Chapter 12.CR, Problem 9CR

To determine

(a)

To calculate:

The sum of squared errors, SSE.

Answer to Problem 9CR

Solution:

The required SSE is 16.7246.

Explanation of Solution

Given Information:

The following data is collected on the number of years of post high-school education and the annual incomes of eight people ten years after graduation from high school.

Thread Count	150	200	225	250	275	300	350	400
Price (in Dollars)	18	21	25	28	30	31	35	45

The least Squares regression line is the line for which the average variation from the data is the smallest, also called the line of best fit, given by

$\widehat{y}=b_0+b_{1}x$.

Where $b_1$ is the slope of the least-squares regression line for paired data from a sample,

And $b_0$ is the $y$-intercept for the regression line.

Formula used:

The equation of least-squares regression line is given by,

$\widehat{y}=b_0+b_{1}x$

Where $b_1$ is the slope of the least-squares regression line given as,

$b_1=\frac{n{\displaystyle \sum x_i{y}_i}-\left({\displaystyle \sum x_i}\right)\left({\displaystyle \sum y_i}\right)}{n{\displaystyle \sum x_i^2-{\left({\displaystyle \sum x_i}\right)}^2}}$

And $b_0$ is y-intercept given as,

$b_0=\frac{{\displaystyle \sum y_i}}{n}-b_1\frac{{\displaystyle \sum x_i}}{n}$

Where n is the number of data pairs in the sample,

$x_i$ is the $i^{th}$ value of the explanatory variable,

And $y_i$ is the $i^{th}$ value of response variable.

The sum of squared errors (SSE) for a regression line is calculated as,

$\text{SSE}={{\displaystyle \sum \left(y_i-{\widehat{y}}_i\right)}}^2$

Where, $y_i$ is the $i^{th}$ value of response variable,

And ${\widehat{y}}_i$ is the predicted value of $y_i$, using the least-squares regression model.

Calculation:

Thread Count	Price(in Dollars)	$x_i{y}_i$	$x_i^2$	$y_i^2$
150	18	2700	22500	324
200	21	4200	40000	441
225	25	5625	50625	625
250	28	7000	62500	784
275	30	8250	75625	900
300	31	9300	90000	961
350	35	12250	122500	1225
400	45	18000	160000	2025
${\displaystyle \sum x_i=2150}$	${\displaystyle \sum y_i=}233$	${\displaystyle \sum x_i{y}_i=}67325$	${\displaystyle \sum x_i^2=}623750$	${\displaystyle \sum y_i^2=}7285$

Let $x_i$ be the thread counts of various bed sheets,

And $y_i$ be the price of the bed sheets.

The slope of the least-squares regression line is calculated as,

$b_1=\frac{n{\displaystyle \sum x_i{y}_i}-\left({\displaystyle \sum x_i}\right)\left({\displaystyle \sum y_i}\right)}{n{\displaystyle \sum x_i^2-{\left({\displaystyle \sum x_i}\right)}^2}}$

Where, ${\displaystyle \sum x_i}=x_1+x_2+\mathrm{....}+x_8$.

Substitute 150 for $x_1$, 200 for $x_2$ ….., 400 for $x_8$ in the above formula.

$\begin{array}{rll}{\displaystyle \sum x_i}& =& 150+200+\mathrm{.....}+350+400\\ & =& 2150\end{array}$

Proceed in the same manner to calculate ${\displaystyle \sum y_i},{\displaystyle \sum x_i{y}_i},{\displaystyle \sum x_i^2}\text{and}{\displaystyle \sum y_i^2}$ for the rest of the data and refer table for the rest of the values calculated.

$\begin{array}{rll}{\displaystyle \sum y_i}& =& 18+21+\mathrm{.......}+34+45\\ & =& 233\end{array}$

$\begin{array}{rll}{\displaystyle \sum x_i{y}_i}& =& 2700+4200+\mathrm{.......}+12250+18000\\ & =& 67325\end{array}$

$\begin{array}{rll}{\displaystyle \sum x_i^2}& =& 22500+40000+\mathrm{.......}+122500+160000\\ & =& 623750\end{array}$

$\begin{array}{rll}{\displaystyle \sum y_i^2}& =& 324+441+625+\mathrm{.......}+1225+2025\\ & =& 7285\end{array}$

The slope of the least-squares regression line is calculated as,

$b_1=\frac{n{\displaystyle \sum x_i{y}_i}-\left({\displaystyle \sum x_i}\right)\left({\displaystyle \sum y_i}\right)}{n{\displaystyle \sum x_i^2-{\left({\displaystyle \sum x_i}\right)}^2}}$

Substitute 2150 for ${\displaystyle \sum x_i}$, 233 for ${\displaystyle \sum y_i}$, 67325 for ${{\displaystyle \sum x_{i}y}}_i$, 623750 for ${\displaystyle \sum x_i^2}$ and 8 for n in the above formula.

