Answer
$t_0\gt t_α$: null hypothesis is rejected.
There is enough evidence to conclude that the treatment is effective in decreasing the value of the response variable.
Work Step by Step
- Mean;
- Dependent.
$d_i=X_i-Y_i$
$d_1=2$
$d_2=-1$
$d_3=5$
$d_4=0$
$d_5=5$
$d ̅=\frac{∑d_i}{n}=2$
$s_d=\sqrt {\frac{∑(d_i-d ̅)^2}{n-1}}=2.550$
$H_0:~µ_d=0$ versus $H_1:~µ_d\gt0$
$t_0=\frac{d ̅ }{\frac{s_d}{\sqrt n}}=\frac{2}{\frac{2.550}{\sqrt 5}}=1.754$
$n=5$, so:
$d.f.=n-1=4$
Right-tailed test:
$t_α=t_{0.10}=1.533$
(According to Table VI, for d.f. = 4 and area in right tail = 0.10)
Since $t_0\gt t_α$, we reject the null hypothesis.
