Chapter 9, Problem 9CRP

(a)

To determine

Construct a scatter diagram for the data.

Answer to Problem 9CRP

The scatter diagram for data is,

Explanation of Solution

Calculation:

The variable x denotes the weight of incoming mail and y denotes the number of employees required to process the mail.

Step by step procedure to obtain scatter plot using MINITAB software is given below:

• Choose Graph > Scatterplot.
• Choose Simple. Click OK.
• In Y variables, enter the column of x.
• In X variables, enter the column of y.
• Click OK.

(b)

To determine

Find the value of $\overline{x}$.

Find the value of $\overline{y}$.

Find the value of b.

Find the equation of the least-squares line.

Construct the line on the scatter diagram.

Answer to Problem 9CRP

The value of $\overline{x}$ is 16.38.

The value of $\overline{y}$ is 10.13.

The value of b is 0.554.

The equation of the least-squares line is $\widehat{y}=1.051+0.554x$.

The scatter plot with regression line is,

Explanation of Solution

Calculation:

The values are ${\displaystyle \sum x}=131$, ${\displaystyle \sum y}=81$, ${\displaystyle \sum x^2}=2,435$, ${\displaystyle \sum y^2}=927$, ${\displaystyle \sum xy}=1,487$ and $n=8$

The value of $\overline{x}$ is,

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{131}{8}\\ =16.375\\ \approx 16.38\end{array}$

Hence, the value of $\overline{x}$ is 16.38.

The value of $\overline{y}$ is,

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum y}}{n}\\ =\frac{81}{8}\\ =10.125\\ \approx 10.13\end{array}$

Hence, the value of $\overline{y}$ is 10.13.

The value of b is,

$\begin{array}{c}b=\frac{n{\displaystyle \sum xy}-\left({\displaystyle \sum x}\right)\left({\displaystyle \sum y}\right)}{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{8\left(1487\right)-\left(131\times 81\right)}{8\left(2435\right)-{\left(131\right)}^2}\\ =\frac{1285}{2319}\\ =0.554118\\ \approx 0.554\end{array}$

Hence, the value of b is 0.554.

The value of a is,

$\begin{array}{c}a=\overline{y}-b\overline{x}\\ =10.125-\left(0.554118\right)16.375\\ =10.125-9.07368\\ =1.051\end{array}$

The value of a is 1.051.

The equation of the least-squares line is,

$\begin{array}{c}\widehat{y}=a+bx\\ =1.051+0.554x\end{array}$

Hence, the equation of the least-squares line is $\widehat{y}=1.051+0.554x$.

Step by step procedure to obtain scatter plot using MINITAB software is given below:

• Choose Graph > Scatterplot.
• Choose With regression. Click OK.
• In Y variables, enter the column of x.
• In X variables, enter the column of y.
• Click OK.

(c)

To determine

Find the sample correlation coefficient r.

Find the value of the coefficient of determination $r^2$.

Mention percentage of the variation in y is explained by the least-squares model.

Answer to Problem 9CRP

The sample correlation coefficient r is 0.913.

The value of the coefficient of determination $r^2$ is 0.833.

The percentage of the variation in y is explained by the least-squares model is 83.3%.

Explanation of Solution

Calculation:

Coefficient of determination$\left(r^2\right)$:

The coefficient of determination $\left(r^2\right)$ determines the percent of the variation in response variable that is explained by the predictor variables. A larger value of $r^2$ indicates that the model is a good fit.

Step by step procedure to obtain correlation using MINITAB software is given below:

• Choose Stat > Basic Statistics > Correlation.
• In Variable, enter the column as x, y.
• Click OK.

Output using MINITAB software is given below:

From MINITAB output, the correlation is 0.913.

Hence, the correlation coefficient r is 0.913.

The value of $r^2$ is,

$\begin{array}{c}{r}^2={\left(0.913\right)}^2\\ =0.833\end{array}$

Hence, the value of the coefficient of determination $r^2$ is 0.833.

About 83.3% of the variation in y (number of employees required to process the mail) is explained by x (weight of incoming mail). Since the value of $r^2$ is large the model is a strong fit for the data.

Hence, the percentage of the variation in y is explained by the least-squares model is 83.3%.

(d)

To determine

Check whether the claim that the population correlation coefficient is positive or not.

Answer to Problem 9CRP

The population correlation coefficient is positive.

Explanation of Solution

Calculation:

Null hypothesis:

$H_0:\rho =0$

Alternative hypothesis:

$H_1:\rho >0$

Test statistic:

The test statistic formula for test correlation r is,

$t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$

Where r is the sample correlation coefficient, n is the sample size with degrees of freedom $d.f.=n-2$.

Substitute r as 0.913, and n as 8 in the test statistic formula.

$\begin{array}{c}t=\frac{0.913\sqrt{8-2}}{\sqrt{1-{\left(0.913\right)}^2}}\\ =\frac{2.2364}{0.4079596}\\ =5.48\end{array}$

The test statistic value is 5.48.

