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. 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. Among 199 patients given standard care, 62.8% were no longer smoking after one month (based on data from “Sustained Care Intervention and Post discharge Smoking Cessation Among Hospitalized Adults,” by Rigottiet al., Journal of the American Medical Association, Vol. 312, No. 7). Construct the two 95% confidence interval estimates of the percentages of success. Compare the results. What do you conclude?

The given claim is about the program that helped the patients stop smoking. There were 198$\left(n_1\right)$ patients who were given sustained care from them 82.8% $\left({\hat{p}}_1\right)$ people were no longer smoking. And also, there were 199$\left(n_2\right)$ patients who were given standard care out of which 62.8% $\left({\hat{p}}_2\right)$were no longer smoking.

The basic requirement of this confidence interval of proportion is that the sample is a simple random sample.

This requirement is assumed to be satisfied.

Assuming fixed trials and a constant rate of success.

It is required to check,$\mathbf{np}⩾\mathbf{5}\mathbf{and}\mathbf{nq}⩾\mathbf{5}$ for each case.

$$\begin{gathered} n_1{\hat{p}}_1=198\times 0.828 \\ =163.94 \\ ⩾5 \end{gathered}$$

$$\begin{gathered} n_1{\hat{q}}_1=198\times \left(1-0.828\right) \\ =198\times 0.272 \\ =53.86 \\ ⩾5 \end{gathered}$$

Also,

$$\begin{gathered} n_2{\hat{p}}_2=199\times 0.628 \\ =124.97 \\ ⩾5 \\ {n}_2{\hat{q}}_2=199\times \left(1-0.628\right) \\ =199\times 0.372 \\ =74.028⩾5 \end{gathered}$$

Thus, the requirements for the test are satisfied.

The confidence interval for proportion is calculated as below,

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

The margin of error can be computed by the formula given below,

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

For 95% confidence interval, the critical value is obtained from standard normal table as 1.96$\left(z_{\frac{0.05}{2}}\right)$ .

The margin of error for sustained care,

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\times \sqrt{\frac{{\hat{p}}_1{\hat{q}}_1}{n_1}} \\ =z_{0.025}\times \sqrt{\frac{0.828\times 0.172}{198}} \\ =1.96\times \sqrt{\frac{0.828\times 0.172}{198}} \\ =0.0525 \end{gathered}$$

Substitute all the values in the formula given below:

$$\begin{gathered} \text{Confidence}\text{interval}={\hat{p}}_1-E<\hat{p}<{\hat{p}}_1+E \\ =0.828-0.0525<\hat{p}<0.828+0.0525 \\ =0.776<\hat{p}<0.881 \end{gathered}$$

The 95% confidence interval for proportion is between 77.6% and 88.1%.

Margin of error for standard care,

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\times \sqrt{\frac{{\hat{p}}_2{\hat{q}}_2}{n_2}} \\ =z_{0.025}\times \sqrt{\frac{0.628\times 0.372}{199}} \\ =1.96\times \sqrt{\frac{0.828\times 0.372}{199}} \\ =0.067 \end{gathered}$$

Substitute all the values in the formula.

$$\begin{gathered} \text{Confidence}\text{interval}={\hat{p}}_2-E<\hat{p}<{\hat{p}}_2+E \\ =0.628-0.067<\hat{p}<0.628+0.067 \\ =0.561<\hat{p}<0.695 \end{gathered}$$

The 95% confidence interval for proportion is between 56.1% and 69.5%.

The intervals do not overlap as the values range differently. As the confidence interval for sustained care has larger positive range, it can be concluded that the success rate for sustained care is greater than standard care.