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Slide 1 Shakeel Nouman M.Phil Statistics Nonparametric Methods and Chi-Square Tests (1) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman.

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Presentation on theme: "Slide 1 Shakeel Nouman M.Phil Statistics Nonparametric Methods and Chi-Square Tests (1) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman."— Presentation transcript:

1 Slide 1 Shakeel Nouman M.Phil Statistics Nonparametric Methods and Chi-Square Tests (1) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

2 Slide 2 Using Statistics The Sign Test The Runs Test - A Test for Randomness The Mann-Whitney U Test The Wilcoxon Signed-Rank Test The Kruskal-Wallis Test - A Nonparametric Alternative to One-Way ANOVA Nonparametric Methods and Chi-Square Tests (1)  Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

3 Slide 3 The Friedman Test for a Randomized Block Design The Spearman Rank Correlation Coefficient A Chi-Square Test for Goodness of Fit Contingency Table Analysis - A Chi- Square Test for Independence A Chi-Square Test for Equality of Proportions Summary and Review of Terms Nonparametric Methods and Chi-Square Tests (2)  Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

4 Slide 4 Parametric Methods Inferences based on assumptions about the nature of the population distribution » Usually: population is normal Types of tests » z-test or t-test » Comparing two population means or proportions » Testing value of population mean or proportion » ANOVA » Testing equality of several population means Parametric Methods Inferences based on assumptions about the nature of the population distribution » Usually: population is normal Types of tests » z-test or t-test » Comparing two population means or proportions » Testing value of population mean or proportion » ANOVA » Testing equality of several population means 14-1 Using Statistics (Parametric Tests) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

5 Slide 5 Nonparametric Tests Distribution-free methods making no assumptions about the population distribution Types of tests » Sign tests » Sign Test: Comparing paired observations » McNemar Test: Comparing qualitative variables » Cox and Stuart Test: Detecting trend » Runs tests » Runs Test: Detecting randomness » Wald-Wolfowitz Test: Comparing two distributions Nonparametric Tests Distribution-free methods making no assumptions about the population distribution Types of tests » Sign tests » Sign Test: Comparing paired observations » McNemar Test: Comparing qualitative variables » Cox and Stuart Test: Detecting trend » Runs tests » Runs Test: Detecting randomness » Wald-Wolfowitz Test: Comparing two distributions Nonparametric Tests Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

6 Slide 6 Nonparametric Tests Ranks tests Mann-Whitney U Test: Comparing two populations Wilcoxon Signed-Rank Test: Paired comparisons Comparing several populations: ANOVA with ranks – Kruskal-Wallis Test – Friedman Test: Repeated measures Spearman Rank Correlation Coefficient Chi-Square Tests Goodness of Fit Testing for independence: Contingency Table Analysis Equality of Proportions Nonparametric Tests Ranks tests Mann-Whitney U Test: Comparing two populations Wilcoxon Signed-Rank Test: Paired comparisons Comparing several populations: ANOVA with ranks – Kruskal-Wallis Test – Friedman Test: Repeated measures Spearman Rank Correlation Coefficient Chi-Square Tests Goodness of Fit Testing for independence: Contingency Table Analysis Equality of Proportions Nonparametric Tests (Continued) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

7 Slide 7 Deal with enumerative (frequency counts) data. Do not deal with specific population parameters, such as the mean or standard deviation. Do not require assumptions about specific population distributions (in particular, the normality assumption). Deal with enumerative (frequency counts) data. Do not deal with specific population parameters, such as the mean or standard deviation. Do not require assumptions about specific population distributions (in particular, the normality assumption). Nonparametric Tests (Continued) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

8 Slide 8 Comparing paired observations Paired observations: X and Y p = P(X>Y) » Two-tailed test H 0 : p = 0.50 H 1 : p  0.50 » Right-tailed testH 0 : p  0.50 H 1 : p  0.50 » Left-tailed testH 0 : p  0.50 H 1 : p  0.50 » Test statistic: T = Number of + signs Comparing paired observations Paired observations: X and Y p = P(X>Y) » Two-tailed test H 0 : p = 0.50 H 1 : p  0.50 » Right-tailed testH 0 : p  0.50 H 1 : p  0.50 » Left-tailed testH 0 : p  0.50 H 1 : p  0.50 » Test statistic: T = Number of + signs 14-2 Sign Test Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

9 Slide 9 Small Sample: Binomial Test For a two-tailed test, find a critical point corresponding as closely as possible to  /2 (C 1 ) and define C 2 as n-C 1. Reject null hypothesis if T  C 1 or T  C 2. For a right-tailed test, reject H 0 if T  C, where C is the value of the binomial distribution with parameters n and p = 0.50 such that the sum of the probabilities of all values less than or equal to C is as close as possible to the chosen level of significance, . For a left-tailed test, reject H 0 if T  C, where C is defined as above. Small Sample: Binomial Test For a two-tailed test, find a critical point corresponding as closely as possible to  /2 (C 1 ) and define C 2 as n-C 1. Reject null hypothesis if T  C 1 or T  C 2. For a right-tailed test, reject H 0 if T  C, where C is the value of the binomial distribution with parameters n and p = 0.50 such that the sum of the probabilities of all values less than or equal to C is as close as possible to the chosen level of significance, . For a left-tailed test, reject H 0 if T  C, where C is defined as above. Sign Test Decision Rule Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

10 Slide 10 Cumulative Binomial Probabilities (n=15, p=0.5) x F(x) 00.00003 10.00049 20.00369 30.01758 40.05923 50.15088 60.30362 70.50000 80.69638 90.84912 100.94077 110.98242 120.99631 130.99951 140.99997 151.00000 CEO Before After Sign 1 34 1+ 2 55 0 3 2 3 1+ 4 24 1+ 5 44 0 6 23 1+ 7 12 1+ 8 54 -1- 9 45 1+ 10 54 -1- 11 34 1+ 12 25 1+ 13 25 1+ 14 23 1+ 15 1 2 1+ 16 32 -1- 17 45 1+ CEO Before After Sign 1 34 1+ 2 55 0 3 2 3 1+ 4 24 1+ 5 44 0 6 23 1+ 7 12 1+ 8 54 -1- 9 45 1+ 10 54 -1- 11 34 1+ 12 25 1+ 13 25 1+ 14 23 1+ 15 1 2 1+ 16 32 -1- 17 45 1+ n = 15 T = 12  0.025 C1=3 C2 = 15-3 = 12 H 0 rejected, since T  C2 C1 Example 14-1 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

11 Slide 11 Example 14-1- Using the Template H 0 : p = 0.5 H 1 : p  0.5 Test Statistic: T = 12 p-value = 0.0352. For a = 0.05, the null hypothesis is rejected since 0.0352 < 0.05. Thus one can conclude that there is a change in attitude toward a CEO following the award of an MBA degree. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

12 Slide 12 A run is a sequence of like elements that are preceded and followed by different elements or no element at all. Case 1: S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E : R = 20 Apparently nonrandom Case 2: SSSSSSSSSS|EEEEEEEEEE : R = 2 Apparently nonrandom Case 3: S|EE|SS|EEE|S|E|SS|E|S|EE|SSS|E : R = 12 Perhaps random Case 1: S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E : R = 20 Apparently nonrandom Case 2: SSSSSSSSSS|EEEEEEEEEE : R = 2 Apparently nonrandom Case 3: S|EE|SS|EEE|S|E|SS|E|S|EE|SSS|E : R = 12 Perhaps random A two-tailed hypothesis test for randomness: H 0 : Observations are generated randomly H 1 : Observations are not generated randomly Test Statistic: R=Number of Runs Reject H 0 at level  if R  C1 or R  C2, as given in Table 8, with total tail probability P(R  C 1 ) + P(R  C 2 ) =   A two-tailed hypothesis test for randomness: H 0 : Observations are generated randomly H 1 : Observations are not generated randomly Test Statistic: R=Number of Runs Reject H 0 at level  if R  C1 or R  C2, as given in Table 8, with total tail probability P(R  C 1 ) + P(R  C 2 ) =   14-3 The Runs Test - A Test for Randomness Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

