Lecture Slides. Elementary Statistics. by Mario F. Triola. and the Triola Statistics Series

Lecture Slides Elementary Statistics Tenth Edition and the Triola Statistics Series by Mario F. Triola Slide 1

Chapter 13 Nonparametric Statistics 13-1 Overview 13-2 Sign Test 13-3 Wilcoxon Signed-Ranks Test for Matched Pairs 13-4 Wilcoxon Rank-Sum Test for Two Independent Samples 13-5 Kruskal-Wallis Test 13-6 Rank Correlation 13-7 Runs Test for Randomness Slide 2

Section 13-1 Overview Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN Slide 3

Overview Definitions Parametric tests have requirements about the nature or shape of the populations involved. Nonparametric tests do not require that samples come from populations with normal distributions or have any other particular distributions. Consequently, nonparametric tests are called distribution-free tests. Slide 4

Advantages of Nonparametric Methods 1. Nonparametric methods can be applied to a wide variety of situations because they do not have the more rigid requirements of the corresponding parametric methods. In particular, nonparametric methods do not require normally distributed populations. 2. Unlike parametric methods, nonparametric methods can often be applied to categorical data, such as the genders of survey respondents. 3. Nonparametric methods usually involve simpler computations than the corresponding parametric methods and are therefore easier to understand and apply. Slide 5

Disadvantages of Nonparametric Methods 1. Nonparametric methods tend to waste information because exact numerical data are often reduced to a qualitative form. 2. Nonparametric tests are not as efficient as parametric tests, so with a nonparametric test we generally need stronger evidence (such as a larger sample or greater differences) before we reject a null hypothesis. Slide 6

Efficiency of Nonparametric Methods Slide 7

Definitions Data are sorted when they are arranged according to some criterion, such as smallest to the largest or best to worst. A rank is a number assigned to an individual sample item according to its order in the sorted list. The first item is assigned a rank of 1, the second is assigned a rank of 2, and so on. Slide 8

Handling Ties in Ranks Find the mean of the ranks involved and assign this mean rank to each of the tied items. Sorted Data 4 5 5 5 10 11 12 12 Preliminary Ranking 1 2 3 4 5 6 7 8 Mean is 3. Mean is 7.5. Rank 1 3 3 3 5 6 7.5 7.5 Slide 9

Section 13-2 Sign Test Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN Slide 10

Key Concept The main objective of this section is to understand the sign test procedure, which involves converting data values to plus and minus signs, then testing for disproportionately more of either sign. Slide 11

Definition Sign Test The sign test is a nonparametric (distribution free) test that uses plus and minus signs to test different claims, including: 1) Claims involving matched pairs of sample data; 2) Claims involving nominal data; 3) Claims about the median of a single population. Slide 12

Basic Concept of the Sign Test The basic idea underlying the sign test is to analyze the frequencies of the plus and minus signs to determine whether they are significantly different. Slide 13

Figure 13-1 Sign Test Procedure Slide 14

Figure 13-1 Sign Test Procedure Slide 15

Figure 13-1 Sign Test Procedure Slide 16

Requirements 1. The sample data have been randomly selected. 2. There is no requirement that the sample data come from a population with a particular distribution, such as a normal distribution. Slide 17

Notation for Sign Test x = the number of times the less frequent sign occurs n = the total number of positive and negative signs combined Slide 18

Test Statistic For n 25: x (the number of times the less frequent sign occurs) For n > 25: z = n 2 (x + 0.5) n Critical values For n 25, critical x values are in Table A-7. For n > 25, critical z values are in Table A-2. n 2 Slide 19

Claims Involving Matched Pairs When using the sign test with data that are matched pairs, we convert the raw data to plus and minus signs as follows: 1. Subtract each value of the second variable from the corresponding value of the first variable. 2. Record only the sign of the difference found in step 1. Exclude ties: that is, any matched pairs in which both values are equal. Slide 20

Key Concept Underlying This Use of the Sign Test If the two sets of data have equal medians, the number of positive signs should be approximately equal to the number of negative signs. Slide 21

Example: Yields of Corn from Different Seeds Use the data in Table 13-3 with a 0.05 significance level to test the claim that there is no difference between the yields from the regular and kiln-dried seed. H 0 : The median of the differences is equal to 0. H 1 : The median of the differences is not equal to 0. α = 0.05 x = minimum(7, 4) = 4 (From Table 13-3, there are 7 negative signs and 4 positive signs.) Critical value = 1 (From Table A-7 where n = 11 and α = 0.05) Slide 23

