Answer
Confidence interval: $-0.1343\lt β_1\lt0.1051$
We are 95% confident that $β_1$ is between -0.1343 and 0.1051.
Work Step by Step
From item (e):
$∑(x_i-x ̅)^2=\sqrt {n-1}s_x=\sqrt {14-1}\times47.330=170.651$
$n=14$, so:
$d.f.=n-2=12$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.179$
(According to Table VI, for d.f. = 12 and area in right tail = 0.025)
$Lower~bound=b_1-t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}=−0.0146-2.179\times\frac{9.375}{170.651}=-0.1343$
$Upper~bound=b_1+t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}=−0.0146+2.179\times\frac{9.375}{170.651}=0.1051$
