Chapter 14, Problem 67SE

Income and Percent Audited. The Transactional Records Access Clearinghouse at Syracuse University reported data showing the odds of an Internal Revenue Service audit. The following table shows the average adjusted gross income reported and the percent of the returns that were audited for 20 selected IRS districts.

1. a. Develop the estimated regression equation that could be used to predict the percent audited given the average adjusted gross income reported.
2. b. At the .05 level of significance, determine whether the adjusted gross income and the percent audited are related.
3. c. Did the estimated regression equation provide a good fit? Explain.
4. d. Use the estimated regression equation developed in part (a) to calculate a 95% confidence interval for the expected percent audited for districts with an average adjusted gross income of $35,000.

To determine

Find the estimated regression equation to predict the percent audited given the average adjusted gross income reported.

Answer to Problem 67SE

The estimated regression equation to predict the percent audited, given the average adjusted gross income reported is as follows:

$\widehat{\text{Percent Audited}}=-0.471+0.000039\text{Adjusted Gross income}$.

Explanation of Solution

Calculation:

The data are related to the adjusted gross income ($) and percent audited for 20 selected IRS districts.

In the given problem, the percent audited is the dependent variable (y) and the adjusted gross income is the independent variable (x).

Regression:

Software procedure:

Step-by-step procedure to obtain the estimated regression equation using EXCEL:

• In Excel sheet, enter Adjusted Gross income and Percent Audited in different columns.
• In Data, select Data Analysis and choose Regression.
• In Input Y Range, select Percent Audited.
• In Input X Range, select Adjusted Gross income.
• Select Labels.
• Click OK.

Output obtained using EXCEL is given below:

Thus, the estimated regression equation to predict the percent audited, given the average adjusted gross income reported is as follows:

$\widehat{\text{Percent Audited}}=-0.471+0.000039\text{Adjusted Gross income}$.

To determine

Use $\alpha =0.05$ to determine whether the adjusted gross income and the percent audited are related.

Answer to Problem 67SE

There is a significant relationship between the adjusted gross income and the percent audited.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

$H_0:{\beta }_1=0$

That is, there is no significant relationship between the adjusted gross income and the percent audited.

Alternative hypothesis:

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

That is, there is a significant relationship between the adjusted gross income and the percent audited.

From the output in Part (a), it is found that the F-test statistic is 4.99.

Level of significance:

The given level of significance is $\alpha =0.05$.

p-value:

From Part (a) in the output, it is found that the p-value is 0.038.

Rejection rule:

If the $p\text{-value}\le \alpha$, then reject the null hypothesis.

Conclusion:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.038\right)<\alpha \left(=0.05\right)$

Thus, the decision is “reject the null hypothesis”.

Therefore, the data provide sufficient evidence to conclude that there is a significant relationship between the adjusted gross income and the percent audited.

Thus, the adjusted gross income and the percent audited are related.

To determine

Explain whether the estimated regression equation provides a good fit to the data.

Answer to Problem 67SE

The estimated regression equation does not provide a good fit to the data.

Explanation of Solution

$R^2$(R-square):

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 the fraction of variability of the response variable (y) that is accounted for by the linear regression model.

In the given output of Part (a), $\text{R square}=21.71\%$.

Thus, the percentage of variation in the observed values of percent audited that is explained by the regression is 21.71%, which indicates that only 21.71% of the variability in percent audited is explained by the variability in the adjusted gross income using the linear regression model.

Thus, the estimated regression equation does not provide a good fit to the data.

To determine

Find a 95% confidence interval for the expected percent audited for districts with an average adjusted gross income of $35,000.

Answer to Problem 67SE

The 95% confidence interval for the expected percent audited for districts with an average adjusted gross income of $35,000 is $\left(0.7841\%,\text{1}\text{.0039\%}\right)$.

Explanation of Solution

Calculation:

The estimate of standard deviation of $\widehat{y}*$ is given as follows:

$s_{\widehat{y}*}=s\sqrt{\frac{1}{n}+\frac{{\left(x*-\overline{x}\right)}^2}{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}}$

From Part (a), the estimated regression equation is as follows:

$\widehat{\text{Percent Audited}}=-0.471+0.000039\text{Adjusted Gross income}$.

Also, the mean square error (MSE) is 0.0436.

