Non-Parametric Tests of Significance, Second Set of Exercises Vartanian: SW 540

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1 Non-Parametric Tests of Significance, Second Set of Exercises Vartanian: SW You are examining the rankings of players on two rugby teams. Is there a significant difference in the two teams? Use a 5% test. Sample 1: Sample 2: You are examining the rankings of relative crime levels of residents in two neighborhoods. Is there a significant difference in the two groups? Use a 5% test. Sample 1: Sample 2: You are examining the rankings of cities in two areas of the country for air pollution. Is there a difference in the two? Use a 1% test. Sample 1: 50 observations Sample 2: 50 observations Number of runs=22 4. You are examining the socio-economic status (an ordinal scale variable) for those people in high paying and low paying jobs. Is there a difference in socio-economic status for the two groups? Use a one-tailed, 5% test. Socio-Economic Level Low Paying Jobs High Paying Jobs N Cumul. % N Cumul. % Lower-Lower Upper-Lower Lower-Middle Upper-Middle Lower-Upper Upper-Upper D:\WP60\LECT1.PHD\PROBSETS\nonparpro.b2.wpd Page 1

2 Total You are examining the socio-economic status (an ordinal scale variable) for those people in high paying and low paying jobs. Is there a difference in socio-economic status for the two groups? Use a one-tailed, 5% test. Socio-Economic Level Low Paying Jobs High Paying Jobs N Cumul. % N Cumul. % Lower-Lower Upper-Lower Lower-Middle Upper-Middle Lower-Upper Upper-Upper Total You are using a matched pairs experiment to determine if a particular type of treatment is effective for the people who get a little crazed at the thought of a final exam. You cannot assume normality within this slightly strange population. The following tables gives anxiety levels (an interval scale variable) after the experiment with a control for previous levels of anxiety (this is how the individuals were matched). Pair No. Group 1: Control Group 2: Treatment Rank of Negative Rank D:\WP60\LECT1.PHD\PROBSETS\nonparpro.b2.wpd Page 2

3 Total 7. You are using a matched pairs experiment to determine if a particular type of treatment is effective for the alcoholics who wish to stop drinking. You cannot assume normality within this population. The following tables gives the number of drinks (an interval scale variable) after the experiment with a control for previous levels of drinking (this is how the individuals were matched). Pair No. Group 1: Control Group 2: Treatment Rank of Negative Rank Total 8. You are examining two ordinal level variables to determine the correlation between these variables. You also wish to determine if there is a significant relationship between the variables. Use both methods to determine the correlation coefficient and the level of significance. School Student Depression Levels Faculty Depression Levels Univ. of Wisconsin 1 2 Univ. of Michigan 2 3 Univ. of California 7 6 Penn State 5 4 Harvard 8 1 Penn 6 5 Yale 4 7 Univ of Virginia 3 8 D:\WP60\LECT1.PHD\PROBSETS\nonparpro.b2.wpd Page 3

4 Answers 1. There are two ways of running this test: the runs test and the Mann-Whitney test. We ll do both tests. A. Runs Test If we use A s for the first sample and B s for the second sample, we d have the following number of runs: AAAAABBBBBBBAA. So the number of runs=3 Is this significant? Look in table E. In the 7 X 7 case, you would need 4 or fewer runs to reject the null hypothesis. Here, you would reject the null hypothesis for a 5% test. B. Mann-Whitney Test If we first compare the second group to the first, we d get U= =14 U = =35 Use the lower of the U and U values and look this up in table F. A value of U=14, with an N 1 =7 and N 2 =7 corresponds to a.104 level of significance. For a 5% test, we would accept the null hypothesis that states that there is no difference between the groups. 2. You can run both the runs and Mann-Whitney tests. A. Runs test ABABABABABABAB A total of 14 runs. Look in table E, where the critical value is 4. Because the total number of runs is larger than the critical value, accept the null hypothesis. B. The Mann-Whitney Test First, compare the second sample to the first U= =21 U = =28 Use the lower of these two values, U=21 and use table F and you get a value of.355. You will thus accept the null hypothesis at the.05 level. D:\WP60\LECT1.PHD\PROBSETS\nonparpro.b2.wpd Page 4

