Answer
(c)
Work Step by Step
$1-\alpha=0.95$, and $df=n-1=4-1=3$, thus according to the table, $t_{\alpha/2}=3.182$.
Thus the confidence interval is: $\overline{x}\pm t_{\alpha/2}\frac{\sigma}{\sqrt n}$, which here is: $0.63\pm 3.182\frac{0.01}{\sqrt{4}}$, thus the interval is: $(0.614,0.646)$, thus the answer is (c).