According to the regression equation $\widehat{y}*=-0.471+0.000039x$, the value of $\widehat{y}*$ for $\text{adjusted gross income}=\$35,000$ is as follows:

$\begin{array}{c}\widehat{y}*=-0.471+0.000039\left(35,000\right)\\ =-0.471+1.365\\ =0.894\end{array}$

Thus, the possible value of the dependent variable y when $x=\$35,000$ is 0.894.

The standard error of the estimate is obtained as follows:

$\begin{array}{c}s=\sqrt{\text{MSE}}\\ =\sqrt{0.0436}\\ =0.2088\end{array}$

Thus, the standard error of the estimate is 0.2088.

It is known that for a sample of size n, the mean of a random variable x can be obtained as follows:

$\overline{x}=\frac{{\displaystyle \sum _1^n{x}_i}}{n}$

Thus, the mean of the random variable x is obtained as follows:

$\begin{array}{c}\overline{x}=\frac{\left(\begin{array}{l}36,664+39,845+34,886+32,512+34,531+35,995+37,799+\\ 33,876+30,513+30,174+30,060+37,153+34,918+33,291+\\ 31,504+29,199+33,072+30,859+32,566+34,296\end{array}\right)}{20}\\ =\frac{672,713}{20}\\ =33,636\end{array}$

The mean of the random variable x is $\overline{x}=33,636$.

The value of ${\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}$ is calculated as follows:

$x_i$	$x_i-\overline{x}$	${\left(x_i-\overline{x}\right)}^2$
36,664	3,028	9,168,784
38,845	5,209	27,133,681
34,886	1,250	1,562,500
32,512	–1,124	1,263,376
34,531	895	801,025
35,995	2,359	5,564,881
37,799	4,163	17,330,569
33,876	240	57,600
30,513	–3,123	9,753,129
30,174	–3,462	11,985,444
30,060	–3,576	12,787,776
37,153	3,517	12,369,289
34,918	1,282	1,643,524
33,291	–345	119,025
31,504	–2,132	4,545,424
29,199	–4,437	19,686,969
33,072	–564	318,096
30,859	–2,777	7,711,729
32,566	–1,070	1,144,900
34,296	660	435,600
		${\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}=145,383,321$

For the adjusted gross income of (35,000), the standard deviation of $\widehat{y}*$ is obtained as follows:

$\begin{array}{c}{s}_{\widehat{y}*}=s\sqrt{\frac{1}{n}+\frac{{\left(x*-\overline{x}\right)}^2}{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}}\\ =\left(0.2088\right)\sqrt{\frac{1}{20}+\frac{{\left(35,000-33,636\right)}^2}{145,383,321}}\\ =\left(0.2088\right)\sqrt{\frac{1}{20}+\frac{1,860,496}{145,383,321}}\\ =\left(0.2088\right)\sqrt{0.0628}\end{array}$

     $\begin{array}{c}=\left(0.2088\right)\left(0.2506\right)\\ \approx 0.0523\end{array}$

Thus, the standard deviation of $\widehat{y}*$ for the adjusted gross income of 35,000 is 0.0523.

The confidence interval for the expected value of y $\left(E\left(y*\right)\right)$ is $\widehat{y}*\pm t_{\frac{\alpha }{2}}{s}_{\widehat{y}*}$, where the confidence coefficient is $1-\alpha$ and $t_{\frac{\alpha }{2}}$ is the value of t distribution with the degrees of freedom of $n-2$ for a sample of size n.

Degrees of freedom:

For a sample of size n, the degrees of freedom is given as $n-2$.

In this given problem, for a sample size 20, the degrees of freedom is as follows:

$\begin{array}{c}df=n-2\\ =20-2\\ =18\end{array}$

Thus, the degrees of freedom is 18.

Level of significance:

The given level of significance is $\alpha =0.05$.

For both tails distribution:

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

Form Table 2 of “t Distribution” in Appendix B, it is found that the value of t test statistic with the level of significance 0.025 and degrees of freedom 18 is $t_{0.025,18}=2.101$.

Therefore, the required confidence interval is obtained as follows:

$\begin{array}{c}\widehat{y}*\pm t_{\frac{\alpha }{2}}{s}_{\widehat{y}*}=0.894\pm \left(2.101\right)\left(0.0523\right)\\ =0.894\pm 0.1099\\ =\left(0.7841,\text{ 1}\text{.0039}\right)\end{array}$

Thus, the 95% confidence interval for the expected percent audited for districts with an average adjusted gross income of $35,000 is $\left(0.7841\%,\text{1}\text{.0039\%}\right)$.