Answer
Confidence interval: $0.6420\lt y\lt1.1560$
We are 95% confident that the nicotine content of a randomly selected cigarette whose tar content is 12 mg is between 0.6420 and 1.1560 mg.
Work Step by Step
From problem 16 from Section 14.1:
$s_e=0.1121$
$∑(x_i-x ̅)^2=24.366^2=593.702$
$x ̅=\frac{5+16+24+24+8+9+9+24+15+18+17+5+10}{13}=14.15$
$n=13$, so:
$d.f.=n-2=11$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.201$
(According to Table VI, for d.f. = 11 and area in right tail = 0.025)
$Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {1+\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=0.899-2.201\times0.1121\sqrt {1+\frac{1}{13}+\frac{(12-14.15)^2}{593.702}}=0.6420$
$Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {1+\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=0.899+2.201\times0.1121\sqrt {1+\frac{1}{13}+\frac{(12-14.15)^2}{593.702}}=1.1560$