Answer
$T\lt T_\frac{α}{2}$: null hypothesis is rejected.
There is enough evidence to conclude that the median length of stay for
employer referrals is different from those referred by the criminal justice
system.
Work Step by Step
$H_0:M_D=0$ versus $M_D\ne0$
$D_i=X_i-107~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Rank$
$D_1=X_1-107=80-107=-27~~~~~~~~~~~~~~~~~-11$
$D_2=X_2-107=108-107=1~~~~~~~~~~~~~~~~~~~~~+1$
$D_3=X_3-107=95-107=-12~~~~~~~~~~~~~~~~~~-7$
$D_4=X_4-107=107-107=0~~~~~~~~~~~~~~~~~~~discard$
$D_5=X_5-107=89-107=-18~~~~~~~~~~~~~~~~~~-9$
$D_6=X_6-107=100-107=-7~~~~~~~~~~~~~~~~~~-4$
$D_7=X_7-107=85-107=-22~~~~~~~~~~~~~~~~~-10$
$D_8=X_8-107=102-107=-5~~~~~~~~~~~~~~~~~~-3$
$D_9=X_9-107=115-107=8~~~~~~~~~~~~~~~~~~~~+5.5$
$D_{10}=X_{10}-107=109-107=2~~~~~~~~~~~~~~~~~~~+2$
$D_{11}=X_{11}-107=99-107=-8~~~~~~~~~~~~~~~~-5.5$
$D_{12}=X_{12}-107=94-107=-13~~~~~~~~~~~~~~~~-8$
$n=11$
Two-tailed test.
$T_+=1+5.5+2=8.5$
$|T_-|=|-11-7-9-4-10-3-5.5-8|=57.5$
$T_+\lt |T_-|$. So:
Test statistic: $T=T_+=8.5$
Critical value: $T_\frac{α}{2}=10$
(According to table XII, for $n=11$ and $\frac{α}{2}=0.025$)
Since $T\lt T_\frac{α}{2}$, we reject the null hypothesis.
