Answer
Confidence interval: $2.2852\lt ŷ\lt4.3838$
We are 95% confident that the GPA of a randomly selected student who chooses a seat in the fifth row is between 2.2852 and 4.3838.
Work Step by Step
$s_e=0.5102$ (item (b))
$∑(x_i-x ̅)^2=13.948^2=194.547$ (item (f))
$x ̅=\frac{1+2+2+2+2+3+3+4+4+4+5+5+5+5+5+5+5+5+6+6+6+6+6+7+7+7+7+7+7+7+7+8+8+8+9+9+9+11}{38}=5.6579$
$n=38$, so:
$d.f.=n-2=36$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.028$
(According to Table VI, for d.f. = 7 and area in right tail = 0.025)
$Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {1+\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=3.3345-2.028\times0.5102\sqrt {1+\frac{1}{38}+\frac{(5-5.6579)^2}{194.547}}=2.2852$
$Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {1+\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=3.3345+2.028\times0.5102\sqrt {1+\frac{1}{38}+\frac{(5-5.6579)^2}{194.547}}=4.3838$
