91影视

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
Alternative Hypothesis\(P\)-value

\(\begin{array}{l}{H_a}:{\mu _1} - {\mu _2} > {\Delta _0}{P_0}(W \ge w)\\{H_a}:{\mu _1} - {\mu _2} < {\Delta _0}{P_0}(W \le w) = {P_0}(W \ge m(m + n + 1) - w)\\{H_a}:{\mu _1} - {\mu _2} \ne {\Delta _0}2{Y_0}(W \ge \max \{ w,m(m + n + 1) - w)\end{array}\)

Use Table\(A.14\)to determine\({P^2}\)-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

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

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

The alternative hypothesis is two-sided; thus, the-value is

\(2{P_0}\left( {W \ge \max \{ w,m(m + n + 1) - w) = 2{P_0}(W \ge 37)} \right.\)

Using the table in appendix, for\(\alpha = 0.05\), since\(w = 37 < {W_{m,n,\alpha /2}} = 61\)

the corresponding\(P\)-value is larger than\(0.05\).

Hence, do not reject null hypothesis at given significance level, the mean burning time of two types of wood is the same.