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 the reaction time to the blue stimulus is different from the reaction time to the red stimulus.
Work Step by Step
$d_i=B_i-R_i$
$d_1=0.174$
$d_2=0.074$
$d_3=0.299$
$d_4=-0.135$
$d_5=0.229$
$d_6=-0.083$
$d ̅=\frac{∑d_i}{n}=0.093$
$s_d=\sqrt {\frac{∑(d_i-d ̅)^2}{n-1}}=0.1737$
$H_0:~µ_d=0$ versus $H_1:~µ_d\ne0$
$t_0=\frac{d ̅ }{\frac{s_d}{\sqrt n}}=\frac{0.093}{\frac{0.1737}{\sqrt {6}}}=1.311$
$n=6$, so:
$d.f.=n-1=5$
Two-tailed test:
$t_{\frac{α}{2}}=t_{0.005}=4.032$
(According to Table VI, for d.f. = 5 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.
