91影视

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,$\mathbf{np}猢�\mathbf{5}\mathbf{and}\mathbf{nq}猢�\mathbf{5}$ for each case.

$$\begin{gathered} n{\hat{p}}_1=198脳0.258 \\ =51.084 \\ 猢�5 \end{gathered}$$

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

Also,

$$\begin{gathered} n{\hat{p}}_2=198脳0.409 \\ =80.982 \\ 猢�5 \end{gathered}$$

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

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,

$E=z_{\frac{伪}{2}}脳\sqrt{\frac{\hat{p}\hat{q}}{n}}$

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,

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

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,

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

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.