13 Slide 13 Table 8: Number of Runs (r) (n 1,n 2 )11121314151617181920. (10,10)0.5860.7580.8720.9490.9810.9960.9991.0001.0001.000 Table 8: Number of Runs (r) (n 1,n 2 )11121314151617181920. (10,10)0.5860.7580.8720.9490.9810.9960.9991.0001.0001.000 Case 1: n 1 = 10 n 2 = 10 R= 20 p-value  0 Case 2: n 1 = 10 n 2 = 10 R = 2 p-value  0 Case 3: n 1 = 10 n 2 = 10 R= 12 p-value    PR          F(11)] = (2)(1-0.586) = (2)(0.414) = 0.828 H 0 not rejected Case 1: n 1 = 10 n 2 = 10 R= 20 p-value  0 Case 2: n 1 = 10 n 2 = 10 R = 2 p-value  0 Case 3: n 1 = 10 n 2 = 10 R= 12 p-value    PR          F(11)] = (2)(1-0.586) = (2)(0.414) = 0.828 H 0 not rejected Runs Test: Examples Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

14 Slide 14 Large-Sample Runs Test: Using the Normal Approximation Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

15 Slide 15 Example 14-2: n 1 = 27 n 2 = 26 R = 15 H 0 should be rejected at any common level of significance. Large-Sample Runs Test: Example 14-2 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

16 Slide 16 Large-Sample Runs Test: Example 14-2 – Using the Template Note:The computed p-value using the template is 0.0005 as compared to the manually computed value of 0.0006. The value of 0.0005 is more accurate. Note: The computed p-value using the template is 0.0005 as compared to the manually computed value of 0.0006. The value of 0.0005 is more accurate. Reject the null hypothesis that the residuals are random. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

17 Slide 17 The null and alternative hypotheses for the Wald-Wolfowitz test: H 0 : The two populations have the same distribution H 1 : The two populations have different distributions The test statistic: R = Number of Runs in the sequence of samples, when the data from both samples have been sorted The null and alternative hypotheses for the Wald-Wolfowitz test: H 0 : The two populations have the same distribution H 1 : The two populations have different distributions The test statistic: R = Number of Runs in the sequence of samples, when the data from both samples have been sorted Salesperson A:35443950 48296075 4966 Salesperson B:17231324 33211816 32 Salesperson A:35443950 48296075 4966 Salesperson B:17231324 33211816 32 Using the Runs Test to Compare Two Population Distributions (Means): the Wald-Wolfowitz Test Example 14-3: Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

18 Slide 18 Table Number of Runs (r) (n 1,n 2 )2345. (9,10) 0.0000.0000.0020.004... Table Number of Runs (r) (n 1,n 2 )2345. (9,10) 0.0000.0000.0020.004... Sales SalesSalesPerson SalesPerson(Sorted)(Sorted)Runs 35A13 B 44A16 B 39A17 B 48A21 B 60A24 B 1 75A29 A 2 49A32 B 66A33 B 3 17B35 A 23B39 A 13B44 A 24B48 A 33B49 A 21B50 A 18B60 A 16B66 A 32B75 A 4 Sales SalesSalesPerson SalesPerson(Sorted)(Sorted)Runs 35A13 B 44A16 B 39A17 B 48A21 B 60A24 B 1 75A29 A 2 49A32 B 66A33 B 3 17B35 A 23B39 A 13B44 A 24B48 A 33B49 A 21B50 A 18B60 A 16B66 A 32B75 A 4 n 1 = 10 n 2 = 9 R= 4 p-value  P   H 0 may be rejected n 1 = 10 n 2 = 9 R= 4 p-value  P   H 0 may be rejected The Wald-Wolfowitz Test: Example 14-3 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

19 Slide 19 Ranks tests Mann-Whitney U Test: Comparing two populations Wilcoxon Signed-Rank Test: Paired comparisons Comparing several populations: ANOVA with ranks Kruskal-Wallis Test Friedman Test: Repeated measures Ranks tests Mann-Whitney U Test: Comparing two populations Wilcoxon Signed-Rank Test: Paired comparisons Comparing several populations: ANOVA with ranks Kruskal-Wallis Test Friedman Test: Repeated measures Ranks Tests Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

20 Slide 20 The null and alternative hypotheses: H 0 : The distributions of two populations are identical H 1 : The two population distributions are not identical The Mann-Whitney U statistic: where n 1 is the sample size from population 1 and n 2 is the sample size from population 2. The null and alternative hypotheses: H 0 : The distributions of two populations are identical H 1 : The two population distributions are not identical The Mann-Whitney U statistic: where n 1 is the sample size from population 1 and n 2 is the sample size from population 2. 14-4 The Mann-Whitney U Test (Comparing Two Populations) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

21 Slide 21 Cumulative Distribution Function of the Mann-Whitney U Statistic n 2 =6 n 1 =6 u. 40.0130 50.0206 60.0325. Rank ModelTimeRankSum A35 5 A38 8 A4010 A4212 A4111 A36 652 B29 2 B27 1 B30 3 B33 4 B39 9 B37 726 Rank ModelTimeRankSum A35 5 A38 8 A4010 A4212 A4111 A36 652 B29 2 B27 1 B30 3 B33 4 B39 9 B37 726 P( u  5) The Mann-Whitney U Test: Example 14-4 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

22 Slide 22 Example 14-5: Large-Sample Mann-Whitney U Test Score Rank Score Program Rank Sum 85120.020.0 87121.041.0 92127.068.0 98130.098.0 90126.0124.0 88123.0147.0 75117.0164.0 72113.5177.5 6016.5184.0 93128.0212.0 88123.0235.0 89125.0260.0 96129.0289.0 73115.0304.0 6218.5312.5 Score Rank Score Program Rank Sum 85120.020.0 87121.041.0 92127.068.0 98130.098.0 90126.0124.0 88123.0147.0 75117.0164.0 72113.5177.5 6016.5184.0 93128.0212.0 88123.0235.0 89125.0260.0 96129.0289.0 73115.0304.0 6218.5312.5 Score Rank Score Program Rank Sum 65210.0 10.0 5724.0 14.0 74216.0 30.0 4322.0 32.0 3921.0 33.0 88223.0 56.0 6228.5 64.5 69211.0 75.5 70212.0 87.5 72213.5101.0 5925.0106.0 6026.5112.5 80218.0130.5 83219.0149.5 5023.0152.5 Score Rank Score Program Rank Sum 65210.0 10.0 5724.0 14.0 74216.0 30.0 4322.0 32.0 3921.0 33.0 88223.0 56.0 6228.5 64.5 69211.0 75.5 70212.0 87.5 72213.5101.0 5925.0106.0 6026.5112.5 80218.0130.5 83219.0149.5 5023.0152.5 Since the test statistic is z = -3.32, the p-value  0.0005, and H 0 is rejected. Since the test statistic is z = -3.32, the p-value  0.0005, and H 0 is rejected. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

23 Slide 23 Example 14-5: Large-Sample Mann-Whitney U Test – Using the Template Since the test statistic is z = - 3.32, the p-value  0.0005, and H 0 is rejected. That is, the LC (Learning Curve) program is more effective. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

24 Slide 24 The null and alternative hypotheses: H 0 : The median difference between populations are 1 and 2 is zero H 1 : The median difference between populations are 1 and 2 is not zero Find the difference between the ranks for each pair, D = x 1 -x 2, and then rank the absolute values of the differences. The Wilcoxon T statistic is the smaller of the sums of the positive ranks and the sum of the negative ranks: For small samples, a left-tailed test is used, using the values in Appendix C, Table 10. The large-sample test statistic: The null and alternative hypotheses: H 0 : The median difference between populations are 1 and 2 is zero H 1 : The median difference between populations are 1 and 2 is not zero Find the difference between the ranks for each pair, D = x 1 -x 2, and then rank the absolute values of the differences. The Wilcoxon T statistic is the smaller of the sums of the positive ranks and the sum of the negative ranks: For small samples, a left-tailed test is used, using the values in Appendix C, Table 10. The large-sample test statistic: 14-5 The Wilcoxon Signed- Ranks Test (Paired Ranks) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

