91影视

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.

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.

$H_1:p_1>p_2$

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:

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

Hence,

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

The test statistic is equal to:

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

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.

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.

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 $\left(伪\right)$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.

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

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

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.

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.