Question 15.6: Refer to Example 5 and test the hypothesis H0: F = G against......
Refer to Example 5 and test the hypothesis H_{0}\colon F=G against the alternatives H_{A}^{\prime\prime}\colon F\neq G and H_{A}\colon F\gt G.
In Example 5 we saw that R_{X}=26\ (\mathrm{and~}R_{Y}=19) Since m = 5 and n = 4, relation (20) gives: U = 11.
U=R_{X}-\frac{m(m+1)}{2}=m n+\frac{n(n+1)}{2}-R_{Y}.~~~~~~~~~~~~~~~~(20)
From the tables cited above, we have: P(U ≤ 2) = P(U ≥ 17) = 0.032. So, C_{1}=2,\ C_{2}=17,\operatorname{and}\ H_{0} is rejected in favor of H_{A}^{\prime\prime} at the level of significance 0.064. From the same tables, we have that P(U ≤ 3) = 0.056, so that C = 3, and \textstyle H_{0} is rejected in favor of \textstyle H_{A} at level of significance 0.056. The results stated in (23) are formulated as a lemma below.
E U={\frac{m n}{2}},\quad V\! a r(U)={\frac{m n(m+n+1)}{12}}.\qquad\qquad\qquad(23)
LEMMA 2
With U defined by (19), the relations in (23) hold true under H_0.
U=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n}u(X_{i}-Y_{j}).\qquad\qquad\qquad\qquad\qquad(19)LEMMA 3
With U, EU, and Var(U) defined, respectively, by (19) and (23),