25 Slide 25 Sold Sold Rank Rank Rank (1) (2) D=x 1 -x 2 ABS(D) ABS(D)(D>0) (D<0) 564016169.09.00 4870-222212.00.012 10060404015.015.00 857015158.08.00 22814147.07.00 4440442.02.00 3545-10106.00.06 287212111.011.00 5260-885.00.05 7770773.53.50 8990-111.00.01 10100**** 6585-202010.00.010 9061292913.013.00 7040303014.014.00 3326773.53.50 Sum:8634 Sold Sold Rank Rank Rank (1) (2) D=x 1 -x 2 ABS(D) ABS(D)(D>0) (D<0) 564016169.09.00 4870-222212.00.012 10060404015.015.00 857015158.08.00 22814147.07.00 4440442.02.00 3545-10106.00.06 287212111.011.00 5260-885.00.05 7770773.53.50 8990-111.00.01 10100**** 6585-202010.00.010 9061292913.013.00 7040303014.014.00 3326773.53.50 Sum:8634 T=34 n=15 P=0.0530 P=0.02525 P=0.0120 P=0.00516 H 0 is not rejected (Note the arithmetic error in the text for store 13) T=34 n=15 P=0.0530 P=0.02525 P=0.0120 P=0.00516 H 0 is not rejected (Note the arithmetic error in the text for store 13) Example 14-6 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

26 Slide 26 Hourly Rank Rank Rank Messages Md 0 D=x 1 -x 2 ABS(D) ABS(D) (D>0) (D<0) 151149221.01.00.0 144149-552.00.02.0 123149-262613.00.013.0 178149292915.015.00.0 105149-444423.00.023.0 112149-373720.00.020.0 140149-994.00.04.0 167149181810.010.00.0 177149282814.014.00.0 185149363619.019.00.0 129149-202011.00.011.0 16014911116.06.00.0 110149-393921.00.021.0 170149212112.012.00.0 198149494925.025.00.0 16514916168.08.00.0 109149-404022.00.022.0 118149-313116.50.016.5 155149663.03.00.0 102149-474724.00.024.0 16414915157.07.00.0 180149313116.516.50.0 139149-10105.00.05.0 16614917179.09.00.0 82149333318.018.00.0 Sum:163.5161.5 Hourly Rank Rank Rank Messages Md 0 D=x 1 -x 2 ABS(D) ABS(D) (D>0) (D<0) 151149221.01.00.0 144149-552.00.02.0 123149-262613.00.013.0 178149292915.015.00.0 105149-444423.00.023.0 112149-373720.00.020.0 140149-994.00.04.0 167149181810.010.00.0 177149282814.014.00.0 185149363619.019.00.0 129149-202011.00.011.0 16014911116.06.00.0 110149-393921.00.021.0 170149212112.012.00.0 198149494925.025.00.0 16514916168.08.00.0 109149-404022.00.022.0 118149-313116.50.016.5 155149663.03.00.0 102149-474724.00.024.0 16414915157.07.00.0 180149313116.516.50.0 139149-10105.00.05.0 16614917179.09.00.0 82149333318.018.00.0 Sum:163.5161.5 Example 14-7 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

27 Slide 27 Example 14-7 using the Template Note 1: Note 1: You should enter the claimed value of the mean (median) in every used row of the second column of data. In this case it is 149. Note 2: Note 2: In order for the large sample approximations to be computed you will need to change n > 25 to n >= 25 in cells M13 and M14. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

28 Slide 28 The Kruskal-Wallis hypothesis test: H 0 : All k populations have the same distribution H 1 : Not all k populations have the same distribution The Kruskal-Wallis test statistic: If each n j > 5, then H is approximately distributed as a  2. The Kruskal-Wallis hypothesis test: H 0 : All k populations have the same distribution H 1 : Not all k populations have the same distribution The Kruskal-Wallis test statistic: If each n j > 5, then H is approximately distributed as a  2. 14-6 The Kruskal-Wallis Test - A Nonparametric Alternative to One-Way ANOVA Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

29 Slide 29 SoftwareTimeRank Group RankSum 14514190 13810256 15616325 16017 14715 16518 2308 24011 2287 24413 2255 24212 3224 3193 3151 3319 3276 3172 SoftwareTimeRank Group RankSum 14514190 13810256 15616325 16017 14715 16518 2308 24011 2287 24413 2255 24212 3224 3193 3151 3319 3276 3172  2 (2,0.005) =10.5966, so H 0 is rejected. Example 14-8: The Kruskal- Wallis Test Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

30 Slide 30 Example 14-8: The Kruskal-Wallis Test – Using the Template Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

31 Slide 31 If the null hypothesis in the Kruskal-Wallis test is rejected, then we may wish, in addition, compare each pair of populations to determine which are different and which are the same. Further Analysis (Pairwise Comparisons of Average Ranks) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

32 Slide 32 Pairwise Comparisons: Example 14-8 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

33 Slide 33 The Friedman test is a nonparametric version of the randomized block design ANOVA. Sometimes this design is referred to as a two-way ANOVA with one item per cell because it is possible to view the blocks as one factor and the treatment levels as the other factor. The test is based on ranks. 14-7 The Friedman Test for a Randomized Block Design The Friedman hypothesis test: H 0 : The distributions of the k treatment populations are identical H 1 : Not all k distribution are identical The Friedman test statistic: The degrees of freedom for the chi-square distribution is (k – 1). The Friedman hypothesis test: H 0 : The distributions of the k treatment populations are identical H 1 : Not all k distribution are identical The Friedman test statistic: The degrees of freedom for the chi-square distribution is (k – 1). Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

34 Slide 34 Example 14-10 – using the Template Note: Note: The p-value is small relative to a significance level of a = 0.05, so one should conclude that there is evidence that not all three low- budget cruise lines are equally preferred by the frequent cruiser population Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

35 Slide 35 The Spearman Rank Correlation Coefficient is the simple correlation coefficient calculated from variables converted to ranks from their original values. 14-8 The Spearman Rank Correlation Coefficient Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

36 Slide 36 Using Statistics The Sign Test The Runs Test - A Test for Randomness The Mann-Whitney U Test The Wilcoxon Signed-Rank Test The Kruskal-Wallis Test - A Nonparametric Alternative to One-Way ANOVA Nonparametric Methods and Chi-Square Tests (1)  Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

37 Slide 37 The Friedman Test for a Randomized Block Design The Spearman Rank Correlation Coefficient A Chi-Square Test for Goodness of Fit Contingency Table Analysis - A Chi- Square Test for Independence A Chi-Square Test for Equality of Proportions Summary and Review of Terms Nonparametric Methods and Chi-Square Tests (2)  Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

38 Slide 38 Parametric Methods Inferences based on assumptions about the nature of the population distribution » Usually: population is normal Types of tests » z-test or t-test » Comparing two population means or proportions » Testing value of population mean or proportion » ANOVA » Testing equality of several population means Parametric Methods Inferences based on assumptions about the nature of the population distribution » Usually: population is normal Types of tests » z-test or t-test » Comparing two population means or proportions » Testing value of population mean or proportion » ANOVA » Testing equality of several population means 14-1 Using Statistics (Parametric Tests) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

39 Slide 39 Nonparametric Tests Distribution-free methods making no assumptions about the population distribution Types of tests » Sign tests » Sign Test: Comparing paired observations » McNemar Test: Comparing qualitative variables » Cox and Stuart Test: Detecting trend » Runs tests » Runs Test: Detecting randomness » Wald-Wolfowitz Test: Comparing two distributions Nonparametric Tests Distribution-free methods making no assumptions about the population distribution Types of tests » Sign tests » Sign Test: Comparing paired observations » McNemar Test: Comparing qualitative variables » Cox and Stuart Test: Detecting trend » Runs tests » Runs Test: Detecting randomness » Wald-Wolfowitz Test: Comparing two distributions Nonparametric Tests Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

