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.

Measured Results vs. Reported Results The same study cited in the preceding exercise produced these results after six months for the 198 patients given sustained care: 25.8% were no longer smoking, and these results were biochemically confirmed, but 40.9% of these patients

reported that they were no longer smoking. Construct the two 95% confidence intervals. Compare the results. What do you conclude?

The given claim is about the program that helped the patients to stop smoking. There were 198(n) patients who were given sustained care, 25.8% $\left({\hat{p}}_1\right)$ did not smoke and it was biochemically confirmed. And 40.9% $\left({\hat{p}}_2\right)$of these patients reported that they no longer smoke.

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 constant rate of success.

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

$$\begin{gathered} n{\hat{p}}_1=198\times 0.258 \\ =51.084 \\ ⩾5 \end{gathered}$$

$$\begin{gathered} n{\hat{q}}_1=198\times \left(1-0.258\right) \\ =146.92 \\ ⩾5 \end{gathered}$$

Also,

$$\begin{gathered} n{\hat{p}}_2=198\times 0.409 \\ =80.982 \\ ⩾5 \end{gathered}$$

$$\begin{gathered} n{\hat{q}}_2=198\times \left(1-0.409\right) \\ =117.02 \\ ⩾5 \end{gathered}$$

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$\left(z_{\frac{0.05}{2}}\right)$ .

Margin of error for sustained care no longer smoking are,

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\times \sqrt{\frac{{\hat{p}}_1{\hat{q}}_1}{n_1}} \\ =1.96\times \sqrt{\frac{0.258\times 0.742}{198}} \\ =0.061 \end{gathered}$$

Substitute all the values in the formula.

$$\begin{gathered} \text{Confidence}\text{interval}={\hat{p}}_1-E<\hat{p}<{\hat{p}}_1+E \\ =0.258-0.061<\hat{p}<0.258+0.061 \\ =0.197<\hat{p}<0.319 \end{gathered}$$

The 95% confidence interval for proportion is between 19.7% and 31.9%.

Margin of error for standard care,

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\times \sqrt{\frac{{\hat{p}}_2{\hat{q}}_2}{n_2}} \\ =1.96\times \sqrt{\frac{0.409\times 0.591}{198}} \\ =0.0685 \end{gathered}$$

Substitute all the values in the formula.

$$\begin{gathered} \text{Confidence}\text{interval}={\hat{p}}_2-E<\hat{p}<{\hat{p}}_2+E \\ =0.409-0.068<\hat{p}<0.409+0.068 \\ =0.340<\hat{p}<0.477 \end{gathered}$$

The 95% confidence interval for proportion is between 34.0% and 47.7%.

The values range differently, as they are non-overlapping.

It can be concluded that the proportion of patients confirmed no longer smoking biomedically was higher.