Question

A company wants to test whether the new software is better than the old software. The following table record the efficiency of the old software and the new software (the larger the better). Old New 0.122 0.151 0.123 0.167 0.132 0.173 0.169 0.261 0.262 0.298 (a) Calculate the Mann Whitney U statistics for comparing these two sets of data. (b) Is there a significant increase in the efficiency with the new software at a = 0.05 level? Use Table 8 in the textbook to find the p-value and make a decision.

Answer 1

Objective: To test whether the new software is better than the old software

To test: H₀: There is no difference between the New and the Old software H_a: The New software is better than the Old software

The Mann Whitney U statistic is given by the formula:

$U_{Old}=n_{1}n_{2}+\frac{n_{1}(n_{1}+1)}{2}-R_{1}$ and

$U_{New}=n_{1}n_{2}+\frac{n_{2}(n_{2}+1)}{2}-R_{2}$

where, $n_{1},n_{2}$ are the no. of observations in the group Old and New respectively; and $R_{1},R_{2}$ are the sum of the ranks assigned to the sorted observations.

The critical / rejection region of the test is given by $U_{Test}\leq U_{Critical}$

(a)

Sorted Ranks Observations Old New Old New Old New 0.122 0.151 0.122 1 0.123 0.167 0.123 2 0.132 0.173 0.132 3 0.169 0.261 0.151 4 0.262 0.167 5 0.298 0.169 6 0.173 7 0.261 8 0.262 9 0.298 10 Sum of Ranks 12 43

Substituting the values,

$U_{Old}=n_{1}n_{2}+\frac{n_{1}(n_{1}+1)}{2}-R_{1}$

  $=(4)(6)+\frac{4(4+1)}{2}-12$

= 24 + 10 - 12

= 22

$U_{New}=n_{1}n_{2}+\frac{n_{2}(n_{2}+1)}{2}-R_{2}$

$=(4)(6)+\frac{6(6+1)}{2}-43$

= 24 + 21 - 43

= 2

Next, the smaller of the two values (which supports the alternate / research hypothesis) of the test statistic obtained is compared with the critical value of the test statistic.The Critical value can be obtained from the critical value table for one tailed test:

п, 6 7 8 9 2 3 5 10 11 12 13 14 15 16 17 18 19 20 4 3 0 4 1 5 0 1 2 4 6 0 2 31 5 7 7 0 2 6 8 11 4. 3 5 8 10 13 15 1 9 1 6 9 12 15 18 21 7 10 14 17 20 24 27 10 1 4 8 12 16 19 23 9 13 17 22 26 30 34 38 42 6 11 15 19 24 28 712 16 21 26 31 7 13 18 23 28 33 39 44 50 55 60 11 1 5 27 31 34 12 2 5 13 2 33 38 42 47 51 14 3 37 41 46 51 56 61 15 3 66 72 8 1319 24 30 36 9 15 20 26 32 39 9 16 22 28 35 41 4 10 17 23 30 37 44 51 58 65 72 79 4 11 18 25 32 39 47 16 3 42 48 54 60 65 71 77 83 17 3 45 51 57 64 70 76 83 89 95 109 18 3 48 55 61 68 75 82 88 95 102 19 87 94 101 109 116 123 20 54 62 69 76 84 92 100 107 114 123 130 138 O O

(b) Here, since, the observed value of the test statistic U = 2 < 3, lies in the rejection region, we may reject the null hypothesis at 5% level. Based on the given data, we may conclude that there is an increase in the efficiency with the new software.