Answer
$-t_{\frac{α}{2}}\lt t_0\lt t_{\frac{α}{2}}$: null hypothesis is not rejected.
There is not enough evidence to conclude that $µ_d\ne0$.
Work Step by Step
$H_0:~µ_d=0$ versus $H_1:~µ_d\ne0$
$t_0=\frac{d ̅ }{\frac{s_d}{\sqrt n}}=\frac{-0.2}{\frac{0.526}{\sqrt 7}}=-1.006$
$n=7$, so:
$d.f.=n-1=6$
Two-tailed test:
$t_{\frac{α}{2}}=t_{0.005}=3.707$
(According to Table VI, for d.f. = 6 and area in right tail = 0.005)
Also, $-t_{\frac{α}{2}}=-3.707$
Since $-t_{\frac{α}{2}}\lt t_0\lt t_{\frac{α}{2}}$, we do not reject the null hypothesis.
