Answer
Confidence interval: $2.2008\lt β_1\lt3.0692$
We are 95% confident that as the age increases by 1 year the boy height increases between 2.2008 and 3.0692 inches.
Work Step by Step
$s_e=2.450$ (item (b))
To find $s_x$:
In MINITAB, enter the Age values in C1.
Select Stat -> Basic Statistics -> Display Descriptive Statistics
In Variables enter C1 and click in Statistics.
Select "Standard deviation". Click OK.
Click Ok.
$s_x=2.7198$
$\sqrt {∑(x_i-x ̅)^2}=\sqrt {n-1}s_x=\sqrt {20-1}\times2.7198=11.855$ (see note on page 684)
$n=20$, so:
$d.f.=n-2=18$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.306$
(According to Table VI, for d.f. = 18 and area in right tail = 0.025)
$Lower~bound=b_1-t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}=2.635-2.101\times\frac{2.450}{11.855}=2.2008$
$Upper~bound=b_1+t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}=2.635+2.101\times\frac{2.450}{11.855}=3.0692$
