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# Nonparametric methods and chi square tests (1)

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### Nonparametric methods and chi square tests (1)

1. 1. Nonparametric Methods and Chi-Square Tests (1) Slide 1 Shakeel Nouman M.Phil Statistics Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
2. 2. Slide 2 14 Nonparametric Methods and Chi-Square Tests (1) • 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) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
3. 3. Slide 3 14 Nonparametric Methods and Chi-Square Tests (2) • 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 (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
4. 4. 14-1 Using Statistics (Parametric Tests) 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 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
5. 5. Nonparametric Tests 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 Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
6. 6. Nonparametric Tests (Continued) 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 Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
7. 7. Nonparametric Tests (Continued) • • • 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). Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
8. 8. 14-2 Sign Test Slide 8 • Comparing paired observations Paired observations: X and Y p = P(X>Y) » Two-tailed test H0: p = 0.50 H1: p0.50 » Right-tailed test H0: p  0.50 H1: p0.50 » Left-tailed test H0: p  0.50 H1: p 0.50 » Test statistic: T = Number of + signs Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
9. 9. Sign Test Decision Rule Slide 9 • Small Sample: Binomial Test For a two-tailed test, find a critical point corresponding as closely as possible to /2 (C1) and define C2 as n-C1. Reject null hypothesis if T  C1or T  C2. For a right-tailed test, reject H0 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 H0 if T  C, where C is defined as above. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
10. 10. Example 14-1 CEO Before After 1 3 4 1 2 5 5 0 3 2 3 1 4 2 4 1 5 4 4 0 6 2 3 1 7 1 2 1 8 5 4 -1 9 4 5 1 10 5 4 -1 11 3 4 1 12 2 5 1 13 2 5 1 14 2 3 1 15 1 2 1 16 3 2 -1 17 4 5 1 Sign + + + + + + + + + + + + n = 15 T = 12   0.025 C1=3 C2 = 15-3 = 12 H0 rejected, since T  C2 Slide 10 Cumulative Binomial Probabilities (n=15, p=0.5) x F(x) 0 0.00003 1 0.00049 2 0.00369 3 0.01758 4 0.05923 5 0.15088 6 0.30362 7 0.50000 8 0.69638 9 0.84912 10 0.94077 11 0.98242 12 0.99631 13 0.99951 14 0.99997 15 1.00000 C1 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
11. 11. Example 14-1- Using the Template Slide 11 H0: p = 0.5 H1: 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. 12. 14-3 The Runs Test - A Test for Randomness 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 Case 2: SSSSSSSSSS|EEEEEEEEEE Case 3: S|EE|SS|EEE|S|E|SS|E|S|EE|SSS|E : R = 20 Apparently nonrandom : R = 2 Apparently nonrandom : R = 12 Perhaps random A two-tailed hypothesis test for randomness: H0: Observations are generated randomly H1: Observations are not generated randomly Test Statistic: R=Number of Runs Reject H0 at level if R C1 or R C2, as given in Table 8, with total tail probability P(R C1) + P(R C2) =  Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
13. 13. Runs Test: Examples Table 8: (n1,n2) 11 12 (10,10) 0.586 0.758 0.872 0.949 0.981 0.996 0.999 1.000 1.000 1.000 . . . 13 Number of Runs (r) 14 15 16 17 Slide 13 18 19 20 Case 1: n1 = 10 n2 = 10 R= 20 p-value 0 Case 2: n1 = 10 n2 = 10 R = 2 p-value  0 Case 3: n1 = 10 n2 = 10 R= 12 p-value R  P F(11)] = (2)(1-0.586) = (2)(0.414) = 0.828 H0 not rejected Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
14. 14. Large-Sample Runs Test: Using the Normal Approximation Slide 14 The mean of the normal distribution of the number of runs: 2n n E ( R)  1 n n 1 2 1 2 The standard deviation:   R 2n n ( 2n n  n  n ) ( n  n ) ( n  n  1) 1 2 1 2 1 2 2 1 2 1 2 The standard normal test statistic: R  E ( R) z  R Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
15. 15. Large-Sample Runs Test: Example 14-2 Slide 15 Example 14-2: n1 = 27 n2 = 26 R = 15 2n n ( 2)( 27)( 26) E ( R)  1 2  1   1  26.49  1  27.49 n n ( 27  26) 1 2 2n n ( 2n n  n  n ) 1 2 1 2 1 2  ( 2)( 27)( 26)(( 2)( 27)( 26)  27  26))   R ( n  n ) 2 ( n  n  1) ( 27  26) 2 ( 27  26  1) 1 2 1 2 1896804   12.986  3.604 146068 R  E ( R ) 15  27.49 z   3.47  3.604 R p - value = 2(1 - .9997) = 0.0006 H0 should be rejected at any common level of significance. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
