Final answer:
The 90% confidence interval for estimating the proportion of adult Seattle residents that do not believe they can contract an STI is (0.173, 0.207), suggesting the true proportion is between 17.3% and 20.7%.
Step-by-step explanation:
To construct a 90% confidence interval estimating the proportion of adult Seattle residents that do not believe they can contract a sexually transmitted infection (STI), we will use the formula for a confidence interval for a population proportion. The formula for a confidence interval is given by:
P± Z*sqrt(P(1-P)/n)
where:
- P is the sample proportion,
- n is the sample size, and
- Z* is the z-score corresponding to the desired confidence level.
For this survey conducted by the University of Washington School of Medicine:
- P = 266/1400 = 0.19,
- n = 1400, and
- Z* (for 90% confidence) is approximately 1.645
Plugging these values into the formula gives us:
Confidence Interval = 0.19 ± 1.645 * sqrt(0.19(0.81)/1400)
Now, we need to calculate the error margin:
Error Margin = 1.645 * sqrt(0.19*0.81/1400) ≈ 1.645 * sqrt(0.1539/1400) ≈ 1.645 * sqrt(0.0001099) ≈ 1.645 * 0.010488 ≈ 0.017253
Therefore, the 90% confidence interval is:
0.19 ± 0.017 ≈ (0.173, 0.207)
Rounded to three decimal places, the confidence interval is (0.173, 0.207).
This interval suggests that with 90% confidence, the true proportion of adult Seattle residents that do not believe they can contract an STI is between 17.3% and 20.7%.