Chapter 7.5, Problem 26E

Federal Income Tax Returns. The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return. The mean amount of deductions for this population of taxpayers was $16,642. Assume the standard deviation is σ = $2400.

1. a. What is the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for each of the following sample sizes: 30, 50, 100, and 400?
2. b. What is the advantage of a larger sample size when attempting to estimate the population mean?

To determine

Find the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for the sample sizes: 30, 50, 100, and 400.

Answer to Problem 26E

The probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for the sample size 30 is 0.3544.

The probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 50 is 0.4448.

The probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 100 is 0.5934.

The probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 400 is 0.9050.

Explanation of Solution

Calculation:

The given information is that the standard deviation is $\sigma =\$2,400$, and the mean amount of deductions for this population of taxpayers is $16,642.

Sampling distribution of $\overline{x}$:

The probability distribution of all possible values of the sample mean $\overline{x}$ is termed as the sampling distribution of $\overline{x}$.

• The expected value of $\overline{x}$ is $E\left(\overline{x}\right)=\mu$. Here, $\mu$ represents the population mean.
• The standard deviation of $\overline{x}$ is as follows:

For finite population, ${\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}\left(\frac{\sigma }{\sqrt{n}}\right)$. Here, N represents population size, n represents sample size, and $\sigma$ represents population standard deviation.

  For infinite population, ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

• When the population size (N) is infinite or finite and $\frac{n}{N}\le 0.05$, then the standard deviation of $\overline{x}$ is ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

The expected value of $\overline{x}$ is as follows:

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

Thus, the expected value of $\overline{x}$ is $16,642.

For sample size 30:

The standard deviation is ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

Substitute $\sigma$ as $2,400 and n as 30 in the formula ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{2,400}{\sqrt{30}}\\ =\frac{2,400}{5.4772}\\ =438.18\end{array}$

Thus, the standard deviation of $\overline{x}$ is $438.18.

Central limit theorem:

For a simple random sample of size n drawn from a population, the sampling distribution of the sample mean $\overline{x}$ is approximately normal when the sample size is larger.

Therefore, the sampling distribution of the sample mean $\overline{x}$ is approximately normal with $\mu =\$16,642$ and ${\sigma }_{\overline{x}}=\$438.18$.

For sample size 30:

The probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 30 is denoted as $P\left(\overline{x}\le \mu \pm 200\right)$.

$\begin{array}{c}P\left(\mu -200\le \overline{x}\le \mu +200\right)=P\left[\frac{-200}{{\sigma }_{\overline{x}}}\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le \frac{200}{{\sigma }_{\overline{x}}}\right]\\ =P\left(\frac{-200}{438.18}\le z\le \frac{200}{438.18}\right)\\ =P\left(-0.46\le z\le 0.46\right)\\ =1-2P\left(z\le 0.46\right)\end{array}$

Use Table 1: Cumulative probabilities for the standard normal distribution to find probability.

• Locate the value 0.4 in the first column.
• Locate the value 0.06 in the first row corresponding to the value 0.4 in the first column.
• The intersecting value of row and column is 0.6772.

$\begin{array}{c}P\left(\mu -200\le \overline{x}\le \mu +200\right)=1-2\left(0.6772\right)\\ =1-1.3544\\ =0.3544\end{array}$

Thus, the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 30 is 0.3544.

For sample size 50:

Substitute $\sigma$ as $2,400 and n as 50 in the formula ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{2,400}{\sqrt{50}}\\ =\frac{2,400}{7.0711}\\ =339.41\end{array}$

Thus, the standard deviation of $\overline{x}$ is $339.41.

The sampling distribution of the sample mean $\overline{x}$ is approximately normal with $\mu =\$16,642$ and ${\sigma }_{\overline{x}}=\$339.41$.

The probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 50 is denoted as $P\left(\overline{x}\le \mu \pm 200\right)$.

