Answer
Confidence interval: $-0.1929\lt µ_d\lt0.3789$
We are 99% confident that the population mean difference is between -0.1929 and 0.3789.
Work Step by Step
$n=6$, so:
$d.f.=n-1=5$
$level~of~confidence=(1-α).100$%
$99$% $=(1-α).100$%
$0.99=1-α$
$α=0.01$
$t_{\frac{α}{2}}=t_{0.005}=4.032$
(According to Table VI, for d.f. = 5 and area in right tail = 0.005)
$Lower~bound=d ̅-t_{\frac{α}{2}}.\frac{s_d}{\sqrt n}=0.093-4.032\times\frac{0.1737}{\sqrt {6}}=-0.1929$
$Upper~bound=d ̅+t_{\frac{α}{2}}.\frac{s_d}{\sqrt n}=0.093+4.032\times\frac{0.1737}{\sqrt {6}}=0.3789$