16. 16. Large-Sample Runs Test: Example 14-2 – Using the Template Slide 16 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. 17. Using the Runs Test to Compare Two Population Distributions (Means): the Wald-Wolfowitz Test Slide 17 The null and alternative hypotheses for the Wald-Wolfowitz test: H0: The two populations have the same distribution H1: 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 Example 14-3: Salesperson A: Salesperson B: 35 44 17 23 39 13 50 24 48 33 29 21 60 18 75 16 49 32 66 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
18. 18. The Wald-Wolfowitz Test: Example 14-3 Sales 35 44 39 48 60 75 49 66 17 23 13 24 33 21 18 16 32 Sales Person A A A A A A A A B B B B B B B B B Sales (Sorted) 13 16 17 21 24 29 32 33 35 39 44 48 49 50 60 66 75 Sales Person (Sorted) B B B B B A B B A A A A A A A A A Runs 1 2 3 Slide 18 n1 = 10 n2 = 9 R= 4 p-value PR  4 H0 may be rejected Table Number of Runs (r) (n1,n2) 2 3 4 5 . . . (9,10) 0.000 0.000 0.002 0.004 ... 4 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
19. 19. Ranks Tests 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 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
20. 20. 14-4 The Mann-Whitney U Test (Comparing Two Populations) Slide 20 The null and alternative hypotheses: H0: The distributions of two populations are identical H1: The two population distributions are not identical The Mann-Whitney U statistic: n1 ( n1  1) U  n1 n2   R1 2 R 1   Ranks from sample 1 where n1 is the sample size from population 1 and n2 is the sample size from population 2. n1 n2 n1 n2 (n1  n2  1) E [U ]  U  2 12 U  E [U ] The large - sample test statistic: z   U Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
21. 21. The Mann-Whitney U Test: Example 14-4 Model A A A A A A B B B B B B Time 35 38 40 42 41 36 29 27 30 33 39 37 Rank 5 8 10 12 11 6 2 1 3 4 9 7 Rank Sum U  n1 n 2  n1 ( n1  1) 2 (6)(6 + 1) = (6)(6) + 5 52 26 Slide 21  R1  52 2 Cumulative Distribution Function of the Mann-Whitney U Statistic n2=6 n1=6 u . . . 4 0.0130 P(u 5) 5 0.0206 6 0.0325 . . . Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
22. 22. Example 14-5: Large-Sample Mann-Whitney U Test Score Rank Score Program Rank Sum 85 1 20.0 20.0 87 1 21.0 41.0 92 1 27.0 68.0 98 1 30.0 98.0 90 1 26.0 124.0 88 1 23.0 147.0 75 1 17.0 164.0 72 1 13.5 177.5 60 1 6.5 184.0 93 1 28.0 212.0 88 1 23.0 235.0 89 1 25.0 260.0 96 1 29.0 289.0 73 1 15.0 304.0 62 1 8.5 312.5 Score Rank Score Program Rank Sum 65 2 10.0 10.0 57 2 4.0 14.0 74 2 16.0 30.0 43 2 2.0 32.0 39 2 1.0 33.0 88 2 23.0 56.0 62 2 8.5 64.5 69 2 11.0 75.5 70 2 12.0 87.5 72 2 13.5 101.0 59 2 5.0 106.0 60 2 6.5 112.5 80 2 18.0 130.5 83 2 19.0 149.5 50 2 3.0 152.5 U  n1n2  Slide 22 n1 ( n1  1)  R1 2 (15)(15  1)  (15)(15)   312 .5  32 .5 2 n1n2 (15)(15) E [U ]  = = 112.5 2 2 n1n2 ( n1  n2  1) U  12 (15)(15)(15  15  1)   24 .109 12 U  E [U ] 32 .5  112 .5 z    3.32 U 24 .109 Since the test statistic is z = -3.32, the p-value 0.0005,Tests (1) HShakeelrejected. Statistics Govt. College University Lahore, Statistical Nonparametric Methods and Chi-Square and By 0 is Nouman M.Phil Officer
23. 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 H0 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. 24. 14-5 The Wilcoxon SignedRanks Test (Paired Ranks) Slide 24 The null and alternative hypotheses: H0: The median difference between populations are 1 and 2 is zero H1: The median difference between populations are 1 and 2 is not zero Find the difference between the ranks for each pair, D = x1 -x2, 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: T  min  (  ),  (  ) For small samples, a left-tailed test is used, using the values in Appendix C, Table 10. E[T ]  n ( n  1) 4 The large-sample test statistic: n ( n  1)( 2 n  1) T  24 z T  E[T ] T Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
25. 25. Example 14-6 Sold Sold (1) (2) Rank Rank Rank D=x1-x2 ABS(D) ABS(D) (D>0) (D<0) 56 48 100 85 22 44 35 28 52 77 89 10 65 90 70 33 16 -22 40 15 14 4 -10 21 -8 7 -1 0 -20 29 30 7 40 70 60 70 8 40 45 7 60 70 90 10 85 61 40 26 16 22 40 15 14 4 10 21 8 7 1 * 20 29 30 7 9.0 12.0 15.0 8.0 7.0 2.0 6.0 11.0 5.0 3.5 1.0 * 10.0 13.0 14.0 3.5 9.0 0.0 15.0 8.0 7.0 2.0 0.0 11.0 0.0 3.5 0.0 * 0.0 13.0 14.0 3.5 0 12 0 0 0 0 6 0 5 0 1 * 10 0 0 0 Sum: 86 Slide 25 T=34 n=15 P=0.05 30 P=0.02525 P=0.01 20 P=0.00516 34 H0 is not rejected (Note the arithmetic error in the text for store 13) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
