Question

14.93 Corvette Prices. Following are the age and price data for Corvettes from Exercise 14.23.

x	6	6	6	2	2	5	4	5	1	4
y	290	280	295	425	384	315	355	328	425	325

a. Obtain a point estimate for the mean price of all 4-year-old Corvettes.

b. Determine a $90\%$ confidence interval for the mean price of all 4-year-old Corvettes.

c. Find the predicted price of a 4-year-old Corvette.
d. Determine a $90\%$ prediction interval for the price of a 4 -year-old Corvette.
e. Draw graphs similar to those in Fig. 14.11 on page 576 , showing
f. Why is the prediction interval wider than the confidence interval?

Short answer

(a) A point estimate for the mean price of all 4-year-old Corvettes is ${\hat{y}}_p=344.99$

(b) A $90\%$confidence interval for the mean price of all 4-year-old Corvettes is between $\$336.6$to $\$353.38$.

(c) The predicted price of a 4- year-old Corvette is ${\hat{y}}_p=344.99$

(d) The predicted interval for the price of a 4- year -old Corvette is $\$31720$to $\$37278$

(e) The graphs similar to those in Fig. 14.11 on page 576 as:

(f) The error in the estimate is due to the sample regression line's estimation of the population regression line, and the error in the prediction is due to the mistake in estimating the mean price plus price variance.

Step-by-step solution

Part (a) Step 1: Given information

To find the predicted price of a $4$year-old Corvette. The age and price data for Corvettes are:

x	6	6	6	2	2	5	4	5	1	4
y	290	280	295	425	384	315	355	328	425	325

Part (a) Step 2: Explanation

Table computation as:

x	y	$xy$	$x^2$	$y^2$
6	290	1740	36	84100
6	280	1680	36	78400
6	295	1770	36	87025
2	425	850	4	180625
2	384	768	4	147456
5	315	1575	25	99225
4	355	1420	16	126025
5	328	1640	25	107584
1	425	425	1	180625
4	325	1300	16	105625
$\sum x_i=41$	$\sum y_i=3422$	$\sum x_i{y}_i=13168$	$\sum {x_i}^2=199$	$\sum {y_i}^2=1196690$

Part (a) Step 3: Explanation

Let, $S_{xy}=\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n$
$=13168-\left(41\right)\left(3422\right)/10$
$=13168-140302/10$
$=13168-14030.2$
$=-862.2$

Then,

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

$=199-(41{)}^2/10$

$=199-1681/10$

$=199-168.1$

$=30.9$

Part (a) Step 4: Explanation

Here, $SST$ stands for total sum of squares.
$S_{yy}=\left(\sum y_i^2\right)-{\left(\sum y_i\right)}^2/n$

$=1196690-(3422{)}^2/10$
$=1196690-11710084/10$
$=1196690-1171008.4$
$=25681.6$

The $SSR$regression sum of squares is equal to,

$SSR=\frac{S_{x{y}^2}}{S_{xx}}$
$=\frac{(-862.2{)}^2}{30.9}$
$=\frac{743388.84}{30.9}$
$=24057.8913$

Then,

$SSE=SST-SSR$
$=25681.6-24057.8913$
$=1623.70874$

Part (a) Step 5: Explanation

The slope of the regression line is determined as:
$b_i=\frac{S_{xy}}{S_{xx}}$
$=\frac{-862.2}{30.9}$
$=-27.9029$
The standard error of the estimate is determined as:
$S_e=\sqrt{\frac{SSE}{n-2}}$
$=\sqrt{\frac{1623.70874}{10-2}}$
$=14.24652913$
$\approx 14.246$

Part (a) Step 6: Explanation

Determining the value of the$y$-intercept is as follows:
$\begin{array}{r}{b}_0=\frac{1}{n}\left(\sum y_i-b_i\sum x_i\right)\end{array}$
$=\frac{1}{10}(3422-(-27.9\left)\right(41\left)\right)$
$=\frac{1}{10}(4565.9)$
$=456.59$
As a result, the regression equation is ${\hat{y}}_p=436.59-27.9x_p$.
By inputting the value of $x_p=4$ in the regression equation, the formula for determining the value of the point estimation is achieved.
${\hat{y}}_p=436.59-27.9x_p$
$=456.59-27.9\left(4\right)$
$=344.99$
As a result, the point estimate as ${\hat{y}}_p=344.99$.

