Chapter 15 - Section 15.5 - Assess Your Understanding - Applying the Concepts - Page 768: 9

$w_{\frac{α}{2}}\lt T\lt w_{1-\frac{α}{2}}$: null hypothesis is not rejected. There is not enough evidence to conclude that the median sociability score for women is different from the median sociability score for men.

$H_0:M_{women}=M_{men}$ versus $H_1:M_{women}\ne M_{men}$ $Score~~~~~~Sample~~~~~~Rank$ $~~~~3~~~~~~~~~~women~~~~~~~~~~~1$ $~~~~6~~~~~~~~~~women~~~~~~~~~~~2$ $~~~~7~~~~~~~~~~women~~~~~~~~~~3.5$ $~~~~7~~~~~~~~~~women~~~~~~~~~~3.5$ $~~~~8~~~~~~~~~~~~men~~~~~~~~~~~~~~5$ $~~~~9~~~~~~~~~~women~~~~~~~~~~6.5$ $~~~~9~~~~~~~~~~~~men~~~~~~~~~~~~~6.5$ $~~~10~~~~~~~~~women~~~~~~~~~~8.5$ $~~~10~~~~~~~~~~~men~~~~~~~~~~~~~8.5$ $~~~11~~~~~~~~~women~~~~~~~~~~10$ $~~~12~~~~~~~~~women~~~~~~~~~11.5$ $~~~12~~~~~~~~~~~men~~~~~~~~~~~~11.5$ $~~~13~~~~~~~~~women~~~~~~~~~14.5$ $~~~13~~~~~~~~~~~men~~~~~~~~~~~~14.5$ $~~~13~~~~~~~~~~~men~~~~~~~~~~~~14.5$ $~~~13~~~~~~~~~~~men~~~~~~~~~~~~14.5$ $~~~14~~~~~~~~~women~~~~~~~~~17.5$ $~~~14~~~~~~~~~~~men~~~~~~~~~~~~17.5$ $~~~15~~~~~~~~~women~~~~~~~~~20.5$ $~~~15~~~~~~~~~women~~~~~~~~~20.5$ $~~~15~~~~~~~~~~~men~~~~~~~~~~~~20.5$ $~~~15~~~~~~~~~~~men~~~~~~~~~~~~20.5$ $~~~16~~~~~~~~~women~~~~~~~~~~~23$ $~~~17~~~~~~~~~women~~~~~~~~~24.5$ $~~~17~~~~~~~~~~~men~~~~~~~~~~~~24.5$ $~~~18~~~~~~~~~women~~~~~~~~~26.5$ $~~~18~~~~~~~~~~~men~~~~~~~~~~~~26.5$ $~~~20~~~~~~~~~~~men~~~~~~~~~~~~~~28$ Small-sample case: $S=1+2+3.5+3.5+6.5+8.5+10+11.5+14.5+17.5+20.5+20.5+23+24.5+26.5=193.5$ $T=S-\frac{n_1(n_1+1)}{2}=193.5-\frac{15(15+1)}{2}=193.5-120=73.5$ Critical values: $w_{\frac{α}{2}}=w_{0.025}=55$ (According to table XIII, for $n_1=15$, $n_2=13$ and $\frac{α}{2}=0.025$) $w_{1-\frac{α}{2}}=n_1n_2-w_{\frac{α}{2}}$ $w_{0.975}=15\times13-55=140$ Since $w_{\frac{α}{2}}\lt T\lt w_{1-\frac{α}{2}}$, we do not reject the null hypothesis.