In each of Exercises 14.64-14.69, apply Procedure 14.2 an page 567 to find and interpret a confidence interval, at the specified confidence level for the slope of the population regression line that relates rite response variable to the predicter variable.

Study Time and Score. Refer to Exercise 14.63; $99\%$.

STEP 1: $\alpha =0.01$for a $99\%$confidence interval. Given that $\mathrm{n}=8$,

$\mathrm{df}=n-2$

$=8-2$

$=6$

Because of technology, $t_{\omega 2}=t_{0.01/2}=t_{0.005}=3.708$

STEP 2:

For ${\beta }_1$, the formula for finding the end points of the confidence interval is

$b_1\pm t_{\alpha 2}\times \frac{s_e}{\sqrt{S_{xx}}}$

$S_{xy}=\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n$

$=9519-\left(117\right)\left(660\right)/8$

$=9519-77220/8$

$=9519-9652.5$

$=-133.5$

$S_{xx}=\sum x_i^2-{\left(\sum x_i\right)}^2/n$

$=1869-(117{)}^2/8$

$=1869-13689/8$

$=1869-1711.125$

$=157.875$

The entire amount of square SST is calculated as follows:

$S_{yy}=\sum y_i^2-{\left(\sum y_i\right)}^2/n$

$=54638-(660{)}^2/8$

$=54638-435600/8$

$=54638-54450$

$=188$

SSR stands for regression sum of squares.

$SSR=\frac{{S_{xy}}^2}{S_{xx}}$

$=\frac{(-133.5{)}^2}{157.875}$

$=\frac{17822.25}{157.875}$

$=112.888361$

SSE $=$SST $-$SSR

$=188-112.888361$

$=75.11163895$

The slope of the regression line is calculated using the formula,

$b_1=\frac{S_{xy}}{S_{xx}}$

$=\frac{-133.5}{157.875}$

$=-0.845605701$

The standard error of the estimate is calculated using this formula:

$s_4=\sqrt{\frac{SSE}{n-2}}$

$=\sqrt{\frac{75.11163895}{8-2}}$

$=3.538164283$

$=3.54$

We have $b=-0.845605701$

$s_e=3.54$

$S\_\left\{z\right\}=157.875,$

So, $=0.845605701\pm 3.708\times \frac{3.54}{\sqrt{157,875}}$

Or $-0.845605701\pm 1.04414554$, or $-1.89\$to\$0.20$

STEP 3:

Interretation: We can be $99\%$certain that the change in mean test score per hour studied for students in introductory calculus courses is between - $1.89$ and $0.20$ points.