Answer
$T\gt T_\frac{α}{2}$: null hypothesis is not rejected.
There is not enough evidence to conclude that an individual’s height and arm span are different.
Work Step by Step
$H_0:M_D=0$ versus $M_D\ne0$
Let the "height" values to be the X and the "arm spam" values to be the Y.
$D_i=X_i-Y_i~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Rank$
$D_1=X_1-Y_1=59.5-62=-2.5~~~~~~~~~~~~~~~-6$
$D_2=X_2-Y_2=69-65.5=3.5~~~~~~~~~~~~~~~~+9.5$
$D_3=X_3-Y_3=77-76=1~~~~~~~~~~~~~~~~~~~~~~~~+3$
$D_4=X_4-Y_4=59.5-63=-3.5~~~~~~~~~~~~~-9.5$
$D_5=X_5-Y_5=74.5-74=0.5~~~~~~~~~~~~~~~~+1.5$
$D_6=X_6-Y_6=63-66=-3~~~~~~~~~~~~~~~~~~~-7.5$
$D_7=X_7-Y_7=61.5-61=0.5~~~~~~~~~~~~~~~~+1.5$
$D_8=X_8-Y_8=67.5-69=-1.5~~~~~~~~~~~~~~~-4$
$D_9=X_9-Y_9=73-70=3~~~~~~~~~~~~~~~~~~~~~~~+7.5$
$D_{10}=X_{10}-Y_{10}=69-71=-2~~~~~~~~~~~~~~~~-5$
$n=10$
Two-tailed test.
$T_+=9.5+3+1.5+1.5+7.5=23$
$|T_-|=|-6-9.5-7.5-4-5|=32$
$T_+\lt |T_-|$. So:
Test statistic: $T=T_+=23$
Critical value: $T_\frac{α}{2}=8$
(According to table XII, for $n=10$ and $\frac{α}{2}=0.025$)
Since $T\gt T_\frac{α}{2}$, we do not reject the null hypothesis.
