# Let t be the smaller of like signed ranks that is t

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Let T be the smaller of like-signed ranks. That is T is either the same of the positive ranks or the sum of the negative ranks, whichever is smaller. Statistical tables gives the critical value of T at the -levels of significance. This table is reffered for a small value of n (n<25) thus, if the observed value of T is smaller than the critical value, the null hypothesis is rejected at - level of significance. When n is larger than 25, a normal approximation is used to test the hypothesis. Here, the observed value of T is assumed to follow a normal distribution with ( 1) ( ) 4 ( 1)(2 1) ( ) 24 n n E T n n n SD T Thus, ( ) ( ) T E T Z SD T Is approximately normally distributed with mean zero and unit standard deviation. Example - 3 A company wishes to determine whether employee attitudes might be improved by a series of weekly meetings at which the management presents films and lectures dealing with the company and its products. As an experiment, the company selects a groups of 30 employees, who are tested by administering a questionnaire before and after a series of such meetings. One of the questions concerned the quality of the company’s produ cts and the management wished to find out whether employee attitudes towards their products have improved. The results obtained form this sample are given in Table . Test the null hypothesis that there is no significant impact of the weekly meetings on the employee attitudes towards its products.

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115 Table 14.1 Difference in scores of 30 employees before and after the weekly meeting. Employee No. Difference Score d Rank of d Rank with less frequent sign 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 -2 1 0 1 -2 0 4 4 1 3 -1 5 -3 5 3 -1 1 5 8 2 0 2 -3 2 1 4 8 2 -3 1 -11.5 4.5 - 4.5 -11.5 - 21.0 21.0 4.5 17.0 -4.5 24.0 -17.0 24.0 17.0 -4.5 4.5 24.0 26.5 11.5 - 11.5 -17.0 11.5 4.5 21.0 26.5 11.5 -17.0 4.5 11.5 11.5 4.5 17.0 4.5 17.0 17.0 T = 83.0 d = (after score before score) Solution The null hypothesis to be tested is : there is no impact of meetings on employee attitudes. As the difference score of three employees is zero, the sample size for analysis is 27. The T value here is the sum of the negative ranks, as the negative differences are fewer in number. From table, we have T=83.0. For the given data, we get
116 ( 1) 27 28 ( ) 189.0 4 4 ( 1)(2 1) ( ) 41.62 24 n n E T n n n SD T Therefore, ( ) 83 189.0 2.55 ( ) 41.62 T E T Z SD T   Referring to the normal probability table, we find that the calculated value of Z lies in the rejection region at 5% level of significance. Hence the null hypothesis is rejected, i.e., there is an employees attitudes after the weekly meetings The procedure involved in the Wilcoxon matched pairs signed ranks test is summarized as follows; i. For each pair of observed data, determine the signed difference i d between the two scores. The pairs with 0 i d are deleted from the analysis.

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