M12_BERE8380_12_SE_C12.9.qxd 2/21/11 3:56 PM Page 1 12.9 Friedman Rank Test: Nonparametric Analysis for the Randomized Block Design 1 12.9 Friedman Rank Test: Nonparametric Analysis for the Randomized Block Design When analyzing a randomized block design, sometimes the data consist of only the ranks within each block. Other times, you cannot assume that the data from each of the c groups are from normally distributed populations. In these situations, you can use the Friedman rank test. You use the Friedman rank test to determine whether c groups have been selected from populations having equal medians. That is, you test = = Á = H0: M.1 M.2 M.c against the alternative = Á H1: Not all M.j are equal (where j 1, 2, , c). To conduct the test, you replace the data values in each of the r independent blocks with the corresponding ranks, so that you assign rank 1 to the smallest value in the block and rank c to the largest. If any values in a block are tied, you assign them the mean of the ranks that they would otherwise have been assigned. Thus, Rij is the rank (from 1 to c) associated with the jth group in the ith block. Equation (12.13) defines the test statistic for the Friedman rank test. FRIEDMAN RANK TEST FOR DIFFERENCES AMONG C MEDIANS c = 12 2 - + FR + a R.j 3r(c 1) (12.13) rc(c 1) j=1 where 2 = = Á R.j square of the total of the ranks for group j ( j 1, 2, , c) r = number of blocks c = number of groups As the number of blocks gets large (i.e., greater than 5), you can approximate the test sta- - tistic FR by using the chi-square distribution with c 1 degrees of freedom. Thus, for any se- lected level of significance a, you reject the null hypothesis if the computed value of FR is 2 - greater than xU, the upper-tail critical value for the chi-square distribution having c 1 degrees of freedom, as shown in Figure 12.22. That is, 7 2 Reject H0 if FR xa; otherwise, do not reject H0. FIGURE 12.22 Determining the Rejection Region for the Friedman Test 1 – α α 0 χ2 Region of Region of Nonrejection Rejection Critical Value M12_BERE8380_12_SE_C12.9.qxd 2/21/11 3:56 PM Page 2 2 CHAPTER 12 Chi-Square Tests and Nonparametric Tests The critical values from the chi-square distribution are given in Table E.4. To illustrate the Friedman rank test, return to the fast-food chain study from Section 11.2, in which six raters (blocks) evaluated four restaurants (groups). The results of the experiment are displayed in Table 12.21, along with some summary computations. If you cannot make the assumption that the service ratings are normally distributed for each restaurant, the Friedman rank test is more appropriate than the F test. The null hypothesis is that the median service ratings for the four restaurants are equal. The alternative hypothesis is that at least one of the restaurants differs from at least one of the others: = = = H0: M.1 M.2 M.3 M.4 H1: Not all the medians are equal. TABLE 12.21 Restaurant Converting Data to Ranks Within Blocks A B C D Blocks of Raters Rating Rank Rating Rank Rating Rank Rating Rank 1 70 2.0 61 1.0 82 4.0 74 3.0 2 77 3.0 75 1.0 88 4.0 76 2.0 3 76 2.0 67 1.0 90 4.0 80 3.0 4 80 3.0 63 1.0 96 4.0 76 2.0 5 84 2.5 66 1.0 92 4.0 84 2.5 6 78 2.0 68 1.0 98 4.0 86 3.0 Rank total 14.5 6.0 24.0 15.5 Table 12.21 provides the 24 service ratings from Table 11.7 (see FFChain ), along with the ranks assigned within each block. From Table 12.21, you compute the following rank totals for each group: Rank Totals: = = = = R.1 14.5 R.2 6.0 R.3 24.0 R.4 15.5 Equation (12.14) provides a check on the rankings. CHECKING THE RANKINGS IN THE FRIEDMAN TEST rc(c + 1) R + R + R + R = (12.14) .1 .2 .3 .4 2 For the data in Table 12.21, (6)(4)(5) 14.5 + 6 + 24 + 15.5 = 2 60 = 60 Using Equation (12.13), c = 12 2 - + FR + a R.j 3r(c 1) rc(c 1) j=1 12 = 14.52 + 6.02 + 24.02 + 15.52 - (3)(6)(5) (6)(4)(5)3 4 12 = (1,062.5) - 90 = 16.25 a 120 b b r M12_BERE8380_12_SE_C12.9.qxd 2/21/11 3:56 PM Page 3 12.9 Friedman Rank Test: Nonparametric Analysis for the Randomized Block Design 3 = 7 Because FR 16.25 7.815, the upper-tail critical value of the chi-square distribution with c - 1 = 3 degrees of freedom (see Table E.4), or using the Excel or Minitab results of Figure 12.22, because the p-value = 0.001 6 0.05, you reject the null hypothesis at the a = 0.05 level. You conclude that there are significant differences (as perceived by the raters) in the median service ratings at the four restaurants. Minitab labels the Friedman test statistic as S (which is equivalent to the statistic FR). If there are ties in the rankings, as is the case here, Minitab provides an adjustment to the test FIGURE 12.22 Excel and Minitab results for the Friedman rank test for differences among the four medians for the fast-food chain study M12_BERE8380_12_SE_C12.9.qxd 2/21/11 3:56 PM Page 4 4 CHAPTER 12 Chi-Square Tests and Nonparametric Tests statistic S, along with an adjusted p-value. This adjustment has a minimal impact on these results. The following assumptions are needed to use the Friedman rank test: • The r blocks are independent so that the values in one block have no influence on the val- ues in any other block. • The underlying variable is continuous. • The data constitute at least an ordinal scale of measurement within each of the r blocks. • There is no interaction between the r blocks and the c treatment levels. • The c populations have the same variability. • The c populations have the same shape. The Friedman rank test makes less stringent assumptions than does the randomized block F test. If you ignore the last two assumptions (variability and shape), you could still use the Friedman rank test to determine whether at least one of the populations differs from the other populations in some characteristic—either central tendency, variation, or shape. On the other hand, the randomized block F test requires that the level of measurement is an interval or ratio scale and that the c samples are from underlying normal populations hav- ing equal variances. Both the randomized block F test and the Friedman test assume that there is no interaction between the treatments and the blocks. When the more stringent assumptions of the randomized block F test hold, you should se- lect it over the Friedman test because it has more power to detect significant differences among the groups. However, if the assumptions of the randomized block F test are inappropriate, you should use the Friedman rank test. Problems for Section 12.9 LEARNING THE BASICS four characteristics: taste, aroma, richness, and acidity. The following table displays the summated ratings—accumulated 12.105 What is the upper-tail critical value when testing over all four characteristics. for the equality of the medians in six populations using a = 0.10? Brand 12.106 For Problem 12.105: Expert A B C D a. State the decision rule for testing the null hypothesis that all six groups have equal population medians. C.C. 24 26 25 22 = S.E. 27 27 26 24 b. What is your statistical decision if FR 11.56? E.G. 19 22 20 16 B.L. 24 27 25 23 APPLYING THE CONCEPTS C.M. 22 25 22 21 C.N. 26 27 24 24 12.107 Nine experts rated four brands of Colombian G.N. 27 26 22 23 coffee in a taste-testing experiment (see Coffee ). A R.M. 25 27 24 21 rating on a 7-point scale (1 = extremely unpleasing, 7 = P. V. 22 23 20 19 = extremely pleasing) is given for each of the following M12_BERE8380_12_SE_C12.9.qxd 2/21/11 3:56 PM Page 5 12.9 Friedman Rank Test: Nonparametric Analysis for the Randomized Block Design 5 a. At the 0.05 level of significance, is there evidence of a 12.111 The data in the Concrete2 file represent the com- difference in the median summated ratings of the four pressive strength in thousands of pounds per square inch of brands of Colombian coffee? 40 samples of concrete taken 2, 7, and 28 days after pouring. b. Are there any differences in the results of (a) from those Source: O. Carrillo-Gamboa and R. F. Gunst, “Measurement- of Problem 11.23 on page 437? Discuss. Error-Model Collinearities,” Technometrics, 34, 1992, pp. 12.108 Which cell phone service has the highest rating? 454–464. The data in CellRating represent the mean ratings for a. At the 0.05 level of significance, is there evidence of a Verizon, AT&T, T-Mobile, and Sprint in 19 different cities.