Question

Critical Thinking. In Exercises 17–28, use the data and confidence level to construct a confidence interval estimate of p, then address the given question.

Medication UsageIn a survey of 3005 adults aged 57 through 85 years, it was found that 81.7% of them used at least one prescription medication (based on data from “Use of Prescription and Over-the-Counter Medications and Dietary Supplements Among Older Adults in the United States,” by Qato et al.,Journal of the American Medical Association,Vol. 300, No. 24).

a.How many of the 3005 subjects used at least one prescription medication?

b.Construct a 90% confidence interval estimate of thepercentageof adults aged 57 through 85 years who use at least one prescription medication.

c.What do the results tell us about the proportion of college students who use at least one prescription medication?

Short answer

a.The number of adults aged 57 through 85 years who use at least one prescription medication is equal to 2455.

b. The 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication is equal to (80.54%, 82.86%).

c. As the survey involved only adults aged 57 through 85 years, the above results do not suggest anything about the proportion of college students who use at least one prescription medication.

Step-by-step solution

Given information

The medication usage of adults aged 57 through 85 years is recorded. In a sample of 3005 adults aged 57 through 85 years, 81.7% of them use at least one prescription medication.

Conversion of proportion to a number

a.

The number of adults aged 57 through 85 years who use at least one prescription medication is equal to:

$$\begin{gathered} 81.7\%\text{of}3005=\frac{81.7}{100}\times 3005 \\ =2455 \end{gathered}$$

Therefore, the number of adults aged 57 through 85 years who use at least one prescription medication is equal to 2455.

Expression of the confidence interval

The confidence interval has the following expression:

$\hat{p}-E<p<\hat{p}+E$

Here, E is the margin of error and has the following formula:

$E=z_{\frac{\alpha }{2}}\times \sqrt{\frac{\hat{p}\hat{q}}{n}}$where

$\hat{p}$is the sample proportion of adults aged 57 through 85 years who use at least one prescription medication.

$\hat{q}$is the sample proportion of adults aged 57 through 85 years who do not use any prescription medication.

n is the sample size

$z_{\frac{\alpha }{2}}$is the one-tailed critical value of z

Compute the critical value

The confidence level is given to be equal to 90%. Thus, the corresponding level of significance is equal to 0.10.

Now,$\alpha =0.10$

The value of $z_{\frac{\alpha }{2}}$form the standard normal table is equal to 1.645.

Compute the margin of error

The margin of error is computed as shown below:

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\times \sqrt{\frac{\hat{p}\hat{q}}{n}} \\ =1.645\times \sqrt{\frac{0.817\times 0.183}{3005}} \\ =0.0116 \end{gathered}$$

Compute the confidence interval

b.

The value of the confidence interval is computed as follows:

$$\begin{gathered} \hat{p}-E<p<\hat{p}+E \\ 0.817-0.0116<E<0.817+0.0116 \\ 0.8054<p<0.8286 \\ 80.54\%<p<82.86\% \end{gathered}$$

The 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication is equal to (80.54%, 82.86%).

Subjects targeted by the results

c.

The above results do not suggest anything about the proportion of college students who use at least one prescription medication because the sample consisted of only adults aged 57 through 85 years.