To construct a 99% confidence interval for the population proportion of adults who believe in UFOs, follow these steps.

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Step 1: Define Variables

• Sample size $n$ = 22,642,264
• Number of successes $x$ = 729,729 (adults who believe in UFOs)
• Sample proportion $\hat{p}$ = $\frac{x}{n}$
  $\hat{p} = \frac{729,729}{22,642,264} \approx 0.0322$

Step 2: Find the Critical Value $Z_{\alpha/2}$

For a 99% confidence interval, the critical value $Z_{\alpha/2}$ is approximately 2.576 (from the standard normal distribution).

Step 3: Calculate the Standard Error

The standard error $SE$ for a population proportion is:

$SE = \sqrt{ \frac{\hat{p}(1 - \hat{p})}{n} }$

$SE = \sqrt{ \frac{0.0322 \times (1 - 0.0322)}{22,642,264} } \approx 0.000036$

Step 4: Compute the Confidence Interval

The formula for the confidence interval is:

$\hat{p} \pm Z_{\alpha/2} \times SE$

$0.0322 \pm 2.576 \times 0.000036$

$0.0322 \pm 0.0000927$

Step 5: Final Interval

$(0.0321, 0.0323)$

Thus, the 99% confidence interval for the proportion of adults who believe in UFOs is approximately:

$(0.0321, 0.0323)$

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Would you like more details or have any questions about these calculations?

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5 Related Questions

1. How does the choice of confidence level affect the width of a confidence interval?
2. What would the confidence interval look like if the sample size was much smaller?
3. How can we interpret this confidence interval in real-world terms?
4. What is the margin of error, and how is it connected to the confidence interval?
5. Why do we assume a normal distribution when calculating confidence intervals for proportions?

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Tip: Larger sample sizes result in narrower confidence intervals because the standard error becomes smaller.