26. 26. Example 14-7 Hourly Messages 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 Rank D=x1-x2 ABS(D) ABS(D) 2 -5 -26 29 -44 -37 -9 18 28 36 -20 11 -39 21 49 16 -40 -31 6 -47 15 31 -10 17 33 2 5 26 29 44 37 9 18 28 36 20 11 39 21 49 16 40 31 6 47 15 31 10 17 33 Rank (D>0) Rank (D<0) 1.0 2.0 13.0 15.0 23.0 20.0 4.0 10.0 14.0 19.0 11.0 6.0 21.0 12.0 25.0 8.0 22.0 16.5 3.0 24.0 7.0 16.5 5.0 9.0 18.0 1.0 0.0 0.0 15.0 0.0 0.0 0.0 10.0 14.0 19.0 0.0 6.0 0.0 12.0 25.0 8.0 0.0 0.0 3.0 0.0 7.0 16.5 0.0 9.0 18.0 0.0 2.0 13.0 0.0 23.0 20.0 4.0 0.0 0.0 0.0 11.0 0.0 21.0 0.0 0.0 0.0 22.0 16.5 0.0 24.0 0.0 0.0 5.0 0.0 0.0 Sum: 151 144 123 178 105 112 140 167 177 185 129 160 110 170 198 165 109 118 155 102 164 180 139 166 82 Md0 163.5 161.5 Slide 26 E[ T ]  n ( n  1) T   (25)(25 + 1) = 4 4 n ( n  1)( 2 n  1) = 162.5 24 25( 25  1)(( 2 )( 25)  1) 24  33150  37 .165 24 The large - sample test statistic: z  T  E[ T ] T  163.5  162 .5  0.027 37 .165 H 0 cannot be rejected Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
27. 27. Example 14-7 using the Template Slide 27 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: 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. 28. 14-6 The Kruskal-Wallis Test - A Nonparametric Alternative to One-Way ANOVA Slide 28 The Kruskal-Wallis hypothesis test: H0: All k populations have the same distribution H1: Not all k populations have the same distribution The Kruskal-Wallis test statistic: 2 12  k Rj  H  3(n  1) n(n  1)   n j   j 1  2 If each nj > 5, then H is approximately distributed as a  . Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
29. 29. Example 14-8: The Kruskal-Wallis Test SoftwareTime Rank Group RankSum 1 45 14 1 90 1 38 10 2 56 1 56 16 3 25 1 60 17 1 47 15 1 65 18 2 30 8 2 40 11 2 28 7 2 44 13 2 25 5 2 42 12 3 22 4 3 19 3 3 15 1 3 31 9 3 27 6 3 17 2 Slide 29  k R2  12 j  H   j 1   3( n  1) n ( n  1)  nj  12  902 562 252        3(18  1) 18(18  1)  6 6 6   12   11861  57     342   6   12 .3625 2(2,0.005)=10.5966, so H0 is rejected. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
30. 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. 31. Further Analysis (Pairwise Comparisons of Average Ranks) 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. The pairwise comparison test statistic: D  Ri  R j where R i is the mean of the ranks of the observations from population i. The critical point for the paired comparisons:  n(n  1)  1 1  2 C KW  (   , k 1 )    ni  n j    12  Reject if D > C KW Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
32. 32. Pairwise Comparisons: Example 14-8 C KW Slide 32 Critical Point: n(n  1)  1 1  2  (   ,k 1 )   12  ni  n j     18(18  1)  1 1  ( 9.21034)    12  6 6  87.49823  9.35 90  15 6 56 R2   9.33 6 25 R3   4.17 6 R1  D1,2  15  9.33  5.67 D1,3  15  4.17  10.83 *** D2,3  9.33  4.17  516 . Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
33. 33. 14-7 The Friedman Test for a Randomized Block Design 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. The Friedman hypothesis test: H0: The distributions of the k treatment populations are identical H1: Not all k distribution are identical The Friedman test statistic: 12    R  3n(k  1) nk (k  1) 2 k j 1 2 j 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. 34. Example 14-10 – using the Template Slide 34 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 lowbudget 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. 35. 14-8 The Spearman Rank Correlation Coefficient Slide 35 The Spearman Rank Correlation Coefficient is the simple correlation coefficient calculated from variables converted to ranks from their original values. The Spearman Rank Correlation Coefficient (assuming no ties): n 2 6  di rs  1  i 1 where d = R(x ) - R(y ) 2 i i i n ( n  1) Null and alternative hypotheses: H 0:  s = 0 H1:  s  0 Critical values for small sample tests from Appendix C, Table 11 Large sample test statistic: z = rs ( n  1) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
36. 36. Slide 36 14 Nonparametric Methods and Chi-Square Tests (1) • 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) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
37. 37. Slide 37 14 Nonparametric Methods and Chi-Square Tests (2) • 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 (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
38. 38. 14-1 Using Statistics (Parametric Tests) 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 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
39. 39. Nonparametric Tests 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 Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
40. 40. Nonparametric Tests (Continued) 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 Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
41. 41. Nonparametric Tests (Continued) • • • 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). Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
