Answer
Confidence interval: $-0.79\lt µ_d\lt0.45$
We are 98% confident that the population mean difference is between -0.79 and 0.45.
Work Step by Step
$n=6$, so:
$d.f.=n-1=5$
$level~of~confidence=(1-α).100$%
$98$% $=(1-α).100$%
$0.98=1-α$
$α=0.02$
$t_{\frac{α}{2}}=t_{0.01}=3.365$
(According to Table VI, for d.f. = 5 and area in right tail = 0.025)
$Lower~bound=d ̅-t_{\frac{α}{2}}.\frac{s_d}{\sqrt n}=−0.167-3.365\times\frac{0.4502}{\sqrt {6}}=-0.79$
$Upper~bound=d ̅+t_{\frac{α}{2}}.\frac{s_d}{\sqrt n}=−0.167+3.365\times\frac{0.4502}{\sqrt {6}}=0.45$