Example: Yields of Corn from Different Seeds Use the data in Table 13-3 with a 0.05 significance level to test the claim that there is no difference between the yields from the regular and kiln-dried seed. H 0 : The median of the differences is equal to 0. H 1 : The median of the differences is not equal to 0. With a test statistic of x = 4 and a critical value of 1, we fail to reject the null hypothesis of no difference. There is not sufficient evidence to warrant rejection of the claim that the median of the differences is equal to 0. Slide 24

Claims Involving Nominal Data The nature of nominal data limits the calculations that are possible, but we can identify the proportion of the sample data that belong to a particular category. Then we can test claims about the corresponding population proportion p. Slide 25

Example: Gender Selection Of the 325 babies born to parents using the XSORT method of gender selection, 295 were girls. Use the sign test and a 0.05 significance level to test the claim that this method of gender selection has no effect. The procedures are for cases in which n > 25. Note that the only requirement is that the sample data are randomly selected. H 0 : p = 0.5 (the proportion of girls is 0.5) H 1 : p 0.5 Slide 26

Example: Gender Selection Of the 325 babies born to parents using the XSORT method of gender selection, 295 were girls. Use the sign test and a 0.05 significance level to test the claim that this method of gender selection has no effect. Denoting girls by the positive sign (+) and boys by the negative sign ( ), we have 295 positive signs and 30 negative signs. Test statistic x = minimum(295, 30) = 30 The test involves two tails. Slide 27

Example: Gender Selection Of the 325 babies born to parents using the XSORT method of gender selection, 295 were girls. Use the sign test and a 0.05 significance level to test the claim that this method of gender selection has no effect. With α = 0.05 in a two-tailed test, the critical values are z = ± 1.96. The test statistic z = -14.64 is less than -1.96. We reject the null hypothesis that p = 0.5. There is sufficient evidence to warrant rejection of the claim that the method of gender selection has no effect. Slide 29

Claims About the Median of a Single Population The negative and positive signs are based on the claimed value of the median. Slide 31

Example: Body Temperature Use the temperatures for 12:00 A.M. on Day 2 in Data Set 2 in Appendix B. Use the sign test to test the claim that the median is less than 98.6 F. There are 68 subjects with temperatures below 98.6 F, 23 subjects with temperatures above 98.6 F, and 15 subjects with temperatures equal to 98.6 F. H 0 : Median is equal to 98.6 F. H 1 : Median is less than 98.6 F. Since the claim is that the median is less than 98.6 F. the test involves only the left tail. Slide 32

Example: Body Temperature Use the temperatures for 12:00 A.M. on Day 2 in Data Set 2 in Appendix B. Use the sign test to test the claim that the median is less than 98.6 F. Discard the 15 zeros. Use ( ) to denote the 68 temperatures below 98.6 F, and use ( + ) to denote the 23 temperatures above 98.6 F. So n = 91 and x = 23 Slide 33

Example: Body Temperature Use the temperatures for 12:00 A.M. on Day 2 in Data Set 2 in Appendix B. Use the sign test to test the claim that the median is less than 98.6 F. z = (x + 0.5) n 2 n 2 z = (23 + 0.5) 91 2 91 2 = 4.61 Slide 34

Example: Body Temperature Use the temperatures for 12:00 A.M. on Day 2 in Data Set 2 in Appendix B. Use the sign test to test the claim that the median is less than 98.6 F. We use Table A-2 to get the critical z value of 1.645. The test statistic of z = 4.61 falls into the critical region. We reject the null hypothesis. We support the claim that the median body temperature of healthy adults is less than 98.6 F. Slide 35

Example: Body Temperature Use the temperatures for 12:00 A.M. on Day 2 in Data Set 2 in Appendix B. Use the sign test to test the claim that the median is less than 98.6 F. Figure 13.3 Slide 36

Recap In this section we have discussed: Sign tests where data are assigned plus or minus signs and then tested to see if the number of plus and minus signs is equal. Sign tests can be performed on claims involving: Matched pairs Nominal data The median of a single population Slide 37

Section 13-3 Wilcoxon Signed-Ranks Test for Matched Pairs Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN Slide 38

Key Concept The Wilcoxon signed-ranks test uses ranks of sample data consisting of matched pairs. This test is used with a null hypothesis that the population of differences from the matched pairs has a median equal to zero. Slide 39

