Answer
$t_0\lt -t_α$: null hypothesis is not rejected.
There is not enough evidence to conclude that the treatment is effective in increasing the value of the response variable.
Work Step by Step
- Mean;
- Dependent.
$d_i=X_i-Y_i$
$d_1=-2$
$d_2=2$
$d_3=-5$
$d_4=-3$
$d_5=0$
$d ̅=\frac{∑d_i}{n}=-1.6$
$s_d=\sqrt {\frac{∑(d_i-d ̅)^2}{n-1}}=2.702$
$H_0:~µ_d=0$ versus $H_1:~µ_d\lt0$
$t_0=\frac{d ̅ }{\frac{s_d}{\sqrt n}}=\frac{-1.6}{\frac{2.702}{\sqrt 5}}=-1.324$
$n=5$, so:
$d.f.=n-1=4$
Left-tailed test:
$t_α=t_{0.05}=2.132$
(According to Table VI, for d.f. = 4 and area in right tail = 0.05)
So, $-t_α=-2.132$
Since $t_0\gt -t_α$, we do not reject the null hypothesis.
