Chapter 8.4, Problem 43E

A Phoenix Wealth Management/Harris Interactive survey of 1500 individuals with net worth of $1 million or more provided a variety of statistics on wealthy people (BusinessWeek, September 22, 2003). The previous three-year period had been bad for the stock market, which motivated some of the questions asked.

1. a. The survey reported that 53% of the respondents lost 25% or more of their portfolio value over the past three years. Develop a 95% confidence interval for the proportion of wealthy people who lost 25% or more of their portfolio value over the past three years.
2. b. The survey reported that 31% of the respondents feel they have to save more for retirement to make up for what they lost. Develop a 95% confidence interval for the population proportion.
3. c. Five percent of the respondents gave $25,000 or more to charity over the previous year. Develop a 95% confidence interval for the proportion who gave $25,000 or more to charity.
4. d. Compare the margin of error for the interval estimates in parts (a), (b), and (c). How is the margin of error related to $\bar{p}$? When the same sample is being used to estimate a variety of proportions, which of the proportions should be used to choose the planning value p*? Why do you think p* = .50 is often used in these cases?

To determine

Find the 95% confidence interval for the proportion of wealthy people who lost 25% or more of their portfolio over the past three years.

Answer to Problem 43E

The 95% confidence interval for the proportion of wealthy people who lost 25% or more of their portfolio over the past three years is (0.5047, 0.5553).

Explanation of Solution

Calculation:

A total of 1,500 individuals were surveyed. The survey reported that about 53% of the respondents lost 25% or more of their portfolio over the past three years. That is, $\overline{p}=0.53$.

The formula for margin of error is given below:

$\text{Margin of error}=z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}$

The value of $z_{\frac{\alpha }{2}}$ is obtained as follows:

$\begin{array}{c}\alpha =1-0.95\\ =0.05\end{array}$

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

From “Table 8.1: Values of $z_{\frac{\alpha }{2}}$ for the most commonly used confidence levels”, the value of $z_{\frac{\alpha }{2}}$ for 95% is 1.96.

The margin of error is calculated as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.53\left(1-0.53\right)}{1,500}}\\ =1.96\sqrt{\frac{0.2491}{1,500}}\\ =1.96\times 0.01289\\ =0.0253\end{array}$

Thus, the value of margin of error is 0.0253.

Confidence interval:

The 95% confidence interval for population proportion is calculated as follows:

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=\overline{p}\pm \text{margin of error}\\ =0.53\pm 0.0253\\ =\left(0.53-0.0253,0.53+0.0253\right)\\ =\left(0.5047,0.5553\right)\end{array}$

Thus, the 95% confidence interval for the proportion of wealthy people who lost 25% or more of their portfolio over the past three years is (0.5047, 0.5553).

To determine

Find the 95% confidence interval for the population proportion.

Answer to Problem 43E

The 95% confidence interval for the population proportion is (0.2866, 0.3334).

Explanation of Solution

Calculation:

The survey results that 31% of the respondents feel they have to save more for the retirement to make up for what they lost. That is, $\overline{p}=0.31$.

The value of $z_{\frac{\alpha }{2}}$ is obtained as follows:

$\begin{array}{c}\alpha =1-0.95\\ =0.05\end{array}$

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

From “Table 8.1: Values of $z_{\frac{\alpha }{2}}$ for the most commonly used confidence levels”, the value of $z_{\frac{\alpha }{2}}$ for 95% is 1.96.

Margin of error:

The value of margin of error is calculated as follows:

$\begin{array}{c}\text{Margin of error,}E=1.96\sqrt{\frac{0.31\left(1-0.31\right)}{1,500}}\\ =1.96\sqrt{\frac{0.2139}{1,500}}\\ =1.96\times 0.01194\\ =0.0234\end{array}$

Thus, the value of margin of error is 0.0234.

Confidence interval:

The 95% confidence interval for population proportion is calculated as follows:

$\begin{array}{c}\text{Confidence interval=}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}\\ =\overline{p}\pm \text{Margin of error}\\ =0.31\pm 0.0234\\ =\left(0.31-0.0234,0.31+0.0234\right)\\ =\left(0.2866,0.3334\right)\end{array}$

Thus, the 95% confidence interval for the population proportion is (0.2866, 0.3334).

To determine

Find the 95% confidence interval for the proportion of respondents who gave $25,000 or more to charity.

Answer to Problem 43E

The 95% confidence interval for the proportion of respondents who gave $25,000 or more to charity is (0.039, 0.061).

Explanation of Solution

Calculation:

The survey reported that 5% of the respondents gave $25,000 or more to charity. That is, $\overline{p}=0.05$.

The value of margin of error is calculated as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.05\left(1-0.05\right)}{1,500}}\\ =1.96\sqrt{\frac{0.0475}{1,500}}\\ =1.96\times 0.005627\\ =0.0110\end{array}$

Thus, the value of margin of error is 0.0110.

Confidence interval:

The 95% confidence interval for population proportion is calculated as follows:

$\begin{array}{c}\text{Confidence interval}=\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}\\ =\overline{p}\pm \text{Margin of error}\\ =0.05\pm 0.0110\\ =\left(0.05-0.0110,0.05+0.0110\right)\\ =\left(0.039,0.0610\right)\end{array}$

Thus, the 95% confidence interval for the proportion of respondents who gave $25,000 or more to charity is (0.039, 0.061).

To determine

Compare the margin of errors in parts (a), (b), and (c).

Explain the margin of error related to sample proportion.

Identify the proportion that should be used to choose the planning value $p\ast$.

Explain why $p\ast =0.05$ is often used in this case.

Explanation of Solution

The margin of errors for part (a) is 0.0253, for part (b) is 0.0234, and for part (c) is 0.3334.

From the results, it can be observed that the margin of error decreases as the $\overline{p}$ value decreases.

Here, the margin of error for the entire interval is less than the given value (say 0.05). Therefore, an estimate of the largest proportion is taken as the planning value $p\ast$. That is, $p\ast =0.53$. The planning value $p\ast =0.5$ ensures that the margin of error for the interval estimate will be small.