Answer
Confidence interval: $2.489\lt y\lt6.071$
We are 95% confident that the mean value of $y$ when $x=7$ is between 2.489 and 6.071.
Work Step by Step
From problem 7 from Section 14.1:
$s_e=0.8944$
$∑(x_i-x ̅)^2=(\sqrt {5-1}\times1.581)^2=9.998$
$x ̅=0$ (Problem 9d, 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)
$Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=4.28-3.182\times0.8944\sqrt {\frac{1}{5}+\frac{(1.4-0)^2}{9.998}}=2.489$
$Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=4.28+3.182\times0.8944\sqrt {\frac{1}{5}+\frac{(1.4-0)^2}{9.998}}=6.071$
