In a survey funded by the University of Washington School of Medicine, 176 of 1,100 adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI). Construct a 99\% confidence interval estimating the proportion of adult Seattle residents that do not believe they can contract an STI. Round the answers to three decimal places.
Preliminary:
a. Is it safe to assume that $n \leq 5 \%$ of all residents in Seattle?
Yes
No
b. Verify $n \widehat{p}(1-\hat{p}) \geq 10$. Round your answer to one decimal place,
\[
n \widehat{p}(1-\widehat{p})=
\]
Confidence Interval: What is the $99 \%$ confidence interval to estimate the population proportion? Round your answer to three decimal places.
\[
< p<
\]
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Answer
Final Answer: The 99% confidence interval for the proportion of adult Seattle residents that do not believe they can contract an STI is \(\boxed{0.132 < p < 0.188}\). The conditions for constructing this confidence interval are met. The sample size is less than 5% of the population size (\(\boxed{True}\)), and the number of successes and failures in the sample are both at least 10 (\(\boxed{True}\)).
Step 1 ：First, calculate the sample proportion (p̂), which is the number of successes (people who do not believe they can contract an STI) divided by the total number of trials (total number of people surveyed). In this case, p̂ = 176 / 1100 = 0.16.
Step 2 ：Next, calculate the standard error of the proportion, which is the square root of [ p̂(1 - p̂) / n ]. In this case, the standard error = sqrt[ 0.16(1 - 0.16) / 1100 ] = 0.011.
Step 3 ：Then, use the Z-score for a 99% confidence interval, which is approximately 2.576, to calculate the margin of error, which is the Z-score multiplied by the standard error. In this case, the margin of error = 2.576 * 0.011 = 0.028.
Step 4 ：Finally, calculate the confidence interval by subtracting and adding the margin of error from/to the sample proportion. In this case, the confidence interval is 0.16 - 0.028 = 0.132 and 0.16 + 0.028 = 0.188.
Step 5 ：Check the conditions for constructing a confidence interval for a proportion. The first condition is that the sample size should be less than 5% of the population size. The second condition is that the number of successes and failures in the sample should both be at least 10. In this case, both conditions are met.
Step 6 ：Final Answer: The 99% confidence interval for the proportion of adult Seattle residents that do not believe they can contract an STI is \(\boxed{0.132 < p < 0.188}\). The conditions for constructing this confidence interval are met. The sample size is less than 5% of the population size (\(\boxed{True}\)), and the number of successes and failures in the sample are both at least 10 (\(\boxed{True}\)).
