Comparing Two Numerical Response Populations: Paired DataSubject ReuseThe ScatterplotPutting the `R' in RPDOther Ways to Form PairsForming Pairs from Adjacent TrialsWhen is it Valid to do a Paired Data Analysis?An Extended ExampleComputingSummaryPractice ProblemsSolutions to Practice ProblemsHomework ProblemsChapter 20Comparing Two Numerical ResponsePopulations: Paired DataThis chapter is an extension of Chapter 16. In Chapter 16 we considered populations in whicheach population member or trial yields two dichotomous respons es. In the current chapter eachpopulation member or trial yi el ds two numbers. In other ways, however, this chapter also extendsthe work we did in Chapters 17–19.20.1 Subject ReuseI will introduce you to the idea of subject reuse with an artificial s tudy of drug therapy for tensionheadaches. We will compare two different ways to design a study. I cannot use a real scientificstudy to make my comparisons because, to my knowledge, medical researchers select a design anduse it. They do not investigate a medical i ssue twice, with two different designs, just to make mehappy!I am interested in studying drug therapies fo r a fairly mild health ai lment, tension h eadaches.As you will see sh ortly, it is imp ortant t hat I have chosen an ai lment that is bo th nonlethal andrecurrent. I want to compare two drug therapies for the treatment of a tension headache. I willrefer to the t wo therapies as drug A (treatment 1 and populat ion 1) and drug B (treatment 2 andpopulation 2).We n eed a response that is a number. Each subject is given the foll owing instructions:The next tim e you experience a tension headache, take the drug we have given to you.Wait 20 minutes. Writ e down your assess ment of your pain on a scale from 0 (no pain)to 10 (worst pain ever).How can I study this? Goin g all the way b ack to Chapter 1, I can use a completely randomizeddesign. Followin g Chapter 19, I can perform population-based inference on the d at a I obtain frommy completely randomi zed design. In particular, I can compare the mean of population 1 (drug A),µ1, to the mean of population 2 (drug B), µ2. I can estimate µ1− µ2with confidence and test the513Table 20.1: Artificial data from a CRD on headache pain , sorted within each treatment.Position: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16Drug A: 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9Drug B: 0 1 2 2 3 3 3 4 4 4 5 5 6 7 7 8null hypoth es is that µ1= µ2. In o rder to choose an alternative, we need more information aboutthe drugs. Three scenarios come to mi nd, listed below:• Drug A is a placebo and drug B supposedly is beneficial. In this situat ion, remembering thatsmaller responses are preferred to larger responses, my alternative would be >.• Drug A is the extra-strength version of drug B. In this situation , my alternative would be <.• Drugs A and B are different active drugs. In this situation , my alternative would be 6=.Suppose now that I h ave 3 2 subjects available for study and I am willing to pretend that theyare a random sample from my sup erpop ulation of interest. I decide to use a balanced design. Thus ,I will use the online randomizer to assign 16 subjects to each treatment.The artificial d at a for my CRD on the 32 subjects is given in Table 20.1. The data have beenseparated by treatments and sorted within each treatment. You can verify the following values ofsummary statistics (or trust me if you don’t need additional practice on these computations):¯x = 5.500, s1= 2.366, ¯y = 4.000, s2= 2.251 and n1= n2= 16.Next, I calculates2p=(2.366)2+ (2.251)22= 5.3325 and sp=√5.3325 = 2.309.The 95% confidence interval estimate of µ1− µ2is (see Formula 19.9 on page 501):(5.50 − 4.00) ± 2.04 2(2.309)q2/16 = 1.50 ± 2.042(0.8164) = 1.50 ± 1.67 = [−0.17, 3.17].This interval is inconclusive b ecause it contains both positive and negative numbers. For futurereference, note that the half-width of this interval is 1.67.For a test o f hypotheses, from Equation 19.11 on page 502, the observed value of t he teststatistic i st = 1.50/0.8164 = 1.837.With the h el p of our website calculator,http://stattrek.com/online-calculator/t-distribution.aspx,514we find that the area under the t-curve with df = 1 6 + 16 −2 = 30 to the right of 1.837 is equal to0.0381. Thus, the approximate P-value for the alternative > is 0.0381 and the approximate P-valuefor the alternative 6= is 2(0.0381) = 0.0762.Let’s look at the dat a in Table 20.1 again. In the drug A row, two subjects gave a respon se of2—not much pain—and two gave a response of 9—a great deal of pain. In words, for drug A thereis a large amount of subject-to-subject variatio n. The same is true for drug B. The idea behind therandomized pairs design (RPD) is to attempt to reduce this subject-to-subject variation.I mentioned above that it is important that tens ion headaches are no nlethal and recurrent. Re-currence is important because if each subject has a h eadache (which is necessary in th e CRD forus to obtain a response from each subject) then the subject will have a second headache. The RPDwe learn about below will use responses from two headaches per sub ject, compared to the CRDwhich loo ked at one headache per subject. Nonlethal is important because—and I don’t mean tobe insensitive—in order to have a second headache the subject must s urvive the first one.Admittedly, I am ignoring studies that would involve looking at 3, 4, 5 or more headaches persubject. I must draw t he line somewhere!You can now see the reason for the term subject reus e. We reuse each subject and , thus, obtaintwo respon ses per subject. And, somewhat obviously, because our goal is to compare the twotreatments, for each subject we obtain a response from both treatments. T hus, for example, subjectSally gives us two numbers: her pain with drug A and her pain wit h drug B.My next step is to provide you with artificial headache pain data from an RPD. My go al is tocompare my RPD to my CRD for th e artificial headache pain study. What is a fair way to do this?Well, my CRD had 32 sub jects, wit h one response per subject, yielding a total of 32 observations.I cou ld have 32 subjects in my RPD, but that would yield 32 × 2 = 64 observations. T his strikesme as an unfair comparison. Thus, instead, my RPD below has only 16 subjects; with each subjectgiving two responses, I wil l have a total of 16×2 = 32 observations, the same as I had in my CRD.In fact, my RPD has exactly the same 32
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