Answer
$t_0\lt -t_α$: null hypothesis is rejected.
There is enough evidence to conclude that the clarity of the lake is improving.
Work Step by Step
$d_i=X_i-Y_i$
$d_1=-14$
$d_2=-2$
$d_3=-7$
$d_4=2$
$d_5=2$
$d_6=-12$
$d_7=-2$
$d_8=-8$
$d ̅=\frac{∑d_i}{n}=-5.125$
$s_d=\sqrt {\frac{∑(d_i-d ̅)^2}{n-1}}=6.081$
$H_0:~µ_d=0$ versus $H_1:~µ_d\lt0$
$t_0=\frac{d ̅ }{\frac{s_d}{\sqrt n}}=\frac{-5.125}{\frac{6.081}{\sqrt 8}}=-2.384$
$n=8$, so:
$d.f.=n-1=7$
Left-tailed test:
$t_α=t_{0.05}=1.895$
(According to Table VI, for d.f. = 7 and area in right tail = 0.05)
So, $-t_α=-1.895$
Since $t_0\lt -t_α$, we reject the null hypothesis.
