Chapter 12, Problem 69SE

a.

To determine

Find the interval estimate for the slope of the population regression.

Answer to Problem 69SE

The 95% confidence interval for the slope of the population regression is $\underset{\_}{0.888452\le {\beta }_1\le 1.085948}$.

Explanation of Solution

Given info:

The data represents the values of the variables height in feet and price in dollars for a sample of warehouses.

Calculation:

Linear regression model:

In a linear equation $y=b_0+b_1{x}_i$ the constant $b_1$ be the slope and $b_0$ be the y-intercept and x is the independent variable and y is the independent variable.

A linear regression model is given as $\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x$ where $\widehat{y}$ be the predicted values of response variable and x be the predictor variable. The ${\widehat{\beta }}_1$ be the estimate of slope and ${\widehat{\beta }}_0$ be the estimate of intercept of the line.

Regression:

Software procedure:

Step by step procedure to obtain regression equation using MINITAB software is given as,

• Choose Stat > Regression > Fit Regression Line.
• In Response (Y), enter the column of Price.
• In Predictor (X), enter the column of Height.
• Click OK.

Output using MINITAB software is given below:

Thus, the regression line for the variables sale price $\left(y\right)$ and height $\left(x\right)$ is $\underset{\_}{\stackrel{⌢}{y}=23.77+0.9872x}$.

Therefore, the slope coefficient of the regression equation is $b_1={\widehat{\beta }}_1=0.9872$.

Confidence interval:

The general formula for the confidence interval for the slope of the regression line is,

$CI={\widehat{\beta }}_1\pm t_{a/2,\left(n-2\right)}\times s_{{\widehat{\beta }}_1}$

Where, ${\widehat{\beta }}_1$ be the slope of the sample regression line, $s_{{\widehat{\beta }}_1}$ be the estimate of error standard deviation of slope coefficient.

From the MINITAB output, the estimate of error standard deviation of slope coefficient is $s_{{\widehat{\beta }}_1}=0.0468$.

Since, the level of confidence is not specified. The prior confidence level 95% can be used.

Critical value:

For 95% confidence level,

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Degrees of freedom:

The sample size is $n=19$

The degrees of freedom is,

$\begin{array}{c}d.f=n-2\\ =19-2\\ =17\end{array}$

From Table A.5 of the t-distribution in Appendix A, the critical value corresponding to the right tail area 0.025 and 17 degrees of freedom is 2.110.

Thus, the critical value is $\left(t_{0.025,17}\right)=2.110$.

The 95% confidence interval is,

$\begin{array}{c}C.I={\widehat{\beta }}_1-\left(t_{a/2}\times s_{{\widehat{\beta }}_1}\right)\le {\beta }_1\le {\widehat{\beta }}_1+\left(t_{a/2}\times s_{{\widehat{\beta }}_1}\right)\\ =\left(0.9872-\left(2.110\times 0.0468\right)\le {\beta }_1\le 0.9872+\left(2.110\times 0.0468\right)\right)\\ =\left(0.9872\pm 0.098748\right)\\ \simeq \left(0.888452,1.085948\right)\end{array}$

Thus, the 95% confidence interval for the slope of the population regression is $\underset{\_}{0.888452\le {\beta }_1\le 1.085948}$.

Interpretation:

There is 95% confident, that the expected change in sale price associated with 1 foot increase in height lies between $0.888452 and $1.085948.

c.

To determine

Find the interval estimate for the true mean sale price of all warehouses with 25 ft truss height.

Answer to Problem 69SE

The 95% specified confidence interval for the true mean sale price of all warehouses with 25 ft truss height is $\underset{\_}{\left(47.730,49.172\right)}$.

Explanation of Solution

Calculation:

Here, the regression equation is $\stackrel{⌢}{y}=23.77+0.9872x$. Where y represents the variable sale price and x represents the variable height.

Expected sale price when the height is 25 feet:

The expected sale price with 25 ft height ware houses is obtained as follows:

$\begin{array}{c}{\mu }_{\widehat{y}}=23.77+0.9872x\\ =23.77+0.9872\times 25\\ =48.45\end{array}$

Thus, the expected sale price with 25 ft height ware houses is 48.45.

