Answer
Confidence interval: $-0.52\lt y\lt1.24$
We are 90% confident that the mean value of $y$ when $x=1.8$ is between -0.52 and 1.24.
Work Step by Step
From problem 8 from Section 14.1:
$s_e=0.5164$
$∑(x_i-x ̅)^2=(\sqrt {5-1}\times1.581)^2=9.998$
$x ̅=0$ (Problem 10d, section 4.2)
$n=5$, so:
$d.f.=n-2=3$
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.90=1-α$
$α=0.1$
$t_{\frac{α}{2}}=t_{0.05}=2.353$
(According to Table VI, for d.f. = 3 and area in right tail = 0.05)
$Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=0.36-2.353\times0.5164\sqrt {\frac{1}{5}+\frac{(1.8-0)^2}{9.998}}=-0.52$
$Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=0.36+2.353\times0.5164\sqrt {\frac{1}{5}+\frac{(1.8-0)^2}{9.998}}=1.24$