40 Slide 40 Nonparametric Tests Ranks tests Mann-Whitney U Test: Comparing two populations Wilcoxon Signed-Rank Test: Paired comparisons Comparing several populations: ANOVA with ranks – Kruskal-Wallis Test – Friedman Test: Repeated measures Spearman Rank Correlation Coefficient Chi-Square Tests Goodness of Fit Testing for independence: Contingency Table Analysis Equality of Proportions Nonparametric Tests Ranks tests Mann-Whitney U Test: Comparing two populations Wilcoxon Signed-Rank Test: Paired comparisons Comparing several populations: ANOVA with ranks – Kruskal-Wallis Test – Friedman Test: Repeated measures Spearman Rank Correlation Coefficient Chi-Square Tests Goodness of Fit Testing for independence: Contingency Table Analysis Equality of Proportions Nonparametric Tests (Continued) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

41 Slide 41 Deal with enumerative (frequency counts) data. Do not deal with specific population parameters, such as the mean or standard deviation. Do not require assumptions about specific population distributions (in particular, the normality assumption). Deal with enumerative (frequency counts) data. Do not deal with specific population parameters, such as the mean or standard deviation. Do not require assumptions about specific population distributions (in particular, the normality assumption). Nonparametric Tests (Continued) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

42 Slide 42 Comparing paired observations Paired observations: X and Y p = P(X>Y) » Two-tailed test H 0 : p = 0.50 H 1 : p  0.50 » Right-tailed testH 0 : p  0.50 H 1 : p  0.50 » Left-tailed testH 0 : p  0.50 H 1 : p  0.50 » Test statistic: T = Number of + signs Comparing paired observations Paired observations: X and Y p = P(X>Y) » Two-tailed test H 0 : p = 0.50 H 1 : p  0.50 » Right-tailed testH 0 : p  0.50 H 1 : p  0.50 » Left-tailed testH 0 : p  0.50 H 1 : p  0.50 » Test statistic: T = Number of + signs 14-2 Sign Test Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

43 Slide 43 Small Sample: Binomial Test For a two-tailed test, find a critical point corresponding as closely as possible to  /2 (C 1 ) and define C 2 as n-C 1. Reject null hypothesis if T  C 1 or T  C 2. For a right-tailed test, reject H 0 if T  C, where C is the value of the binomial distribution with parameters n and p = 0.50 such that the sum of the probabilities of all values less than or equal to C is as close as possible to the chosen level of significance, . For a left-tailed test, reject H 0 if T  C, where C is defined as above. Small Sample: Binomial Test For a two-tailed test, find a critical point corresponding as closely as possible to  /2 (C 1 ) and define C 2 as n-C 1. Reject null hypothesis if T  C 1 or T  C 2. For a right-tailed test, reject H 0 if T  C, where C is the value of the binomial distribution with parameters n and p = 0.50 such that the sum of the probabilities of all values less than or equal to C is as close as possible to the chosen level of significance, . For a left-tailed test, reject H 0 if T  C, where C is defined as above. Sign Test Decision Rule Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

44 Slide 44 Cumulative Binomial Probabilities (n=15, p=0.5) x F(x) 00.00003 10.00049 20.00369 30.01758 40.05923 50.15088 60.30362 70.50000 80.69638 90.84912 100.94077 110.98242 120.99631 130.99951 140.99997 151.00000 Cumulative Binomial Probabilities (n=15, p=0.5) x F(x) 00.00003 10.00049 20.00369 30.01758 40.05923 50.15088 60.30362 70.50000 80.69638 90.84912 100.94077 110.98242 120.99631 130.99951 140.99997 151.00000 CEO Before After Sign 1 34 1+ 2 55 0 3 2 3 1+ 4 24 1+ 5 44 0 6 23 1+ 7 12 1+ 8 54 -1- 9 45 1+ 10 54 -1- 11 34 1+ 12 25 1+ 13 25 1+ 14 23 1+ 15 1 2 1+ 16 32 -1- 17 45 1+ CEO Before After Sign 1 34 1+ 2 55 0 3 2 3 1+ 4 24 1+ 5 44 0 6 23 1+ 7 12 1+ 8 54 -1- 9 45 1+ 10 54 -1- 11 34 1+ 12 25 1+ 13 25 1+ 14 23 1+ 15 1 2 1+ 16 32 -1- 17 45 1+ n = 15 T = 12  0.025 C1=3 C2 = 15-3 = 12 H 0 rejected, since T  C2 n = 15 T = 12  0.025 C1=3 C2 = 15-3 = 12 H 0 rejected, since T  C2 C1 Example 14-1 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

45 Slide 45 Example 14-1- Using the Template H 0 : p = 0.5 H 1 : p  Test Statistic: T = 12 p-value = 0.0352. For  = 0.05, the null hypothesis is rejected since 0.0352 < 0.05. Thus one can conclude that there is a change in attitude toward a CEO following the award of an MBA degree. H 0 : p = 0.5 H 1 : p  Test Statistic: T = 12 p-value = 0.0352. For  = 0.05, the null hypothesis is rejected since 0.0352 < 0.05. Thus one can conclude that there is a change in attitude toward a CEO following the award of an MBA degree. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

46 Slide 46 A run is a sequence of like elements that are preceded and followed by different elements or no element at all. Case 1: S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E : R = 20 Apparently nonrandom Case 2: SSSSSSSSSS|EEEEEEEEEE : R = 2 Apparently nonrandom Case 3: S|EE|SS|EEE|S|E|SS|E|S|EE|SSS|E : R = 12 Perhaps random Case 1: S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E|S|E : R = 20 Apparently nonrandom Case 2: SSSSSSSSSS|EEEEEEEEEE : R = 2 Apparently nonrandom Case 3: S|EE|SS|EEE|S|E|SS|E|S|EE|SSS|E : R = 12 Perhaps random A two-tailed hypothesis test for randomness: H 0 : Observations are generated randomly H 1 : Observations are not generated randomly Test Statistic: R=Number of Runs Reject H 0 at level  if R  C1 or R  C2, as given in Table 8, with total tail probability P(R  C 1 ) + P(R  C 2 ) =  A two-tailed hypothesis test for randomness: H 0 : Observations are generated randomly H 1 : Observations are not generated randomly Test Statistic: R=Number of Runs Reject H 0 at level  if R  C1 or R  C2, as given in Table 8, with total tail probability P(R  C 1 ) + P(R  C 2 ) =  14-3 The Runs Test - A Test for Randomness Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

47 Slide 47 Table 8: Number of Runs (r) (n 1,n 2 )11121314151617181920. (10,10)0.5860.7580.8720.9490.9810.9960.9991.0001.0001.000 Table 8: Number of Runs (r) (n 1,n 2 )11121314151617181920. (10,10)0.5860.7580.8720.9490.9810.9960.9991.0001.0001.000 Case 1: n 1 = 10 n 2 = 10 R= 20 p-value  0 Case 2: n 1 = 10 n 2 = 10 R = 2 p-value  0 Case 3: n 1 = 10 n 2 = 10 R= 12 p-value  P  R  F(11)] = (2)(1-0.586) = (2)(0.414) = 0.828 H 0 not rejected Case 1: n 1 = 10 n 2 = 10 R= 20 p-value  0 Case 2: n 1 = 10 n 2 = 10 R = 2 p-value  0 Case 3: n 1 = 10 n 2 = 10 R= 12 p-value  P  R  F(11)] = (2)(1-0.586) = (2)(0.414) = 0.828 H 0 not rejected Runs Test: Examples Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

48 Slide 48 Large-Sample Runs Test: Using the Normal Approximation Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

49 Slide 49 Example 14-2: n 1 = 27 n 2 = 26 R = 15 H 0 should be rejected at any common level of significance. Large-Sample Runs Test: Example 14-2 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

