Chapter 8, Problem 63SE

Credit Card Ownership. Credit card ownership varies across age groups. In 2018, CreditCards.com estimated that the percentage of people who own at least one credit card is 67% in the 18–24 age group, 83% in the 25–34 age group, 76% in the 35–49 age group, and 78% in the 50+ age group. Suppose these estimates are based on 455 randomly selected people from each age group.

1. a. Construct a 95% confidence interval for the proportion of people in each of these age groups who owns at least one credit card.
2. b. Assuming the same sample size will be used in each age group, how large would the sample need to be to ensure that the margin of error is .03 or less for each of the four confidence intervals?

To determine

Find the 95% confidence interval for the population proportion of people in each of the given age groups who own at least one credit card.

Answer to Problem 63SE

The 95% confidence interval for the proportion of people in 18-24 age group who owns at least one credit card is $\underset{\_}{\left(0.6269,0.7131\right)}$.

The 95% confidence interval for the proportion of people in 25-34 age group who owns at least one credit card is $\underset{\_}{\left(0.7955,0.8645\right)}$.

The 95% confidence interval for the proportion of people in 35-49 age group who owns at least one credit card is $\underset{\_}{\left(0.7208,0.7992\right)}$.

The 95% confidence interval for the proportion of people in 50+ age group who owns at least one credit card is $\underset{\_}{\left(0.742,0.818\right)}$.

Explanation of Solution

Calculation:

The study is based on people from different age groups who own at least one credit card. There are 4 different age groups and in each group, 455 people are randomly selected.

For 18-24 age group, the percentage of people who owns at least one credit card is 67%.

For 25-34 age group, the percentage of people who owns at least one credit card is 83%.

For 35-49 age group, the percentage of people who owns at least one credit card is 76%.

For 50+ age group, the percentage of people who owns at least one credit card is 78%.

Margin of error for18-24 age group:

The formula for margin of error is as follows:

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

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

The value of margin of error is obtained as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.67\left(1-0.67\right)}{455}}\\ =1.96\sqrt{\frac{0.2211}{455}}\\ =1.96\times 0.022\\ =0.0431\end{array}$

Confidence interval:

The 95% confidence interval for the proportion of people in 18-24 age group who owns at least one credit card is obtained as follows:

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.67\pm 0.0431\\ =\left(0.67-0.0431,0.67+0.0431\right)\\ =\left(0.6269,0.7131\right)\end{array}$

Thus, the 95% confidence interval for the proportion of people in 18-24 age group who owns at least one credit card is $\underset{\_}{\left(0.6269,0.7131\right)}$.

Margin of error for25-34 age group:

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

The value of margin of error is obtained as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.83\left(1-0.83\right)}{455}}\\ =1.96\sqrt{\frac{0.1411}{455}}\\ =1.96\times 0.0176\\ =0.0345\end{array}$

Confidence interval:

The 95% confidence interval for the proportion of people in 25-34 age group who owns at least one credit card is obtained as follows:

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.83\pm 0.0345\\ =\left(0.83-0.0345,0.83+0.0345\right)\\ =\left(0.7955,0.8645\right)\end{array}$

Thus, the 95% confidence interval for the proportion of people in 25-34 age group who owns at least one credit card is $\underset{\_}{\left(0.7955,0.8645\right)}$.

Margin of error for 35-49 age group:

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

The value of margin of error is obtained as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.76\left(1-0.76\right)}{455}}\\ =1.96\sqrt{\frac{0.1824}{455}}\\ =1.96\times 0.0200\\ =0.0392\end{array}$

Confidence interval:

The 95% confidence interval for the proportion of people in 35-49 age group who owns at least one credit card is obtained as follows:

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.76\pm 0.0392\\ =\left(0.76-0.0392,0.76+0.0392\right)\\ =\left(0.7208,0.7992\right)\end{array}$

Thus, the 95% confidence interval for the proportion of people in 35-49 age group who owns at least one credit card is $\underset{\_}{\left(0.7208,0.7992\right)}$.

Margin of error for 50+ age group:

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

The value of margin of error is obtained as follows:

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.78\left(1-0.78\right)}{455}}\\ =1.96\sqrt{\frac{0.1716}{455}}\\ =1.96\times 0.0194\\ =0.0380\end{array}$

Confidence interval:

The 95% confidence interval for the proportion of people in 50+ age group who owns at least one credit card is obtained as follows:

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.78\pm 0.0380\\ =\left(0.78-0.0380,0.78+0.0380\right)\\ =\left(0.742,0.818\right)\end{array}$

Thus, the 95% confidence interval for the proportion of people in 50+ age group who owns at least one credit card is $\underset{\_}{\left(0.742,0.818\right)}$.

To determine

Find the number of sample needed for each of the four confidence intervals.

Answer to Problem 63SE

The number of sample needed for each of the four confidence intervals is 489.

Explanation of Solution

Calculation:

The margin of error is 0.03 and the value of $p\ast$ is 0.67.

The formula for sample size for an interval estimate of a population proportion is as follows:

$n=\frac{{\left(z_{\frac{\alpha }{2}}\right)}^{2}p\ast \left(1-p\ast \right)}{E^2}$ .

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

The required sample size is as follows:

$\begin{array}{c}n=\frac{{\left(1.96\right)}^2\left(0.67\right)\left(1-0.67\right)}{{\left(0.03\right)}^2}\\ =\frac{3.8416\times 0.2211}{0.0009}\\ =\frac{0.8494}{0.0009}\\ =943.75\\ \approx 944\end{array}$

Thus, the sample size is 944.

Hence, the number of sample needed for each of the four confidence intervals 489.