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 $95\mathrm{\%}$ 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 $(p1-p2)\pm z\ast \sqrt{\frac{p1(1-p1)}{n1}+\frac{p2(1-p2)}{n2}}$, 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 $(0.75-0.795)\pm 1.96\ast \sqrt{\frac{0.75(1-0.75)}{1000}+\frac{0.795(1-0.795)}{1123}}$.

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 $\boxed{{\displaystyle (-0.0809,-0.0095)}}$.