Answer
Confidence interval: $36.64\lt y\lt37.39$
We are 95% confident that the mean length of the right tibia of all rats 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)
$Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=37.0137-2.262\times0.494\sqrt {\frac{1}{11}+\frac{(25.83-25.34)^2}{10.837}}=36.64$
$Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=37.0137+2.262\times0.494\sqrt {\frac{1}{11}+\frac{(25.83-25.34)^2}{10.837}}=37.39$
