Question 15.7: Refer to Example 25 in Chapter 1 (see also Example 4 here), ......

Here m=n=15,\,{\ z_{0.05}}=1.645,\,{\mathrm{and}}\,z_{0.025=1.96}. Then:

C_{1}\simeq112.50-1.96\times24.105=112.50-47.2458\simeq65.254,

C_{2}\simeq112.50+47.2458\simeq159.746,

C\ \;\simeq\;112.50-1.645\times24.105\simeq112.50-39.653=72.847,

C^{\prime}\simeq112.50+39.653=152.153.

Next, comparing all 15 × 15 pairs in Table 15.3, we get the observed value of U = 185.
Therefore the hypothesis \textstyle H_{0} is rejected in favor of {H}_{A}^{\prime\prime}, since U = 185 > 159.746 = C_{2}; the null hypothesis is also rejected in favor of {H}_{A}^{\prime}, since U = 185 > 152.153 = C^{\prime}{\mathrm{;}} but the null hypothesis is not rejected when the alternative is H_{A}, because U=185\nless72.847=C.