Who prepares your tax return? Refer to the Behavioral Research and Accounting (January 2015) study on income tax compliance, Exercise 5.50 (p. 321). Recall that in a sample of 270 U.S. adult workers, the researchers found that 37% prepare their own tax return.

a. Construct a 99% confidence interval for the true proportion of all U.S. adult workers who prepare their own tax return.

b. Suppose an IRS tax consultant claims that 50% of all U.S. adult workers prepare their own tax return. Make an inference about this claim.

c. According to the researchers, about 70% of the sampled workers were recruited from a shopping mall (where they were reimbursed $5 for their time) and about 30% were full-time workers enrolled in a professional graduate degree program. How might this information impact the inference you made in part b?

Referring to the Behavioral Research and Accounting (January 2015) study on income tax compliance, exercise 5.50.

There are 270 U.S adult workers, the researchers found that 37% prepare their own tax returns.

a.

The sample proportion is the point estimator of the population proportion p.

The sample proportion is,

$\begin{array}{rcl}\hat{\mathrm{p}}& =& 37\%\\ & =& 0.37\end{array}$

Then the level of$100\left(1-\mathrm{\alpha }\right)\%$ confidence interval for p (proportion) is,

$\hat{\mathrm{p}}\pm {\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{\hat{\mathrm{p}}\left(1-\hat{\mathrm{p}}\right)}{\mathrm{n}}}$

For a 99% confidence interval, the value of$\frac{\mathrm{\alpha }}{2}$ is,

$$\begin{gathered} 100\left(1-\mathrm{\alpha }\right)\%=99\% \\ \left(1-\mathrm{\alpha }\right)=0.99 \end{gathered}$$

For $\mathrm{\alpha }=0.01\mathrm{and}\frac{\mathrm{\alpha }}{2}=0.005$,

The 99% confidence interval is,

$\begin{array}{rcl}\hat{\mathrm{p}}\pm {\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{\hat{\mathrm{p}}\left(1-\hat{\mathrm{p}}\right)}{\mathrm{n}}}& =& 0.37\pm 2.576\sqrt{\frac{0.37\left(1-0.37\right)}{270}}\left[\mathrm{From}\mathrm{Standard}\mathrm{Normal}\mathrm{Table}\right]\\ & =& \left(0.37\pm 2.576\sqrt{0.000863}\right)\\ & =& \left(0.37\pm 0.0757\right)\\ & =& \left(0.2943,0.4457\right)\end{array}$

Therefore, the 99% confidence interval for the true proportion of all U.S adult workers is$\left(0.2943,0.4457\right)$.

b.

The claim population proportion is 0.5, in part (a) the 99% confidence interval for the true proportion of all U.S adult workers is$\left(0.2943,0.4457\right)$.

So, the claim population proportion does not belong to the confidence interval $\left(0.2943,0.4457\right)$.

So, at a 1% level of significance, the claim is not true.

c.

According to the researchers, about 70% of the sampled workers were recruited from a shopping malland about 30% were full-time workers enrolled in a professional graduate degree program, from the given information the value of the sample proportion does not change, that is why the confidence interval will remain the same.

So, this information does not impact the inference in part (b).