0.042).9.4 Sign TestSign TestsThe paired t-test assumes that differences are normallydistributed.If this is not true, the Central Limit Theorem can be used tojustify that the t-test is still valid, provided that the samplesizes are large enough. (As usual, how large is large enoughdepends on the character of the nonnormality in thepopulation.)The sign test is a nonparametric test that does not depend ona normal assumption for the population of differences.To carry out a sign test, we ignore the magnitude ofdifferences and just record whether each difference is positiveor negative.Sign Tests (cont.)P-values are then computed from a binomial distribution withp = 0.5.Technically, the sign test is not testing equality of populationmeans.Instead, a sign test is testing if the differences are equallylikely to be positive versus either a nondirectional ordirectional alternative.ExampleThe compound mCPP is thought to be a hunger supressant.In an experiment, nine obese men had their weight change(kg) recored after each of two two-week periods, once whentaking a placebo and once when taking mCPP.There was a two week “washout period” between measurementperiods.ExampleNegative values indicate a weight loss.A negative difference indicates more weight was lost withmCPP than with the placebo.Subject mCPPP Placebo difference1 0.0 −1.1 1.12 −1.1 0.5 −1.63 −1.6 0.5 −2.14 −0.3 0.0 −0.35 −1.1 −0.5 −0.66 −0.9 1.3 −2.27 −0.5 −1.4 0.98 0.7 0.0 0.79 −1.2 −0.8 −0.4CalculationIn the data, six of nine men lost more weight while usingmCPP than when using the placebo. Is this differencesignificant (here consider a one-sided test)?If mCPP had absolutely no effect, then we would expect thechanges in weight to be random and either treatment wouldbe equally likely to appear better for each individual.With this null assumption, the number of individuals that losemore weight with mCPP than with a placebo is a binomialrandom variable with n = 9 and p = 0.5.If mCPP has an effect, we would expect the proportion ofmen who lose more weight with the drug than with a placeboto be higher than 0.5.Calculation (cont.)The test statistic is the number of negative differences, 6.The p-value is the probability of obtaining 6 or more successesin 9 independent trials with success probability p = 0.5.Here is how to compute this in R.> p1 = sum(dbinom(6:9, 9, 0.5))> p1[1] 0.2539063or> p1 = 1-pbinom(5,9,.5)> p1[1] 0.2539063The sign test: test statisticN+= number of positives, N−= number of negatives1nondirectional (µ16= µ2)Bs= max(N+, N−)2µ1> µ2Bs= N+3µ1< µ2Bs= N−Calculating the p-value (two-sided)p = 2 ∗ IP{Y ≥ Bs},where Y ∼ B(n, 0.5)Calculating the p-value (one-sided)Two steps:1Check directionality—see if the data deviate from H0in thedirection specified by HA:1If not, the p-value is greater than .502If so, proceed to Step 2.2p = IP{Y ≥ Bs}, whereY ∼ B(n, 0.5)Example: rat experiment8 rats were given a drug. Hemoglobin content of blood wasmeasured before and after the drug. Where d = ybefore− yafter.Test whether the drug has an impact on hemoglobin content.data summary:N−= 1 and N+= 7.Chalkboard CalculationsThe sign test: What if there are ties?Tie: y1= y2for some subject. Then d = 0 for this subject. Nosign!!Exclude all zeros, and decrease the sample size accordingly.idea: each pair whose difference is zero is ignored; such pairsare regarded as providing no evidence against H0in


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UW-Madison STAT 371 - Chapter 9 - Comparison of Paired Samples

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