Part (b) Step 1: Given information

To determine a $90\%$ confidence interval for the mean price of all $4$-year-old Corvettes.

Part (b) Step 2: Explanation

Here, $a=0.10$for a $90\%$confidence interval.
Since $n=10$,
$df=n-2$
$=10-2$
$=8$
From calculation, $t_{a/2}=t_{0.10/2}=t_{0.05}=1.860$
The formula for computing the confidence interval end points for the conditional mean of the response variable is as follows:
${\hat{y}}_p\pm t_{a/2}{s}_e\sqrt{\frac{1}{n}+\frac{{\left(x_p-\sum x_i/n\right)}^2}{S_{xx}}}$
We have, $x_p=4$
${\hat{y}}_p=324.99$
$S_e=14.246$
$S_{xx}=30.9$

Hence,

$344.99\pm 1.86(14.246)\sqrt{\frac{1}{10}+\frac{(4-41/10{)}^2}{30.9}}$

$344.99\pm 26.49756\sqrt{0.1+0.000323625}$

$344.99\pm 8.39281$, or $336.6$to $353.38$

$336.6$to $353.38$is the average price of all $4$-year-old corvettes, according to $90\%$confidence.

Part (c) Step 1: Given information

To find the predicted price of a $4$-year-old Corvette.

Part (c) Step 2: Explanation

Let, determine the predicted price of a $4$-year old corvette.

And the regression equation is, ${\hat{y}}_p=456.59-27.9x_p$

Substituting the value of $x_p=4$into the regression equation yields the expected value.

${\hat{y}}_p=456.59-27.9x_p$
$=456.59-27.9\left(4\right)$
$=344.99$
As a result, the predicted value is ${\hat{y}}_p=344.99$

Part (d) Step 1: Given information

To determine a $90\%$prediction interval for the price of all $4$-year old corvettes.

Part (d) Step 2: Explanation

Let, $a=0.10$for a $90\%$ confidence interval.
Since $n=10$,
$df=n-2$
$=10-2$
$=8$
From calculation, $t_{a/2}=t_{0.10/2}=t_{0.05}=1.860$
Computing the confidence interval end points for the conditional mean of the response variable is as follows:
${\hat{y}}_p\pm t_{a/2}{s}_e\sqrt{\frac{1}{n}+\frac{{\left(x_p-\sum x_i/n\right)}^2}{S_{xx}}}$
Since,$x_p=4$
${\hat{y}}_p=344.99$

$S_e=14.246$
$S_{xx}=30.9$

Hence, $344.99\pm 1.86(14.246)\sqrt{1+\frac{1}{10}+\frac{(4-41/10{)}^2}{30.9}}$

$344.99\pm 26.49756\sqrt{1+0.1+0.000323625}$
Also,
$344.99\pm 27.79496318,$or $317.2$to $372.78$
As a result, $90\%$ confident that the price of $4$-year old corvette will be is somewhere between $31720$to $37278$.

Part (e) Step 1: Given information

To draw the graph similar to a $90\%$confidence interval and $90\%$ prediction interval.

Part (e) Step 2: Explanation

Let, $90\%$prediction interval graph as follows:

Then, $90\%$confidence interval graph as follows:

As a result, a graph is generated with a $90\%$confidence interval and a $90\%$prediction interval.

Part (f) Step 1: Given information

To explain the prediction interval is wider than confidence interval.

Part (f) Step 2: Explanation

Because of the following, the prediction interval is larger than the confidence interval:
The error in estimating the mean price of all $4$-year-old Corvettes is attributable only to the sample regression line being used to estimate the population regression line.
The inaccuracy in predicting the price of a $4$-year-old Corvette, on the other hand, is related to the error in guessing the mean price plus the range in $4$-year-old Corvette prices.
As a result, the error in the estimate is due to the sample regression line's estimation of the population regression line, and the error in the prediction is due to the mistake in estimating the mean price plus price variance.