Answer
Confidence interval: $0.1283\lt β_1\lt0.2955$
We are 95% confident that as the temperature increases by 1 °F the number of chirps per second increases between 0.12831 and 0.2955.
Work Step by Step
From item (d):
$\sqrt {∑(x_i-x ̅)^2}=\sqrt {n-1}s_x=\sqrt {15-1}\times6.7073=25.0964$
$n=15$, so:
$d.f.=n-2=13$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.160$
(According to Table VI, for d.f. = 13 and area in right tail = 0.025)
$Lower~bound=b_1-t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}=0.2119-2.160\times\frac{0.9715}{25.0964}=0.1283$
$Upper~bound=b_1+t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}=0.2119+2.160\times\frac{0.9715}{25.0964}=0.2955$
