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The point estimate for the population proportion of "no-shows" (denoted as \(\hat{p}\)) is the midpoint of the confidence interval. 

The given confidence interval is (0.028, 0.055). To find the point estimate:

\[
\hat{p} = \frac{0.028 + 0.055}{2} = \frac{0.083}{2} = 0.0415
\]

So, the point estimate \(\hat{p}\) for the population proportion of "no-shows" is 0.0415 or 4.15%.

Do you have any questions or would you like more details? Here are some related questions:

1. How is the confidence interval for a population proportion constructed?
2. What does a 90% confidence level mean in the context of this problem?
3. What formula is used to calculate the margin of error for a population proportion?
4. How does the sample size affect the width of the confidence interval?
5. What would the point estimate be if the confidence interval were wider, say (0.025, 0.060)?

**Tip**: The point estimate of a proportion is always at the center of the confidence interval, regardless of the confidence level.

Based on a random sample of 143 passengers, a 90% confidence interval for the proportion of "no-shows" on the day of the flight was determined to be (0.028, 0.055). What is the point estimate for the population proportion of all passenger "no-shows" on the day of the flight?

Find the point estimate for the population proportion of 'no-shows' from a 90% confidence interval (0.028, 0.055) based on a sample of 143 passengers. Learn the statistical approach to estimate the population proportion using confidence intervals.

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