5 3. You only have enough information for the runs test. A. Runs Expect number of runs or the mean= 2 * 50* = The standard deviation = ó r ' 2(50(50(2(50(50&50&50) (50%50) 2 (50%50&1) ' (10,000)((99) ' 24.75'4.97 Z=(22-51)/4.97 = -29/4.97= Since the critical value is 2.57 or 2.58, you will reject the null. 4. Use the Kolmogorov-Smirnov Test. D= since this is the greatest difference in the cumulative percentages for the two groups. You will then use a chi-square test to determine significance. 2 '4((.280) 2 ( 127( '4(.078( 127% '.312(78.27' at 2 Dfs, for the.001 level of significance the C.V. = Therefore reject the null hypothesis. 5. Again, use the Kolmogorov-Smirnov Test. D= '4((.462) 2 ( 155( '4(.213( 155% '.852(95.68' D:\WP60\LECT1.PHD\PROBSETS\nonparpro.b2.wpd Page 5

6 at 2 Dfs, for the.05 level of significance, the C.V.= Therefore, reject the null hypothesis that the two groups are the same. 6. Pair No. Group 1: Control Group 2: Treatment Rank of Negative Rank Total The value of 21.5 is the sum of the positive differences and the value of 15.5 is the total of the negative differences. The T value is 15.5 since it is the lower of the two scores (21.5 and 15.5). The critical value for this test (from table H) is 4 for a two-tailed, 5% test. Therefore, accept the null hypothesis. 7. Pair No. Group 1: Control Group 2: Treatment Rank of Negative Rank Total D:\WP60\LECT1.PHD\PROBSETS\nonparpro.b2.wpd Page 6

7 You ll use the value of 14 for your T score, with a critical value of 4 for a two-tailed, 5% test, or a onetailed,.025 test. Either way, you ll accept the null hypothesis. 8. Use Spearman s r and Kendall s tau to determine correlations between these two ordinal level variables. First, we ll determine Spearman s r. Use the formula: N 2 6j D i i'1 r s '1& N(N 2 &1) School Student Faculty D D 2 Univ. of Wisconsin Univ. of Michigan Univ. of California Penn State Harvard Penn Yale Univ of Virginia D 2 i '88. r s '1& 6(88. 8(63 '1&1.048'&.048 We could then determine significance by using the Z formula for r s. Z' r s &0 1/ N&1 Z' &.048 ' &.048. This Z score is not significant. Therefore, you will fail to reject the null 1/ 7.38 '&.127 hypothesis. D:\WP60\LECT1.PHD\PROBSETS\nonparpro.b2.wpd Page 7

8 For Kendall s tau, we have the following: School Student Faculty Univ. of Wisconsin 1 2 Univ. of Michigan 2 3 Univ. of California 7 6 Penn State 5 4 Harvard 8 1 Penn 6 5 Yale 4 7 Univ of Virginia 3 8 Put these in order by student depression levels: Univ. of Wisconsin a 1 2 Univ. of Michigan b 2 3 Univ of Virginia c 3 8 Yale d 4 7 Penn State e 5 4 Penn f 6 5 Univ. of California g 7 6 Harvard h 8 1 We can then determine S: (a,b)=+1 (b,c)=+1 (c,d)= -1 (d,e)= -1 (e,f)=+1 (f,g)=+1 (g,h)=-1 (a,c)=+1 (b,d)=+1 (c,e)= -1 (d,f)= -1 (e,g)=+1 (f,h)=-1 (a,d)=+1 (b,e)=+1 (c,f)= -1 (d,g)= -1 (e,h)=-1 (a,e)=+1 (b,f)=+1 (c,g)= -1 (d,h)= -1 (a,f)=+1 (b,g)=+1 (c,h)= -1 (a,g)=+1 (b,h)=-1 (a,h)= S=0. To then determine the correlation coefficient, use the formula: S ô a ' 1 2 N(N&1) The denominator is.5*8*7 = 28. The numerator is 0. The correlation coefficient is thus equal to 0. D:\WP60\LECT1.PHD\PROBSETS\nonparpro.b2.wpd Page 8

9 To then determine significance, use Z' S ó where óis: ó 2 s ' 1 18 N(N&1)(2N%5). ó 2 ' 1 18 (8(7(21'65.33 ó'8.08. Z=0/8.08 = 0. Insignificant. D:\WP60\LECT1.PHD\PROBSETS\nonparpro.b2.wpd Page 9