$\begin{array}{c}P\left(\mu -200\le \overline{x}\le \mu +200\right)=P\left[\frac{-200}{{\sigma }_{\overline{x}}}\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le \frac{200}{{\sigma }_{\overline{x}}}\right]\\ =P\left(\frac{-200}{339.41}\le z\le \frac{200}{339.41}\right)\\ =P\left(-0.59\le z\le 0.59\right)\\ =2P\left(z\le 0.59\right)-1\end{array}$

Use Table 1: Cumulative probabilities for the standard normal distribution to find probability.

• Locate the value 0.5 in the first column.
• Locate the value 0.09 in the first row corresponding to the value 0.5 in the first column.
• The intersecting value of row and column is 0.7224.

$\begin{array}{c}P\left(\mu -200\le \overline{x}\le \mu +200\right)=2\left(0.7224\right)-1\\ =1.4448-1\\ =0.4448\end{array}$

Thus, the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 50 is 0.4448.

For sample size 100:

Substitute $\sigma$ as $2,400 and n as 100 in the formula ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{2,400}{\sqrt{100}}\\ =\frac{2,400}{10}\\ =240\end{array}$

Thus, the standard deviation of $\overline{x}$ is $240.

The sampling distribution of the sample mean $\overline{x}$ is approximately normal with $\mu =\$16,642$ and ${\sigma }_{\overline{x}}=\$240$.

The probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 100 is denoted as $P\left(\overline{x}\le \mu \pm 200\right)$.

$\begin{array}{c}P\left(\mu -200\le \overline{x}\le \mu +200\right)=P\left[\frac{-200}{{\sigma }_{\overline{x}}}\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le \frac{200}{{\sigma }_{\overline{x}}}\right]\\ =P\left(\frac{-200}{240}\le z\le \frac{200}{240}\right)\\ =P\left(-0.83\le z\le 0.83\right)\\ =2P\left(z\le 0.83\right)-1\end{array}$

Use Table 1: Cumulative probabilities for the standard normal distribution to find probability.

• Locate the value 0.8 in the first column.
• Locate the value 0.03 in the first row corresponding to the value 0.8 in the first column.
• The intersecting value of row and column is 0.7967.

$\begin{array}{c}P\left(\mu -200\le \overline{x}\le \mu +200\right)=2\left(0.7967\right)-1\\ =1.5934-1\\ =0.5934\end{array}$

Thus, the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 100 is 0.5934.

For sample size 400:

Substitute $\sigma$ as $2,400 and n as 400 in the formula ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{2,400}{\sqrt{400}}\\ =\frac{2,400}{20}\\ =120\end{array}$

Thus, the standard deviation of $\overline{x}$ is $120.

The sampling distribution of the sample mean $\overline{x}$ is approximately normal with $\mu =\$16,642$ and ${\sigma }_{\overline{x}}=\$120$.

The probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 400 is denoted as $P\left(\overline{x}\le \mu \pm 200\right)$.

$\begin{array}{c}P\left(\mu -200\le \overline{x}\le \mu +200\right)=P\left[\frac{-200}{{\sigma }_{\overline{x}}}\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le \frac{200}{{\sigma }_{\overline{x}}}\right]\\ =P\left(\frac{-200}{120}\le z\le \frac{200}{120}\right)\\ =P\left(-1.67\le z\le 1.67\right)\\ =2P\left(z\le 1.67\right)-1\end{array}$

Use Table 1: Cumulative probabilities for the standard normal distribution to find probability.

• Locate the value 1.6 in the first column.
• Locate the value 0.07 in the first row corresponding to the value 1.6 in the first column.
• The intersecting value of row and column is 0.9525.

$\begin{array}{c}P\left(\mu -200\le \overline{x}\le \mu +200\right)=2\left(0.9525\right)-1\\ =1.905-1\\ =0.9050\end{array}$

Thus, the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 400 is 0.9050.

To determine

Explain the advantage of a larger sample size when attempting to estimate the population mean.

Explanation of Solution

In general, the probability of the sample mean will lie in a specified distance of the population mean when the sample size increases.

In this case, the probability of being within $\pm 200$ of $\mu$ is from 0.3544 for sample size 30 to 0.905 for sample size 400.