Definition The Wilcoxon signed-ranks test is a nonparametric test that uses ranks of sample data consisting of matched pairs. It is used to test the null hypothesis that the population of differences has a median of zero. H 0 : The matched pairs have differences that come from a population with a median equal to zero. H 1 : The matched pairs have differences that come from a population with a nonzero median. Slide 40

Wilcoxon Signed-Ranks Test Requirements 1. The data consist of matched pairs that have been randomly selected. 2. The population of differences (found from the pairs of data) has a distribution that is approximately symmetric, meaning that the left half of its histogram is roughly a mirror image of its right half. (There is no requirement that the data have a normal distribution.) Slide 41

Notation T = the smaller of the following two sums: 1. The sum of the absolute values of the negative ranks of the nonzero differences d 2. The sum of the positive ranks of the nonzero differences d Slide 42

Test Statistic for the Wilcoxon Signed-Ranks Test for Matched Pairs For n 30, the test statistic is T. For n > 30, the test statistic is z = T n(n + 1) 4 n(n +1) (2n +1) 24 Slide 43

Critical Values for the Wilcoxon Signed-Ranks Test for Matched Pairs For n 30, the critical T value is found in Table A-8. For n > 30, the critical z values are found in Table A-2. Slide 44

Procedure for Finding the Value of the Test Statistic Step 1: For each pair of data, find the difference d by subtracting the second value from the first. Keep the signs, but discard any pairs for which d = 0. Step 2: Ignore the signs of the differences, then sort the differences from lowest to highest and replace the differences by the corresponding rank value. When differences have the same numerical value, assign to them the mean of the ranks involved in the tie. Step 3: Attach to each rank the sign difference from which it came. That is, insert those signs that were ignored in step 2. Step 4: Find the sum of the absolute values of the negative ranks. Also find the sum of the positive ranks. Slide 45

Procedure for Finding the Value of the Test Statistic Step 5: Let T be the smaller of the two sums found in Step 4. Either sum could be used, but for a simplified procedure we arbitrarily select the smaller of the two sums. Step 6: Let n be the number of pairs of data for which the difference d is not 0. Step 7: Determine the test statistic and critical values based on the sample size, as shown above. Step 8: When forming the conclusion, reject the null hypothesis if the sample data lead to a test statistic that is in the critical region - that is, the test statistic is less than or equal to the critical value(s). Otherwise, fail to reject the null hypothesis. Slide 46

Example: Does the Type of Seed Affect Corn Growth? Use the data in Table 13-4 with the Wilcoxon signed-ranks test and 0.05 significance level to test the claim that there is no difference between the yields from the regular and kiln-dried seed. H 0 : There is no difference between the times of the first and second trials. H 1 : There is a difference between the times of the first and second trials. Slide 48

Example: Does the Type of Seed Affect Corn Growth? The ranks of differences in row four of the table are found by ranking the absolute differences, handling ties by assigning the mean of the ranks. The signed ranks in row five of the table are found by attaching the sign of the differences to the ranks. The differences in row three of the table are found by computing the first time second time. Slide 49

Example: Does the Type of Seed Affect Corn Growth? Calculate the Test Statistic Step 1: In Table 13-4, the row of differences is obtained by computing this difference for each pair of data: d = yield from regular seed yield from kiln-dried seed Step 2: Ignoring their signs, we rank the absolute differences from lowest to highest. Step 3: The bottom row of Table 13-4 is created by attaching to each rank the sign of the corresponding differences. Slide 50

Example: Does the Type of Seed Affect Corn Growth? Calculate the Test Statistic Step 3 (cont.): If there really is no difference between the yields from the two types of seed (as in the null hypothesis), we expect the sum of the positive ranks to be approximately equal to the sum of the absolute values of the negative ranks. Step 4: We now find the sum of the absolute values of the negative ranks, and we also find the sum of the positive ranks. Slide 51

Example: Does the Type of Seed Affect Corn Growth? Calculate the Test Statistic Step 4 (cont.): Sum of absolute values of negative ranks: 51 (from 10 + 9 + 8 + 6 + 5 + 11 + 2) Sum of positive ranks: 15 (from 1 + 3 + 4 + 7) Step 5: Letting T be the smaller of the two sums found in Step 4, we find that T = 15. Step 6: Letting n be the number of pairs of data for which the difference d is not 0, we have n = 11. Slide 52

