Question 18

In a survey funded by the UW school of medicine, 750 of 1,000 adult Seattle residents said they did not believe they could contract a sexually transmitted infection (STI) and 893 of 1,123 Denver residents said they did not believe they could contract an (STI). Construct a $95 \%$ confidence interval of the difference in proportions of Seattle and Denver residents who do not believe they could contract a STI. (Use a $z$ score of 1.96 for your computations.)
\[
\begin{array}{l}
(.0809, .0095) \\
(-.0809,-.0095) \\
(-.0809, .0095) \\
(-.0095, .0809)
\end{array}
\]
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Step 1 ：Given that 750 out of 1000 Seattle residents and 893 out of 1123 Denver residents do not believe they could contract a STI, we can calculate the proportions for each city. For Seattle, the proportion (p1) is \(\frac{750}{1000} = 0.75\). For Denver, the proportion (p2) is \(\frac{893}{1123} = 0.795\).

Step 2 ：We are asked to construct a 95% confidence interval for the difference in these proportions. The formula for this is \((p1 - p2) \pm z * \sqrt{\frac{p1(1-p1)}{n1} + \frac{p2(1-p2)}{n2}}\), where n1 and n2 are the sizes of the two groups, and z is the z-score corresponding to the desired level of confidence.

Step 3 ：In this case, n1 is the number of Seattle residents surveyed (1000), n2 is the number of Denver residents surveyed (1123), and z is the z-score for a 95% confidence interval (1.96).

Step 4 ：Substituting these values into the formula, we get \((0.75 - 0.795) \pm 1.96 * \sqrt{\frac{0.75(1-0.75)}{1000} + \frac{0.795(1-0.795)}{1123}}\).

Step 5 ：Solving this gives us a confidence interval of (-0.0809, -0.0095).

Step 6 ：So, the 95% confidence interval for the difference in proportions of Seattle and Denver residents who do not believe they could contract a STI is \(\boxed{(-0.0809, -0.0095)}\).