91影视

a) Toverify that the value of Minitab's test statistic is correct,

The following table represents required data to compute test statistic value.

The test statistic value is

\({\rm{w = 41}}\)

which confirms the output.

b)

When testing null hypothesis

\({H_0}:{\mu _1} - {\mu _2} = {\Delta _0}\)

versus one of the alternative hypothesis, one could use test statistic value

\(w = \sum\limits_{i = 1}^m {{r_i}} \)

where\({r_i}\)is rank of\(\left( {{x_i} - {\Delta _0}} \right)\)in the combined sample of\(m + n{\left( {x - {\Delta _0}} \right)^\prime }{\rm{s}}\)and\({y^\prime }\)s.

The\(P\)-value depends on alternative hypothesis

\(\begin{array}{*{20}{l}}{{\rm{ Alternative Hypothesis }}}&{P{\rm{ - value }}}\\{{H_a}:{\mu _1} - {\mu _2} > {\Delta _0}}&{{P_0}(W \ge w)}\\{{H_a}:{\mu _1} - {\mu _2} < {\Delta _0}}&{{P_0}(W \le w) = {P_0}(W \ge m(m + n + 1) - w)}\\{{H_a}:{\mu _1} - {\mu _2} \ne {\Delta _0}}&{2{P_0}(W \ge \max \{ w,m(m + n + 1) - w)}\end{array}\)

Use Table\(\Lambda .14\)to determine\(P\)-values (use closest value to corresponding significance level to make conclusions).

The null hypotheses of interest, are

\({H_0}:{\mu _1} - {\mu _2} = 0\),versus alternative hypothesis

\({H_a}:{\mu _1} - {\mu _2} < 0,\)

The alternative hypothesis is lower-sided; thus, the\(P\)-value is

\(\begin{array}{l}{P_0}(W \le w) = {P_0}(W \ge m(m + n + 1) - w)\\{P_0}(W \le w) = {P_0}(W \ge 95).\end{array}\)

Using the table in appendix, for\(\alpha = 0.05,m = 8,n = 8,P\)value is

\({P_0}(W \ge 95) = 0.027,\)

the corresponding \(P\)-value is less than\(0.05\). Hence, reject null hypothesis at given significance level, the true average level for exposed infants appears to exceed the unexposed infants by more than \(25\). Do not reject null hypothesis at significance level \(0.01\).