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 there is a difference in the measurement of the muzzle velocity between device A and device B.
Work Step by Step
$d_i=A_i-B_i$
$d_1=0.6$
$d_2=-0.2$
$d_3=-0.2$
$d_4=0.2$
$d_5=-0.2$
$d_6=0.8$
$d_7=0.1$
$d_8=-0.1$
$d_9=1.1$
$d_{10}=-0.3$
$d_{11}=-0.4$
$d_{12}=0$
$d ̅=\frac{∑d_i}{n}=0.1167$
$s_d=\sqrt {\frac{∑(d_i-d ̅)^2}{n-1}}=0.4745$
$H_0:~µ_d=0$ versus $H_1:~µ_d\ne0$
$t_0=\frac{d ̅ }{\frac{s_d}{\sqrt n}}=\frac{0.1167}{\frac{0.4745}{\sqrt {12}}}=0.852$
$n=12$, so:
$d.f.=n-1=11$
Two-tailed test:
$t_{\frac{α}{2}}=t_{0.005}=3.106$
(According to Table VI, for d.f. = 6 and area in right tail = 0.005)
Also, $-t_{\frac{α}{2}}=-3.106$
Since $-t_{\frac{α}{2}}\lt t_0\lt t_{\frac{α}{2}}$, we do not reject the null hypothesis.
