The Sum of Ranks TestRanksThe Hypotheses for the Sum of Ranks TestStep 2: The Test Statistic and Its Sampling DistributionStep 3: The Three Rules for Calculating the P-valueOrdinal DataComputingSummaryPractice ProblemsSolutions to Practice ProblemsHomework ProblemsChapter 6The Sum of Ranks TestThus far in these Course Notes we have considered CRDs with a numerical response. In Chapter 5we learned how to perform a statistical test o f h ypotheses to investigate whether the Skeptic’sArgument is correct. Every test of hypotheses has a test statistic; in Chapter 5 we chose the tes tstatistic U which has observed value u = ¯x − ¯y. For rather obvious reasons, this test using U isreferred to a test of means or a test of comparing means.We learned in Chapter 1 that the mean is a popular way to summarize a list of numbers. Thus,it is not surprising to learn that comparing means, by subtraction, is a popular way to compare twotreatments and, h ence, the test of Chapter 5 seems sensible. But we also learned i n Chapter 1 thatthe median is another popular way to summarize a list of numbers. Thus, you might guess thatanother popular choice for a test statistic would be the one whose observed value is v = ˜x − ˜y. Ifyou make this guess, you would be wron g , but close to the truth.Recall from Chapter 1 that the distinction between the mean and the median can be viewed asthe di stinction between focusing on arithmetic versus position. The median, recall, is the numberat the center position of a sorted listed—for an odd sample size—or the average of the values atthe two center positions—for an even sample s ize. Thus, the value v in the previous paragraphcompares two sorted lists by comparing the nu mbers in their center po sitions. This comparisonignores a great deal of information! In those situations in which, for wh at ever reasons, we preferto focus on positions rather than arithmetic, it turns out that using ranks, defined below, is superiorto us ing m edians in order to compare two sets of nu mbers.In this chapter we will consider an option to using U: i.e., we will present a test that comparesthe two sets of data b y comparing their ranks. When we study power in a later chapter, we will seethat sometimes the test that compares ranks is better than the test that compares means. The lastsection of this chapter presents an addition al advantage of using a test b ased on ranks; it can beused when the response is ordinal, but not numerical.6.1 RanksWe begin by do ing something that seems quite odd: We combine t he data from t he two treatm entsinto one set of d ata and then we sort the n = n1+ n2response values. For example, for Dawn’s117Tabl e 6.1: Dawn’s 20 sorted response values, with ranks.Position: 1 2 3 4 5 6 7 8 9 10Response: 0 1 1 1 2 3 3 3 3 4Rank: 1 3 3 3 5 7.5 7.5 7.5 7.5 10.5Position: 11 12 13 14 15 16 17 18 19 20Response: 4 5 5 5 6 6 6 7 7 8Rank: 10.5 1 3 13 13 16 16 16 18. 5 18.5 20study of h er cat, the 20 sorted response values are given in Table 6.1. You can verify these numbersfrom the data presented in Table 1.3 in Chapter 1, but I recommend that you just trust me on this.We note t hat Dawn’s 20 numb ers consist of nine distinct values. Her number of distinct values issmaller than 20 because several o f the responses are tied; for example, four responses are tied withthe value 3. Going back t o Chapter 1, we talk about the 20 posi tions in the list i n Table 6.1. A sexamples: position 1 has the response 0; position 20 has the response 8; and positions 6–9 all havethe response 3.If the n numbers in our list are all distinct, then the rank of each response is its position. Thisis referred to as the no-ties situation and it makes all of the computations below much s impler.Sadly, in practice, data with ties are common place. Whenever there are ties, all tied responsesreceive the same rank, which is equal to the mean of their positions. Thus, for example, all fourof the responses equal to 3 receive the rank of 7.5 because t hey occupy positions 6 through 9 andthe mean of 6, 7, 8 and 9 is 7. 5. It is tediou s to sum these four numbers to find their mean; hereis a shortcut that always works: sim ply compute the mean of the smallest (first) and largest (last)positions in the l ist. For example, to find the mean of 6, 7, 8 and 9, simply calculate (6+9)/2 = 7.5.When we consider ordinal data in Section 6.5 we will have occasion to find the mean of43, 44, . . . , and 75.Summing these 33 numb ers is much more tedious than simply computi ng (43 + 75)/2 = 59.Finally, for any responses in the list that is not tied with another—responses 0, 2 and 8 i n Dawn’sdata—its rank equals its position.The basic idea of our test based on ranks is that we analyze the ranks, not the responses. Forexample, I have retyped Table 6.1 in Table 6.2 (dropp ing t he two Position rows) with the addedfeature that th e responses from treatment 1 (chicken) and their ranks are in bold face type. For thetest statistic U we performed arithmetic on the responses to obtain the means for each treatmentand then we s ubtracted. We do the same arithmetic now, but we use the ranks instead of theresponses. For example, let R1denote the sum of the ranks for treatment 1 and let r1denote itsobserved value. For Dawn’s d at a we get:r1= 3 + 7.5 + 10.5 + 13 + 13 + 16 + 16 + 16 + 18.5 + 20 = 133.5.118Tabl e 6.2: Dawn’s 20 sort ed responses, with ranks. The respon ses from treatment 1, and theirranks, are in bold-faced typ e.Response: 0 1 1 1 2 3 3 3 3 4Rank: 1 3 3 3 5 7.5 7.5 7.5 7.5 1 0.5Response: 4 5 5 5 6 6 6 7 7 8Rank: 10.5 1 3 13 13 16 16 16 18.5 18.5 20Similarly, let R2denote t h e sum of the ranks for treatment 2 and let r2denote i ts observed value.For Dawn’s data we get:r2= 1 + 3 + 3 + 5 + 7 .5 + 7.5 + 7.5 + 10.5 + 13 + 18.5 = 76 .5.In order to compare t h e treatments’ ranks d escrip tively, we calculate the mean of the ranks for eachtreatment:¯r1= r1/n1= 13 3.5/10 = 13.35 and ¯r2= r2/n2= 76.5/10 = 7.65,which show that, based on ranks, the responses on treatment 1 are larger than the responses ontreatment 2 .The next obvious step is that we definev = ¯r1− ¯r2= r1/n1− r2/n2,to be the observed value of the test statistic V . I say t hat this step is obvious because it is analogousto our definition of u = ¯x − ¯y; …
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