Answer
$T\gt T_α$: null hypothesis is not rejected.
There is not enough evidence to conclude that the exercise program is effective.
Work Step by Step
$H_0:M_D=0$ versus $M_D\gt0$
Let the "before" to be the X and the "after" values to be the Y.
$D_i=X_i-Y_i~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Rank$
$D_1=X_1-Y_1=136-128=8~~~~~~~~~~~~~~~~+5$
$D_2=X_2-Y_2=120-111=9~~~~~~~~~~~~~~~+6.5$
$D_3=X_3-Y_3=129-129=0~~~~~~~~~~~~~~~discard$
$D_4=X_4-Y_4=143-148=-5~~~~~~~~~~~~-3.5$
$D_5=X_5-Y_5=115-110=5~~~~~~~~~~~~~~~+3.5$
$D_6=X_6-Y_6=113-112=1~~~~~~~~~~~~~~~+1.5$
$D_7=X_7-Y_7=89-98=-9~~~~~~~~~~~~~~~~-6.5$
$D_8=X_8-Y_8=122-103=19~~~~~~~~~~~~~+8.5$
$D_9=X_9-Y_9=102-103=-1~~~~~~~~~~~~-1.5$
$D_{10}=X_{10}-Y_{10}=122-103=19~~~~~~~~~+8.5$
$n=9$
Right-tailed test.
Test statistic: $T=|T_-|=|-3.5-6.5-1.5|=11.5$
Critical value: $T_α=8$
(According to table XII, for $n=9$ and $α=0.05$)
Since $T\gt T_α$, we do not reject the null hypothesis.