Example: Does the Type of Seed Affect Corn Growth? Calculate the Test Statistic Step Step 7: 7: Because n = 11, we have n 30, so we use a test statistic of T = 15. From Table A-8, the critical T = 11 (using n = 11 and α = 0.05 in two tails). Step 8: The test statistic T = 15 is not less than or equal to the critical value of 11, so we fail to reject the null hypothesis. It appears that there is no difference between yields from regular seed and kiln-dried seed. Slide 53

Recap In this section we have discussed: The Wilcoxon signed-ranks test which uses matched pairs. The hypothesis is that the matched pairs have differences that come from a population with a median equal to zero. Slide 54

Section 13-4 Wilcoxon Rank-Sum Test for Two Independent Samples Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN Slide 55

Key Concept The Wilcoxon signed-ranks test (Section 13-3) involves matched pairs of data. The Wilcoxon rank-sum test of this section involves two independent samples that are not related or somehow matched or paired. Slide 56

Definition The Wilcoxon rank-sum test is a nonparametric test that uses ranks of sample data from two independent populations. It is used to test the null hypothesis that the two independent samples come from populations with equal medians. H 0 : The two samples come from populations with equal medians. H 1 : The two samples come from populations with different medians. Slide 57

Basic Concept If two samples are drawn from identical populations and the individual values are all ranked as one combined collection of values, then the high and low ranks should fall evenly between the two samples. Slide 58

Requirements 1. There are two independent samples of randomly selected data. 2. Each of the two samples has more than 10 values. 3. There is no requirement that the two populations have a normal distribution or any other particular distribution. Slide 59

n 1 = size of Sample 1 n 2 = size of Sample 2 Notation for the Wilcoxon Rank-Sum Test R 1 = sum of ranks for Sample 1 R 2 = sum of ranks for Sample 2 R = same as R 1 (sum of ranks for Sample 1) µ R = mean of the sample R values that is expected when the two populations have equal medians σ R = standard deviation of the sample R values that is expected with two populations having equal medians Slide 60

Test Statistic for the Wilcoxon Rank-Sum Test where µ R z = = R µ R σ R n 1 (n 1 + n 2 + 1) 2 σ R = n 1 n 2 (n 1 + n 2 + 1) 12 n 1 = size of the sample from which the rank sum R is found n 2 = size of the other sample R = sum of ranks of the sample with size n 1 Slide 61

Critical Values for the Wilcoxon Rank-Sum Test Critical values can be found in Table A-2 (because the test statistic is based on the normal distribution). Slide 62

Procedure for Finding the Value of the Test Statistic 1. Temporarily combine the two samples into one big sample, then replace each sample value with its rank. 2. Find the sum of the ranks for either one of the two samples. 3. Calculate the value of the z test statistic as shown in the previous slide, where either sample can be used as Sample 1. Slide 63

Example: BMI of Men and Women The data in Table 13-5 are from Data Set 1 in Appendix B and use only the first 13 sample values for men and the first 12 sample values for women. The numbers in parentheses are their ranks beginning with a rank of 1 assigned to the lowest value of 17.7. R 1 and R 2 at the bottom denote the sum of ranks. Slide 64

Example: BMI of Men and Women Use the data in Table 13-5 with the Wilcoxon rank-sum test and a 0.05 significance level to test the claim that the median BMI of men is equal to the median BMI of women. The requirements of having two independent and random samples and each having more than 10 values are met. H 0 : Men and women have BMI values with equal medians H 1 : Men and women have BMI values with medians that are not equal Slide 65

Example: BMI of Men and Women Use the data in Table 13-5 with the Wilcoxon rank-sum test and a 0.05 significance level to test the claim that the median BMI of men is equal to the median BMI of women. Procedures. 1. Rank all 25 BMI measurements combined. This is done in Table 13-5. 2. Find the sum of the ranks of either one of the samples. For men the sum of ranks is R = 11.5 + 9 + 14 + + 15.5 = 187 Slide 66

Example: BMI of Men and Women Procedures (cont.). 3. Calculate the value of the z test statistic. µ R n ( n + n + 1) 13(13 + 12 + 1) 2 2 1 1 2 = = = 169 σ R n n ( n + n + 1) (13)(12)(13 + 12 + 1) 12 12 1 2 1 2 = = = 18.385 z R µ R 187 169 = = = σ 18.385 R 0.98 Slide 67

