Chapter 8.4, Problem 43E

a.

To determine

Find the 95% confidence interval for the proportion of wealthy people who lost 25% or more of their portfolio value 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 value over the past three years is (0.8799, 0.9201).

Explanation of Solution

Calculation:

The PWMI survey conducted a survey on individuals with a net worth of $1 million or more. It provided a variety of statistics on wealthy people for a sample of 1,500 individual. It is given that 53% of the respondents lost 25% or more of their portfolio value over the past 3 years.

The sampling distribution of $\overline{p}$ is the population proportion p.

The formula for margin of error is as follows:

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

The value of $\overline{p}$ is 0.53.

From “Table 8.1 value of $z_{\frac{\alpha }{2}}$”, the value of $z_{\frac{\alpha }{2}}$ for 95% is 1.96.

The value of margin of error is calculated as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.53\left(1-0.53\right)}{1,500}}\\ =1.96\sqrt{\frac{0.2491}{1,500}}\\ =1.96\times 0.0129\\ =0.0253\end{array}$

Thus, the value of margin of error is 0.0253.

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

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

Thus, the 95% confidence interval for population proportion is $\underset{\_}{\left(0.5047,0.5553\right)}$.

b.

To determine

Compute the 95% confidence interval for the proportion.

Answer to Problem 43E

The 95% confidence interval for the proportion of is $\underset{\_}{\left(0.2867,0.3333\right)}$.

Explanation of Solution

Calculation:

The survey results that 31% of respondents feel they have to save more for retirement to make up for what they lost.

The value of $\overline{p}$ is 0.31.

From “Table 8.1 value of $z_{\frac{\alpha }{2}}$”, the value of $z_{\frac{\alpha }{2}}$ for 95% is 1.96.

The value of margin of error is obtained as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.31\left(1-0.31\right)}{1,500}}\\ =1.96\sqrt{\frac{0.2139}{1,500}}\\ =1.96\times 0.0119\\ =0.0233\end{array}$

Thus, the value of margin of error is 0.0233.

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

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.31\pm 0.0233\\ =\left(0.31-0.0233,0.31+0.0233\right)\\ =\left(0.2867,0.3333\right)\end{array}$

Thus, the 95% confidence interval for population proportion is $\underset{\_}{\left(0.2867,0.3333\right)}$.

c.

To determine

Calculate the 95% confidence interval for the proportion that gave $25,000 or more to charity.

Answer to Problem 43E

The 95% confidence interval for the proportion that gave $25,000 or more to charity is $\underset{\_}{\left(0.039,0.061\right)}$.

Explanation of Solution

Calculation:

The survey results that 5% of respondents gave $25,000 or more to charity over the previous year.

The value of $\overline{p}$ is 0.05.

From “Table 8.1 value of $z_{\frac{\alpha }{2}}$”, the value of $z_{\frac{\alpha }{2}}$ for 95% is 1.96.

The value of margin of error is obtained as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.05\left(1-0.05\right)}{1,500}}\\ =1.96\sqrt{\frac{0.0475}{1,500}}\\ =1.96\times 0.005\\ =0.0110\end{array}$

Thus, the value of margin of error is 0.0110.

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

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.05\pm 0.0110\\ =\left(0.05-0.0110,0.05+0.0110\right)\\ =\left(0.039,0.061\right)\end{array}$

Thus, the 95% confidence interval for population proportion is $\underset{\_}{\left(0.039,0.061\right)}$.

d.

To determine

Compare the margin of error in Parts (a), (b), and (c).

Explain the margin of error related to sample proportion.

Find the proportion that is used to choose the planning value when the same sample is used to estimate a variety of proportions.

Explain why one would think $p*=0.50$ is often used in these cases.

Explanation of Solution

The margin of error for Part (a) is 0.0253, for Part (b) is 0.0233, and for Part (c) is 0.0110.

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

An estimate of the largest proportion should be used as a planning value if the estimates must be less than a given value.

A planning value of $p*=0.50$ would guarantee small margin of error for all the interval estimates.