Answer
Confidence interval: $0.8272\lt y\lt0.9708$
We are 95% confident that the mean nicotine content of all cigarettes whose
tar content is 12 mg is between 0.8272 and 0.9708 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)
$Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=0.899-2.201\times0.1121\sqrt {\frac{1}{13}+\frac{(12-14.15)^2}{593.702}}=0.8272$
$Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=0.899+2.201\times0.1121\sqrt {\frac{1}{13}+\frac{(12-14.15)^2}{593.702}}=0.9708$
