Question

Testing Claims About Proportions. In Exercises 7–22, 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.

Smoking Cessation Programs Among 198 smokers who underwent a “sustained care” program, 51 were no longer smoking after six months. Among 199 smokers who underwent a “standard care” program, 30 were no longer smoking after six months (based on data from “Sustained Care Intervention and Postdischarge Smoking Cessation Among Hospitalized Adults,” by Rigotti et al., Journal of the American Medical Association, Vol. 312, No. 7). We want to use a 0.01 significance level to test the claim that the rate of success for smoking cessation is greater with the sustained care program.

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

c. Does the difference between the two programs have practical significance?

Short answer

a. There is sufficient evidence to support the claim that therate of success for smoking cessation is greater with the sustained care program.

b. The 98% confidence interval is equal to (0.013, 0.200), and it suggests that the claim stating that therate of success for smoking cessation is greater with the sustained care program is true.

c. The difference between the success rates of the two programs is considerable and appears to have practical significance.

Step-by-step solution

Given information

In a sample of 198 smokers who underwent a sustained care program, 51 were no longer smoking after six months. In another sample of 199 smokers who underwent a standard care program, 30 were no longer smoking after six months.

The level of significance to test the hypothesis is 0.01.

Describe the hypotheses to be tested

Null hypothesis: The proportion of smokers who underwent the sustained care program and left smoking after six months is equal to the proportion of smokers who underwent the standard care program and left smoking after six months.

$H_0:p_1=p_2$

Alternate hypothesis: The proportion of smokers who underwent the sustained care program and left smoking after six months is greater than the proportion of smokers who underwent the standard care program and quit smoking after six months.

$H_1:p_1>p_2$

Calculate the sample statistics

Let $n_1$denote the sample size corresponding to the smokers who took the sustained care program.

Let $n_2$denote the sample size corresponding to the smokers who took the standard care program.

Assume that $x_1$and $x_2$are the number of smokers who left smoking after six months when they took the sustained care program and the standard care program, respectively.

Let${\hat{p}}_1$ be the sample success rate of the sustained care program.

Thus,

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{51}{198} \\ =0.258 \end{gathered}$$

$$\begin{gathered} {\hat{q}}_1=1-{\hat{p}}_1 \\ =0.742 \end{gathered}$$

Let ${\hat{p}}_2$be the sample success rate of the standard care program.

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{30}{199} \\ =0.151 \end{gathered}$$

$$\begin{gathered} {\hat{q}}_2=1-{\hat{p}}_2 \\ =0.849 \end{gathered}$$

The value of the pooled sample proportion is equal to:

$$\begin{gathered} \overline{p}=\frac{x_1+x_2}{n_1+n_2} \\ =\frac{51+30}{198+199} \\ =0.204 \end{gathered}$$

Hence,

$$\begin{gathered} \overline{q}=1-\overline{p} \\ =1-0.204 \\ =0.796 \end{gathered}$$

Compute the value of the test statistic

The test statistic is equal to:

$$\begin{gathered} z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\frac{\overline{p}\overline{q}}{n_1}+\frac{\overline{p}\overline{q}}{n_2}}} \\ =\frac{\left(0.258-0.151\right)-0}{\sqrt{\frac{\left(0.204\right)\left(0.796\right)}{198}+\frac{\left(0.204\right)\left(0.796\right)}{199}}} \\ =2.641 \end{gathered}$$

Referring to the standard normal distribution table, the critical value of z corresponding to for a right-tailed test is equal to 2.33.

Referring to the standard normal distribution table, the corresponding p-value is equal to 0.0041.

Here, the value of the test statistic is greater than the critical value, and the p-value is less than 0.01.

Therefore, reject the null hypothesis at a 0.01 significance level.

Conclusion of the test

a.

There is sufficient evidence to support the claim that the success rateof the sustained care program is greater than the success rate of the standard care program.

Describe the confidence interval

If the level of significance for a two-tailed test is equal to 0.01, then the corresponding confidence level to construct the confidence interval is equal to 98%.

Hence, the level of significance $\left(\alpha \right)$for the confidence interval method is equal to 0.02.

The expression of the confidence interval is as follows:

$\left({\hat{p}}_1-{\hat{p}}_2\right)-E<p_1-p_2<\left({\hat{p}}_1-{\hat{p}}_2\right)+E$

Here, E is the margin of error.

Calculate the margin of error

The value of the margin of error is computed as follows:

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\sqrt{\frac{{\hat{p}}_1{\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2{\hat{q}}_2}{n_2}} \\ =2.33\times \sqrt{\frac{\left(0.258\right)\left(0.742\right)}{198}+\frac{\left(0.151\right)\left(0.849\right)}{199}} \\ =0.0935 \end{gathered}$$

Construct the confidence interval.

b.

Substituting the required values, the following confidence interval is obtained:

$$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)-E<p_1-p_2<\left({\hat{p}}_1-{\hat{p}}_2\right)+E \\ (0.258-0.151)-0.0935<p_1-p_2<(0.258-0.151)+0.0935 \\ 0.013<p_1-p_2<0.200 \end{gathered}$$

Thus, the 99% confidence interval is equal to (0.013, 0.200).

This confidence interval does not contain zero that means there is a significant difference between the two population proportions of smokers who left smoking after six months.

Therefore, the confidence interval suggests that there is sufficient evidence to support the claim that therate of success of the sustained care program is greater as compared to that of the standard care program.

Compare the success rates

c.

Here, the sample success rate of the sustained care program and the standard care program are 25.8% and 15.1%, respectively.

Thus, the success rate of the sustained care program appears to be greater than that of the standard care program.

The sustained care program gives better results and should be preferred over the standard care program.