50 Slide 50 Large-Sample Runs Test: Example 14-2 – Using the Template Note:The computed p- value using the template is 0.0005 as compared to the manually computed value of 0.0006. The value of 0.0005 is more accurate. Note: The computed p- value using the template is 0.0005 as compared to the manually computed value of 0.0006. The value of 0.0005 is more accurate. Reject the null hypothesis that the residuals are random. Note:The computed p- value using the template is 0.0005 as compared to the manually computed value of 0.0006. The value of 0.0005 is more accurate. Note: The computed p- value using the template is 0.0005 as compared to the manually computed value of 0.0006. The value of 0.0005 is more accurate. Reject the null hypothesis that the residuals are random. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

51 Slide 51 The null and alternative hypotheses for the Wald-Wolfowitz test: H 0 : The two populations have the same distribution H 1 : The two populations have different distributions The test statistic: R = Number of Runs in the sequence of samples, when the data from both samples have been sorted The null and alternative hypotheses for the Wald-Wolfowitz test: H 0 : The two populations have the same distribution H 1 : The two populations have different distributions The test statistic: R = Number of Runs in the sequence of samples, when the data from both samples have been sorted Salesperson A:35443950 48296075 4966 Salesperson B:17231324 33211816 32 Salesperson A:35443950 48296075 4966 Salesperson B:17231324 33211816 32 Using the Runs Test to Compare Two Population Distributions (Means): the Wald-Wolfowitz Test Example 14-3: Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

52 Slide 52 Table Number of Runs (r) (n 1,n 2 )2345. (9,10) 0.0000.0000.0020.004... Table Number of Runs (r) (n 1,n 2 )2345. (9,10) 0.0000.0000.0020.004... Sales SalesSalesPerson SalesPerson(Sorted)(Sorted)Runs 35A13 B 44A16 B 39A17 B 48A21 B 60A24 B 1 75A29 A 2 49A32 B 66A33 B 3 17B35 A 23B39 A 13B44 A 24B48 A 33B49 A 21B50 A 18B60 A 16B66 A 32B75 A 4 Sales SalesSalesPerson SalesPerson(Sorted)(Sorted)Runs 35A13 B 44A16 B 39A17 B 48A21 B 60A24 B 1 75A29 A 2 49A32 B 66A33 B 3 17B35 A 23B39 A 13B44 A 24B48 A 33B49 A 21B50 A 18B60 A 16B66 A 32B75 A 4 n 1 = 10 n 2 = 9 R= 4 p-value  P  R  H 0 may be rejected n 1 = 10 n 2 = 9 R= 4 p-value  P  R  H 0 may be rejected The Wald-Wolfowitz Test: Example 14-3 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

53 Slide 53 Ranks tests Mann-Whitney U Test: Comparing two populations Wilcoxon Signed-Rank Test: Paired comparisons Comparing several populations: ANOVA with ranks Kruskal-Wallis Test Friedman Test: Repeated measures Ranks tests Mann-Whitney U Test: Comparing two populations Wilcoxon Signed-Rank Test: Paired comparisons Comparing several populations: ANOVA with ranks Kruskal-Wallis Test Friedman Test: Repeated measures Ranks Tests Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

54 Slide 54 The null and alternative hypotheses: H 0 : The distributions of two populations are identical H 1 : The two population distributions are not identical The Mann-Whitney U statistic: where n 1 is the sample size from population 1 and n 2 is the sample size from population 2. The null and alternative hypotheses: H 0 : The distributions of two populations are identical H 1 : The two population distributions are not identical The Mann-Whitney U statistic: where n 1 is the sample size from population 1 and n 2 is the sample size from population 2. 14-4 The Mann-Whitney U Test (Comparing Two Populations) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

55 Slide 55 Cumulative Distribution Function of the Mann-Whitney U Statistic n 2 =6 n 1 =6 u. 40.0130 50.0206 60.0325. Rank ModelTimeRankSum A35 5 A38 8 A4010 A4212 A4111 A36 652 B29 2 B27 1 B30 3 B33 4 B39 9 B37 726 Rank ModelTimeRankSum A35 5 A38 8 A4010 A4212 A4111 A36 652 B29 2 B27 1 B30 3 B33 4 B39 9 B37 726  The Mann-Whitney U Test: Example 14-4 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

56 Slide 56 Example 14-5: Large-Sample Mann-Whitney U Test Score Rank Score Program Rank Sum 85120.020.0 87121.041.0 92127.068.0 98130.098.0 90126.0124.0 88123.0147.0 75117.0164.0 72113.5177.5 6016.5184.0 93128.0212.0 88123.0235.0 89125.0260.0 96129.0289.0 73115.0304.0 6218.5312.5 Score Rank Score Program Rank Sum 85120.020.0 87121.041.0 92127.068.0 98130.098.0 90126.0124.0 88123.0147.0 75117.0164.0 72113.5177.5 6016.5184.0 93128.0212.0 88123.0235.0 89125.0260.0 96129.0289.0 73115.0304.0 6218.5312.5 Score Rank Score Program Rank Sum 65210.0 10.0 5724.0 14.0 74216.0 30.0 4322.0 32.0 3921.0 33.0 88223.0 56.0 6228.5 64.5 69211.0 75.5 70212.0 87.5 72213.5101.0 5925.0106.0 6026.5112.5 80218.0130.5 83219.0149.5 5023.0152.5 Score Rank Score Program Rank Sum 65210.0 10.0 5724.0 14.0 74216.0 30.0 4322.0 32.0 3921.0 33.0 88223.0 56.0 6228.5 64.5 69211.0 75.5 70212.0 87.5 72213.5101.0 5925.0106.0 6026.5112.5 80218.0130.5 83219.0149.5 5023.0152.5 Since the test statistic is z = -3.32, the p-value  0.0005, and H 0 is rejected. Since the test statistic is z = -3.32, the p-value  0.0005, and H 0 is rejected. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

57 Slide 57 Example 14-5: Large-Sample Mann-Whitney U Test – Using the Template Since the test statistic is z = - 3.32, the p-value  0.0005, and H 0 is rejected. That is, the LC (Learning Curve) program is more effective. Since the test statistic is z = - 3.32, the p-value  0.0005, and H 0 is rejected. That is, the LC (Learning Curve) program is more effective. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

58 Slide 58 The null and alternative hypotheses: H 0 : The median difference between populations are 1 and 2 is zero H 1 : The median difference between populations are 1 and 2 is not zero Find the difference between the ranks for each pair, D = x 1 -x 2, and then rank the absolute values of the differences. The Wilcoxon T statistic is the smaller of the sums of the positive ranks and the sum of the negative ranks: For small samples, a left-tailed test is used, using the values in Appendix C, Table 10. The large-sample test statistic: The null and alternative hypotheses: H 0 : The median difference between populations are 1 and 2 is zero H 1 : The median difference between populations are 1 and 2 is not zero Find the difference between the ranks for each pair, D = x 1 -x 2, and then rank the absolute values of the differences. The Wilcoxon T statistic is the smaller of the sums of the positive ranks and the sum of the negative ranks: For small samples, a left-tailed test is used, using the values in Appendix C, Table 10. The large-sample test statistic: 14-5 The Wilcoxon Signed- Ranks Test (Paired Ranks) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

59 Slide 59 Sold Sold Rank Rank Rank (1) (2) D=x 1 -x 2 ABS(D) ABS(D)(D>0) (D<0) 564016169.09.00 4870-222212.00.012 10060404015.015.00 857015158.08.00 22814147.07.00 4440442.02.00 3545-10106.00.06 287212111.011.00 5260-885.00.05 7770773.53.50 8990-111.00.01 10100**** 6585-202010.00.010 9061292913.013.00 7040303014.014.00 3326773.53.50 Sum:8634 Sold Sold Rank Rank Rank (1) (2) D=x 1 -x 2 ABS(D) ABS(D)(D>0) (D<0) 564016169.09.00 4870-222212.00.012 10060404015.015.00 857015158.08.00 22814147.07.00 4440442.02.00 3545-10106.00.06 287212111.011.00 5260-885.00.05 7770773.53.50 8990-111.00.01 10100**** 6585-202010.00.010 9061292913.013.00 7040303014.014.00 3326773.53.50 Sum:8634 T=34 n=15 P=0.0530 P=0.02525 P=0.0120 P=0.00516 H 0 is not rejected (Note the arithmetic error in the text for store 13) T=34 n=15 P=0.0530 P=0.02525 P=0.0120 P=0.00516 H 0 is not rejected (Note the arithmetic error in the text for store 13) Example 14-6 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

