91Ó°ÊÓ

The given data is

Regression $t-$interval

For a $99\%$confidence interval,

$\mathrm{Î\pm }=0.01$

$n=8$

$df=n-2$

$$\begin{gathered} =8-2 \\ =6 \end{gathered}$$

$t_{\mathrm{Î\pm }/2}=t_{0.01/2}$

$$\begin{gathered} =t_{0.005} \\ =3.708 \end{gathered}$$

The formula for calculating the end points of the confidence interval for ${\mathrm{β}}_1$are

localid="1652464634241" $b_1Â\pm t_{a/2}\frac{s_e}{\sqrt{S_{xx}}}$

Computation table

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

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

$=9519-77220/8$

$$\begin{gathered} =9519-9652.5 \\ =-133.5 \end{gathered}$$

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

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

$=1869-13689/8$

$$\begin{gathered} =1869-1711.125 \\ =157.875 \end{gathered}$$

The total sum of square SST is

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

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

$=54638-435600/8$

$$\begin{gathered} =54638-54450 \\ =188 \end{gathered}$$

The regression sum of squares SSR is

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

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

$=\frac{17822.25}{157.875}$

$=112.888361$

$SSE=SST-SSR$

$$\begin{gathered} =118-112.888361 \\ =75.11163895 \end{gathered}$$

The slope of the regression line is

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

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

$=-0.845605701$

The standard error of the estimate is

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

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

$$\begin{gathered} =3.538164283 \\ ≈3.54 \end{gathered}$$

Substituting the values

$-0.845605701Â\pm 3.708×\frac{3.54}{\sqrt{157.875}}$

localid="1652464398861" $-0.845605701Â\pm 1.04414554,Or-1.89to0.20$

We may be $99\%$positive that the change in mean test score per hour studied for students in beginning calculus courses is between $-1.89$and$0.20$points.