Confidence interval:

The general formula for the $\left(1-\alpha \right)$% confidence interval for the conditional mean at $x=x_p$ is,

$CI={\mu }_{\widehat{y}}\pm t_{\left(\frac{\alpha }{2},n-2\right)}s\sqrt{\frac{1}{n}+\frac{{\left(x_p-\overline{x}\right)}^2}{S_{\text{xx}}}}$

Where, ${\widehat{y}}_p$ be the point estimate for the conditional mean of the response variable at $x=x_p$, and $S_{\text{xx}}={\displaystyle \sum _i{x}_i^2}-\frac{{\left({\displaystyle \sum _i{x}_i}\right)}^2}{n}$.

From the MINITAB output in part (a), the value of the standard error of the estimate is $s=1.41590$.

The value of $S_{\text{xx}}$ is obtained as follows:

i	Truss height x	$x_i^2$
1	12	144
2	14	196
3	14	196
4	15	225
5	15	225
6	16	256
7	18	324
8	22	484
9	22	484
10	24	576
11	24	576
12	26	676
13	26	676
14	27	729
15	28	784
16	30	900
17	30	900
18	33	1089
19	36	1296
Total	${\displaystyle \sum x_i}=432$	${\displaystyle \sum _i{x}_i^2}=10,736$

Thus, the total of truss height is ${\displaystyle \sum x_i}=432$.

The mean truss height is,

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x_i}}{n}\\ =\frac{432}{19}\\ =22.737\end{array}$

Thus, the mean truss height is $\underset{\_}{\overline{x}=22.737}$.

Covariance term $S_{\text{xx}}$:

The value of $S_{\text{xx}}$ is,

$\begin{array}{c}{S}_{\text{xx}}={\displaystyle \sum _i{x}_i^2}-\frac{{\left({\displaystyle \sum _i{x}_i}\right)}^2}{n}\\ =10,736-\frac{{432}^2}{19}\\ =10,736-9,822.316\\ =913.684\end{array}$

Thus, the covariance term $S_{\text{xx}}$ is 913.684.

Since, the level of confidence is not specified. The prior confidence level 95% can be used.

Critical value:

For 95% confidence level,

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Degrees of freedom:

The sample size is $n=19$

The degrees of freedom is,

$\begin{array}{c}d.f=n-2\\ =19-2\\ =17\end{array}$

From Table A.5 of the t-distribution in Appendix A, the critical value corresponding to the right tail area 0.025 and 17 degrees of freedom is 2.110.

Thus, the critical value is $\left(t_{{}_{0.025,17}}\right)=2.110$.

The 95% confidence interval is,

$\begin{array}{c}C.I={\mu }_{\widehat{y}}-t_{\left(\frac{\alpha }{2},n-2\right)}s\sqrt{\frac{1}{n}+\frac{{\left(x_p-\overline{x}\right)}^2}{S_{\text{xx}}}}\le {\mu }_y\le {\mu }_{\widehat{y}}+t_{\left(\frac{\alpha }{2},n-2\right)}s\sqrt{\frac{1}{n}+\frac{{\left(x_p-\overline{x}\right)}^2}{S_{\text{xx}}}}\\ =\left(48.45\pm 2.110\times 1.141590\sqrt{\frac{1}{19}+\frac{{\left(25-22.737\right)}^2}{913.684}}\right)\\ =\left(48.45\pm 0.721\right)\\ \simeq \left(47.730,49.172\right)\end{array}$

Thus, the 95% specified confidence interval for the true mean of all warehouses with 25 ft truss height is $\underset{\_}{\left(47.730,49.172\right)}$.

Interpretation:

There is 95% specified confidence interval for the true mean of all warehouses with 25 ft truss height lies between $47.730 and $49.172.

d.

To determine

Find the prediction interval of sale price for a single warehouse of truss height 25 ft.

Compare the width of the prediction interval with the confidence interval obtained in part (a).

Answer to Problem 69SE

The 95% prediction interval of sale price for a single warehouse of truss height 25 ft is $\underset{\_}{\left(45.377,51.523\right)}$.

The prediction interval is wider than the confidence interval.

Explanation of Solution

Calculation:

Here, the regression equation is $\stackrel{⌢}{y}=23.77+0.9872x$. Where y represents the variable sale price and x represents the variable height.

