Chapter 10, Problem 48E

(a)

Section 1:

To determine

To graph: A scatterplot that shows the increase in GPA over time by hand and check whether the linear increase seems reasonable.

Explanation of Solution

Graph: Plot the data from the provided table with GPA on y-axis and Year on x-axis.

Hence, the obtained graph is shown below:

Interpretation: From the scatterplot, it can be seen that there is a linear dependency between GPA and year, which infers that the linear increase appears reasonable.

Section 2:

To determine

To graph: A scatterplot that shows the increase in GPA over time by using software and check whether the linear increase seems reasonable.

Explanation of Solution

Graph: Construct a scatterplot using excel as follows:

Step 1: Enter the data in Excel.

Step 2: Select the data. Click on Insert $\to$ Scatterplot $\to$ Scatter.

Hence, the obtained graph is shown below:

Interpretation: From the scatterplot, it can be seen that there is a linear dependency between GPA and year, so it can be said that by using software the results are same.

(b)

Section 1:

To determine

To find: The least square regression line for predicting GPA from year by hand.

Answer to Problem 48E

Solution: The regression line is

$\text{GPA}=-19.06+0.011\text{Years}$

Explanation of Solution

Calculation: Compute the value of $b_0\text{ and }{b}_1$ using formula as follows:

$\begin{array}{c}{b}_1=\frac{n{\displaystyle \sum xy-{\displaystyle \sum x}{\displaystyle \sum y}}}{n{\displaystyle \sum {\left(x\right)}^2}-{\left({\displaystyle \sum x}\right)}^2}\\ b_0=\overline{y}-b_1\overline{x}\end{array}$

Now, compute ${\displaystyle \sum x,{\displaystyle \sum y},{\displaystyle \sum xy},{\displaystyle \sum {\left(x\right)}^2\text{ and }{\left({\displaystyle \sum x}\right)}^2}}$ using the provided data, which are as follows:

Year (x)	GPA (y)	xy	x2
1992	2.85	5677.2	3968064
1996	2.90	5788.4	3984016
2002	2.97	5945.9	4008004
2007	3.01	6041.1	4028049
${\displaystyle \sum x}=7997$	${\displaystyle \sum y}=11.73$	${\displaystyle \sum xy}=23452.6$	${\displaystyle \sum {\left(x\right)}^2=15988133}$

Now, compute the value of $b_1$ as follows:

$\begin{array}{c}{b}_1=\frac{n{\displaystyle \sum xy-{\displaystyle \sum x}{\displaystyle \sum y}}}{n{\displaystyle \sum {\left(x\right)}^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{\left(4\times 23452.6\right)-\left(7997\times 11.73\right)}{\left(4\times 15988133\right)-{\left(7997\right)}^2}\\ =\frac{5.59}{523}\\ =0.011\end{array}$

Compute the value of $\overline{x}\text{ and }\overline{y}$ as follows:

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{7997}{4}\\ =1999.25\end{array}$

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum y}}{n}\\ =\frac{11.73}{4}\\ =2.93\end{array}$

Now, compute the value of $b_0$ as follows:

$\begin{array}{c}{b}_0=\overline{y}-b_1\overline{x}\\ =2.93-\left(0.011\times 1999.25\right)\\ =-19.06\end{array}$

Hence, the obtained regression equation is

$\text{GPA}=-19.06+0.011\text{Years}$

Interpretation: Therefore, it can be concluded from the obtained regression equation that the GPA increases by 0.011 times with the increase in year.

Section 2:

To determine

To graph: A scatterplot that shows the fitting of least-squares regression line by hand.

Explanation of Solution

Calculation: Obtain points for plotting on the graph using the regression equation as follows:

Let $\text{Year}\left(x\right)=1992$, then the GPA will be obtained as:

$\begin{array}{c}\text{GPA}=-19.06+0.011\text{Years}\\ =-19.06+\left(0.011\times 1992\right)\\ =2.852\end{array}$

Let $\text{Year}\left(x\right)=1996$, then the GPA will be obtained as:

$\begin{array}{c}\text{GPA}=-19.06+0.011\text{Years}\\ =-19.06+\left(0.011\times 1996\right)\\ =2.896\end{array}$

Let $\text{Year}\left(x\right)=2002$, then the GPA will be obtained as:

$\begin{array}{c}\text{GPA}=-19.06+0.011\text{Years}\\ =-19.06+\left(0.011\times 2002\right)\\ =2.962\end{array}$

Let $\text{Year}\left(x\right)=2007$, then the GPA will be obtained as:

$\begin{array}{c}\text{GPA}=-19.06+0.011\text{Years}\\ =-19.06+\left(0.011\times 2007\right)\\ =3.017\end{array}$

Now, plot these points of GPA on $y-$ axis and Year on $x-$ axis as shown below:

Graph:

Interpretation: Therefore, it can be said that all the points lie on the regression line, so it can be concluded that the line is a good fit.

