To develop a 95% confidence interval for the population mean rating, we use the following formula:

$CI = \bar{x} \pm z \left(\frac{s}{\sqrt{n}}\right)$

Where:

• $\bar{x}$ = sample mean
• $z$ = z-score for 95% confidence level (1.96 for a two-tailed test)
• $s$ = sample standard deviation
• $n$ = sample size (which is 50)

Step 1: Calculate the sample mean ($\bar{x}$)

$\bar{x} = \frac{\sum x_i}{n}$

Step 2: Calculate the sample standard deviation ($s$)

$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

Step 3: Compute the margin of error

$\text{Margin of Error} = z \times \frac{s}{\sqrt{n}}$

Step 4: Construct the confidence interval

$CI = \left( \bar{x} - \text{Margin of Error}, \bar{x} + \text{Margin of Error} \right)$

Let me calculate these values for you.The sample mean ($\bar{x}$) is 6.32, and the sample standard deviation ($s$) is 3.01.

Using a 95% confidence level, the confidence interval estimate for the population mean rating is:

$(5.48, 7.16)$

This means we are 95% confident that the true mean rating for Miami International Airport lies between 5.48 and 7.16.

Would you like more details on any step, or do you have any questions?

Related questions:

1. What factors can influence the width of a confidence interval?
2. How does the sample size affect the confidence interval?
3. How is the z-score determined for different confidence levels?
4. What are the assumptions underlying the confidence interval calculation?
5. How would the confidence interval change if the sample had more variability?

Tip:

A larger sample size reduces the margin of error, providing a more precise estimate of the population mean.