42. 42. 14-2 Sign Test Slide 42 • Comparing paired observations Paired observations: X and Y p = P(X>Y) » Two-tailed test H0: p = 0.50 H1: p0.50 » Right-tailed test H0: p  0.50 H1: p0.50 » Left-tailed test H0: p  0.50 H1: p 0.50 » Test statistic: T = Number of + signs Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
43. 43. Sign Test Decision Rule Slide 43 • Small Sample: Binomial Test For a two-tailed test, find a critical point corresponding as closely as possible to /2 (C1) and define C2 as n-C1. Reject null hypothesis if T  C1or T  C2. For a right-tailed test, reject H0 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 H0 if T  C, where C is defined as above. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
44. 44. Example 14-1 CEO 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Before After 3 4 5 5 2 3 2 4 4 4 2 3 1 2 5 4 4 5 5 4 3 4 2 5 2 5 2 3 1 2 3 2 4 5 Sign 1 0 1 1 0 1 1 -1 1 -1 1 1 1 1 1 -1 1 + + + + + + + + + + + + n = 15 T = 12   0.025 C1=3 C2 = 15-3 = 12 C1 H0 rejected, since T  C2 Slide 44 Cumulative Binomial Probabilities (n=15, p=0.5) x F(x) 0 0.00003 1 0.00049 2 0.00369 3 0.01758 4 0.05923 5 0.15088 6 0.30362 7 0.50000 8 0.69638 9 0.84912 10 0.94077 11 0.98242 12 0.99631 13 0.99951 14 0.99997 15 1.00000 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
45. 45. Example 14-1- Using the Template Slide 45 H0: p = 0.5 H1: p  5 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. 46. 14-3 The Runs Test - A Test for Randomness 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 Case 2: SSSSSSSSSS|EEEEEEEEEE Case 3: S|EE|SS|EEE|S|E|SS|E|S|EE|SSS|E : R = 20 Apparently nonrandom : R = 2 Apparently nonrandom : R = 12 Perhaps random A two-tailed hypothesis test for randomness: H0: Observations are generated randomly H1: Observations are not generated randomly Test Statistic: R=Number of Runs Reject H0 at level  if R  C1 or R  C2, as given in Table 8, with total tail probability P(R  C1) + P(R  C2) =  Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
47. 47. Runs Test: Examples Table 8: (n1,n2) 11 12 (10,10) 0.586 0.758 . . . Number of Runs (r) 13 14 0.872 0.949 Slide 47 15 16 17 18 19 20 0.981 0.996 0.999 1.000 1.000 1.000 Case 1: n1 = 10 n2 = 10 R= 20 p-value0 Case 2: n1 = 10 n2 = 10 R = 2 p-value 0 Case 3: n1 = 10 n2 = 10 R= 12 p-value PR  11F(11)] = (2)(1-0.586) = (2)(0.414) = 0.828 H0 not rejected Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
48. 48. Large-Sample Runs Test: Using the Normal Approximation Slide 48 The mean of the normal distribution of the number of runs: 2n n E ( R)  1 n n 1 2 1 2 The standard deviation:   R 2n n ( 2n n  n  n ) ( n  n ) ( n  n  1) 1 2 1 2 1 2 2 1 2 1 2 The standard normal test statistic: R  E ( R) z  R Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
49. 49. Large-Sample Runs Test: Example 14-2 Slide 49 Example 14-2: n1 = 27 n2 = 26 R = 15 2n n ( 2)( 27)( 26) E ( R)  1 2  1   1  26.49  1  27.49 n n ( 27  26) 1 2 2n n ( 2n n  n  n ) 1 2 1 2 1 2  ( 2)( 27)( 26)(( 2)( 27)( 26)  27  26))   R ( n  n ) 2 ( n  n  1) ( 27  26) 2 ( 27  26  1) 1 2 1 2 1896804   12.986  3.604 146068 R  E ( R ) 15  27.49 z   3.47  3.604 R p - value = 2(1 - .9997) = 0.0006 H0 should be rejected at any common level of significance. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
50. 50. Large-Sample Runs Test: Example 14-2 – Using the Template Slide 50 Note: The computed pvalue 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. 51. Using the Runs Test to Compare Two Population Distributions (Means): the Wald-Wolfowitz Test Slide 51 The null and alternative hypotheses for the Wald-Wolfowitz test: H0: The two populations have the same distribution H1: 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 Example 14-3: Salesperson A: Salesperson B: 35 17 44 23 39 13 50 24 48 33 29 21 60 18 75 16 49 32 66 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
52. 52. The Wald-Wolfowitz Test: Example 14-3 Sales 35 44 39 48 60 75 49 66 17 23 13 24 33 21 18 16 32 Sales Person A A A A A A A A B B B B B B B B B Sales (Sorted) 13 16 17 21 24 29 32 33 35 39 44 48 49 50 60 66 75 Sales Person (Sorted) B B B B B A B B A A A A A A A A A Runs 1 2 3 Slide 52 n1 = 10 n2 = 9 R= 4 p-value PR  4 H0 may be rejected Table Number of Runs (r) (n1,n2) 2 3 4 5 . . . (9,10) 0.000 0.000 0.002 0.004 ... 4 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
53. 53. Ranks Tests 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 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