From part (c), the the expected sale price with 25 ft height ware houses is ${\mu }_{\widehat{y}}=48.45$.

Prediction interval for a single future value:

Prediction interval is used to predict a single value of the focus variable that is to be observed at some future time. In other words it can be said that the prediction interval gives a single future value rather than estimating the mean value of the variable.

The general formula for $\left(1-\alpha \right)$% prediction interval for the conditional mean at $x=x_p$ is,

$P.I={\widehat{y}}_p\pm t_{\frac{\alpha }{2}}s\sqrt{1+\frac{1}{n}+\frac{{\left(x_p-{\displaystyle \sum _i\frac{x_i}{n}}\right)}^2}{S_{\text{xx}}}}$

where ${\widehat{y}}_p$ be the predicted value of the response variable at $x=x_p$ and $S_{\text{xx}}={\displaystyle \sum _i{x}_i^2}-\frac{{\left({\displaystyle \sum _i{x}_i}\right)}^2}{n}$

From the MINITAB output in part (a), the value of the standard error of the estimate is $s=1.41590$.

From part (c), the truss height is $\overline{x}=22.737$ and the covariance term is $S_{\text{xx}}=913.684$.

Since, the level of confidence is not specified. The prior confidence level 95% can be used.

Critical value:

For 95% confidence level,

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Degrees of freedom:

The sample size is $n=19$

The degrees of freedom is,

$\begin{array}{c}d.f=n-2\\ =19-2\\ =17\end{array}$

From Table A.5 of the t-distribution in Appendix A, the critical value corresponding to the right tail area 0.025 and 17 degrees of freedom is 2.110.

Thus, the critical value is $\left(t_{{}_{0.025,17}}\right)=2.110$.

The 95% prediction interval is,

$\begin{array}{c}P.I={\widehat{y}}_p\pm t_{\frac{\alpha }{2}}s\sqrt{1+\frac{1}{n}+\frac{{\left(x_p-\overline{x}\right)}^2}{S_{\text{xx}}}}\\ =48.45\pm \left(2.110\right)\left(1.41590\right)\sqrt{1+\frac{1}{9}+\frac{{\left(25-22.737\right)}^2}{913.684}}\\ =48.45\pm 3.073\\ \simeq \left(45.377,51.523\right)\end{array}$

Thus, the 95% prediction interval of sale price for a single warehouse of truss height 25 ft is $\underset{\_}{\left(45.377,51.523\right)}$.

Interpretation:

For repeated samples, there is 95% confident that the sale price for a single warehouse of truss height 25 ft lies between $45.377 and $51.523.

Comparison:

The 95% prediction interval of sale price for a single warehouse of truss height 25 ft is $\underset{\_}{\left(45.377,51.523\right)}$.

Width of the prediction interval:

The width of the 95% prediction interval is,

$\begin{array}{c}{\text{Width}}_1=\text{Upper}\text{bound}-\text{Lower}\text{bound}\\ \text{=51}\text{.523}-45.377\\ =6.146\end{array}$

Thus, the width of the 95% prediction interval is 6.146.

The 95% specified confidence interval for the true mean of all warehouses with 25 ft truss height is $\underset{\_}{\left(47.730,49.172\right)}$.

Width of the confidence interval:

The width of the 95% confidence interval is,

$\begin{array}{c}{\text{Width}}_2=\text{Upper}\text{bound}-\text{Lower}\text{bound}\\ \text{=49}\text{.172}-47.730\\ =1.442\end{array}$

Thus, the width of the 95% confidence interval is 1.442.

From, the obtained two widths it is observed that the width of the prediction interval is typically larger than the width of the confidence interval.

Thus, the prediction interval is wider than the confidence interval.

e.

To determine

Compare the width of the 95% prediction interval of sale price of ware houses for 25 ft truss height and for 30 ft truss height.

Answer to Problem 69SE

The 95% prediction interval of sale price of ware houses for 30 ft truss height will be wider than the sale price of ware houses for 25 ft truss height.

Explanation of Solution

Calculation:

Here, the regression equation is $\stackrel{⌢}{y}=23.77+0.9872x$. Where y represents the variable sale price and x represents the variable height.

