Answer
Confidence interval: $10.838\lt y\lt12.836$
We are 95% confident that the mean value of $y$ when $x=7$ is between 10.838 and 12.836.
Work Step by Step
From problem 5 from Section 14.1:
$s_e=0.5315$
$∑(x_i-x ̅)^2=(\sqrt {5-1}\times2.073)^2=17.1893$
$x ̅=5.4$ (Problem 7d, section 4.2)
$n=5$, so:
$d.f.=n-2=3$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=3.182$
(According to Table VI, for d.f. = 3 and area in right tail = 0.025)
$Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=11.837-3.182\times0.5315\sqrt {\frac{1}{7}+\frac{(7-5.4)^2}{17.1893}}=10.838$
$Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=11.837+3.182\times0.5315\sqrt {\frac{1}{5}+\frac{(7-5.4)^2}{17.1893}}=12.836$
