Question

14.97 Study Time and Score. Following are the data on total hours studied over 2 weeks and test score at the end of the 2 weeks from Exercise 14.27.

x	10	15	12	20	8	16	14	22
y	91	81	84	74	85	80	84	80

a. Determine a point estimate for the mean test score of all beginning calculus students who study for $15$hours.
b. Find a $99\%$ confidence interval for the mean test score of all beginning calculus students who study for $15$ hours.
c. Find the predicted test score of a beginning calculus student who studies for $15$ hours.
d. Determine a $99\%$ prediction interval for the test score of a beginning calculus student who studies for $15$hours.

Short answer

(a) A point estimate for the mean test score is ${\hat{y}}_p=82.18$

(b) A $99\%$confidence interval for the mean test score is between $77.53$ and $86.84$

(c) The predicted test score is ${\hat{y}}_p=82.18$.

(d) A $99\%$prediction interval for the test score is $68.26$and $96.10$.

Step-by-step solution

Part (a) Step 1: Given information

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

Part (a) Step 2: Explanation

The table is calculated as follows:

x	y	$xy$	$x^2$	$y^2$
10	92	920	100	8464
15	81	1215	225	6561
12	84	1008	144	7056
20	74	1480	400	5476
8	85	680	64	7225
16	80	1280	256	6400
14	84	1176	196	7056
22	80	1760	484	6400
$\sum x_i=117$	$\sum y_i=660$	$\sum x_i{y}_i=9519$	$\sum {x_i}^2=1869$	$\sum {y_i}^2=54638$

Part (a) Step 3: Explanation

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

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

Part (a) Step 4: Explanation

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

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

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

$=188-112.888361$

$=75.11163895$

Part (a) Step 5: Explanation

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:
$b_1=\frac{S_w}{S_w}$
$=\frac{-133.5}{157.875}$
$=-0.845605701$

Part (a) Step 6: Explanation

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

$b_0=\frac{1}{n}\left(\sum y_i-b_1\sum x_i\right)$
$=\frac{1}{8}(660+0.8456(117\left)\right)$
$=\frac{1}{8}(758.9352)$
$\approx 94.8669$
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 ${\hat{y}}_p=82.18$.

Part (b) Step 1: Given information

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

Part (b) Step 2: Explanation

Let, $\alpha =0.01$for a $99\%$confidence interval.

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

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

$=t_{0,005}$

$=3.708$

Part (b) Step 3: Explanation

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

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

$82.18\pm 4.657360702$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 $86.84$.

Part (c) Step 1: Given information

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

Part (c) Step 2: Explanation

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 ${\hat{y}}_p=82.18$.

Part (d) Step 1: Given information

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

Part (d) Step 2: Explanation

Let,$\alpha =0.01$for a $99\%$confidence interval.

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

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

$=t_{0,005}$

$=3.708$

Part (d) Step 3: Explanation

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

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

$82.18\pm 13.92807544$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 $96.10$.