Example: BMI of Men and Women Use the data in Table 13-5 with the Wilcoxon rank-sum test and a 0.05 significance level to test the claim that the median BMI of men is equal to the median BMI of women. A large positive value of z would indicate that the higher ranks are found disproportionately in Sample 1, and a large negative value of z would indicate that Sample 1 had a disproportionate share of lower ranks. Slide 68

Example: BMI of Men and Women Use the data in Table 13-5 with the Wilcoxon rank-sum test and a 0.05 significance level to test the claim that the median BMI of men is equal to the median BMI of women. We have a two tailed test (with α = 0.05), so the critical values are 1.96 and 1.96. The test statistic of 0.98 does not fall within the critical region, so we fail to reject the null hypothesis that men and women have BMI values with equal medians. It appears that BMI values of men and women are basically the same. Slide 69

Example: BMI of Men and Women The preceding example used only 13 of the 40 sample BMI values for men listed in Data Set 1 in Appendix B, and it used only 12 of the 40 BMI values for women. Do the results change if we use all 40 sample values for both men and women? The null and alternative hypotheses are the same. Slide 70

Example: BMI of Men and Women In the Minitab display below ETA1 and ETA2 denote the medians of the first and second samples, respectively. The rank sum for men is W = 1727.5 The P-value is 0.3032 (or 0.3031 after adjustment for ties). Minitab Slide 71

Example: BMI of Men and Women Because the P-value is greater than α = 0.05, we fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that men and women have BMI values with equal medians. Minitab Slide 72

Recap In this section we have discussed: The Wilcoxon Rank-Sum Test for Two Independent Samples. TW9 It is used to test the null hypothesis that the two independent samples come from populations with equal medians. Slide 73

Slide 73 TW9 period at end of sentence Tom Wegleitner; 24/5/2006

Section 13-5 Kruskal-Wallis Test Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN Slide 74

Key Concept This section introduces the Kruskal- Wallis test, which uses ranks of data from three or more independent samples to test the null hypothesis that the samples come from populations with equal medians. Slide 75

Kruskal-Wallis Test Definition. The Kruskal-Wallis test (also called the H test) is a nonparametric test that uses ranks of sample data from three or more independent populations. It is used to test the null hypothesis that the independent samples come from populations with the equal medians. H 0 : The samples come from populations with equal medians. H 1 : The samples come from populations with medians that are not all equal. Slide 76

Kruskal-Wallis Test We compute the test statistic H, which has a distribution that can be approximated by the chi-square (χ 2 ) distribution as long as each sample has at least 5 observations. When we use the chi-square distribution in this context, the number of degrees of freedom is k 1, where k is the number of samples. Slide 77

Kruskal-Wallis Test Requirements 1. We have at least three independent samples, all of which are randomly selected. 2. Each sample has at least 5 observations. 3. There is no requirement that the populations have a normal distribution or any other particular distribution. Slide 78

N k Kruskal-Wallis Test Notation = total number of observations in all observations combined = number of samples R 1 = sum of ranks for Sample 1 n 1 = number of observations in Sample 1 For Sample 2, the sum of ranks is R 2 and the number of observations is n 2, and similar notation is used for the other samples. Slide 79

Kruskal-Wallis Test H Test Statistic 2 R 2 1 2 R 2 k 12 R = + +... + 3( N + 1) N( N + 1) n n n 1 2 k 1. Test is right-tailed. Critical Values 2. df = k 1 (Because the test statistic H can be approximated by the χ 2 distribution, use Table A-4). Slide 80

Procedure for Finding the Value of the Test Statistic H 1 Temporarily combine all samples into one big sample and assign a rank to each sample value. 2. For each sample, find the sum of the ranks and find the sample size. 3. Calculate H by using the results of Step 2 and the notation and test statistic given on the preceding slide. Slide 81

Procedure for Finding the Value of the Test Statistic H The test statistic H is basically a measure of the variance of the rank sums R 1, R 2,, R k. If the ranks are distributed evenly among the sample groups, then H should be a relatively small number. If the samples are very different, then the ranks will be excessively low in some groups and high in others, with the net effect that H will be large. Slide 82

Example: Effects of Treatments on Poplar Tree Weights Table 13-6 lists weights of poplar trees given different treatments. (Numbers in parentheses are ranks.) Slide 83

