Answer
Confidence interval: $17.12\lt y\lt17.28$
We are 95% confident that the mean head circumference of children who are 25.75 inches tall is between 17.12 and 17.28 inches.
Work Step by Step
From problem 13 from Section 14.1:
$s_e=0.0954$
$∑(x_i-x ̅)^2=(\sqrt {11-1}×1.094)^2=11.968$
$x ̅=\frac{27.75+24.5+25.5+26+25+27.75+26.5+27+26.75+26.75+27.5}{11}=26.455$
$n=11$, so:
$d.f.=n-2=9$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.262$
(According to Table VI, for d.f. = 9 and area in right tail = 0.025)
$Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=17.20-2.262\times0.0954\sqrt {\frac{1}{11}+\frac{(25.75-26.455)^2}{11.968}}=17.12$
$Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=17.20+2.262\times0.0954\sqrt {\frac{1}{11}+\frac{(25.75-26.455)^2}{11.968}}=17.28$
