91Ó°ÊÓ

Use the data and confidence level to construct a confidence interval estimate of \(p,\) then address the given question. 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 standand care, \(62.8 \%\) were no longer smoking after one month (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). Construct the two \(95 \%\) confidence interval estimates of the percentages of success. Compare the results. What do you conclude?

Step by step solution

01

- Identify sample proportions and sizes

For sustained care: - Sample size (\(n_1\))= 198 - Sample proportion (\(\hat{p}_1\)) = 0.828 For standard care: - Sample size (\(n_2\))= 199 - Sample proportion (\(\hat{p}_2\)) = 0.628

02

- Determine the standard error for each proportion

Use the formula for standard error of a proportion: \( SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \) For sustained care: \(SE_1 = \sqrt{ \frac{0.828(1-0.828)}{198} } \approx 0.027 \) For standard care: \(SE_2 = \sqrt{ \frac{0.628(1-0.628)}{199} } \approx 0.034 \)

03

- Determine the Z-score for a 95% confidence interval

For a 95% confidence interval, the Z-score is 1.96.

04

- Calculate the margin of error for each proportion

Use the formula for margin of error: \( ME = Z \times SE \) For sustained care: \(ME_1 = 1.96 \times 0.027 \approx 0.053 \) For standard care: \(ME_2 = 1.96 \times 0.034 \approx 0.067 \)

05

- Construct the confidence intervals

For sustained care: \(CI_1 = \hat{p}_1 \pm ME_1 \) \(CI_1 = 0.828 \pm 0.053 \) So, \(CI_1 = [0.775, 0.881] \) For standard care: \(CI_2 = \hat{p}_2 \pm ME_2 \) \(CI_2 = 0.628 \pm 0.067 \) So, \(CI_2 = [0.561, 0.695] \)

06

- Compare the confidence intervals and draw conclusions

Compare the intervals: - Sustained care CI: \([0.775, 0.881] \) - Standard care CI: \([0.561, 0.695] \) The confidence intervals do not overlap, indicating that sustained care appears to be more effective in helping patients stop smoking after one month.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion

A sample proportion is the fraction of individuals in a sample with a particular characteristic. In this exercise, we use it to find the percentage of patients who stopped smoking. For each care type, we calculate the sample proportion by dividing the number of successes by the total sample size.
For sustained care: \ \ \ \ Sample size (\(n_1\) = 198) and number of successes (\(82.8%\( of 198\textup )) gives the sample proportion \ \ (\textup \ Sample proportion (\( \hat{p}_1 \) = \ \frac{\textup \ 82.8}{100}) = 0.828\)
For standard care: \ Sample size (\textbefore (# 199\textup}} \ \) and number of successes (\textup 62.8%\textup