91Ó°ÊÓ

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

$\mathrm{df}=n-2$

$=8-2$

$=6$

Because of technology, $t_{\mathrm{Ó\neg }2}=t_{0.01/2}=t_{0.005}=3.708$

STEP 2:

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

$b_1Â\pm t_{\mathrm{Î\pm }2}×\frac{s_e}{\sqrt{S_{xx}}}$

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

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

$=9519-77220/8$

$=9519-9652.5$

$=-133.5$

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

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

$=1869-13689/8$

$=1869-1711.125$

$=157.875$

The entire amount of square SST is calculated as follows:

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

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

$=54638-435600/8$

$=54638-54450$

$=188$

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, $=0.845605701Â\pm 3.708×\frac{3.54}{\sqrt{157,875}}$

Or $-0.845605701Â\pm 1.04414554$, 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.