Answer
Confidence interval: $47.1095\lt y\lt49.1805$
We are 90% confident that the mean height of a 7-year-old boy is between 47.1095 and 49.1805 inches.
Work Step by Step
$s_e=2.450$ (item (b))
$∑(x_i-x ̅)^2=11.855^2=140.541$ (item (d))
$x ̅=\frac{2+2+2+3+3+4+4+5+5+5+6+7+7+8+8+8+9+9+10+10}{20}=5.85$
$n=20$, so:
$d.f.=n-2=18$
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.90=1-α$
$α=0.1$
$t_{\frac{α}{2}}=t_{0.05}=1.734$
(According to Table VI, for d.f. = 18 and area in right tail = 0.05)
$Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=48.145-1.734\times2.450\sqrt {\frac{1}{20}+\frac{(7-5.85)^2}{140.541}}=47.1095$
$Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=48.145+1.734\times2.450\sqrt {\frac{1}{20}+\frac{(7-5.85)^2}{140.541}}=49.1805$