Example: Effects of Treatments on Poplar Tree Weights Use the data in Table 13-6 with the Kruskal-Wallis test to test the claim that the four samples come from populations with equal medians. Are requirements met? There are three or more independent and random samples. Each sample size is 5. (Requirement is at least 5.) H 0 : The populations of poplar tree weights from the four treatments have equal medians. H 1 : The four population medians are not all equal. Slide 84

Example: Effects of Treatments on Poplar Tree Weights Use the data in Table 13-6 with the Kruskal-Wallis test to test the claim that the four samples come from populations with equal medians. The following statistics come from Table 13-6: n 1 = 5, n 2 = 5, n 3 = 5, n 4 = 5 N = 20 R 1 = 45, R 2 = 37.5, R 3 = 42.5, R 4 = 85 Slide 85

Example: Effects of Treatments on Poplar Tree Weights Use the data in Table 13-6 with the Kruskal-Wallis test to test the claim that the four samples come from populations with equal medians. Evaluate the test statistic.. H 12 R R R = + +... + 3( N + 1) N( N 1) n n n 2 2 2 1 2 k + 1 2 k 2 2 2 2 12 45 37.5 42.5 85 = + + + 3(20 + 1) 20(20 + 1) 5 5 5 5 = 8.214 Slide 86

Example: Effects of Treatments on Poplar Tree Weights Use the data in Table 13-6 with the Kruskal-Wallis test to test the claim that the four samples come from populations with equal medians. Find the critical value.. Because each sample has at least five observations, the distribution of H is approximately a chi-square distribution. df = k 1 = 4 1 = 3 α = 0.05 From Table A-4 the critical value = 7.815. Slide 87

Example: Effects of Treatments on Poplar Tree Weights Use the data in Table 13-6 with the Kruskal-Wallis test to test the claim that the four samples come from populations with equal medians. The test statistic 8.214 is in the critical region, so we reject the null hypothesis of equal medians. At least one of the medians appears to be different from the others. Slide 88

Recap In this section we have discussed: The Kruskal-Wallis Test is the nonparametric equivalent of ANOVA. It tests the hypothesis that three or more populations have equal means. The populations do not have to be normally distributed. Slide 89

Section 13-6 Rank Correlation Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN Slide 90

Key Concept This section describes the nonparametric method of rank correlation, which uses paired data to test for an association between two variables. In Chapter 10 we used paired sample data to compute values for the linear correlation coefficient r, but in this section we use ranks as a the basis for computing the rank correlation coefficient r s. Slide 91

Rank Correlation Definition The rank correlation test (or Spearman s rank correlation test) is a non-parametric test that uses ranks of sample data consisting of matched pairs. It is used to test for an association between two variables, so the null and alternative hypotheses are as follows (where ρ s denotes the rank correlation coefficient for the entire population): H o : ρ s = 0 (There is no correlation between the two variables.) H 1 : ρ s 0 (There is a correlation between the two variables.) Slide 92

Advantages Rank correlation has these advantages over the parametric methods discussed in Chapter 10: 1. The nonparametric method of rank correlation can be used in a wider variety of circumstances than the parametric method of linear correlation. With rank correlation, we can analyze paired data that are ranks or can be converted to ranks. 2. Rank correlation can be used to detect some (not all) relationships that are not linear. Slide 93

Disadvantages A disadvantage of rank correlation is its efficiency rating of 0.91, as described in Section 13-1. This efficiency rating shows that with all other circumstances being equal, the nonparametric approach of rank correlation requires 100 pairs of sample data to achieve the same results as only 91 pairs of sample observations analyzed through parametric methods, assuming that the stricter requirements of the parametric approach are met. Slide 94

Figure 13-4 Rank Correlation for Testing H 0 : ρ s = 0 Slide 95

Figure 13-4 Rank Correlation for Testing H 0 : ρ s = 0 Slide 96

Requirements 1. The sample paired data have been randomly selected. 2. Unlike the parametric methods of Section 10-2, there is no requirement that the sample pairs of data have a bivariate normal distribution. There is no requirement of a normal distribution for any population. Slide 97

Notation r s = rank correlation coefficient for sample paired data (r s is a sample statistic) ρ s = rank correlation coefficient for all the population data (ρ s is a population parameter) n = number of pairs of data d = difference between ranks for the two values within a pair Slide 98

Rank Correlation Test Statistic No ties: After converting the data in each sample to ranks, if there are no ties among ranks for either variable, the exact value of the test statistic can be calculated using this formula: r s = 1 6Σd n n Ties: After converting the data in each sample to ranks, if either variable has ties among its ranks, the exact value of the test statistic rs can be found by using Formula 10-1 with the ranks: r s = 2 2 ( 1) nσxy ( Σx)( Σy) n( Σx ) ( Σx) n( Σy ) ( Σy) 2 2 2 2 Slide 99

