Chapter 6, Problem 8P

a.

To determine

Obtain the sampling distribution of mean $\overline{x}$ for itemized deductions for the population of taxpayers.

Answer to Problem 8P

The sampling distribution of the mean $\overline{x}$ for itemized deductions for the random sample of 30 taxpayers with adjusted gross income between $30,000 and $60,000 is normal with mean $E(\overline{x})=16,642$ and standard deviation ${\sigma }_{\overline{x}}=438.18$.

The sampling distribution of the mean $\overline{x}$ for itemized deductions for the random sample of 50 taxpayers with adjusted gross income between $30,000 and $60,000 is normal with mean $E(\overline{x})=16,642$ and standard deviation ${\sigma }_{\overline{x}}=339.41$.

The sampling distribution of the mean $\overline{x}$ for itemized deductions for the random sample of 100 taxpayers with adjusted gross income between $30,000 and $60,000 is normal with mean $E(\overline{x})=16,642$ and standard deviation ${\sigma }_{\overline{x}}=240$.

The sampling distribution of the mean $\overline{x}$ for itemized deductions for the random sample of 400 taxpayers with adjusted gross income between $30,000 and $60,000 is normal with mean $E(\overline{x})=16,642$ and standard deviation ${\sigma }_{\overline{x}}=120$.

Explanation of Solution

It is given that 33% of taxpayers in the year 2010 with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return. The mean amount of deductions for the given population of taxpayers is $\mu =16,642$ and standard deviation is $\sigma =2400$.

The sampling distribution of the mean is approximately normal when the sample size is large regardless of the underlying distribution of the population from where it is taken.

The mean of the $\overline{x}$ is $E(\overline{x})=\mu$  and standard deviation of $\overline{x}$ is ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

In this context, $\overline{x}$ is the mean amount of itemized deductions for the given population of taxpayers.

For the random sample of 30 taxpayers, the mean and standard deviation is calculated as,

$\begin{array}{c}E(\overline{x})=\mu \\ =16,642\end{array}$

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{2400}{\sqrt{30}}\\ =438.18\end{array}$

Thus, for the random sample of 30 taxpayers with adjusted gross income between $30,000 and $60,000, the sampling distribution of the mean $\overline{x}$ for itemized deductions is normal with mean $E(\overline{x})=16,642$ and standard deviation ${\sigma }_{\overline{x}}=438.18$.

For the random sample of 50 taxpayers, the mean and standard deviation is calculated as,

$\begin{array}{c}E(\overline{x})=\mu \\ =16,642\end{array}$

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{2400}{\sqrt{50}}\\ =339.41\end{array}$

Thus, for the random sample of 50 taxpayers with adjusted gross income between $30,000 and $60,000, the sampling distribution of the mean $\overline{x}$ for itemized deductions is normal with mean $E(\overline{x})=16,642$ and standard deviation ${\sigma }_{\overline{x}}=339.41$

For the random sample of 100 taxpayers, the mean and standard deviation is calculated as,

$\begin{array}{c}E(\overline{x})=\mu \\ =16,642\end{array}$

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{2400}{\sqrt{100}}\\ =240\end{array}$

Thus, for the random sample of 100 taxpayers with adjusted gross income between $30,000 and $60,000, the sampling distribution of the mean $\overline{x}$ for itemized deductions is normal with mean $E(\overline{x})=16,642$ and standard deviation ${\sigma }_{\overline{x}}=240$

For the random sample of 400 taxpayers, the mean and standard deviation is calculated as,

$\begin{array}{c}E(\overline{x})=\mu \\ =16,642\end{array}$

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{2400}{\sqrt{400}}\\ =120\end{array}$

Thus, for the random sample of 400 taxpayers with adjusted gross income between $30,000 and $60,000, the sampling distribution of the mean $\overline{x}$ for itemized deductions is normal with mean $E(\overline{x})=16,642$ and standard deviation ${\sigma }_{\overline{x}}=120$

b.

To determine

Explain the advantage of using larger sample size.

Explanation of Solution

From part (a), it can be observed that when sample size increases the standard error of the sample mean decreases.

Thus, larger sample reduces the standard error which results in a more accurate estimate of the population mean.