91Ó°ÊÓ

To determine a point estimate for the mean test score of all beginning calculus students who study for $15$ hours.

The table is calculated as follows:

Since, $S_{xy}=∑x_i{y}_i-\left(∑x_i\right)\left(∑y_i\right)/n$
$=9519-\left(117\right)\left(660\right)/8$
$=9519-77220/8$
$=9519-9652.5$
$=-133.5$
Then,

$S_{xx}=∑x_i^2-{\left(∑x_i\right)}^2/n$
$=1869-(117{)}^2/8$
$=1869-13689/8$
$=1869-1711.125$
$=157.875$

The total $SST$sum of square is calculated as follows:

$S_w=∑y_i^2-{\left(∑y_i\right)}^2/n$
$=54638-(660{)}^2/8$
$=54638-435600/8$
$=54638-54450$
$=188$
The $\mathrm{SSR}$regression sum of squares is calculated as follows:

$SSR=\frac{S_m^2}{S_m}$
$=\frac{(-133.5{)}^2}{157.875}$
$=\frac{17822.25}{157.875}$
$=112.888361$
$SSE=SST-SSR$

$=188-112.888361$

$=75.11163895$

Determining the standard error of the estimate as follows:

$s_e=\sqrt{\frac{SSE}{n-2}}$
$=\sqrt{\frac{75.11163895}{8-2}}$

$=3.538164283$
Determining the slope of the regression line as follows:
$b_1=\frac{S_w}{S_w}$
$=\frac{-133.5}{157.875}$
$=-0.845605701$

Determine the value of $y-$intercept as follows:

$b_0=\frac{1}{n}\left(∑y_i-b_1∑x_i\right)$
$=\frac{1}{8}(660+0.8456(117\left)\right)$
$=\frac{1}{8}(758.9352)$
$≈94.8669$
As a result, the regression equation is ${\hat{y}}_p=94.8669-0.8456x_p$.
Substituting the value of $x_p=15$ into the regression equation yields the method for determining the value of the point estimate.
${\hat{y}}_p=94.8669-0.8456x_p$
$=94.8669-0.8456\left(15\right)$

$=82.1829$
As a result, the point estimate is ${\hat{y}}_p=82.18$.

To find a $99\%$ confidence interval for the mean test score of all beginning calculus students who study for $15$hours.

Let, $\mathrm{Î\pm }=0.01$for a $99\%$confidence interval.

Since,$n=8$
$df=n-2$
$=8-2$
$=6$
According to calculation:

$t_{a2}=t_{0,01/2}$

$=t_{0,005}$

$=3.708$

Computing the confidence interval end points for the conditional mean of the response variable is as follows:
${\hat{y}}_pÂ\pm t_{\mathrm{Î\pm }\mathrm{Î\copyright }}{s}_e\sqrt{\frac{1}{n}+\frac{{\left(x_p-∑x_i/n\right)}^2}{{\mathrm{S}}_x}}$
Note that: $x_p=15$
${\hat{y}}_p=82.18$
$s_e=3.54$
$S_{xx}=157.875$
Hence,

$82.18Â\pm 3.708(3.54)\sqrt{\frac{1}{8}+\frac{(15-117/8{)}^2}{157.875}}$
$82.18Â\pm 13.12632\sqrt{0.125890736}$
Also,

$82.18Â\pm 4.657360702$Or $77.53$to$86.84$

As a result, a $99\%$confident that the mean test score of all beginning calculus students who study for $15$hours is somewhere between $77.53$and $86.84$.

To find the predicted test score of a beginning calculus student who studies for $15$ hours.

Let, the regression equation is ${\hat{y}}_p=94.8669-0.8456x_p$
By substituting the value of $x_p=15$ in the regression equation, the projected value is obtained.
${\hat{y}}_p=94.8669-0.8456x_p$
$=94.8669-0.8456\left(15\right)$
$=82.1829$
As a result, the point estimate is ${\hat{y}}_p=82.18$.

To determine a $99\%$prediction interval for the test score of a beginning calculus student who studies for $15$ hours.

Let,$\mathrm{Î\pm }=0.01$for a $99\%$confidence interval.

Since, $n=8$.
$df=n-2$
$=8-2$
$=6$
According to calculation:

$t_{a2}=t_{0,01/2}$

$=t_{0,005}$

$=3.708$

Determining the end points of the prediction interval for the response variable's value is as follows:
${\hat{y}}_pÂ\pm t_{{\mathrm{Î\pm }}_2{s}_p}\sqrt{1+\frac{1}{n}+\frac{{\left(x_p-∑x_i/n\right)}^2}{{\mathrm{S}}_x}}$
Note that: $x_p=15$
${\hat{y}}_p=82.18$
$s_e=3.54$
$S_{xx}=157.875$
Hence,

$82.18Â\pm 3.708(3.54)\sqrt{1+\frac{1}{8}+\frac{(15-117/8{)}^2}{157.875}}$
$82.18Â\pm 13.12632\sqrt{1+0.125890736}$
Also,

$82.18Â\pm 13.92807544$Or $68.26$to $96.10$
As a result, $99\%$certain that the test score of all beginning calculus students who study for $15$hours will be somewhere between $68.26$and $96.10$.