54. 54. 14-4 The Mann-Whitney U Test (Comparing Two Populations) Slide 54 The null and alternative hypotheses: H0: The distributions of two populations are identical H1: The two population distributions are not identical The Mann-Whitney U statistic: n1 ( n1  1) U  n1 n2   R1 R 1   Ranks from sample 1 2 where n1 is the sample size from population 1 and n2 is the sample size from population 2. n1 n2 n1 n2 (n1  n2  1) E [U ]  U  2 12 U  E [U ] The large - sample test statistic: z  U Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
55. 55. The Mann-Whitney U Test: Example 14-4 Model A A A A A A B B B B B B Time 35 38 40 42 41 36 29 27 30 33 39 37 Rank 5 8 10 12 11 6 2 1 3 4 9 7 Rank Sum U  n1 n 2  n1 ( n1  1) 2 (6)(6 + 1) = (6)(6) + 5 52 26 Slide 55  R1  52 2 Cumulative Distribution Function of the Mann-Whitney U Statistic n2=6 n1=6 u . . . 4 0.0130 Pu5 5 0.0206 6 0.0325 . . . Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
56. 56. Example 14-5: Large-Sample Mann-Whitney U Test Score Score Program Rank 85 1 20.0 87 1 21.0 92 1 27.0 98 1 30.0 90 1 26.0 88 1 23.0 75 1 17.0 72 1 13.5 60 1 6.5 93 1 28.0 88 1 23.0 89 1 25.0 96 1 29.0 73 1 15.0 62 1 8.5 Rank Sum 20.0 41.0 68.0 98.0 124.0 147.0 164.0 177.5 184.0 212.0 235.0 260.0 289.0 304.0 312.5 Score Score Program Rank 65 2 10.0 57 2 4.0 74 2 16.0 43 2 2.0 39 2 1.0 88 2 23.0 62 2 8.5 69 2 11.0 70 2 12.0 72 2 13.5 59 2 5.0 60 2 6.5 80 2 18.0 83 2 19.0 50 2 3.0 Rank Sum 10.0 14.0 30.0 32.0 33.0 56.0 64.5 75.5 87.5 101.0 106.0 112.5 130.5 149.5 152.5 Since the test statistic is z = -3.32, the p-value  0.0005, and H0 is rejected. U  n1n2  Slide 56 n1 ( n1  1)  R1 2 (15)(15  1)  (15)(15)   312 .5  32 .5 2 n1n2 (15)(15) E [U ]  = = 112.5 2 2 n1n2 ( n1  n2  1) U  12 (15)(15)(15  15  1)   24 .109 12 U  E [U ] 32 .5  112 .5 z    3.32 U 24 .109 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
57. 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 H0 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. 58. 14-5 The Wilcoxon Signed-Ranks Test (Paired Ranks) Slide 58 The null and alternative hypotheses: H0: The median difference between populations are 1 and 2 is zero H1: The median difference between populations are 1 and 2 is not zero Find the difference between the ranks for each pair, D = x1 -x2, 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:  (  ),  (  ) T  min   For small samples, a left-tailed test is used, using the values in Appendix C, Table 10. E[T ]  n ( n  1) 4 The large-sample test statistic: z  T  T  E[T ] n ( n  1)( 2 n  1) 24 T Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
59. 59. Example 14-6 Sold Sold (1) (2) 56 48 100 85 22 44 35 28 52 77 89 10 65 90 70 33 40 70 60 70 8 40 45 7 60 70 90 10 85 61 40 26 Rank Rank Rank D=x1-x2 ABS(D) ABS(D)(D>0) (D<0) 16 -22 40 15 14 4 -10 21 -8 7 -1 0 -20 29 30 7 16 22 40 15 14 4 10 21 8 7 1 * 20 29 30 7 9.0 12.0 15.0 8.0 7.0 2.0 6.0 11.0 5.0 3.5 1.0 * 10.0 13.0 14.0 3.5 9.0 0.0 15.0 8.0 7.0 2.0 0.0 11.0 0.0 3.5 0.0 * 0.0 13.0 14.0 3.5 0 12 0 0 0 0 6 0 5 0 1 * 10 0 0 0 Sum: 86 Slide 59 T=34 P=0.05 P=0.025 P=0.01 P=0.005 n=15 30 25 20 16 34 H0 is not rejected (Note the arithmetic error in the text for store 13) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
60. 60. Example 14-7 Hourly Messages 151 144 123 178 105 112 140 167 177 185 129 160 110 170 198 165 109 118 155 102 164 180 139 166 82 Md0 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 149 Rank Rank D=x1-x2 ABS(D) ABS(D) (D>0) 2 -5 -26 29 -44 -37 -9 18 28 36 -20 11 -39 21 49 16 -40 -31 6 -47 15 31 -10 17 33 2 5 26 29 44 37 9 18 28 36 20 11 39 21 49 16 40 31 6 47 15 31 10 17 33 Rank (D<0) 1.0 2.0 13.0 15.0 23.0 20.0 4.0 10.0 14.0 19.0 11.0 6.0 21.0 12.0 25.0 8.0 22.0 16.5 3.0 24.0 7.0 16.5 5.0 9.0 18.0 1.0 0.0 0.0 15.0 0.0 0.0 0.0 10.0 14.0 19.0 0.0 6.0 0.0 12.0 25.0 8.0 0.0 0.0 3.0 0.0 7.0 16.5 0.0 9.0 18.0 0.0 2.0 13.0 0.0 23.0 20.0 4.0 0.0 0.0 0.0 11.0 0.0 21.0 0.0 0.0 0.0 22.0 16.5 0.0 24.0 0.0 0.0 5.0 0.0 0.0 Sum: 163.5 161.5 Slide 60 E[ T ]  n ( n  1) (25)(25 + 1) = T   = 162.5 4 4 n ( n  1)( 2 n  1) 24 25( 25  1)(( 2 )( 25)  1) 24  33150  37 .165 24 The large - sample test statistic: z  T  E[ T ] T  163.5  162 .5  0.027 37 .165 H 0 cannot be rejected Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
61. 61. Example 14-7 using the Template Slide 61 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: 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. 62. 14-6 The Kruskal-Wallis Test - A Nonparametric Alternative to One-Way ANOVA Slide 62 The Kruskal-Wallis hypothesis test: H0: All k populations have the same distribution H1: Not all k populations have the same distribution The Kruskal-Wallis test statistic: 2 12  k Rj  H   n   3(n  1) n(n  1)  j 1 j  If each nj > 5, then H is approximately distributed as a 2. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
