Chapter 13, Problem 53CE

a.

To determine

Find the regression equation.

Answer to Problem 53CE

The regression equation is $\widehat{y}=26.8054-2.4082x$.

Explanation of Solution

Calculation:

The values of x and y are given.

Regression equation:

Software procedure:

Step-by-step procedure to obtain the ‘Regression equation’ using the MegaStat software:

• In an EXCEL sheet enter the data values of x and y.
• Go to Add-Ins > MegaStat > Correlation/Regression > Regression Analysis.
• Select input range as ‘Sheet1!$B$2:$B$16’ under Y/Dependent variable.
• Select input range ‘Sheet1!$A$2:$A$16’ under X/Independent variables.
• Click on OK.

Output obtained using the MegaStat software is as given below:

From the output, the regression equation is $\widehat{y}=26.8054-2.4082x$,

Where, y is the price per share and x is the dividend.

b.

To determine

Test whether the slope is significant or not.

Answer to Problem 53CE

There is sufficient evidence to conclude that the slope of the regression line is different from zero.

Explanation of Solution

It is given that the regression equation is $\widehat{y}=26.8054-2.4082x$.

The sample size is 30 and the standard error of the slope is 0.3279.

From the regression equation, the estimated slope of the regression line is $b=-2.4082$ and the standard error of $b$ is $s_b=0.3279$.

Let $\beta$ be the slope of the regression line.

The given test hypotheses are as follows:

Null hypothesis:

$H_0:\beta =0$

That is, the slope of the regression line is equal to zero.

Alternate hypothesis:

$H_1:\beta \ne 0$

That is, the slope of the regression line is not equal to zero.

Assume that the level of significance is 0.05.

Test statistic:

The t-test statistic is as follows:

$t=\frac{b-0}{s_b}$,

Where, $b$ is the slope of the computed regression line and $s_b$ is the standard error of $b$.

Thus, the following is obtained:

$\begin{array}{c}t=\frac{b}{s_b}\\ =\frac{-2.4082}{0.3279}\\ =-7.34\end{array}$

Here, the sample size is $n=30$. Thus, the degrees of freedom is as follows:

$\begin{array}{c}n-2=30-2\\ =28\end{array}$

Critical value:

Software procedure:

Step-by-step software procedure to obtain the critical value $t_{\frac{\alpha }{2}}$ using the EXCEL software:

• Open an EXCEL file.
• In cell A1, enter the formula “=T.INV(0.025,28)”.

Output obtained using the EXCEL is given as follows:

From the EXCEL output, the critical value is –2.048 ($-t_{\frac{\alpha }{2}}$).

Decision based on the critical value:

Reject the null hypothesis, if $\left|t\right|>\left|t_{\frac{\alpha }{2}}\right|$. Otherwise, fail to reject H₀.

Conclusion:

The t-calculated value is –7.34 and the critical value is –2.048.

That is, $\left|t\text{-calculated value}\right|\left(=7.34\right)>\left|t_{\frac{\alpha }{2}}\right|\left(=2.048\right)$.

Thus, the null hypothesis is rejected.

Hence, there is sufficient evidence to conclude that the slope of the regression line is different from zero.

c.

To determine

Find and interpret the value of coefficient of determination.

Answer to Problem 53CE

The coefficient of determination is 0.658.

Explanation of Solution

From Part (a), the value of coefficient of determination is 0.658.

Therefore, 65.8% of variation in the ‘selling price’ is explained by ‘the annual dividend’.

d.

To determine

Find the value of correlation coefficient.

Test whether the correlation in the population is greater than zero or not.

Answer to Problem 53CE

The correlation coefficient is 0.811.

There is enough evidence to infer that the correlation in the population is greater than zero.

Explanation of Solution

Calculation:

The correlation coefficient is as follows:

$\begin{array}{c}\text{Correlation coefficient =}\sqrt{r^2}\\ =\sqrt{0.658}\\ =0.811\end{array}$

The given sample size is 30 and correlation is 0.811.

Denote the population correlation as $\rho$.

The hypotheses are given below:

Null hypothesis:

$H_0:\rho \le 0$

That is, the correlation in the population is less than or equal to zero.

Alternative hypothesis:

$H_1:\rho >0$

That is, the correlation in the population is greater than zero.

Test statistic:

The test statistic is as follows:

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

Here, the sample size is 30 and the correlation coefficient is 0.811.

The test statistic is as follows:

$\begin{array}{c}t=\frac{0.811\sqrt{30-2}}{\sqrt{1-{0.811}^2}}\\ =\frac{0.811\times \sqrt{28}}{\sqrt{0.342279}}\\ =7.335\\ \approx 7.34\end{array}$

Degrees of freedom:

$\begin{array}{c}df=n-2\\ =30-2\\ =28\end{array}$

The level of significance is 0.05. Therefore, $1-\alpha =0.95$.

Critical value:

Software procedure:

Step-by-step software procedure to obtain the critical value using the EXCEL software:

• Open an EXCEL file.
• In cell A1, enter the formula “=T.INV (0.95, 28)”.

Output obtained using the EXCEL is given as follows:

From the EXCEL output, the critical value is 1.701.

Conclusion:

The value of test statistic is 7.34 and the critical value is 1.701.

Here, $\left|t\text{-calculated}\right|\left(=7.34\right)>\left|t\text{-critical value}\right|\left(=1.701\right)$.

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that the correlation in the population is greater than zero.