Answer
Confidence interval: $35.83\lt ŷ\lt38.19$
We are 95% confident that the right tibia of a randomly selected rat whose right humerus is 25.83 mm is between 36.64 and 37.39 inches.
Work Step by Step
From problem 14 from Section 14.1:
$s_e=0.494$
$∑(x_i-x ̅)^2=3.292^2=10.837$
$x ̅=\frac{24.80+24.59+24.59+24.29+23.81+24.87+25.90+26.11+26.63+26.31+26.84}{11}=25.34$
$n=11$, so:
$d.f.=n-2=9$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.262$
(According to Table VI, for d.f. = 9 and area in right tail = 0.025)
$Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {1+\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=37.0137-2.262\times0.494\sqrt {1+\frac{1}{11}+\frac{(25.83-25.34)^2}{10.837}}=35.83$
$Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {1+\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=37.0137+2.262\times0.494\sqrt {1+\frac{1}{11}+\frac{(25.83-25.34)^2}{10.837}}=38.19$
