To determine the minimum sample size needed to be 99% confident within 5 percentage points of the actual proportion of on-time flights, we need to follow these steps:
1. Identify the Confidence Level and Corresponding Z-Score:
The confidence level is 99%. The Z-score associated with a 99% confidence level is 2.576.
2. Define the Desired Margin of Error:
The margin of error is 5 percentage points, which can be written as 0.05.
3. Estimate the Population Proportion:
Without prior knowledge of the actual proportion, a conservative estimate is to use 0.5. This provides the maximum possible sample size needed to ensure the margin of error is not exceeded.
4. Use the Sample Size Formula for Proportions:
The formula to calculate the minimum sample size [tex]\(n\)[/tex] is given by:
[tex]\[
n = \frac{{Z^2 \cdot p \cdot (1 - p)}}{{E^2}}
\][/tex]
where:
- [tex]\( Z \)[/tex] is the Z-score (2.576)
- [tex]\( p \)[/tex] is the estimated population proportion (0.5)
- [tex]\( E \)[/tex] is the margin of error (0.05)
5. Calculate the Sample Size:
Plugging the values into the formula:
[tex]\[
n = \frac{{2.576^2 \cdot 0.5 \cdot (1 - 0.5)}}{{0.05^2}}
\][/tex]
Simplify the calculations:
[tex]\[
n = \frac{{2.576^2 \cdot 0.5 \cdot 0.5}}{0.0025}
\][/tex]
[tex]\[
n = \frac{{6.6357 \cdot 0.25}}{{0.0025}}
\][/tex]
[tex]\[
n = \frac{{1.658925}}{{0.0025}}
\][/tex]
[tex]\[
n = 663.57
\][/tex]
6. Round Up to the Nearest Whole Number:
Since the sample size must be a whole number, we round 663.57 up to 664.
Therefore, the minimum sample size needed to be 99% confident within 5 percentage points of the actual proportion of on-time flights is 664.