63. 63. Example 14-8: The Kruskal-Wallis Test SoftwareTime Rank Group RankSum 1 45 14 1 90 1 38 10 2 56 1 56 16 3 25 1 60 17 1 47 15 1 65 18 2 30 8 2 40 11 2 28 7 2 44 13 2 25 5 2 42 12 3 22 4 3 19 3 3 15 1 3 31 9 3 27 6 3 17 2 Slide 63  k R2  12 j  H   j 1   3( n  1) n ( n  1)  nj  12  902 562 252        3(18  1) 18(18  1)  6 6 6   12   11861  57     342   6   12 .3625 2(2,0.005)=10.5966, so H0 is rejected. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
64. 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. 65. Further Analysis (Pairwise Comparisons of Average Ranks) 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. The pairwise comparison test statistic: D  Ri  R j where R i is the mean of the ranks of the observations from population i. The critical point for the paired comparisons:  n(n  1)  1 1  2 C KW  (   , k 1 )    ni  n j    12  Reject if D > C KW Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
66. 66. Pairwise Comparisons: Example 14-8 C KW Slide 66 Critical Point: n(n  1)  1 1  2  (   ,k 1 )   12  ni  n j     18(18  1)  1 1  ( 9.21034)    12  6 6  87.49823  9.35 90  15 6 56 R2   9.33 6 25 R3   4.17 6 R1  D1,2  15  9.33  5.67 D1,3  15  4.17  10.83 *** D2,3  9.33  4.17  516 . Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
67. 67. 14-7 The Friedman Test for a Randomized Block Design 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. The Friedman hypothesis test: H0: The distributions of the k treatment populations are identical H1: Not all k distribution are identical The Friedman test statistic: 12    R  3n(k  1) nk (k  1) The degrees of freedom for the chi-square distribution is (k – 1). 2 k j 1 2 j Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
68. 68. Example 14-10 – using the Template Slide 68 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. 69. 14-8 The Spearman Rank Correlation Coefficient Slide 69 The Spearman Rank Correlation Coefficient is the simple correlation coefficient calculated from variables converted to ranks from their original values. The Spearman Rank Correlation Coefficient (assuming no ties): n 2 6  di rs  1  i 1 where d = R(x ) - R(y ) 2 i i i n ( n  1) Null and alternative hypotheses: H 0:  s = 0 H1:  s  0 Critical values for small sample tests from Appendix C, Table 11 Large sample test statistic: z = rs ( n  1) Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
70. 70. Spearman Rank Correlation Coefficient: Example 14-11 MMI S&P100 220 151 218 150 216 148 217 149 215 147 213 146 219 152 236 165 237 162 235 161 R-MMI R-S&P Diff Diffsq 7 6 1 1 5 5 0 0 3 3 0 0 4 4 0 0 2 2 0 0 1 1 0 0 6 7 -1 1 9 10 -1 1 10 9 1 1 8 8 0 0 Sum: Slide 70 Table 11: =0.005 n . . . 7 -----8 0.881 9 0.833 10 0.794 11 0.818 . . . 4 n   di 6 64 4 1  1  i 1 1 9758794H    99 nn 1 11 1 rejected Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
71. 71. Spearman Rank Correlation Coefficient: Example 14-11 Using the Template Slide 71 Note: The pvalues 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. 72. 14-9 A Chi-Square Test for Goodness of Fit  Slide 72 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: (Oi  Ei ) 2 2   Ei i 1 k  Compare chi-statistic with critical values from the chi-square distribution (with k-1 degrees of freedom) to test the null hypothesis Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
73. 73. Example 14-12: Goodness-of-Fit Test for the Multinomial Distribution Slide 73 The null and alternative hypotheses: H0: The probabilities of occurrence of events E1, E2...,Ek are given by p1,p2,...,pk H1: The probabilities of the k events are not as specified in the null hypothesis Assuming equal probabilities, p1= p2 = p3 = p4 =0.25 and n=80 Preference Tan Brown Maroon Black Total Observed 12 40 8 20 80 Expected(np) 20 20 20 20 80 (O-E) -8 20 -12 0 0 k ( Oi  E i ) 2    i 1 Ei 2  ( 8 ) 20 2  ( 20 ) 2  ( 12 ) 20 20 2  ( 0) 2 20  30.4   2  11.3449 ( 0.01, 3) H 0 is rejected at the 0.01 level. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
74. 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. 75. Goodness-of-Fit for the Normal Distribution: Example 14-13 p(z<-1) p(-1<z<-0.44) p(-0.44<z<0) p(0<z<0.44) p(0.44<z<14) p(z>1) = 0.1587 = 0.1713 = 0.1700 = 0.1700 = 0.1713 = 0.1587 Partitioning the Standard Normal Distribution 0.1700 0.1700 0.4 0.1713 0.1713 0.3 f(z) 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. Slide 75 0.2 0.1587 0.1587 0.1 z 0.0 -5 -1 0 1 5 -0.44 0.44 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. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