$\begin{array}{rll}{b}_1& =& \frac{\left(8\times 67325\right)-\left(233\times 2150\right)}{8\left(623750\right)-{\left(2150\right)}^2}\\ & =& 0.1024\end{array}$

The y-intercept of regression line is calculated as,

$b_0=\frac{{\displaystyle \sum y_i}}{n}-b_1\frac{{\displaystyle \sum x_i}}{n}$

Substitute 2150 for ${\displaystyle \sum x_i}$, 233 for ${\displaystyle \sum y_i}$, 8 for n and 0.1024for $b_1$.

$\begin{array}{rll}{b}_0& =& \frac{233}{8}-\left(0.1024\right)\frac{2150}{8}\\ & =& 1.6050\end{array}$

The equation of least-squares regression line is given by,

$\widehat{y}=b_0+b_{1}x$

Substitute 28.6514 for $b_0$ and 0.1024 for $b_1$ in the above formula.

$\widehat{y}=1.6050+0.1024x$.

Number of years $x_i$	Annual income $y_i$	Predicted value ${\widehat{y}}_i$	$y_i-{\widehat{y}}_i$	${\left(y_i-{\widehat{y}}_i\right)}^2$
150	18	16.965	1.035	1.071225
200	21	22.085	-1.085	1.177225
225	25	24.645	0.355	0.126025
250	28	27.205	0.795	0.632025
275	30	29.765	0.235	0.055225
300	31	32.325	-1.325	1.755625
350	35	37.445	-2.445	5.978025
400	45	42.565	2.435	5.929225

The predicted values are calculated as,

$\widehat{y}=1.6050+0.1024x$

The predicted value $y_1$ is calculated as,

${\widehat{y}}_1=1.6050+0.1024x_1$

Substitute 150 for $x_1$ in the above formula.

$\begin{array}{rll}{\widehat{y}}_1& =& 1.6050+0.1024\left(150\right)\\ & =& 16.965\end{array}$

Proceed in the same manner to calculate ${\widehat{y}}_1$ for the rest of the data and refer table for the rest of the values calculated.

The residual is calculated as, $y_i-{\widehat{y}}_i$,

Substitute 18 for $y_1$ and 16.965for ${\widehat{y}}_1$.

$\begin{array}{rll}{y}_1-{\widehat{y}}_1& =& 18-16.965\\ & =& 1.035\end{array}$

Square both sides of the equation.

$\begin{array}{rll}{\left(y_1-{\widehat{y}}_1\right)}^2& =& {\left(1.035\right)}^2\\ & =& 1.0712\end{array}$

Proceed in the same manner to calculate ${\left(y_i-{\widehat{y}}_i\right)}^2$ for all the $1\le i\le n$ for the rest data and refer table for the rest of the ${\left(y_i-{\widehat{y}}_i\right)}^2$ values calculated. Then the value of ${\displaystyle \sum {\left(y_i-{\widehat{y}}_i\right)}^2}$ is calculated as,

$\begin{array}{rll}\text{SSE}& =& {\displaystyle \sum {\left(y_i-{\widehat{y}}_i\right)}^2}\\ & =& 1.0712+1.177+0.1260+\mathrm{......}+5.92922\\ & =& 16.7246\end{array}$

Conclusion:

Thus, the SSE is 16.7246

To determine

(b)

To calculate:

The standard error of estimate, $S_e$.

Answer to Problem 9CR

Solution:

The required standard error of estimate is $1.6696$.

Explanation of Solution

Given Information:

The following data is collected on the number of years of post high-school education and the annual incomes of eight people ten years after graduation from high school.

