Answer
Confidence interval: $16.24\lt y\lt17.12$
We are 90% confident that the mean number of chirps per second when the temperature is 80.2°F is between 16.24 and 17.12.
Work Step by Step
$s_e=0.9715$ (item (b))
$∑(x_i-x ̅)^2=25.0964^2=629.8293$ (item (f))
$x ̅=\frac{88.6+93.3+80.6+69.7+69.4+79.6+80.6+76.3+71.6+84.3+75.2+82.0+83.3+82.6+83.5}{15}=80.04$
$n=15$, so:
$d.f.=n-2=13$
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.90=1-α$
$α=0.1$
$t_{\frac{α}{2}}=t_{0.05}=1.771$
(According to Table VI, for d.f. = 13 and area in right tail = 0.05)
$Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=16.68-1.771\times0.9715\sqrt {\frac{1}{15}+\frac{(80.2-80.04)^2}{629.8293}}=16.24$
$Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=16.68+1.771\times0.9715\sqrt {\frac{1}{15}+\frac{(80.2-80.04)^2}{629.8293}}=17.12$
