91影视

When testing null hypothesis

\({H_0}:\mu = {\mu _0}\)

Versus one of the alternative hypothesis, one could use test static value

\({s_ + }\)= the sum of the ranks associated with positive\(\left( {{x_i} = {\mu _0}} \right)'s\)

The P-value depends on the alternative hypothesis

Alternative hypothesis P-values

\(\begin{array}{l}{H_0}:\mu > {\mu _0}\\{H_0}:\mu < {\mu _0}\\{H_0}:\mu \ne {\mu _0}\end{array}\) \(\begin{array}{l}{P_0}\left( {{S_ + } \ge {s_ + }} \right)\\{P_0}\left( {{S_ + } \le {s_ + }} \right) = {P_0}\left( {{S_ + } \ge \frac{{n\left( {n + 1} \right)}}{2} - {s_ + }} \right)\\2{P_0}\left( {{S_ + } \ge \max \left\{ {{s_ + },\frac{{n\left( {n + 1} \right)}}{2} - {s_ + }} \right\}} \right)\end{array}\)

The test of interest is

\({H_0}:\mu = 30\)

Versus alternative hypothesis

\({H_a}:\mu \ne 30\)

The following table represents values required to compute the test statistic value and corresponding P-value. Value 鈥�-1鈥� represents 鈥�-鈥� as a sing (negative difference), and value 1 is 鈥�-+鈥�.

The test statistic value is,

\({s_ + }\)= the sum of the ranks associated with positive\(\left( {x\_i - \mu \_0} \right)'s\)

\(\begin{array}{l} = 3 + 1 + 9 + ... + 15 + 2\\ = 39\end{array}\)

The corresponding P-value is,

\(\begin{array}{l}{P_0}\left( {{S_ + } \le {s_ + }} \right) = {P_0}\left( {{S_ + } \ge \frac{{n\left( {n + 1} \right)}}{2} - {s_ + }} \right)\\ = {P_0}\left( {{S_ + } \ge \frac{{15\left( {14 + 1} \right)}}{2} - 39} \right)\\ = {P_0}\left( {{S_ + } \ge 2} \right)\end{array}\)

In the table in the appendix one could find only particular values ; thus, the null hypothesis should be rejected at\(\alpha = 0.1\), when

\({s_ + } \le \frac{{15.\left( {15 + 1} \right)}}{2} - 83 = 37\)

For n = 15 where 83 in the number from the table, so, since

\({s_ + } = 39 \ge 37\)

Do not reject null hypothesis

Hence, do not reject null hypothesis.