91影视

To define:

\({S_ + } = \)the number of averages,\(\frac{{{X_ - }i + {X_ - }j}}{2},i \le j\)larger than\({\mu _0}\).
At significance level\(\alpha \), with a rejection criteria

\({s_ + } \ge c\)and \({s_ + } \le \frac{{n(n + 1)}}{2} - c\).

when testing null hypothesis\({H_0}:\mu = {\mu _0}\)versus alternative\({H_a}:\mu \ne {\mu _0}\)using Wilcoxon signed-rank test, a\(100(1 - \alpha )\)confidence interval for\(\mu \)is

\(\left( {{{\bar x}_{(n(n + 1)/2 - c + 1)}},{{\bar x}_{(c)}}} \right)\).

Use Appendix Table A.15 to determine values of\(c\).
For\(n = 4\), and the Table\(A.13\), two possible significance levels would be\(P\left( {{S_ + } \ge c} \right) = 0.125\)

a\(75\% \)confidence interval\((2 \cdot 0.125 = 0.25)\), and

\(P\left( {{S_ + } \ge c} \right) = 0.062\)

a\(87.6\)\(\% \)confidence interval\((2 \cdot 0.062 = 0.142)\), where one performs two-tailed test. From the Table A.15, a confidence interval is

\(\begin{array}{l}\left( {{{\bar x}_{(n(n + 1)/2 - c + 1)}},{{\bar x}_{(c)}}} \right) = \left( {{{\bar x}_{(4 \cdot (4 + 1)/2 - 10 + 1)}},{{\bar x}_{(10)}}} \right)\\ = \left( {{{\bar x}_{(1)}},{{\bar x}_{(10)}}} \right)\\ = (0.045,0.177)\end{array}\)

One should compute all averages and find\({1^{st}}\)and\({10^{{\rm{th }}}}\)ordered value which are easy to find.

Note: \(\frac{{n(n + 1)}}{2} = 10.\)