Math Problem Statement

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)

Solution

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 p1p_1 be the proportion of Seattle residents who do not believe they can contract an STI: p1=7501000=0.75p_1 = \frac{750}{1000} = 0.75
    • Let p2p_2 be the proportion of Denver residents who do not believe they can contract an STI: p2=89311230.794p_2 = \frac{893}{1123} \approx 0.794
  2. Calculate the Difference in Proportions: p1^p2=0.750.794=0.044^\hat{p_1} - \hat{p_2 = 0.75 - 0.794 = -0.044}

  3. Calculate the Standard Error (SE) of the Difference: SE=p1(1p1)n1+p2(1p2)n2SE = \sqrt{\frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}} Where:

    • n1=1000n_1 = 1000 (Seattle residents)
    • n2=1123n_2 = 1123 (Denver residents)

    Substituting the values: SE=0.75(10.75)1000+0.794(10.794)1123SE = \sqrt{\frac{0.75(1 - 0.75)}{1000} + \frac{0.794(1 - 0.794)}{1123}} =0.75×0.251000+0.794×0.2061123= \sqrt{\frac{0.75 \times 0.25}{1000} + \frac{0.794 \times 0.206}{1123}} =0.18751000+0.1632441123= \sqrt{\frac{0.1875}{1000} + \frac{0.163244}{1123}} =0.0001875+0.0001452270.0003327270.0182= \sqrt{0.0001875 + 0.000145227} \approx \sqrt{0.000332727} \approx 0.0182

  4. Calculate the Margin of Error (ME): ME=z×SE=1.96×0.01820.0357ME = z \times SE = 1.96 \times 0.0182 \approx 0.0357

  5. Construct the Confidence Interval: CI=(p1^p2^)±MECI = (\hat{p_1} - \hat{p_2}) \pm ME CI=0.044±0.0357CI = -0.044 \pm 0.0357 =(0.0797,0.0083)= (-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.

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Math Problem Analysis

Mathematical Concepts

Statistics
Proportions
Confidence Intervals

Formulas

p1 = x1/n1, p2 = x2/n2
Difference in Proportions = p1 - p2
Standard Error (SE) = sqrt[(p1(1 - p1)/n1) + (p2(1 - p2)/n2)]
Margin of Error (ME) = z * SE
Confidence Interval = (Difference in Proportions) ± ME

Theorems

Central Limit Theorem
Proportions Difference Theorem

Suitable Grade Level

Grades 11-12