Answer
Confidence interval: $-0.1791\lt β_1\lt-0.0307$
We are 95% confident that $β_1$ is between -0.1791 and -0.0307
Work Step by Step
From item (d):
$\sqrt {∑(x_i-x ̅)^2}=\sqrt {n-1}s_x=\sqrt {38-1}\times2.293=13.948$
$n=38$, so:
$d.f.=n-2=36$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.028$
(According to Table VI, for d.f. = 36 and area in right tail = 0.025)
$Lower~bound=b_1-t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}=-0.1049-2.028\times\frac{0.5102}{13.948}=-0.1791$
$Upper~bound=b_1+t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}=-0.1049+2.028\times\frac{0.5102}{13.948}=-0.0307$
