Answer
$T\gt T_α$, we do not reject the null hypothesis.
There is not enough evidence to conclude that the exercise program increases flexibility.
Work Step by Step
$H_0:M_D=0$ versus $M_D\lt0$
Let the "Before" values to be the X and the "After" values to be the Y.
$D_i=X_i-Y_i~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Rank$
$D_1=X_1-Y_1=18.5-19.25=-0.75~~~~~~~~-4.5$
$D_2=X_2-Y_2=21.5-21.75=-0.25~~~~~~~~-1.5$
$D_3=X_3-Y_3=16.5-16.5=0~~~~~~~~~~~~~~~~~~discard$
$D_4=X_4-Y_4=21-20.25=0.75~~~~~~~~~~~~~~~+4.5$
$D_5=X_5-Y_5=20-22.25=-2.25~~~~~~~~~~~~-10$
$D_6=X_6-Y_6=15-16=-1~~~~~~~~~~~~~~~~~~~~~~~-6$
$D_7=X_7-Y_7=19.75-19.5=0.25~~~~~~~~~~~~+1.5$
$D_8=X_8-Y_8=15.75-17=-1.25~~~~~~~~~~~~-7.5$
$D_9=X_9-Y_9=18-19.25=-1.25~~~~~~~~~~~~-7.5$
$D_{10}=X_{10}-Y_{10}=22-19.5=2.5~~~~~~~~~~~~~~~+11$
$D_{11}=X_{11}-Y_{11}=15-16.5=-1.5~~~~~~~~~~~~~-9$
$D_{12}=X_{12}-Y_{12}=20.5-20=0.5~~~~~~~~~~~~~+3$
$n=11$
Left-tailed test.
Test statistic: $T=T_+=4.5+1.5+11+3=20$
Critical value: $T_α=13$
(According to table XII, for $n=11$ and $α=0.05$)
Since $T\gt T_α$, we do not reject the null hypothesis.