Thread Count	150	200	225	250	275	300	350	400
Price (in Dollars)	18	21	25	28	30	31	35	45

Formula used:

The standard error of estimate, which is used to measure by how much the sample data points deviate from regression line is given by,

$\begin{array}{rll}{S}_e& =& \sqrt{\frac{{\displaystyle \sum {\left(y_i-{\widehat{y}}_i\right)}^2}}{n-2}}\\ & =& \sqrt{\frac{\text{SSE}}{n-2}}\end{array}$

Where, $y_i$ is the $i^{th}$ value of response variable,

${\widehat{y}}_i$ is the predicted value of $y_i$, using the least-squares regression model,

n is the number of data pairs in the sample,

And SSE is the sum of squared errors.

Calculation:

The standard error of estimate is calculated as,

$\begin{array}{rll}{S}_e& =& \sqrt{\frac{{\displaystyle \sum {\left(y_i-{\widehat{y}}_i\right)}^2}}{n-2}}\\ & =& \sqrt{\frac{\text{SSE}}{n-2}}\end{array}$

Substitute 5868.153 for SSE and 8 for n in the above formula.

$\begin{array}{rll}{S}_e& =& \sqrt{\frac{16.7246}{8-2}}\\ & =& 1.6696\end{array}$

Conclusion:

Thus, the standard error of estimate is $1.6696$.

To determine

(c)

The 95% prediction interval for the price of 350-thread count sheets.

Answer to Problem 9CR

Solution:

The required prediction interval is. $\left(33.0776,41.8124\right)$.

Explanation of Solution

Given Information:

The following data is collected on the number of years of post high-school education and the annual incomes of eight people ten years after graduation from high school.

Thread Count	150	200	225	250	275	300	350	400
Price (in Dollars)	18	21	25	28	30	31	35	45

Formula used:

The margin of error of a prediction interval for an individual y-value is calculated as,

$E=t_{\alpha /2}{S}_e\sqrt{1+\frac{1}{n}+\frac{n{\left(x_0-\overline{x}\right)}^2}{n\left({\displaystyle \sum x_i^2}\right)-{\left({\displaystyle \sum x_i}\right)}^2}}$

With degree of freedom $df=n-2$.

Where, $x_i$ of response variable,

$x_0$ is the fixed value,

$S_e$ is the standard error of estimate,

n is the number of data pairs in the sample,

SSE is the sum of squared errors,

And $t$ -distribution is applied with degree of freedom of $df=n-2$

Then the prediction interval for an individual y-value is,

$\left(\widehat{y}-E,\widehat{y}+E\right)$.

Calculation:

It is given that the level of prediction is 0.95 then the level of significance is calculated as,

$\begin{array}{rll}\alpha & =& 1-0.95\\ & =& 0.05\end{array}$

Then,

$t_{\alpha /2}=2.447$

The mean of the number of years of post high school education is calculated as,

$\overline{x}=\frac{{\displaystyle \sum x_i}}{n}$

Substitute 2150 for ${\displaystyle \sum x_i}$ and 8 for n in the above formula of mean.

$\begin{array}{rll}\overline{x}& =& \frac{2150}{8}\\ & =& 268.75\end{array}$

The margin of error of a prediction interval for an individual y-value is calculated as,

$E=t_{\alpha /2}{S}_e\sqrt{1+\frac{1}{n}+\frac{n{\left(x_0-\overline{x}\right)}^2}{n\left({\displaystyle \sum x_i^2}\right)-{\left({\displaystyle \sum x_i}\right)}^2}}$

Substitute 2.447 for $t_{\alpha /2}$, 1.6696 for $S_e$, 8 for n, 350 for $x_0$, $623750$ for ${\displaystyle \sum x_i^2}$, 2150 for ${\displaystyle \sum x_i}$ and 268.75 for $\overline{x}$ in the above formula.

$\begin{array}{rll}E& =& 2.447\times 1.6696\times \sqrt{1+\frac{1}{8}+\frac{8{\left(350-268.75\right)}^2}{8\left(623750\right)-{\left(2150\right)}^2}}\\ & =& 2.447\times 1.6696\times 1.0690\\ & =& 4.3674\end{array}$

The $\widehat{y}$ is calculated by substituting the value of $x_0$ in the regression line equation.

The regression line is,

$\widehat{y}=1.6050+0.1024x$

Substitute 350 for $x$ in the above formula.

