Answer
Confidence interval: $0.9023\lt β_1\lt1.8777$
We are 99% confident that as the right humerus increases by 1 mm, the right tibia increases between 0.9023 and 1.8777 mm.
Work Step by Step
From item (d):
$∑(x_i-x ̅)^2=\sqrt {n-1}s_x=\sqrt {11-1}\times1.041=3.292$
$n=11$, so:
$d.f.=n-2=9$
$level~of~confidence=(1-α).100$%
$99$% $=(1-α).100$%
$0.99=1-α$
$α=0.01$
$t_{\frac{α}{2}}=t_{0.005}=3.250$
(According to Table VI, for d.f. = 9 and area in right tail = 0.005)
$Lower~bound=b_1-t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}=1.390-3.250\times\frac{0.494}{3.292}=0.9023$
$Upper~bound=b_1+t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}=1.390+3.250\times\frac{0.494}{3.292}=1.8777$
