Answer
$T\geq T_α$: null hypothesis is not rejected.
There is not enough evidence to conclude that the clarity of the lake is improving.
Work Step by Step
$H_0:M_D=0$ versus $M_D\lt0$
Let the "initial depth" values to be the X and the "depth 5 years later" values to be the Y.
$D_i=X_i-Y_i~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Rank$
$D_1=X_1-Y_1=38-52=-14~~~~~~~~~~~~~~~~~~-8$
$D_2=X_2-Y_2=58-60=-2~~~~~~~~~~~~~~~~~~~-2.5$
$D_3=X_3-Y_3=65-72=-7~~~~~~~~~~~~~~~~~~~~-5$
$D_4=X_4-Y_4=74-72=2~~~~~~~~~~~~~~~~~~~~~~+2.5$
$D_5=X_5-Y_5=56-54=2~~~~~~~~~~~~~~~~~~~~~~+2.5$
$D_6=X_6-Y_6=36-48=-12~~~~~~~~~~~~~~~~~~-7$
$D_7=X_7-Y_7=56-58=-2~~~~~~~~~~~~~~~~~~~-2.5$
$D_8=X_8-Y_8=52-60=-8~~~~~~~~~~~~~~~~~~~~-6$
$n=8$
Left-tailed test.
Test statistic: $T=T_+=2.5+2.5=5$
Critical value: $T_α=5$
(According to table XII, for $n=12$ and $α=0.05$)
Since $T\geq T_α$, we do not reject the null hypothesis.
