Answer
Confidence interval: $-0.69\lt µ_d\lt0.29$
We are 95% confident that the population mean difference is between -0.69 and 0.29.
Work Step by Step
$n=7$, so:
$d.f.=n-1=6$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.447$
(According to Table VI, for d.f. = 6 and area in right tail = 0.025)
$Lower~bound=d ̅-t_{\frac{α}{2}}.\frac{s_d}{\sqrt n}=−0.2-2.447\times\frac{0.526}{\sqrt {7}}=-0.69$
$Upper~bound=d ̅+t_{\frac{α}{2}}.\frac{s_d}{\sqrt n}=−0.2+2.447\times\frac{0.526}{\sqrt {7}}=0.29$