From part (c), the truss height is $\overline{x}=22.737$ and the covariance term is $S_{\text{xx}}=913.684$.

Here, the observation $x^{*}=25$ is close to mean value $\overline{x}=22.737$ than the observation $x^{*}=30$.

The general formula to obtain $s_{\widehat{Y}}$ is,

$s_{\widehat{Y}}=s\times \sqrt{\frac{1}{n}+\frac{{\left(x^{*}-\overline{x}\right)}^2}{S_{xx}}}$.

For $x^{*}=30$ the value of $s_{\widehat{Y}}$ is,

$s_{\widehat{Y}\left(30\right)}=s\times \sqrt{\frac{1}{19}+\frac{{\left(30-22.737\right)}^2}{913.684}}$.

For $x^{*}=25$ the value of $s_{\widehat{Y}}$ is,

$s_{\widehat{Y}\left(25\right)}=s\times \sqrt{\frac{1}{19}+\frac{{\left(25-22.737\right)}^2}{913.684}}$.

In the two quantities, the only difference is the values of ${\left(25-22.737\right)}^2$ and ${\left(30-22.737\right)}^2$.

In general, the value of the quantity ${\left(30-22.737\right)}^2$ will be larger than the value of ${\left(25-22.737\right)}^2$. Since, $x^{*}=25$ is close to $\overline{x}=22.737$ than $x^{*}=30$.

Therefore, the value $s_{\widehat{Y}\left(30\right)}$ will be larger than the value of $s_{\widehat{Y}\left(25\right)}$.

Comparison:

Prediction interval:

The general formula for $\left(1-\alpha \right)$% prediction interval for the conditional mean at $x=x_p$ is,

$P.I={\widehat{y}}_p\pm t_{\frac{\alpha }{2}}s\sqrt{1+\frac{1}{n}+\frac{{\left(x_p-{\displaystyle \sum _i\frac{x_i}{n}}\right)}^2}{S_{\text{xx}}}}$

The prediction interval will be wider for large value of $s_{\widehat{Y}}$. Since, the value $s_{\widehat{Y}}$ plays an important role in increasing or decreasing the margin of error.

Here, $s_{\widehat{Y}}$ is higher for $x^{*}=30$.

Thus, the prediction interval is wider for $x^{*}=30$.

Thus, 95% prediction interval of sale price of ware houses for 30 ft truss height will be wider than the sale price of ware houses for 25 ft truss height.

e.

To determine

Find the correlation coefficient between the variables sale price and truss height.

Answer to Problem 69SE

The correlation coefficient between the variables sale price and truss height is 0.9814.

Explanation of Solution

Calculation:

$R^2$(R-squared):

The coefficient of determination ($R^2$) is defined as the proportion of variation in the observed values of the response variable that is explained by the regression. The squared correlation gives fraction of variability of response variable (y) accounted for by the linear regression model.

The general formula to obtain coefficient of variation is,

$R^2=r^2$

From the regression output obtained in part (a), the value of coefficient of determination is 0.9631.

Thus, the coefficient of determination is $\underset{\_}{r^2=0.9631}$.

Correlation coefficient:

Correlation analysis is used to measure the strength of the association between variables. In other words, it can be said that correlation describes the linear association between quantitative variables.

The general formula to calculate correlation coefficient is,

$\begin{array}{c}r=\frac{{\displaystyle \sum _{i=1}^n\left(x_i-\overline{x}\right)\left(y_i-\overline{y}\right)}}{\sqrt{{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}\times {\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}}\\ =\frac{S_{xy}}{\sqrt{S_{xx}\times S_{yy}}}\end{array}$

The coefficient of determination is obtained as follows:

$\begin{array}{c}r=\sqrt{R^2}\\ =\sqrt{0.9631}\\ =0.9814\end{array}$

The sign of the correlation coefficient depends on the sign of the slope coefficient.

Here, ${\widehat{\beta }}_1=0.9872$.

Since, the sign of the slope coefficient is positive. The correlation coefficient is positive.

Thus, the correlation coefficient is 0.9814.

Interpretation:

The strength of the association between the variables sale price and truss height is 0.9814. that is, 1 unit increase in one variable is associated with 98.14% increase in the value of the other variable.