Section 3:

To determine

To find: The least square regression line for predicting GPA from year by software.

Answer to Problem 48E

Solution: The regression line is:

$\text{GPA}=-18.59+0.011\text{Years}$

Explanation of Solution

Calculation: Obtain the regression line using Excel as follows:

Step 1: Click on Data $\to$ Data analysis.

Step 3: Enter Y variable and X variable input range.

Step 4: Click ‘Ok’ to obtain the result.

Hence, the obtained regression line is shown below:

$\text{GPA}=-18.59+0.011\text{Years}$

Interpretation: Therefore, it can be concluded that the regression equation obtained by hand and software are approximately same.

Section 4:

To determine

To graph: A scatterplot which shows the fitting of least-squares regression line by software.

Explanation of Solution

Graph: Construct a scatterplot using excel as follows:

Step 1: Enter the data in Excel.

Step 2: Select the data. Click on Insert $\to$ Scatterplot $\to$ Scatter.

Step 3: Now, put the cursor on the graph and a plus sign appears on the right hand side.

Step 4: Click on the plus sign and check the box for trend line.

Hence, the obtained graph is shown below:

Interpretation: From the scatterplot, it can be seen that there is a linear dependency between GPA and year.

(c)

Section 1:

To determine

To find: The 95% confidence interval for the slope by hand.

Answer to Problem 48E

Solution: The confidence interval is $\underset{\_}{\left(0.008,0.014\right)}$.

Explanation of Solution

Calculation: Now, compute ${\displaystyle \sum {\left(y_i-{\widehat{y}}_i\right)}^2}\text{ and }{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}$ using $\overline{x}=1999.25$ and the provided data is as follows:

Year (x)	GPA (y)	$\widehat{y}$	${\left(y_i-\widehat{y}\right)}^2$	${\left(x_i-\overline{x}\right)}^2$
1992	2.85	2.852	0.000004	52.5625
1996	2.90	2.896	0.000016	10.5625
2002	2.97	2.962	0.000064	7.5625
2007	3.01	3.017	0.000049	60.0625
${\displaystyle \sum x}=7997$	${\displaystyle \sum y}=11.73$		${\displaystyle \sum {\left(y_i-\widehat{y}\right)}^2=0.000133}$	${\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}=130.75$

Now, compute the standard error of slope of regression $\left({\beta }_1\right)$ as follows:

$\begin{array}{c}SE\left({\beta }_1\right)=\frac{\sqrt{\frac{{\displaystyle \sum {\left(y_i-{\widehat{y}}_i\right)}^2}}{n-2}}}{\sqrt{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}}\\ =\frac{\sqrt{\frac{0.000133}{4-2}}}{\sqrt{130.75}}\\ =\frac{0.0082}{11.435}\\ =0.0007\end{array}$

For 5% level of significance and $\left(n-2\right)$ degree of freedom, that is, $\left(4-2\right)=2$ degree of freedom, the tabulated value of $t$ is 4.303.

Compute the confidence interval as follows:

$\begin{array}{c}\text{Confidence interval}=\left[{\beta }_1\pm t_{tab}\left(SE{\beta }_1\right)\right]\\ =\left[\left\{0.011-\left(4.303\times 0.0007\right)\right\},\left\{0.011+\left(4.303\times 0.0007\right)\right\}\right]\\ =\left[\left(0.011-0.003\right),\left(0.011+0.003\right)\right]\\ =\left(0.008,0.014\right)\end{array}$

Interpretation: It can be said with 95% confidence that the GPA will increase between 0.008 and 0.014 over the time.

Section 2:

To determine

To find: The 95% confidence interval for the slope by software.

Answer to Problem 48E

Solution: The confidence interval is $\underset{\_}{\left(0.008,0.014\right)}$.

Explanation of Solution

Calculation: Obtain the regression line using Excel as follows:

Step 1: Click on Data $\to$ Data analysis.

Step 3: Enter Y variable and X variable input range.

Step 4: Click ‘Ok’ to obtain the result.

Hence, the obtained confidence interval for ${\beta }_1$ is shown below:

$\begin{array}{c}\text{Confidence interval}=\left(\text{Lower 95\% of X variable 1, Upper 95\% of X variable 1}\right)\\ =\left(0.008,0.014\right)\end{array}$

Interpretation: Therefore, it can be concluded that the 95% confidence interval obtained by hand matched with the 95% confidence interval obtained by software.