Answer
Confidence interval: $68.93\lt ŷ\lt169.64$
We are 95% confident that the weight of an American black bear that is 154.5 cm long is between 68.93 and 169.64 kg.
Work Step by Step
From problem 18 from Section 14.1:
$s_e=20.8577$ (see 18b: S in the Model Summary)
$∑(x_i-x ̅)^2=(\frac{s_e}{SE~Coef})^2=(\frac{20.8577}{0.541})^2=1486.41$ (see 18c)
$x ̅=\frac{139.0+138.0+139.0+120.5+149.0+141.0+141.0+150.0+166.0+151.5+129.5+150.0}{12}=142.88$
$n=12$, so:
$d.f.=n-2=10$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.228$
(According to Table VI, for d.f. = 10 and area in right tail = 0.025)
$Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {1+\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=119.283-2.228\times20.8577\sqrt {1+\frac{1}{12}+\frac{(154.5-142.88)^2}{1486.41}}=68.93$
$Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {1+\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=119.283+2.228\times20.8577\sqrt {1+\frac{1}{12}+\frac{(154.5-142.88)^2}{1486.41}}=169.64$