Rank Correlation Critical values: If n 30, critical values are found in Table A-9. If n > 30, use Formula 13-1. Formula 13-1 r s = ± z n 11 where the value of z corresponds to the significance level. (For example, if α = 0.05, z 1.96.) Slide 100

Example: Rankings of Colleges Use the data in Table 13-7 to determine if there is a correlation between the student rankings and the rankings of the magazine. Slide 101

Example: Rankings of Colleges Use the data in Table 13-7 to determine if there is a correlation between the student rankings and the rankings of the magazine. H 0 : ρ s = 0 H 1 : ρ s 0 Since neither variable has ties in the ranks: r s 2 6Σd 6(24) = 1 = 1 n( n 2 1) 8(8 2 1) 144 = 1 = 0.714 504 Slide 102

Example: Rankings of Colleges Use the data in Table 13-7 to determine if there is a correlation between the student rankings and the rankings of the magazine. H 0 : ρ s = 0 H 1 : ρ s 0 From Table A-9 the critical values are ±0.738. Because the test statistic of r s = 0.714 does not exceed the critical value, we fail to reject the null hypothesis. There is not sufficient evidence to support a claim of a correlation between the rankings of the students and the magazine. Slide 103

Example: Rankings of Colleges Large Sample Case Assume that the preceding example is expanded by including a total of 40 colleges and that the test statistic r s is found to be 0.300. If the significance level of α = 0.05, what do you conclude about the correlation? Since n = 40 exceeds 30, we find the critical value from Formula 13-1 r s ± z ± 1.96 = = = ± n 1 40 1 0.314 Slide 104

Example: Rankings of Colleges Large Sample Case Assume that the preceding example is expanded by including a total of 40 colleges and that the test statistic r s is found to be 0.300. If the significance level of α = 0.05, what do you conclude about the correlation? The test statistic of r s = 0.300 does not exceed the critical value of 0.314, so we fail to reject the null hypothesis. There is not sufficient evidence to support the claim of a correlation between students and the magazine. Slide 105

Example: Detecting a Nonlinear Pattern The data in Table 13-8 are the numbers of games played and the last scores (in millions) of a Raiders of the Lost Ark pinball game. We expect that there should be an association between the number of games played and the pinball score. H 0 : ρ s = 0 H 1 : ρ s 0 Slide 106

Example: Detecting a Nonlinear Pattern There are no ties among ranks of either list. r s 2 6Σd 6(6) = 1 = 1 n( n 2 1) 9(9 2 1) 36 = 1 = 0.950 720 Slide 107

Example: Detecting a Nonlinear Pattern Since n = 9 is less than 30, use Table A-9 Critical values are ± 0.700 The sample statistic 0.950 exceeds 0.700, so we conclude that there is significant evidence to reject the null hypothesis of no correlation. There appears to be correlation between the number of games played and the score. Slide 108

Example: Detecting a Nonlinear Pattern If the preceding example is done using the methods of Chapter 9, the linear correlation coefficient is r = 0.586. This leads to the conclusion that there is not enough evidence to support the claim of a significant linear correlation, whereas the nonlinear test found that there was enough evidence. The Excel scatter diagram shows that there is a non-linear relationship that the parametric method would not have detected. Excel Slide 109

Recap In this section we have discussed: Rank correlation which is the non-parametric equivalent of testing for correlation described in Chapter 10. It uses ranks of matched pairs to test for association. Sometimes rank correlation can detect nonlinear correlation that the parametric test will not recognize. Slide 110

Section 13-7 Runs Test for Randomness Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN Slide 111

Key Concept This section introduces the runs test for randomness, which can be used to determine whether the sample data in a sequence are in a random order. This test is based on sample data that have two characteristics, and it analyzes runs of those characteristics to determine whether the runs appear to result from some random process, or whether the runs suggest that the order of the data is not random. Slide 112

Runs Test for Randomness Definitions A run is a sequence of data having the same characteristic; the sequence is preceded and followed by data with a different characteristic or by no data at all. The runs test uses the number of runs in a sequence of sample data to test for randomness in the order of the data. Slide 113