60 Slide 60 Hourly Rank Rank Rank Messages Md 0 D=x 1 -x 2 ABS(D) ABS(D) (D>0) (D<0) 151149221.01.00.0 144149-552.00.02.0 123149-262613.00.013.0 178149292915.015.00.0 105149-444423.00.023.0 112149-373720.00.020.0 140149-994.00.04.0 167149181810.010.00.0 177149282814.014.00.0 185149363619.019.00.0 129149-202011.00.011.0 16014911116.06.00.0 110149-393921.00.021.0 170149212112.012.00.0 198149494925.025.00.0 16514916168.08.00.0 109149-404022.00.022.0 118149-313116.50.016.5 155149663.03.00.0 102149-474724.00.024.0 16414915157.07.00.0 180149313116.516.50.0 139149-10105.00.05.0 16614917179.09.00.0 82149333318.018.00.0 Sum:163.5161.5 Hourly Rank Rank Rank Messages Md 0 D=x 1 -x 2 ABS(D) ABS(D) (D>0) (D<0) 151149221.01.00.0 144149-552.00.02.0 123149-262613.00.013.0 178149292915.015.00.0 105149-444423.00.023.0 112149-373720.00.020.0 140149-994.00.04.0 167149181810.010.00.0 177149282814.014.00.0 185149363619.019.00.0 129149-202011.00.011.0 16014911116.06.00.0 110149-393921.00.021.0 170149212112.012.00.0 198149494925.025.00.0 16514916168.08.00.0 109149-404022.00.022.0 118149-313116.50.016.5 155149663.03.00.0 102149-474724.00.024.0 16414915157.07.00.0 180149313116.516.50.0 139149-10105.00.05.0 16614917179.09.00.0 82149333318.018.00.0 Sum:163.5161.5 Example 14-7 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

61 Slide 61 Example 14-7 using the Template Note 1: Note 1: You should enter the claimed value of the mean (median) in every used row of the second column of data. In this case it is 149. Note 2: Note 2: In order for the large sample approximations to be computed you will need to change n > 25 to n >= 25 in cells M13 and M14. Note 1: Note 1: You should enter the claimed value of the mean (median) in every used row of the second column of data. In this case it is 149. Note 2: Note 2: In order for the large sample approximations to be computed you will need to change n > 25 to n >= 25 in cells M13 and M14. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

62 Slide 62 The Kruskal-Wallis hypothesis test: H 0 : All k populations have the same distribution H 1 : Not all k populations have the same distribution The Kruskal-Wallis test statistic: If each n j > 5, then H is approximately distributed as a  2. The Kruskal-Wallis hypothesis test: H 0 : All k populations have the same distribution H 1 : Not all k populations have the same distribution The Kruskal-Wallis test statistic: If each n j > 5, then H is approximately distributed as a  2. 14-6 The Kruskal-Wallis Test - A Nonparametric Alternative to One-Way ANOVA Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

63 Slide 63 SoftwareTimeRank Group RankSum 14514190 13810256 15616325 16017 14715 16518 2308 24011 2287 24413 2255 24212 3224 3193 3151 3319 3276 3172 SoftwareTimeRank Group RankSum 14514190 13810256 15616325 16017 14715 16518 2308 24011 2287 24413 2255 24212 3224 3193 3151 3319 3276 3172  2 (2,0.005) =10.5966, so H 0 is rejected. Example 14-8: The Kruskal- Wallis Test Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

64 Slide 64 Example 14-8: The Kruskal-Wallis Test – Using the Template Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

65 Slide 65 If the null hypothesis in the Kruskal-Wallis test is rejected, then we may wish, in addition, compare each pair of populations to determine which are different and which are the same. Further Analysis (Pairwise Comparisons of Average Ranks) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

66 Slide 66 Pairwise Comparisons: Example 14-8 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

67 Slide 67 The Friedman test is a nonparametric version of the randomized block design ANOVA. Sometimes this design is referred to as a two-way ANOVA with one item per cell because it is possible to view the blocks as one factor and the treatment levels as the other factor. The test is based on ranks. 14-7 The Friedman Test for a Randomized Block Design The Friedman hypothesis test: H 0 : The distributions of the k treatment populations are identical H 1 : Not all k distribution are identical The Friedman test statistic: The degrees of freedom for the chi-square distribution is (k – 1). The Friedman hypothesis test: H 0 : The distributions of the k treatment populations are identical H 1 : Not all k distribution are identical The Friedman test statistic: The degrees of freedom for the chi-square distribution is (k – 1). Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

68 Slide 68 Example 14-10 – using the Template Note: Note: The p- value is small relative to a significance level of  = 0.05, so one should conclude that there is evidence that not all three low-budget cruise lines are equally preferred by the frequent cruiser population Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

69 Slide 69 The Spearman Rank Correlation Coefficient is the simple correlation coefficient calculated from variables converted to ranks from their original values. 14-8 The Spearman Rank Correlation Coefficient Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

70 Slide 70 Table 11:  =0.005 n. 7------ 80.881 90.833 100.794 110.818. Table 11:  =0.005 n. 7------ 80.881 90.833 100.794 110.818.                            rejected  MMIS&P100R-MMIR-S&PDiffDiffsq 220151 7 611 218150 5 500 216148 3 300 217149 4 400 215147 2 200 213146 1 100 219 152 6 7-11 236165 910-11 23716210 911 235161 8 800 Sum:4 MMIS&P100R-MMIR-S&PDiffDiffsq 220151 7 611 218150 5 500 216148 3 300 217149 4 400 215147 2 200 213146 1 100 219 152 6 7-11 236165 910-11 23716210 911 235161 8 800 Sum:4 Spearman Rank Correlation Coefficient: Example 14-11 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

71 Slide 71 Spearman Rank Correlation Coefficient: Example 14-11 Using the Template Note:The p- values in the range J15:J17 will appear only if the sample size is large (n > 30) Note: The p- values in the range J15:J17 will appear only if the sample size is large (n > 30) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

72 Slide 72 l Steps in a chi-square analysis: Formulate null and alternative hypotheses Compute frequencies of occurrence that would be expected if the null hypothesis were true - expected cell counts Note actual, observed cell counts Use differences between expected and actual cell counts to find chi-square statistic: Compare chi-statistic with critical values from the chi- square distribution (with k-1 degrees of freedom) to test the null hypothesis l Steps in a chi-square analysis: Formulate null and alternative hypotheses Compute frequencies of occurrence that would be expected if the null hypothesis were true - expected cell counts Note actual, observed cell counts Use differences between expected and actual cell counts to find chi-square statistic: Compare chi-statistic with critical values from the chi- square distribution (with k-1 degrees of freedom) to test the null hypothesis 14-9 A Chi-Square Test for Goodness of Fit Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

73 Slide 73 The null and alternative hypotheses: H 0 : The probabilities of occurrence of events E 1, E 2...,E k are given by p 1,p 2,...,p k H 1 : The probabilities of the k events are not as specified in the null hypothesis The null and alternative hypotheses: H 0 : The probabilities of occurrence of events E 1, E 2...,E k are given by p 1,p 2,...,p k H 1 : The probabilities of the k events are not as specified in the null hypothesis Assuming equal probabilities, p 1 = p 2 = p 3 = p 4 =0.25 and n=80 PreferenceTanBrownMaroonBlackTotal Observed1240 82080 Expected(np)2020 202080 (O-E)-820-12 00 Assuming equal probabilities, p 1 = p 2 = p 3 = p 4 =0.25 and n=80 PreferenceTanBrownMaroonBlackTotal Observed1240 82080 Expected(np)2020 202080 (O-E)-820-12 00 Example 14-12: Goodness-of-Fit Test for the Multinomial Distribution Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