$\begin{array}{rll}\widehat{y}& =& 1.6050+0.1024\left(350\right)\\ & =& 37.4450\end{array}$

The prediction interval is,

$\begin{array}{rll}\left(\widehat{y}-E,\widehat{y}+E\right)& =& \left(37.4450-4.3674,37.4450+4.3674\right)\\ & =& \left(33.0776,41.8124\right)\end{array}$

Conclusion:

The required prediction interval is. $\left(33.0776,41.8124\right)$

To determine

(d)

The 95% confidence interval for the y-intercept of the regression line.

Answer to Problem 9CR

Solution:

The required confidence interval is $\left(-3.7304,6.9140\right)$.

Explanation of Solution

Given Information:

The following data is collected on the number of years of post high-school education and the annual incomes of eight people ten years after graduation from high school.

Thread Count	150	200	225	250	275	300	350	400
Price (in Dollars)	18	21	25	28	30	31	35	45

Formula Used:

The correlation coefficient measures the linear relationship between a response variable and explanatory variable calculated as,

$r=\frac{n\left({\displaystyle \sum x_i{y}_i}\right)-\left({\displaystyle \sum x_i}\right)\left({\displaystyle \sum y_i}\right)}{\sqrt{n{\displaystyle \sum x_i{}^2-{\left({\displaystyle \sum x_i}\right)}^2}}\sqrt{n{\displaystyle \sum y_i{}^2-{\left({\displaystyle \sum y_i}\right)}^2}}}$

Coefficient of determination measures the proportion of variation in the response variable caused by explanatory variable which is simply the square of r, the correlation coefficient.

The standard error of estimate, $S_e$ which is used to calculate by how much the sample data points deviates from regression line which is calculated as,

$\begin{array}{rll}{S}_e& =& \sqrt{\frac{{\displaystyle \sum {\left(y_i-{\widehat{y}}_i\right)}^2}}{n-2}}\\ & =& \sqrt{\frac{\text{SSE}}{n-2}}\end{array}$

In ANOVA,

Grand Mean is the weighted mean of the $k$ sample means, one from each of the $k$ populations, equivalent to the mean of all the sample data combined, given by

$\overline{\overline{x}}=\frac{{\displaystyle \sum _{i=1}^k{n}_i{\overline{x}}_i}}{{\displaystyle \sum _{i=1}^k{n}_i}}$

Sum of Squares among Treatments (SST) is the measures the variation between the sample means and the grand mean, given by,

$\text{SST =}{\displaystyle \sum _{i=1}^k{n}_i{\left({\overline{x}}_i-\overline{\overline{x}}\right)}^2}$

Sum of Squares for Error (SSE) is the measures the variation in the sample data resulting from the variability within each sample,

$\text{SSE =}{\displaystyle \sum _{j=1}^{n_1}{\left(x_{1j}-{\overline{x}}_1\right)}^2}+{\displaystyle \sum _{j=1}^{n_2}{\left(x_{2j}-{\overline{x}}_2\right)}^2}+\mathrm{...}+{\displaystyle \sum _{j=1}^{n_k}{\left(x_{kj}-{\overline{x}}_k\right)}^2}$

Total Variation, it is the sum of the squared deviations from the grand mean for all of the data values in each sample, given by

$\begin{array}{rllllll}\text{Total Variation=}{\displaystyle \sum _j^{}}& =& 1n_1{\left(x_{1j}-\overline{\overline{x}}\right)}^2+{\displaystyle \sum _j^{}}& =& 1n_2{\left(x_{2j}-\overline{\overline{x}}\right)}^2+\mathrm{...}+{\displaystyle \sum _j^{}}& =& 1n_k{\left(x_{kj}-\overline{\overline{x}}\right)}^2\\ & =& \text{SST + SSE}\end{array}$

Mean Square for Treatments (MST) found by dividing the sum of squares among treatments by its degrees of freedom, given by

$\text{MST=}\frac{\text{SST}}{\text{DFT}}\text{with DFT=}k-1$

Mean Square for Error (MSE) found by dividing the sum of squares for error by its degrees of freedom, given by

$\text{MST=}\frac{\text{SSE}}{\text{DFE}}\text{with DFE=}{n}_T-k$

Test Statistic for an ANOVA Test is used when independent, simple random samples are taken from populations with variances that are unknown and assumed to be equal, where all of the $k$ population distributions are approximately normal, given by,

$F=\frac{\text{MST}}{\text{MSE}}\text{with }d{f}_1=\text{DFT=}k-1\text{ and }d{f}_2=\text{DFE=}{n}_T-k$

Calculation:

To generate the regression table in excel follow the given steps:

1. Under data tab, choose data analytics and then select regression.

2. Select the input Y range and enter the range of the given $y_i$ and select the input X range and enter the range of the given $x_i$ data.

