To construct the 95% confidence interval for the proportion of adult Seattle residents who do not believe they can contract an STI, let's go step-by-step:

Step 1: Identify the necessary information

• Sample proportion ($p$): $\frac{750}{1000} = 0.75$
• Sample size ($n$): $1000$
• Confidence level: 95%

Step 2: Use the confidence interval formula

The formula for the confidence interval is:

$CI = p \pm Z \cdot \sqrt{\frac{p(1-p)}{n}}$

Where:

• $p$ is the sample proportion.
• $Z$ is the critical value for a 95% confidence level ($Z = 1.96$).
• $n$ is the sample size.

Step 3: Compute the standard error

$SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.75(1-0.75)}{1000}} = \sqrt{\frac{0.1875}{1000}} = \sqrt{0.0001875} \approx 0.0137$

Step 4: Compute the margin of error

$ME = Z \cdot SE = 1.96 \cdot 0.0137 \approx 0.0268$

Step 5: Calculate the confidence interval

$CI = 0.75 \pm 0.0268$

Lower bound: $0.75 - 0.0268 = 0.7232$

Upper bound: $0.75 + 0.0268 = 0.7768$

Final Answer:

The 95% confidence interval is approximately:

$(0.723, 0.777)$

Correct answer: (0.723, 0.777)

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Would you like a more detailed breakdown or have further questions?

Related Questions:

1. How is the sample size related to the width of the confidence interval?
2. Why is the $Z$-value for 95% confidence specifically 1.96?
3. How would the confidence interval change with a smaller sample size?
4. What happens to the confidence interval if we increase the confidence level to 99%?
5. Why is it important to construct a confidence interval for survey results?

Tip:

When interpreting confidence intervals, remember they provide a range in which the true population parameter is likely to lie, given the confidence level.