74 Slide 74 Example 14-12: Goodness-of-Fit Test for the Multinomial Distribution using the Template Note: the p- value is 0.0000, so we can reject the null hypothesis at any a level. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

75 Slide 75 50-5 0.4 0.3 0.2 0.1 0.0 z f ( z ) Partitioning the Standard Normal Distribution 1 -0.440.44 0.1700 0.1713 0.1587 0.1700 0.1713 1. Use the table of the standard normal distribution to determine an appropriate partition of the standard normal distribution which gives ranges with approximately equal percentages. p(z<-1) = 0.1587 p(-1<z<-0.44)= 0.1713 p(-0.44<z<0)= 0.1700 p(0<z<0.44)= 0.1700 p(0.44<z<14)= 0.1713 p(z>1) = 0.1587 1. Use the table of the standard normal distribution to determine an appropriate partition of the standard normal distribution which gives ranges with approximately equal percentages. p(z<-1) = 0.1587 p(-1<z<-0.44)= 0.1713 p(-0.44<z<0)= 0.1700 p(0<z<0.44)= 0.1700 p(0.44<z<14)= 0.1713 p(z>1) = 0.1587 2. Given z boundaries, x boundaries can be determined from the inverse standard normal transformation: x =  +  z = 125 + 40z. 3. Compare with the critical value of the  2 distribution with k-3 degrees of freedom. 2. Given z boundaries, x boundaries can be determined from the inverse standard normal transformation: x =  +  z = 125 + 40z. 3. Compare with the critical value of the  2 distribution with k-3 degrees of freedom. Goodness-of-Fit for the Normal Distribution: Example 14-13 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

76 Slide 76 iO i E i O i - E i (O i - E i ) 2 (O i - E i ) 2 / E i 11415.87-1.873.496900.22035 22017.132.878.236910.48085 31617.00-1.001.000000.05882 41917.002.004.000000.23529 51617.13-1.131.276900.07454 61515.87-0.870.756900.04769  2 :1.11755 iO i E i O i - E i (O i - E i ) 2 (O i - E i ) 2 / E i 11415.87-1.873.496900.22035 22017.132.878.236910.48085 31617.00-1.001.000000.05882 41917.002.004.000000.23529 51617.13-1.131.276900.07454 61515.87-0.870.756900.04769  2 :1.11755  2 (0.10,k-3) = 6.5139 > 1.11755  H 0 is not rejected at the 0.10 level Example 14-13: Solution Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

77 Slide 77 Example 14-13: Solution using the Template Note: p-value = 0.8002 > 0.01  H 0 is not rejected at the 0.10 level Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

78 Slide 78 14-9 Contingency Table Analysis: A Chi-Square Test for Independence Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

79 Slide 79 Null and alternative hypotheses: H 0 : The two classification variables are independent of each other H 1 : The two classification variables are not independent Chi-square test statistic for independence: Degrees of freedom: df=(r-1)(c-1) Expected cell count: Null and alternative hypotheses: H 0 : The two classification variables are independent of each other H 1 : The two classification variables are not independent Chi-square test statistic for independence: Degrees of freedom: df=(r-1)(c-1) Expected cell count: A and B are independent if:P(AUB) = P(A)P(B). If the first and second classification categories are independent:E ij = (R i )(C j )/n A and B are independent if:P(AUB) = P(A)P(B). If the first and second classification categories are independent:E ij = (R i )(C j )/n Contingency Table Analysis: A Chi-Square Test for Independence Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

80 Slide 80 ijOEO-E(O-E) 2 (O-E) 2 /E 114228.813.2174.246.0500 121831.2-13.2174.245.5846 21619.2-13.2174.249.0750 223420.813.2174.248.3769  2 : 29.0865 ijOEO-E(O-E) 2 (O-E) 2 /E 114228.813.2174.246.0500 121831.2-13.2174.245.5846 21619.2-13.2174.249.0750 223420.813.2174.248.3769  2 : 29.0865  2 (0.01,(2-1)(2- 1)) =6.63490 H 0 is rejected at the 0.01 level and it is concluded that the two variables are not independent.  2 (0.01,(2-1)(2- 1)) =6.63490 H 0 is rejected at the 0.01 level and it is concluded that the two variables are not independent. Contingency Table Analysis: Example 14-14 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

81 Slide 81 Since p-value = 0.000, H 0 is rejected at the 0.01 level and it is concluded that the two variables are not independent. Contingency Table Analysis: Example 14-14 using the Template Note: When the contingenc y table is a 2x2, one should use the Yates correction. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

82 Slide 82 14-11 Chi-Square Test for Equality of Proportions tests of homogeneity. Tests of equality of proportions across several populations are also called tests of homogeneity. In general, when we compare c populations (or r populations if they are arranged as rows rather than columns in the table), then the Null and alternative hypotheses: H 0 : p 1 = p 2 = p 3 = … = p c H 1 : Not all p i, I = 1, 2, …, c, are equal Chi-square test statistic for equal proportions: Degrees of freedom: df = (r-1)(c-1) Expected cell count: In general, when we compare c populations (or r populations if they are arranged as rows rather than columns in the table), then the Null and alternative hypotheses: H 0 : p 1 = p 2 = p 3 = … = p c H 1 : Not all p i, I = 1, 2, …, c, are equal Chi-square test statistic for equal proportions: Degrees of freedom: df = (r-1)(c-1) Expected cell count: Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

83 Slide 83 14-11 Chi-Square Test for Equality of Proportions - Extension The Median Test Here, the Null and alternative hypotheses are: H 0 : The c populations have the same median H 1 : Not all c populations have the same median Here, the Null and alternative hypotheses are: H 0 : The c populations have the same median H 1 : Not all c populations have the same median Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

84 Slide 84 Chi-Square Test for the Median: Example 14-16 Using the Template Note:The template was used to help compute the test statistic and the p-value for the median test. First you must manually compute the number of values that are above the grand median and the number that is less than or equal to the grand median. Use these values in the template. See Table 14-16 in the text. Note: The template was used to help compute the test statistic and the p-value for the median test. First you must manually compute the number of values that are above the grand median and the number that is less than or equal to the grand median. Use these values in the template. See Table 14-16 in the text. Since the p-value = 0.6703 is very large there is no evidence to reject the null hypothesis. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

85 Slide 85 Table 11:  =0.005 n. 7------ 80.881 90.833 100.794 110.818. Table 11:  =0.005 n. 7------ 80.881 90.833 100.794 110.818.                            rejected  MMIS&P100R-MMIR-S&PDiffDiffsq 220151 7 611 218150 5 500 216148 3 300 217149 4 400 215147 2 200 213146 1 100 219 152 6 7-11 236165 910-11 23716210 911 235161 8 800 Sum:4 MMIS&P100R-MMIR-S&PDiffDiffsq 220151 7 611 218150 5 500 216148 3 300 217149 4 400 215147 2 200 213146 1 100 219 152 6 7-11 236165 910-11 23716210 911 235161 8 800 Sum:4 Spearman Rank Correlation Coefficient: Example 14-11 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

86 Slide 86 Spearman Rank Correlation Coefficient: Example 14-11 Using the Template Note:The p- values in the range J15:J17 will appear only if the sample size is large (n > 30) Note: The p- values in the range J15:J17 will appear only if the sample size is large (n > 30) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

87 Slide 87 l Steps in a chi-square analysis: Formulate null and alternative hypotheses Compute frequencies of occurrence that would be expected if the null hypothesis were true - expected cell counts Note actual, observed cell counts Use differences between expected and actual cell counts to find chi-square statistic: Compare chi-statistic with critical values from the chi- square distribution (with k-1 degrees of freedom) to test the null hypothesis l Steps in a chi-square analysis: Formulate null and alternative hypotheses Compute frequencies of occurrence that would be expected if the null hypothesis were true - expected cell counts Note actual, observed cell counts Use differences between expected and actual cell counts to find chi-square statistic: Compare chi-statistic with critical values from the chi- square distribution (with k-1 degrees of freedom) to test the null hypothesis 14-9 A Chi-Square Test for Goodness of Fit Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

