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.

Smoking Stopped In a program designed to help patients stop smoking, 198 patients were given sustained care, and 82.8% of them were no longer smoking after one month (based on data from “Sustained Care Intervention and Postdischarge Smoking Cessation AmongHospitalized Adults,” by Rigotti et al., Journal of the American Medical Association, Vol. 312, No. 7). Use a 0.01 significance level to test the claim that 80% of patients stop smoking when given sustained care. Does sustained care appear to be effective?

Short answer

Null hypothesis: The proportion of patients who stop smoking when they are given sustained care is equal to 80%.

Alternative hypothesis: The proportion of patients who stop smoking when they are given sustained care is not equal to 80%.

Test statistic: 0.985

Critical value: 2.5758

P-value: 0.3246

The null hypothesis isfailed to reject.

There is not enough evidence to reject the claim that the proportion of patients who stop smoking after they are given sustained care is equal to 80%.

Sustained care appears to be effective because a large proportion of patients (82.8%) have been able to quit smoking after being given sustained care.

Step-by-step solution

Given information

A sample of 198 patients has beengiven sustained care, and 82.8% of them have stopped smoking after one month. It is claimed that 80% of the patients stop smoking after they are given sustained care.

Hypotheses

The null hypothesis is written as follows.

The proportion of patients who stop smoking when they are given sustained care is equal to 80%.

\({H_0}:p = 0.80\).

The alternative hypothesis is written as follows.

The proportion of patients who stop smoking when they are given sustained care is not equal to 80%.

\({H_1}:p \ne 0.80\).

The test is two-tailed.

Sample size, sample proportion,and population proportion

The sample size is n=198.

The sample proportion of patients who stop smoking after they are given sustained care is given as follows.

\(\begin{array}{c}\hat p = 82.8\% \\ = \frac{{82.8}}{{100}}\\ = 0.828\end{array}\).

The sample proportion of patients who stop smoking after they are given sustained care is equal to 0.80.

Test statistic

The value of the test statistic is computed below.

\(\begin{array}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ = \frac{{0.828 - 0.80}}{{\sqrt {\frac{{0.80\left( {1 - 0.80} \right)}}{{198}}} }}\\ = 0.985\end{array}\)

Thus, z=0.985.

Critical value and p-value

Referring to the standard normal table, the critical value of z at\(\alpha = 0.01\)for a two-tailed test is equal to 2.5758.

Referring to the standard normal table, the p-value for the test statistic value of 0.985 is equal to 0.3246.

Asthe p-value is greater than 0.01, the null hypothesis is failed to reject.

Conclusion of the test

There is not enough evidence to reject the claim that the proportion of patients who stop smoking after they are given sustained care is equal to 80%.

A significant percentage of people are able to quit smoking with the help of sustained care.

Thus, sustained care proves to be effective.