Answer
Confidence interval: $-0.31\lt µ_d\lt0.54$
We are 99% confident that the population mean difference is between -0.31 and 0.54.
Work Step by Step
$n=12$, so:
$d.f.=n-1=11$
$level~of~confidence=(1-α).100$%
$99$% $=(1-α).100$%
$0.99=1-α$
$α=0.01$
$t_{\frac{α}{2}}=t_{0.005}=3.106$
(According to Table VI, for d.f. = 11 and area in right tail = 0.005)
$Lower~bound=d ̅-t_{\frac{α}{2}}.\frac{s_d}{\sqrt n}=0.1167-3.106\times\frac{0.4745}{\sqrt {12}}=-0.31$
$Upper~bound=d ̅+t_{\frac{α}{2}}.\frac{s_d}{\sqrt n}=0.1167+3.106\times\frac{0.4745}{\sqrt {12}}=0.54$
