Chapter 11.3, Problem 16E

a.

To determine

Find the value of $b_1$.

Answer to Problem 16E

The slope $b_1$ is –0.154.

Explanation of Solution

Calculation:

The given information is that the sample data consists of 8 values for x and y.

Slope or $b_1$:

$b_1=r\frac{s_y}{s_x}$

where,

r represents the correlation coefficient between x and y.

$s_y$ represents the standard deviation of y.

$s_x$ represents the standard deviation of x.

Software procedure:

Step-by-step procedure to find the mean, standard deviation for x and y values using MINITAB is given below:

• • Choose Stat > Basic Statistics > Display Descriptive Statistics.
• • In Variables enter the columns of x and y.
• • Choose Options Statistics, and select Mean and Standard deviation.
• • Click OK.

Output obtained from MINITAB is given below:

Correlation:

$r=\frac{1}{n-1}{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}$

The table shows the calculation of correlation:

x	y	$x-\overline{x}$	$y-\overline{y}$	$\frac{x-\overline{x}}{s_x}$	$\frac{y-\overline{y}}{s_y}$	$\left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)$
12	13	–0.38	–0.88	–0.0795	–0.888	0.0706
17	14	4.62	0.12	0.96653	0.12109	0.117
3	16	–9.38	2.12	–1.9623	2.13925	–4.1979
17	13	4.62	–0.88	0.96653	–0.888	–0.8583
16	14	3.62	0.12	0.75732	0.12109	0.0917
11	14	–1.38	0.12	–0.2887	0.12109	–0.035
14	13	1.62	–0.88	0.33891	–0.888	–0.301
9	14	–3.38	0.12	–0.7071	0.12109	–0.0856
Total						–5.1985

Thus, the correlation is

$\begin{array}{c}r=\frac{-5.1985}{8-1}\\ =\frac{-5.1985}{7}\\ =-0.743\end{array}$

$b_1=r\frac{s_y}{s_x}$

Substitute r as –0.743, $s_y$ as 0.991 and $s_x$ as 4.78.

$\begin{array}{c}{b}_1=-0.743\left(\frac{0.991}{4.78}\right)\\ =-0.743\left(0.207\right)\\ =-0.154\end{array}$

Thus, the slope $b_1$ is –0.154.

b.

To determine

Find the residual standard deviation $s_e$.

Answer to Problem 16E

The residual standard deviation $s_e$ is 0.717.

Explanation of Solution

Calculation:

Finding the value of the intercept term before find the residual standard deviation:

Intercept or $b_0$:

$b_0=\overline{y}-b_1\overline{x}$

$\overline{y}$ represents the mean of y values.

$\overline{x}$ represents the mean of x values.

$b_1$ represents the slope coefficient.

Substitute $\overline{y}$ as 13.88, $\overline{x}$ as 12.38 and $b_1$ as –0.154.

$\begin{array}{c}{b}_0=13.88-\left(-0.154\right)\left(12.38\right)\\ =13.88+\left(0.154\right)\left(12.38\right)\\ =13.88+1.91\\ =15.79\end{array}$

Thus, the intercept $b_0$ is 15.79.

The residual standard deviation $s_e$ is calculated using the formula,

$s_e=\sqrt{\frac{{{\displaystyle \sum \left(y-\widehat{y}\right)}}^2}{n-2}}$

Where,

${{\displaystyle \sum \left(y-\widehat{y}\right)}}^2$ represents the sum of squares due to error

n represents the sample size.

Thus, the estimated regression equation is $\widehat{y}=15.79-0.154x$

Use the estimated regression equation to find the predicted value of y for each value of x.

x	y	$\widehat{y}$	$y-\widehat{y}$	${\left(y-\widehat{y}\right)}^2$
12	13	13.942	–0.942	0.88736
17	14	13.172	0.828	0.68558
3	16	15.328	0.672	0.45158
17	13	13.172	–0.172	0.02958
16	14	13.326	0.674	0.45428
11	14	14.096	–0.096	0.00922
14	13	13.634	–0.634	0.40196
9	14	14.404	–0.404	0.16322
Total				3.083

Substitute ${{\displaystyle \sum \left(y-\widehat{y}\right)}}^2$ as 3.083 and n as 8.

$\begin{array}{c}{s}_e=\sqrt{\frac{3.083}{8-2}}\\ =\sqrt{\frac{3.083}{6}}\\ =\sqrt{0.514}\\ =0.717\end{array}$

Thus, the residual standard deviation $s_e$ is 0.717.

c.

To determine

Find the sum of squares for x.

Answer to Problem 16E

The sum of squares for x is 159.88.

Explanation of Solution

Calculation:

The table shows the calculation of sum of squares for x

x	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
12	–0.38	0.1444
17	4.62	21.3444
3	–9.38	87.9844
17	4.62	21.3444
16	3.62	13.1044
11	–1.38	1.9044
14	1.62	2.6244
9	–3.38	11.4244
Total		159.88

Thus, the sum of squares for x is 159.88.

d.

To determine

Find the standard error of $b_1,s_b$.

