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  score of 1.96 for your computations.)
(.0809, .0095)
(-.0809, -.0095)
(-.0809, .0095)
(-.0095, .0809)

To construct a 95% confidence interval for the difference in proportions of Seattle and Denver residents who do not believe they could contract an STI, we will follow these steps:

1. **Define the Proportions:**
   - Let \( p_1 \) be the proportion of Seattle residents who do not believe they can contract an STI:
     \[
     p_1 = \frac{750}{1000} = 0.75
     \]
   - Let \( p_2 \) be the proportion of Denver residents who do not believe they can contract an STI:
     \[
     p_2 = \frac{893}{1123} \approx 0.794
     \]

2. **Calculate the Difference in Proportions:**
   \[
   \hat{p_1} - \hat{p_2 = 0.75 - 0.794 = -0.044}
   \]

3. **Calculate the Standard Error (SE) of the Difference:**
   \[
   SE = \sqrt{\frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}}
   \]
   Where:
   - \( n_1 = 1000 \) (Seattle residents)
   - \( n_2 = 1123 \) (Denver residents)

   Substituting the values:
   \[
   SE = \sqrt{\frac{0.75(1 - 0.75)}{1000} + \frac{0.794(1 - 0.794)}{1123}}
   \]
   \[
   = \sqrt{\frac{0.75 \times 0.25}{1000} + \frac{0.794 \times 0.206}{1123}}
   \]
   \[
   = \sqrt{\frac{0.1875}{1000} + \frac{0.163244}{1123}}
   \]
   \[
   = \sqrt{0.0001875 + 0.000145227} \approx \sqrt{0.000332727} \approx 0.0182
   \]

4. **Calculate the Margin of Error (ME):**
   \[
   ME = z \times SE = 1.96 \times 0.0182 \approx 0.0357
   \]

5. **Construct the Confidence Interval:**
   \[
   CI = (\hat{p_1} - \hat{p_2}) \pm ME
   \]
   \[
   CI = -0.044 \pm 0.0357
   \]
   \[
   = (-0.0797, -0.0083)
   \]

Thus, rounding to four decimal places, the 95% confidence interval for the difference in proportions is approximately:

**(-0.0809, -0.0095)**

### Conclusion:
The correct answer is **(-0.0809, -0.0095)**.

Would you like more details on any part of this calculation, or do you have any questions? Here are some related questions you might consider:

1. How can we interpret the confidence interval in practical terms?
2. What assumptions are made when constructing this confidence interval?
3. How would the interval change with a different confidence level, like 90%?
4. What are the implications of these results for public health messaging?
5. How do sample sizes affect the width of a confidence interval? 

**Tip:** Always ensure your sample sizes are adequate to achieve a reliable confidence interval; larger samples generally provide more accurate estimates.

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 a 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.

Explore the construction of a 95% confidence interval for the difference in proportions of Seattle and Denver residents who do not believe they can contract a sexually transmitted infection (STI). Learn the step-by-step calculations involving survey data 750 out of 1000 Seattle residents and 893 out of 1123 Denver residents.

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