76. 76. Example 14-13: Solution i Oi Ei 1 2 3 4 5 6 14 15.87 20 17.13 16 17.00 19 17.00 16 17.13 15 15.87 Slide 76 Oi - Ei (Oi - Ei)2 (Oi - Ei)2/ Ei -1.87 2.87 -1.00 2.00 -1.13 -0.87 3.49690 8.23691 1.00000 4.00000 1.27690 0.75690 2 : 0.22035 0.48085 0.05882 0.23529 0.07454 0.04769 1.11755 2(0.10,k-3)= 6.5139 > 1.11755 H0 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
77. 77. Example 14-13: Solution using the Template Slide 77 Note: p-value = 0.8002 > 0.01 H0 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. 78. 14-9 Contingency Table Analysis: A Chi-Square Test for Independence Slide 78 First Classification Category Second Classification Category 1 2 3 4 5 Column Total 1 O11 O21 O31 O41 O51 2 O12 O22 O32 O42 O52 3 O13 O23 O33 O43 O53 4 O14 O24 O34 O44 O54 5 O15 O25 O35 O45 O55 C1 C2 C3 C4 C5 Row Total R1 R2 R3 R4 R5 n Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
79. 79. Contingency Table Analysis: A Chi-Square Test for Independence Slide 79 A and B are independent if:P(AUB) = P(A)P(B). If the first and second classification categories are independent:Eij = (Ri)(Cj)/ Null and alternative hypotheses: H0: The two classification variables are independent of each other H1: The two classification variables are not independent ( Oij  Eij ) 2    Eij i 1 j 1 r Chi-square test statistic for independence: 2 c Degrees of freedom: df=(r-1)(c-1) Expected cell count: Ri C j Eij  n Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
80. 80. Contingency Table Analysis: Example 14-14 Industry Type Profit Service Nonservice (Expected) (Expected) Total 42 18 60 (Expected) Loss 6 (Expected) (40*48/100)=19.2 Total ij 11 12 21 22 (60*48/100)=28.8 48 O 42 18 6 34 E 28.8 31.2 19.2 20.8 O-E 13.2 -13.2 -13.2 13.2 (O-E)2 174.24 174.24 174.24 174.24 2: Slide 80 2(0.01,(2-1)(21))=6.63490 H0 is rejected at the 0.01 level (60*52/100)=31.2 and 34 40 (40*52/100)=20.8 it is concluded 52 100 that the two variables 2 (O-E) /E Yates corrected  for a 2x2 table: are not 6.0500 2 independent. 5.5846 O  E  0.5 29.0865 2 9.0750 8.3769  2    ij ij Eij  Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
81. 81. Contingency Table Analysis: Slide 81 Example 14-14 using the Template Note: When the contingenc y table is a 2x2, one should use the Yates correction. Since p-value = 0.000, H0 is rejected at the 0.01 level and it is concluded that the two variables are not independent. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
82. 82. 14-11 Chi-Square Test for Equality of Proportions Slide 82 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: H0: p1 = p2 = p3 = … = pc H1: Not all pi, I = 1, 2, …, c, are equal Chi-square test statistic for equal proportions: ( Oij  Eij ) 2 2   Eij i 1 j 1 r c Degrees of freedom: df = (r-1)(c-1) Expected cell count: Ri C j Eij  n Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
83. 83. 14-11 Chi-Square Test for Equality of Proportions - Extension Slide 83 The Median Test Here, the Null and alternative hypotheses are: H0: The c populations have the same median H1: 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. 84. Chi-Square Test for the Median: Example 14-16 Using the Template Slide 84 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. 85. Spearman Rank Correlation Coefficient: Example 14-11 MMI S&P100 220 151 218 150 216 148 217 149 215 147 213 146 219 152 236 165 237 162 235 161 R-MMI R-S&P Diff Diffsq 7 6 1 1 5 5 0 0 3 3 0 0 4 4 0 0 2 2 0 0 1 1 0 0 6 7 -1 1 9 10 -1 1 10 9 1 1 8 8 0 0 Sum: Slide 85 Table 11: =0.005 n . . . 7 -----8 0.881 9 0.833 10 0.794 11 0.818 . . . 4 n   di 6 64 4  1  1  i 1 1 9758794H  rejected  1  99 nn 11 1 Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
86. 86. Spearman Rank Correlation Coefficient: Example 14-11 Using the Template Slide 86 Note: The pvalues 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. 87. 14-9 A Chi-Square Test for Goodness of Fit  Slide 87 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: (Oi  Ei ) 2   Ei i 1 k 2  Compare chi-statistic with critical values from the chisquare distribution (with k-1 degrees of freedom) to test the null hypothesis Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
88. 88. Example 14-12: Goodness-of-Fit Test for the Multinomial Distribution Slide 88 The null and alternative hypotheses: H0: The probabilities of occurrence of events E1, E2...,Ek are given by p1,p2,...,pk H1: The probabilities of the k events are not as specified in the null hypothesis Assuming equal probabilities, p1= p2 = p3 = p4 =0.25 and n=80 Preference Tan Brown Maroon Black Total Observed 12 40 8 20 80 Expected(np) 20 20 20 20 80 (O-E) -8 20 -12 0 0 k ( Oi  E i ) 2    i 1 Ei 2  ( 8 ) 20 2  ( 20 ) 2  ( 12 ) 20 20 2  ( 0) 2 20  30.4   2  11.3449 ( 0.01, 3) H 0 is rejected at the 0.01 level. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
