12.9 friedman rank test: nonparametric analysis for the

12.9 Friedman Rank Test: Nonparametric Analysis for the Randomized Block Design 1

When analyzing a randomized block design, sometimes the data consist of only the rankswithin each block. Other times, you cannot assume that the data from each of the c groupsare from normally distributed populations. In these situations, you can use the Friedmanrank test.

You use the Friedman rank test to determine whether c groups have been selected frompopulations having equal medians. That is, you test

against the alternative

To conduct the test, you replace the data values in each of the r independent blocks withthe corresponding ranks, so that you assign rank 1 to the smallest value in the block and rank cto the largest. If any values in a block are tied, you assign them the mean of the ranks that theywould otherwise have been assigned. Thus, is the rank (from 1 to c) associated with the jthgroup in the ith block.

Equation (12.13) defines the test statistic for the Friedman rank test.

Rij

H1: Not all M.j are equal (where j = 1, 2, Á , c).

H0: M.1 = M.2 =Á

= M.c

FRIEDMAN RANK TEST FOR DIFFERENCES AMONG C MEDIANS

(12.13)

wheresquare of the total of the ranks for group

number of blocksnumber of groupsc =

r =

j ( j = 1, 2, Á , c)R2.j =

FR =

12rc(c + 1)a

c

j = 1R2

.j - 3r(c + 1)

As the number of blocks gets large (i.e., greater than 5), you can approximate the test sta-tistic by using the chi-square distribution with degrees of freedom. Thus, for any se-lected level of significance you reject the null hypothesis if the computed value of isgreater than the upper-tail critical value for the chi-square distribution having degrees of freedom, as shown in Figure 12.22. That is,

otherwise, do not reject H0.

Reject H0 if FR 7 x2a;

c - 1x2U,

FRa,c - 1FR

Region ofRejection

Region ofNonrejection

CriticalValue

0α1 – α

χ2

F I G U R E 1 2 . 2 2Determining theRejection Region for the Friedman Test

12.9 Friedman Rank Test: Nonparametric Analysis for theRandomized Block Design

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CHECKING THE RANKINGS IN THE FRIEDMAN TEST

(12.14)R.1 + R.2 + R.3 + R.4 =

rc(c + 1)

2

Restaurant

A B C D

Blocks ofRaters Rating Rank Rating Rank Rating Rank Rating Rank

1 70 2.0 61 1.0 82 4.0 74 3.02 77 3.0 75 1.0 88 4.0 76 2.03 76 2.0 67 1.0 90 4.0 80 3.04 80 3.0 63 1.0 96 4.0 76 2.05 84 2.5 66 1.0 92 4.0 84 2.56 78 2.0 68 1.0 98 4.0 86 3.0Rank total 14.5 6.0 24.0 15.5

T A B L E 1 2 . 2 1Converting Data toRanks Within Blocks

For the data in Table 12.21,

Using Equation (12.13),

= a 12120b (1,062.5) - 90 = 16.25

= b 12(6)(4)(5)

314.52+ 6.02

+ 24.02+ 15.524 r - (3)(6)(5)

FR =

12rc(c + 1)a

c

j = 1R2

.j - 3r(c + 1)

60 = 60

14.5 + 6 + 24 + 15.5 =

(6)(4)(5)

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 experimentare displayed in Table 12.21, along with some summary computations. If you cannot make theassumption that the service ratings are normally distributed for each restaurant, the Friedmanrank 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 theothers:

H1: Not all the medians are equal.

H0: M.1 = M.2 = M.3 = M.4

Table 12.21 provides the 24 service ratings from Table 11.7 (see ), along with theranks assigned within each block.

From Table 12.21, you compute the following rank totals for each group:Rank Totals:

Equation (12.14) provides a check on the rankings.

R.1 = 14.5 R.2 = 6.0 R.3 = 24.0 R.4 = 15.5

FFChain

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Because the upper-tail critical value of the chi-square distributionwith degrees of freedom (see Table E.4), or using the Excel or Minitab results ofFigure 12.22, because the you reject the null hypothesis at the

level. You conclude that there are significant differences (as perceived by the raters) inthe median service ratings at the four restaurants.

Minitab labels the Friedman test statistic as S (which is equivalent to the statistic ). Ifthere are ties in the rankings, as is the case here, Minitab provides an adjustment to the test

FR

a = 0.05p-value = 0.001 6 0.05,

c - 1 = 3FR = 16.25 7 7.815,

F I G U R E 1 2 . 2 2Excel and Minitab resultsfor the Friedman ranktest for differencesamong the four mediansfor the fast-food chainstudy

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4 CHAPTER 12 Chi-Square Tests and Nonparametric Tests

statistic S, along with an adjusted p-value. This adjustment has a minimal impact on theseresults.

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 blockF test. If you ignore the last two assumptions (variability and shape), you could still use theFriedman rank test to determine whether at least one of the populations differs from the otherpopulations in some characteristic—either central tendency, variation, or shape.

On the other hand, the randomized block F test requires that the level of measurement isan 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 thereis 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 amongthe groups. However, if the assumptions of the randomized block F test are inappropriate, youshould use the Friedman rank test.