3.Choose 95% confidence interval and click OK.

The following table will appear.

Regression Statistics
Multiple R	0.983094904
R Square	0.96647559
Adjusted R Square	0.960888189
Standard Error	1.66955532
Observations	8

ANOVA
	df	SS	MS	F	Significance F
Regression	1	482.1505102	482.1505102	172.9740696	1.19253E-05
Residual	6	16.7244898	2.787414966
Total	7	498.875

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	1.591836735	2.17509095	0.73184835	0.491845191	-3.730419077	6.914092547
Slope	0.10244898	0.007789635	13.15196067	1.19253E-05	0.083388428	0.121509531

RESIDUAL OUTPUT
Observation	Predicted y	Residuals	Standard Residuals
1	16.95918367	1.040816327	0.673359012
2	22.08163265	-1.081632653	-0.699765248
3	24.64285714	0.357142857	0.231054563
4	27.20408163	0.795918367	0.514921598
5	29.76530612	0.234693878	0.151835856
6	32.32653061	-1.326530612	-0.858202663
7	37.44897959	-2.448979592	-1.584374147
8	42.57142857	2.428571429	1.571171029

The confidence interval of the y-intercept can be constructed by adding and subtracting the margin of error to the point estimate by using Microsoft excel.

Referring regression statistics, $R^2$, coefficient of determination is 0.9995 which implies that 99.95% of the response variable is explained by the explanatory variable.

Standard error is the standard error of estimate, $S_e$ calculated by the formula mentioned in concept.

The lower 95% and the upper 95% gives the confidence interval of the y-intercept.

The intercept given in the row of the table above is the $b_0$ of the regression line coming out to be 1.59 and the slope which is 0.1024 given in the table above is $b_1$ of the regression line.

So the regression line is,

$\widehat{y}=1.59+0.1024x$

The lower and the upper endpoints for a 95% confidence interval for the y-intercept of the regression line, ${\beta }_0$ is. $\left(-3.7304,6.9140\right)$

Conclusion:

Thus, the 95% confidence interval for the y-intercept of the regression line is.

To determine

(e)

Construct a 95% confidence interval for the slope of the regression line.

Answer to Problem 9CR

Solution:

The required confidence interval is $\left(0.0833,0.1215\right)$.

Explanation of Solution

Given Information:

The following data is collected on the number of years of post high-school education and the annual incomes of eight people ten years after graduation from high school.

Thread Count	150	200	225	250	275	300	350	400
Price (in Dollars)	18	21	25	28	30	31	35	45

Formula Used:

The correlation coefficient measures the linear relationship between a response variable and explanatory variable calculated as,

$r=\frac{n\left({\displaystyle \sum x_i{y}_i}\right)-\left({\displaystyle \sum x_i}\right)\left({\displaystyle \sum y_i}\right)}{\sqrt{n{\displaystyle \sum x_i{}^2-{\left({\displaystyle \sum x_i}\right)}^2}}\sqrt{n{\displaystyle \sum y_i{}^2-{\left({\displaystyle \sum y_i}\right)}^2}}}$

Coefficient of determination measures the proportion of variation in the response variable caused by explanatory variable which is simply the square of r, the correlation coefficient.

The standard error of estimate, $S_e$ which is used to calculate by how much the sample data points deviates from regression line which is calculated as,

$\begin{array}{rll}{S}_e& =& \sqrt{\frac{{\displaystyle \sum {\left(y_i-{\widehat{y}}_i\right)}^2}}{n-2}}\\ & =& \sqrt{\frac{\text{SSE}}{n-2}}\end{array}$

In ANOVA,

Grand Mean is the weighted mean of the $k$ sample means, one from each of the $k$ populations, equivalent to the mean of all the sample data combined, given by

$\overline{\overline{x}}=\frac{{\displaystyle \sum _{i=1}^k{n}_i{\overline{x}}_i}}{{\displaystyle \sum _{i=1}^k{n}_i}}$