Answer to Problem 16E

The standard error of $b_1$, $s_b$ is 0.057.

Explanation of Solution

Calculation:

The standard error of $b_1$ is calculated by using the formula:

$s_b=\frac{s_e}{\sqrt{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}}$

Where,

$s_e$ represents the residual standard deviation.

${\displaystyle \sum {\left(x-\overline{x}\right)}^2}$ represents the sum of squares due to x.

Substitute ${\displaystyle \sum {\left(x-\overline{x}\right)}^2}$ as 159.88 and $s_e$ as 0.717.

$\begin{array}{c}{s}_b=\frac{0.717}{\sqrt{159.88}}\\ =\frac{0.717}{12.64}\\ =0.057\end{array}$

Thus, the standard error of $b_1$, $s_b$ is 0.057.

e.

To determine

Find the critical value for a 95% confidence interval of ${\beta }_1$.

Answer to Problem 16E

The critical value for a 95% confidence interval of ${\beta }_1$ is 2.447.

Explanation of Solution

Calculation:

Critical value:

Software procedure:

Step-by-step procedure to find the critical value using MINITAB is given below:

• • Choose Graph > Probability Distribution Plot choose View Probability > OK.
• • From Distribution, choose ‘t’ distribution.
• • In Degrees of freedom, enter 6.
• • Click the Shaded Area tab.
• • Choose Probability and Two tail for the region of the curve to shade.
• • Enter the Probability value as 0.05.
• • Click OK.

Output obtained from MINITAB is given below:

Thus, the critical value for a 95% confidence interval of ${\beta }_1$ is 2.447.

f.

To determine

Find the margin of error for a 95% confidence interval of ${\beta }_1$.

Answer to Problem 16E

The margin of error of $b_1$ is 0.139.

Explanation of Solution

Calculation:

The margin of error for a 95% confidence interval of ${\beta }_1$ is calculated using the formula:

$\text{Margin of error}=t_{\frac{\alpha }{2}}\cdot s_{b_1}$

Where,

$t_{\frac{\alpha }{2}}$ represents the two tailed critical value for a given level of significance.

$s_{b_1}$ represents the standard error of $b_1$.

Substitute $t_{\frac{\alpha }{2}}$ as 2.447 and $s_{b_1}$ as 0.057.

$\begin{array}{c}\text{Margin of error}=\left(2.447\right)\left(0.057\right)\\ =0.139\end{array}$

Thus, the margin of error of $b_1$ is 0.139.

g.

To determine

Construct the 95% confidence interval for ${\beta }_1$.

Answer to Problem 16E

The 95% confidence interval for ${\beta }_1$ is $\left(-0.293,-0.015\right)$

Explanation of Solution

Calculation:

The confidence interval for ${\beta }_1$ is calculated by using the formula:

$\text{Confidence interval}=b_1\pm t_{\frac{\alpha }{2}}\cdot s_{b_1}$

Where,

$b_1$ represents the estimated slope coefficient.

$t_{\frac{\alpha }{2}}\cdot s_{b_1}$ represents the margin of error.

$\begin{array}{c}\text{Confidence interval}=-0.154\pm \left(2.447\right)\cdot \left(0.057\right)\\ =-0.154\pm 0.139\\ =-0.293,-0.015\end{array}$

Thus, the 95% confidence interval for ${\beta }_1$ is $\left(-0.293,-0.015\right)$

h.

To determine

Test the significance of ${\beta }_1$ using 5% level of significance.

Answer to Problem 16E

There is a support of evidence to conclude that there is a linear relationship between x and y at 5% level of significance.

Explanation of Solution

Calculation:

The hypotheses used for testing the significance is given below:

Null hypothesis:

$H_0:{\beta }_1=0$

That is, there is no linear relationship between x and y.

Alternate hypothesis:

$H_1:{\beta }_1\ne 0$

That is, there is a linear relationship between x and y.

Test statistic:

$t=\frac{{\widehat{\beta }}_1-{\beta }_1}{s_{b_1}}$

Where,

${\widehat{\beta }}_1$ represents the estimated slope coefficient.

${\beta }_1$ represents the hypothesized value of slope coefficient.

$s_{b_1}$ represents the standard error of slope coefficient.

Substitute ${\widehat{\beta }}_1$ as –0.154, ${\beta }_1$ as 0 and $s_{b_1}$ as 0.057.

$\begin{array}{c}t=\frac{-0.154-0}{0.057}\\ =\frac{-0.154}{0.057}\\ =-2.70\end{array}$

From part e, the critical value is identified as –2.447.

Decision Rule:

If the positive test statistic value is greater than or equal to the positive critical value or less than the negative critical value, then reject the null hypothesis. Otherwise do not reject the null hypothesis.

Conclusion:

The test statistic value is –2.70 and the critical value is –2.447.

The test statistic value is lesser than the critical value.

That is, $-2.70\left(=\text{test statistic}\right)<-2.447\left(=\text{critical value}\right)$

Thus, the null hypothesis is rejected.

Hence, there is a support of evidence to conclude that there is a linear relationship between x and y.