Fundamental Principles of the Run Test Reject randomness if the number of runs is very low or very high. Example: The sequence of genders FFFFFMMMMM is not random because it has only 2 runs, so the number of runs is very low. Example: The sequence of genders FMFMFMFMFM is not random because there are 10 runs, which is very high. It is important to note that the runs test for randomness is based on the order in which the data occur; it is not based on the frequency of the data. Slide 114

Figure 13-5 Procedure for Runs Test for Randomness Slide 115

Figure 13-5 Procedure for Runs Test for Randomness Slide 116

Requirements 1. The sample data are arranged according to some ordering scheme, such as the order in which the sample values were obtained. 2. Each data value can be categorized into one of two separate categories (such as male/female). Slide 117

Notation n 1 = number of elements in the sequence that have one particular characteristic (The characteristic chosen for n 1 is arbitrary.) n 2 = number of elements in the sequence that have the other characteristic G = number of runs Slide 118

Runs Test for Randomness For Small Samples (n 1 20 and n 2 20) and α = 0.05: Test Statistic Test statistic is the number of runs G Critical Values Critical values are found in Table A-10. Slide 119

Runs Test for Randomness For Small Samples (n 1 20 and n 2 20) and α = 0.05: Decision criteria Reject randomness if the number of runs G is: less than or equal to the smaller critical value found in Table A-10. TW16 or greater than or equal to the larger critical value found in Table A-10. Slide 120

Slide 120 TW16 put periods at end of sentences Tom Wegleitner; 24/5/2006

Runs Test for Randomness For Small Samples (n 1 20 and n 2 20) and α = 0.05: Test Statistic µ where z = G σ G 2n n G µ = 1 2 G n + 1 + n2 1 and σ G = (2 n n )(2 n n n n ) 1 2 1 2 1 2 2 1 + 2 1 + 2 ( n n ) ( n n 1) Slide 121

Runs Test for Randomness For Large Samples (n 1 > 20 or n 2 > 20) or α 0.05: Critical Values Critical values of z: Use Table A-2. Slide 122

Example: Small Sample Genders of Bears Listed below are the genders of the first 10 bears from Data Set 6 in Appendix B. Use a 0.05 significance level to test for randomness in the sequence of genders. M M M M F F M M F F Separate the runs as shown below. M M M M F F M M F F 1st run 2nd run 3rd run 4th run Slide 123

Example: Small Sample Genders of Bears M M M M F F M M F F 1st run 2nd run 3rd run 4th run n 1 = total number of males = 6 n 2 = total number of females = 4 G = number of runs = 4 Because n 1 20 and n 2 20 and α = 0.05, the test statistic is G = 4 Slide 124

Example: Small Sample Genders of Bears M M M M F F M M F F 1st run 2nd run 3rd run 4th run From Table A-10, the critical values are 2 and 9. Because G = 4 is not less than or equal to 2, nor is it greater than or equal to 9, we do not reject randomness. It appears the sequence of genders is random. Slide 125

Example: Large Sample Boston Rainfall on Mondays Refer to the rainfall amounts for Boston as listed in Data Set 10 in Appendix B. Is there sufficient evidence to support the claim that rain on Mondays is not random? D D D D R D R D D R D D R D D D R D D R R R D D D D R D R D R R R D R D D D R D D D R D R D D R D D D R H 0 : The sequence is random. H 1 : The sequence is not random. n 1 = number of Ds = 33 n 2 = number or Rs = 19 G = number of runs = 30 Slide 126

Example: Large Sample Boston Rainfall on Mondays Since n 1 > 20, we must calculate z using the formulas: µ = 2 n 1 n 2 2(33)(19) G 1 1 25.115 n + n + = 33 + 19 + = 1 2 σ G = (2 n n )(2 n n n n ) ( n n ) ( n n 1) 1 2 1 2 1 2 2 1 + 2 1 + 2 (2)(33)(19)[(2(33)(19) 33 19] = = 3.306 2 (33 + 19) (33 + 19 1) G µ G 30 25.115 z = = = 1.48 σ 3.306 G Slide 127

Example: Large Sample Boston Rainfall on Mondays The critical values are z = -1.96 and 1.96. The test statistic of z = 1.48 does not fall within the critical region, so we fail to reject the null hypothesis of randomness. The given sequence does appear to be random. Slide 128

Recap In this section we have discussed: The runs test for randomness which can be used to determine whether the sample data in a sequence are in a random order. We reject randomness if the number of runs is very low or very high. Slide 129