Question 1197974: PICTURE FOR PROBLEM https://imagizer.imageshack.com/img923/3996/7mODCn.jpg
Will your flight get you to your destination on time? To the right are a histogram and summary statistics for the percentage of delayed arrivals each month from 2001 thru 2006. Consider these data to be a representative sample of all months. There is no evidence of a time trend. (The correlation of Flights Delayed % with time is r=-0.012.)
b) Find a 99% confidence interval for the true percentage of flights that arrive late.
Answer by onyulee(41)   (Show Source):
You can put this solution on YOUR website! To calculate a 99% confidence interval for the true percentage of flights that arrive late, we'll use the formula for the confidence interval:
\[
\text{Confidence Interval} = \bar{x} \pm z^* \cdot \frac{s}{\sqrt{n}}
\]
Where:
- \(\bar{x}\) = sample mean (percentage of delayed flights)
- \(s\) = standard deviation of the sample
- \(n\) = sample size
- \(z^*\) = critical value for a 99% confidence level (from the standard normal distribution)
### Steps:
1. Identify the sample statistics (\(\bar{x}\), \(s\), and \(n\)) from the given data.
2. Use a \(z^*\)-value of approximately **2.576** for a 99% confidence level.
3. Calculate the margin of error:
\[
\text{Margin of Error} = z^* \cdot \frac{s}{\sqrt{n}}
\]
4. Compute the confidence interval:
\[
(\bar{x} - \text{Margin of Error}, \bar{x} + \text{Margin of Error})
\]
If you provide the sample mean (\(\bar{x}\)), standard deviation (\(s\)), and sample size (\(n\)) from the summary statistics or histogram, I can complete the calculation.
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