Question

Testing Claims About Proportions. In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Medication Usage In a survey of 3005 adults aged 57 through 85 years, it was found that 87.1% of them used at least one prescription medication (based on data from “Use of Prescription Over-the-Counter Medications and Dietary SupplementsAmong Older Adultsin the United States,” by Qato et al., Journal of the American Medical Association,Vol. 300,No. 24). Use a 0.01 significance level to test the claim that more than 3/4 of adults use at least one prescription medication. Does the rate of prescription use among adults appear to be high?

Short answer

The hypotheses are as follows.

$H_0:p=\frac{3}{4}\left(0.75\right)$

$H_1:p>\frac{3}{4}\left(0.75\right)$

The test statistic value is 8.48. The p-value is 0.0001. The decision is to reject the null hypothesis.

There is enough evidence to conclude that more than $\frac{3}{4}$of adults use at least one prescription medication. The rate of prescription use among adults is high.

Step-by-step solution

Given information

The survey of 3005 adults aged between 57 to 85 years is recorded. Therefore, sample size$n=3005$

The sample proportion of adults who has at least one prescribed medication is $\hat{p}=0.871$.

The level of significance $\alpha =0.01$.

The claim states that more than $\frac{3}{4}$of the adults in the population use at least one prescription medication.

Check the requirements

The requirements for the z-test for the claims about the population proportions are as follows.

1. Assume that 3005 adults are selected randomly.
2. They are fixed and independent selections.
3. The requirements and are satisfied as follows.

$$\begin{gathered} np=3005\times \frac{3}{4} \\ =2253.75 \\ ⩾5 \end{gathered}$$

$$\begin{gathered} nq=3005\times \left(1-\frac{3}{4}\right) \\ =751.25 \\ ⩾5 \end{gathered}$$

Thus, the three requirements of normal approximations are satisfied.

State the hypotheses

Let p be the actual proportions of adults in the population with at least one prescription medication.

The hypotheses are as follows.

$H_0:p=\frac{3}{4}\left(0.75\right)$(Null hypothesis)

$H_1:p>\frac{3}{4}\left(0.75\right)$(Alternative hypothesis and original claim)

As the original claim contains a greater-than symbol,it is a right-tailed test.

Compute the test statistic

The sampling distribution of the sample proportions can be approximated by the normal distribution.

The test statistic is given as follows.

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.817-0.75}{\sqrt{\frac{0.75\times 0.25}{3005}}} \\ =8.4820 \end{gathered}$$

The test statistic value is 8.4820.

Find the p-value

For this right-tailed test, the p-value is the area to the right of the test statistic. Using Table A-2 (standard normal table), the cumulative area to the left of 8.48 corresponds to rows 3.50 and above, and its value is 0.9999.

The area to the right of is expressed as follows.

$$\begin{gathered} P\left(Z>8.4820\right)=1-P\left(Z<8.4820\right) \\ =1-0.9999 \\ =0.0001 \end{gathered}$$

Thus, the p-value is 0.0001.

State the decision rule

When the p-value is less than the given level of significance, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.

In this situation, the p-value is lesser than the significance level: $0.0001<0.01$.

Hence, reject the null hypothesis.

State the conclusion

There is enough evidence to support the claim that more than $\frac{3}{4}$of adults use at least one prescription medication.

As more than $\frac{3}{4}$of adults use at least one prescription medication, the rate of prescription use among adults appears to be high.