Problems for Section 12.9LEARNING THE BASICS

12.105 What is the upper-tail critical value when testingfor the equality of the medians in six populations using

12.106 For Problem 12.105:a. State the decision rule for testing the null hypothesis that

all six groups have equal population medians.b. What is your statistical decision if

APPLYING THE CONCEPTS

12.107 Nine experts rated four brands of Colombiancoffee in a taste-testing experiment (see ). Arating on a 7-point scale

is given for each of the following= extremely pleasing)(1 = extremely unpleasing, 7 =

Coffee

FR = 11.56?

a = 0.10?

four characteristics: taste, aroma, richness, and acidity. Thefollowing table displays the summated ratings—accumulatedover all four characteristics.

Brand

Expert A B C D

C.C. 24 26 25 22S.E. 27 27 26 24E.G. 19 22 20 16B.L. 24 27 25 23C.M. 22 25 22 21C.N. 26 27 24 24G.N. 27 26 22 23R.M. 25 27 24 21P.V. 22 23 20 19

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EG12.9 EXCEL GUIDE FOR THE FRIEDMANRANK TEST

Use the worksheets of the Friedman Rank Test workbookas a template for performing the Friedman rank test. For ex-ample, for the Section 12.9 fast-food study example thatcontains six blocks and four groups, open to theFriedman6x4 worksheet.

Friedman worksheets use the RANK(value, set of block

values, order) function to rank the values for each block.This function is used twice in each formula in the rank tablethat begins in row 12, once with order set to 1 (ascendingorder) and then again set to 0 (descending) in a shortcut, torank values in both ascending and descending order. (Thisis done to allow the table to break ties.) The worksheets alsouse the CHIINV(level of significance, degrees of freedom)and CHIDIST( test statistic, degrees of freedom) func-tions to compute the critical value and p-value, respectively.

When you open to a worksheet, enter the data in the tablethat starts in row 3 and enter the level of significance valuein cell B24. The #NA! messages that appear in manycells are not an error and will disappear after you enteryour data.

MG12.9 MINITAB GUIDE FOR THE FRIEDMANRANK TEST

Use Friedman to perform the Freidman rank test. For ex-ample, to perform the Figure 12.22 test for the fast-foodchain study, open to the FFChain worksheet. Select Stat ➔Nonparametrics ➔ Friedman. In the Friedman dialog box:

1. Double-click C3 Rating in the variables list to addRating to the Response box.

2. Double-click C2 Restaurant in the variables list toadd Restaurant to the Treatment box.

3. Double-click C1 Raters in the variables list to addRaters to the Blocks box.

4. Click OK.

x2

a. At the 0.05 level of significance, is there evidence of adifference in the median summated ratings of the fourbrands of Colombian coffee?

b. Are there any differences in the results of (a) from thoseof Problem 11.23 on page 437? Discuss.

12.108 Which cell phone service has the highest rating?The data in represent the mean ratings forVerizon, AT&T, T-Mobile, and Sprint in 19 different cities.Source: Data extracted from “Best Cell-Phone Service,”Consumer Reports, January 2009, pp. 28–32.

a. At the 0.05 level of significance, determine whetherthere is evidence of a difference in the median cell ratingfor the four cell phone services.


12.109 Is there a difference in the prices if you shop as animpulsive shopper, as a savvy shopper, or if you shop at awarehouse club such as Costco, or if you purchase store-brands? To investigate this, a random sample of 10 pur-chases was selected and the prices were compared. (Dataextracted from “Shop Smart and Save Big,” Consumer Re-ports, May 2009, p. 17.) The prices for the products arestored in .a. At the 0.05 level of significance, is there evidence of a

difference between the median price of an impulsiveshopper, a savvy shopper, if you shop at a warehouseclub such as Costco, or if you purchase store-brands?


12.110 Philips Semiconductors is a leading Europeanmanufacturer of integrated circuits. Integrated circuits areproduced on silicon wafers, which are ground to targetthickness early in the production process. The wafers are po-sitioned in various locations on a grinder and kept in placethrough vacuum decompression. One of the goals of processimprovement is to reduce the variability in the thickness ofthe wafers in different positions and in different batches.Data were collected from a sample of 30 batches. In eachbatch, the thickness of the wafers on positions 1 and 2 (outercircle), 18 and 19 (middle circle), and 28 (inner circle) wasmeasured. The results are given in the file.Source: Extracted from K. C. B. Roes and R. J. M. M. Does,“Shewhart-Type Charts in Nonstandard Situations,” Techn-ometrics, 37, 1995, pp. 15–24.

a. At the 0.05 level of significance, is there evidence of adifference in the median thickness of the wafers for thefive positions?


c. Which test is more appropriate for these data, the Fried-man rank test or the randomized block F test? Explain.

Circuits

Shopping2

CellRating

12.111 The data in the file represent the com-pressive strength in thousands of pounds per square inch of40 samples of concrete taken 2, 7, and 28 days after pouring.Source: O. Carrillo-Gamboa and R. F. Gunst, “Measurement-Error-Model Collinearities,” Technometrics, 34, 1992, pp.454–464.

a. At the 0.05 level of significance, is there evidence of adifference in the median compressive strength after 2, 7,and 28 days?


c. Which test is more appropriate for these data, the Fried-man rank test or the randomized block F test? Explain.

Concrete2

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12.9 friedman rank test: nonparametric analysis for the

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