89. 89. Slide 89 Example 14-12: Goodness-of-Fit Test for the Multinomial Distribution using the Template Note: the pvalue 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. 90. Goodness-of-Fit for the Normal Distribution: Example 14-13 p(z<-1) p(-1<z<-0.44) p(-0.44<z<0) p(0<z<0.44) p(0.44<z<14) p(z>1) = 0.1587 = 0.1713 = 0.1700 = 0.1700 = 0.1713 = 0.1587 Partitioning the Standard Normal Distribution 0.1700 0.1700 0.4 0.1713 0.1713 0.3 f(z) 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. Slide 90 0.2 0.1587 0.1587 0.1 z 0.0 -5 -1 0 1 5 -0.44 0.44 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. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
91. 91. Example 14-13: Solution i 1 2 3 4 5 6 Oi 14 20 16 19 16 15 Ei 15.87 17.13 17.00 17.00 17.13 15.87 Slide 91 Oi - Ei (Oi - Ei)2 (Oi - Ei)2/ Ei -1.87 2.87 -1.00 2.00 -1.13 -0.87 3.49690 8.23691 1.00000 4.00000 1.27690 0.75690 2 : 0.22035 0.48085 0.05882 0.23529 0.07454 0.04769 1.11755 2(0.10,k-3)= 6.5139 > 1.11755 H0 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
92. 92. Example 14-13: Solution using the Template Slide 92 Note: p-value = 0.8002 > 0.01 H0 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. 93. 14-9 Contingency Table Analysis: A Chi-Square Test for Independence Slide 93 First Classification Category Second Classification Category 1 2 3 4 5 Column Total 1 O11 O21 O31 O41 O51 2 O12 O22 O32 O42 O52 3 O13 O23 O33 O43 O53 4 O14 O24 O34 O44 O54 5 O15 O25 O35 O45 O55 C1 C2 C3 C4 C5 Row Total R1 R2 R3 R4 R5 n Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
94. 94. Contingency Table Analysis: A Chi-Square Test for Independence Slide 94 A and B are independent if:P(AUB) = P(A)P(B). If the first and second classification categories are independent:Eij = (Ri)(Cj)/ Null and alternative hypotheses: H0: The two classification variables are independent of each other H1: The two classification variables are not independent Chi-square test statistic for independence: ( Oij  Eij ) 2 2   Eij i 1 j 1 r Degrees of freedom: df=(r-1)(c-1) Expected cell count: c Ri C j Eij  n Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
95. 95. Contingency Table Analysis: Example 14-14 2(0.01,(2-1)(2-1))=6.634 90 Industry Type Profit Service Nonservice (Expected) (Expected) Total 42 18 60 (Expected) 6 34 (Expected) (40*48/100)=19.2 (40*52/100)=20.8 Total 48 52 H0 is rejected at the 0.01 level and it is concluded that the two variables are not independent. (60*52/100)=31.2 Loss ij 11 12 21 22 (60*48/100)=28.8 O 42 18 6 34 E 28.8 31.2 19.2 20.8 (O-E)2 174.24 174.24 174.24 174.24 O-E 13.2 -13.2 -13.2 13.2 2: 29.0865 (O-E)2/E 6.0500 5.5846 9.0750 8.3769 Slide 95 40 100 2 Yates corrected  for a 2x2 table: 2 Oij  Eij  0.5 2    Eij   Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
96. 96. Contingency Table Analysis: Slide 96 Example 14-14 using the Template Note: When the contingenc y table is a 2x2, one should use the Yates correction. Since p-value = 0.000, H0 is rejected at the 0.01 level and it is concluded that the two variables are not independent. Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
97. 97. 14-11 Chi-Square Test for Equality of Proportions Slide 97 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: H0: p1 = p2 = p3 = … = pc H1: Not all pi, I = 1, 2, …, c, are equal Chi-square test statistic for equal proportions: ( Oij  Eij ) 2 2   Eij i 1 j 1 r c Degrees of freedom: df = (r-1)(c-1) Expected cell count: Ri C j Eij  n Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
98. 98. 14-11 Chi-Square Test for Equality of Proportions - Extension Slide 98 The Median Test Here, the Null and alternative hypotheses are: H0: The c populations have the same median H1: 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. 99. Chi-Square Test for the Median: Example 14-16 Using the Template Slide 99 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. 100. Name Religion Domicile Contact # E.Mail M.Phil (Statistics) Shakeel Nouman Christian Punjab (Lahore) 0332-4462527. 0321-9898767 sn_gcu@yahoo.com sn_gcu@hotmail.com Slide 100 GC University, . (Degree awarded by GC University) M.Sc (Statistics) Statitical Officer (BS-17) (Economics & Marketing Division) GC University, . (Degree awarded by GC University) Livestock Production Research Institute Bahadurnagar (Okara), Livestock & Dairy Development Department, Govt. of Punjab Nonparametric Methods and Chi-Square Tests (1) By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer