Questions: Edwin conducted a survey to find the percentage of people in an area who smoked regularly. He defined the label "smoking regularly" for males smoking 30 or more cigarettes in a day and for females smoking 20 or more. Out of 635 persons who took part in the survey, 71 are labeled as people who smoke regularly. What is the 90% confidence interval for this population proportion? Answer choices are rounded to the hundredths place. 0.11 to 0.80 0.11 to 0.13 0.09 to 0.80 0.09 to 0.13

Solution Steps

The sample proportion of people who smoke regularly is calculated as follows:

\[ \hat{p} = \frac{x}{n} = \frac{71}{635} \approx 0.1118 \]

where \( x = 71 \) is the number of people who smoke regularly and \( n = 635 \) is the total number of surveyed individuals.

For a 90% confidence interval, the significance level \( \alpha \) is calculated as:

\[ \alpha = 1 - 0.90 = 0.10 \]

The confidence interval for a single population proportion is given by the formula:

\[ \hat{p} \pm z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]

For a 90% confidence level, the critical value \( z \) is approximately \( 1.64 \). Thus, we can compute the margin of error:

\[ \text{Margin of Error} = z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = 1.64 \cdot \sqrt{\frac{0.1118(1 - 0.1118)}{635}} \approx 0.01 \]

Now, we can calculate the confidence interval:

\[ \text{Confidence Interval} = \left( \hat{p} - \text{Margin of Error}, \hat{p} + \text{Margin of Error} \right) = \left( 0.1118 - 0.01, 0.1118 + 0.01 \right) = (0.09, 0.13) \]

Final Answer