88 Slide 88 The null and alternative hypotheses: H 0 : The probabilities of occurrence of events E 1, E 2...,E k are given by p 1,p 2,...,p k H 1 : The probabilities of the k events are not as specified in the null hypothesis The null and alternative hypotheses: H 0 : The probabilities of occurrence of events E 1, E 2...,E k are given by p 1,p 2,...,p k H 1 : The probabilities of the k events are not as specified in the null hypothesis Assuming equal probabilities, p 1 = p 2 = p 3 = p 4 =0.25 and n=80 PreferenceTanBrownMaroonBlackTotal Observed1240 82080 Expected(np)2020 202080 (O-E)-820-12 00 Assuming equal probabilities, p 1 = p 2 = p 3 = p 4 =0.25 and n=80 PreferenceTanBrownMaroonBlackTotal Observed1240 82080 Expected(np)2020 202080 (O-E)-820-12 00 Example 14-12: Goodness-of-Fit Test for the Multinomial Distribution Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

89 Slide 89 Example 14-12: Goodness-of-Fit Test for the Multinomial Distribution using the Template Note: the p- value is 0.0000, so we can reject the null hypothesis at any  level. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

90 Slide 90 50-5 0.4 0.3 0.2 0.1 0.0 z f ( z ) Partitioning the Standard Normal Distribution 1 -0.440.44 0.1700 0.1713 0.1587 0.1700 0.1713 1. Use the table of the standard normal distribution to determine an appropriate partition of the standard normal distribution which gives ranges with approximately equal percentages. p(z<-1) = 0.1587 p(-1<z<-0.44)= 0.1713 p(-0.44<z<0)= 0.1700 p(0<z<0.44)= 0.1700 p(0.44<z<14)= 0.1713 p(z>1) = 0.1587 1. Use the table of the standard normal distribution to determine an appropriate partition of the standard normal distribution which gives ranges with approximately equal percentages. p(z<-1) = 0.1587 p(-1<z<-0.44)= 0.1713 p(-0.44<z<0)= 0.1700 p(0<z<0.44)= 0.1700 p(0.44<z<14)= 0.1713 p(z>1) = 0.1587 2. Given z boundaries, x boundaries can be determined from the inverse standard normal transformation: x =  +  z = 125 + 40z. 3. Compare with the critical value of the  2 distribution with k-3 degrees of freedom. 2. Given z boundaries, x boundaries can be determined from the inverse standard normal transformation: x =  +  z = 125 + 40z. 3. Compare with the critical value of the  2 distribution with k-3 degrees of freedom. Goodness-of-Fit for the Normal Distribution: Example 14-13 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

91 Slide 91 iO i E i O i - E i (O i - E i ) 2 (O i - E i ) 2 / E i 11415.87-1.873.496900.22035 22017.132.878.236910.48085 31617.00-1.001.000000.05882 41917.002.004.000000.23529 51617.13-1.131.276900.07454 61515.87-0.870.756900.04769  2 :1.11755 iO i E i O i - E i (O i - E i ) 2 (O i - E i ) 2 / E i 11415.87-1.873.496900.22035 22017.132.878.236910.48085 31617.00-1.001.000000.05882 41917.002.004.000000.23529 51617.13-1.131.276900.07454 61515.87-0.870.756900.04769  2 :1.11755  2 (0.10,k-3) = 6.5139 > 1.11755  H 0 is not rejected at the 0.10 level Example 14-13: Solution Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

92 Slide 92 Example 14-13: Solution using the Template Note: p-value = 0.8002 > 0.01  H 0 is not rejected at the 0.10 level Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

93 Slide 93 14-9 Contingency Table Analysis: A Chi-Square Test for Independence Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

94 Slide 94 Null and alternative hypotheses: H 0 : The two classification variables are independent of each other H 1 : The two classification variables are not independent Chi-square test statistic for independence: Degrees of freedom: df=(r-1)(c-1) Expected cell count: Null and alternative hypotheses: H 0 : The two classification variables are independent of each other H 1 : The two classification variables are not independent Chi-square test statistic for independence: Degrees of freedom: df=(r-1)(c-1) Expected cell count: A and B are independent if:P(AUB) = P(A)P(B). If the first and second classification categories are independent:E ij = (R i )(C j )/n A and B are independent if:P(AUB) = P(A)P(B). If the first and second classification categories are independent:E ij = (R i )(C j )/n Contingency Table Analysis: A Chi-Square Test for Independence Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

95 Slide 95 ijOEO-E(O-E) 2 (O-E) 2 /E 114228.813.2174.246.0500 121831.2-13.2174.245.5846 21619.2-13.2174.249.0750 223420.813.2174.248.3769  2 : 29.0865 ijOEO-E(O-E) 2 (O-E) 2 /E 114228.813.2174.246.0500 121831.2-13.2174.245.5846 21619.2-13.2174.249.0750 223420.813.2174.248.3769  2 : 29.0865  2 (0.01,(2-1)(2- 1)) =6.63490 H 0 is rejected at the 0.01 level and it is concluded that the two variables are not independent.  2 (0.01,(2-1)(2- 1)) =6.63490 H 0 is rejected at the 0.01 level and it is concluded that the two variables are not independent. Contingency Table Analysis: Example 14-14 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

96 Slide 96 Since p-value = 0.000, H 0 is rejected at the 0.01 level and it is concluded that the two variables are not independent. Contingency Table Analysis: Example 14-14 using the Template Note: When the contingenc y table is a 2x2, one should use the Yates correction. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

97 Slide 97 14-11 Chi-Square Test for Equality of Proportions tests of homogeneity. Tests of equality of proportions across several populations are also called tests of homogeneity. In general, when we compare c populations (or r populations if they are arranged as rows rather than columns in the table), then the Null and alternative hypotheses: H 0 : p 1 = p 2 = p 3 = … = p c H 1 : Not all p i, I = 1, 2, …, c, are equal Chi-square test statistic for equal proportions: Degrees of freedom: df = (r-1)(c-1) Expected cell count: In general, when we compare c populations (or r populations if they are arranged as rows rather than columns in the table), then the Null and alternative hypotheses: H 0 : p 1 = p 2 = p 3 = … = p c H 1 : Not all p i, I = 1, 2, …, c, are equal Chi-square test statistic for equal proportions: Degrees of freedom: df = (r-1)(c-1) Expected cell count: Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

98 Slide 98 14-11 Chi-Square Test for Equality of Proportions - Extension The Median Test Here, the Null and alternative hypotheses are: H 0 : The c populations have the same median H 1 : Not all c populations have the same median Here, the Null and alternative hypotheses are: H 0 : The c populations have the same median H 1 : Not all c populations have the same median Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

99 Slide 99 Chi-Square Test for the Median: Example 14-16 Using the Template Note:The template was used to help compute the test statistic and the p-value for the median test. First you must manually compute the number of values that are above the grand median and the number that is less than or equal to the grand median. Use these values in the template. See Table 14-16 in the text. Note: The template was used to help compute the test statistic and the p-value for the median test. First you must manually compute the number of values that are above the grand median and the number that is less than or equal to the grand median. Use these values in the template. See Table 14-16 in the text. Since the p-value = 0.6703 is very large there is no evidence to reject the null hypothesis. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

100 Slide 100 M.Phil (Statistics) GC University,. (Degree awarded by GC University) M.Sc (Statistics) GC University,. (Degree awarded by GC University) Statitical Officer (BS-17) (Economics & Marketing Division) Livestock Production Research Institute Bahadurnagar (Okara), Livestock & Dairy Development Department, Govt. of Punjab Name Shakeel Nouman Religion Christian Domicile Punjab (Lahore) Contact # 0332-4462527. 0321-9898767 E.Mail sn_gcu@yahoo.comsn_gcu@yahoo.com sn_gcu@hotmail.com Name Shakeel Nouman Religion Christian Domicile Punjab (Lahore) Contact # 0332-4462527. 0321-9898767 E.Mail sn_gcu@yahoo.comsn_gcu@yahoo.com sn_gcu@hotmail.com Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer


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