Answer
Confidence interval: $0.04927\lt β_1\lt0.06579$
We are 90% confident that as the tar level in a cigarette increases by 1 mg the nicotine level increases between 0.04927 and 0.06579 mg.
Work Step by Step
From item (d):
$∑(x_i-x ̅)^2=\sqrt {n-1}s_x=\sqrt {13-1}\times7.034=24.366$
$n=13$, so:
$d.f.=n-2=11$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.90=1-α$
$α=0.1$
$t_{\frac{α}{2}}=t_{0.05}=1.796$
(According to Table VI, for d.f. = 11 and area in right tail = 0.05)
$Lower~bound=b_1-t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}=0.05753-1.796\times\frac{0.1121}{24.366}=0.04927$
$Upper~bound=b_1+t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}=0.05753+1.796\times\frac{0.1121}{24.366}=0.06579$
