Answer
$T\gt T_\frac{α}{2}$: null hypothesis is not rejected.
There is not enough evidence to conclude that the median length of stay is different for males and females.
Work Step by Step
$H_0:M_D=0$ versus $M_D\ne0$
$D_i=X_i-105~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Rank$
$D_1=X_1-105=117-105=12~~~~~~~~~~~~~~~~~~~~+8$
$D_2=X_2-105=111-105=6~~~~~~~~~~~~~~~~~~~~~~+4$
$D_3=X_3-105=81-105=-24~~~~~~~~~~~~~~~~~~-12$
$D_4=X_4-105=110-105=5~~~~~~~~~~~~~~~~~~~~~~+3$
$D_5=X_5-105=103-105=-2~~~~~~~~~~~~~~~~~~~-1$
$D_6=X_6-105=94-105=-11~~~~~~~~~~~~~~~~~~-6.5$
$D_7=X_7-105=92-105=-13~~~~~~~~~~~~~~~~~~~-9$
$D_8=X_8-105=91-105=-14~~~~~~~~~~~~~~~~~~-10$
$D_9=X_9-105=116-105=11~~~~~~~~~~~~~~~~~~~+6.5$
$D_{10}=X_{10}-105=108-105=3~~~~~~~~~~~~~~~~~~~+2$
$D_{11}=X_{11}-105=98-105=-7~~~~~~~~~~~~~~~~~~-5$
$D_{12}=X_{12}-105=84-105=-21~~~~~~~~~~~~~~~~-11$
$n=12$
Two-tailed test.
$T_+=8+4+3+6.5+2=23.5$
$|T_-|=|-12-1-6.5-9-10-5-11|=54.5$
$T_+\lt |T_-|$. So:
Test statistic: $T=T_+=23.5$
Critical value: $T_\frac{α}{2}=13$
(According to table XII, for $n=12$ and $\frac{α}{2}=0.025$)
Since $T\gt T_\frac{α}{2}$, we do not reject the null hypothesis.