The degrees of freedom is,

$\begin{array}{c}d.f.=8-2\\ =6\end{array}$

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• Enter the Degrees of freedom as 6.
• Click the Shaded Area tab.
• Choose X Value and Right Tail, for the region of the curve to shade.
• Enter the X value as 5.48.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.0008.

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 0.0008 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, $0.0008\left(=P\text{-value}\right)<0.01\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is rejected.

Hence, the population correlation coefficient is positive between the weight of incoming mail and number of employees required to process the mail.

(e)

To determine

Find the number of employees should be assigned mail duty that day healthy crunch receives 15 pounds of mail.

Answer to Problem 9CRP

The number of employees should be assigned mail duty that day healthy crunch receives 15 pounds of mail is 9.

Explanation of Solution

Calculation:

From part (b), the equation of the least-squares line is $\widehat{y}=1.051+0.554x$.

Substitute $x=15$ in equation of the least-squares line.

$\begin{array}{c}\widehat{y}=1.051+0.554\left(15\right)\\ =1.051+8.31\\ =9.36\\ \approx 9\end{array}$

Hence, the number of employees should be assigned mail duty that day healthy crunch receives 15 pounds of mail is 9.

(f)

To determine

Verify the values of $S_e$.

Explanation of Solution

Calculation:

The value of $S_e$ is,

$\begin{array}{c}{S}_e=\sqrt{\frac{{\displaystyle \sum y^2-a{\displaystyle \sum y}-b{\displaystyle \sum xy}}}{n-2}}\\ =\sqrt{\frac{927-\left(1.051\times 81\right)-\left(0.554118\times 1487\right)}{8-2}}\\ =\sqrt{2.982589}\\ =1.726\end{array}$

Hence, the value of $S_e$ is verified as 1.726.

(g)

To determine

Find the 95% confidence interval for the number of employees required to process mail for 15 pounds of mail.

Answer to Problem 9CRP

The 95% confidence interval for the number of employees required to process mail for 15 pounds of mail is $4.871<y<13.855$.

Explanation of Solution

Calculation:

Step by step procedure to obtain confidence interval using MINITAB software is given below:

• Choose Stat > Regression > Regression.
• In Response, enter the column containing the response as y.
• In Predictors, enter the columns containing the predictor as x.
• Choose Options.
• In Prediction intervals for new observations, enter the value as 15.
• In Confidence level, enter value as 95.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the confidence interval is $4.871<y<13.855$.

Hence, the 95% confidence interval for the number of sales Dorothy would make in a week during which made 18 visits is $4.871<y<13.855$.

(h)

To determine

Check whether the claim that the slope $\beta$ of the population least-squares line is positive at the 1% level of significance or not.

Answer to Problem 9CRP

The slope $\beta$ of the population least-squares line is positive at the 1% level of significance.

Explanation of Solution

Calculation:

Null hypothesis:

$H_0:\beta =0$

Alternative hypothesis:

$H_1:\beta >0$

Test statistic:

From part (g) MINITAB output, the test statistic value is 5.47.

The degrees of freedom is,

$\begin{array}{c}d.f.=8-2\\ =6\end{array}$

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• Enter the Degrees of freedom as 6.
• Click the Shaded Area tab.
• Choose X Value and Right Tail, for the region of the curve to shade.
• Enter the X value as 5.47.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.0008.

Conclusion:

The P-value is 0.0008 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, $0.0008\left(=P\text{-value}\right)<0.01\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is rejected.

Hence, the slope $\beta$ of the population least-squares line is positive at the 1% level of significance.

(i)

To determine

Find a 80% confidence interval for $\beta$.

Interpret the confidence interval.

Answer to Problem 9CRP

The 80% confidence interval for $\beta$ is $0.408<\beta <0.700$.

Explanation of Solution

Calculation:

Confidence interval for slope:

The confidence interval formula for slope $\beta$ is,

$b-E<\beta <b+E$

Where $E=\frac{t_c{S}_e}{\sqrt{{\displaystyle \sum x^2-\frac{1}{n}{\left({\displaystyle \sum x}\right)}^2}}}$, $t_c$ is the critical value, $S_e$ is the standard error of estimate computed from the sample, n is the sample size with degrees of freedom $d.f.=n-2$.

Critical value:

Use the Appendix II: Tables, Table 6: Critical Values for Student’s t Distribution:

• In d.f. column locate the value 6.
• In the row of two-tail area locate the level of significance $\alpha =0.20$.
• The intersecting value of row and columns is 1.440.

The critical value is $\pm 1.440$.

The margin of error is,

$\begin{array}{c}E=\frac{1.440\times 1.726}{\sqrt{2,435-\frac{1}{8}{\left(131\right)}^2}}\\ =\frac{2.48544}{17.02572}\\ =0.146\end{array}$

The 80% confidence interval for $\beta$ is,

$\begin{array}{c}0.554-0.146<\beta <0.554+0.146\\ 0.408<\beta <0.700\end{array}$

Hence, the 80% confidence interval for $\beta$ is $0.408<\beta <0.700$.

The number of employees required to process the mail increases by an amount that ranges between 0.408 and 0.700, if weight of incoming mail increases by one unit.