Sum of Squares among Treatments (SST) is the measures the variation between the sample means and the grand mean, given by,

$\text{SST=}{\displaystyle \sum _{i=1}^k{n}_i{\left({\overline{x}}_i-\overline{\overline{x}}\right)}^2}$

Sum of Squares for Error (SSE) is the measures the variation in the sample data resulting from the variability within each sample,

$\text{SSE =}{\displaystyle \sum _{j=1}^{n_1}{\left(x_{1j}-{\overline{x}}_1\right)}^2}+{\displaystyle \sum _{j=1}^{n_2}{\left(x_{2j}-{\overline{x}}_2\right)}^2}+\mathrm{...}+{\displaystyle \sum _{j=1}^{n_k}{\left(x_{kj}-{\overline{x}}_k\right)}^2}$

Total Variation, it is the sum of the squared deviations from the grand mean for all of the data values in each sample, given by

$\begin{array}{rllllll}\text{Total Variation=}{\displaystyle \sum _j^{}}& =& 1n_1{\left(x_{1j}-\overline{\overline{x}}\right)}^2+{\displaystyle \sum _j^{}}& =& 1n_2{\left(x_{2j}-\overline{\overline{x}}\right)}^2+\mathrm{...}+{\displaystyle \sum _j^{}}& =& 1n_k{\left(x_{kj}-\overline{\overline{x}}\right)}^2\\ & =& \text{SST + SSE}\end{array}$

Mean Square for Treatments (MST) found by dividing the sum of squares among treatments by its degrees of freedom, given by

$\text{MST=}\frac{\text{SST}}{\text{DFT}}\text{with DFT=}k-1$

Mean Square for Error (MSE) found by dividing the sum of squares for error by its degrees of freedom, given by

$\text{MST=}\frac{\text{SSE}}{\text{DFE}}\text{with DFE=}{n}_T-k$

Test Statistic for an ANOVA Test is used when independent, simple random samples are taken from populations with variances that are unknown and assumed to be equal, where all of the $k$ population distributions are approximately normal, given by,

$F=\frac{\text{MST}}{\text{MSE}}\text{with }d{f}_1=\text{DFT=}k-1\text{ and }d{f}_2=\text{DFE=}{n}_T-k$

Calculation:

To generate the regression table in excel follow the given steps:

1. Under data tab, choose data analytics and then select regression.

2. Select the input Y range and enter the range of the given $y_i$ and select the input X range and enter the range of the given $x_i$ data.

3.Choose 95% confidence interval and click OK.

The following table will appear.

Regression Statistics
Multiple R		0.983094904
R Square		0.96647559
Adjusted R Square		0.960888189
Standard Error		1.66955532
Observations		8
ANOVA
	df		SS				MS			F			Significance F
Regression			1		482.1505102				482.1505102			172.9740696			1.19253E-05
Residual			6		16.7244898				2.787414966
Total			7		498.875
	Coefficients			Standard Error				t Stat			P-value			Lower 95%		Upper 95%
Intercept	1.591836735			2.17509095				0.731848356			0.491845191			-3.730419077		6.914092547
Slope	0.10244898			0.007789635				13.15196067			1.19253E-05			0.083388428		0.121509531
RESIDUAL OUTPUT
Observation			Predicted y			Residuals				Standard Residuals
1			16.95918367			1.040816327				0.673359012
2			22.08163265			-1.081632653				-0.699765248
3			24.64285714			0.357142857				0.231054563
4			27.20408163			0.795918367				0.514921598
5			29.76530612			0.234693878				0.151835856
6			32.32653061			-1.326530612				-0.858202663
7			37.44897959			-2.448979592				-1.584374147
8			42.57142857			2.428571429				1.571171029

The lower 95% and the upper 95% gives the confidence interval of the slope.

The intercept given in the row of the table above is the $b_0$ of the regression line coming out to be 1.59 and the slope which is 0.1024 given in the table above is $b_1$ of the regression line.

So the regression line is,

$\widehat{y}=1.59+0.1024x$

The lower and the upper endpoints for a 95% confidence interval for the slope of the regression line, ${\beta }_1$ is. $\left(0.0833,0.1215\right)$.

Conclusion:

Thus, the 95% confidence interval for the slope of the regression line is $\left(0.0833,0.1215\right)$.