To develop a 95% confidence interval estimate for the population mean rating of Miami International Airport, we need to calculate the sample mean ($\bar{x}$) and the sample standard deviation ($s$). Then, we will use the t-distribution (since the population standard deviation is unknown and the sample size is less than 30) to calculate the confidence interval.

Step-by-Step Solution:

1. Calculate the sample mean ($\bar{x}$): $\bar{x} = \frac{\sum x_i}{n}$ where $x_i$ are the individual ratings and $n = 50$ is the sample size.

2. Calculate the sample standard deviation ($s$): $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}$

3. Determine the critical value from the t-distribution: For a 95% confidence level and 49 degrees of freedom (since $n - 1 = 50 - 1 = 49$), the critical value $t_{0.025, 49}$ can be obtained from the t-distribution table.

4. Compute the margin of error (ME): $\text{ME} = t_{0.025, 49} \times \frac{s}{\sqrt{n}}$

5. Calculate the confidence interval: $\text{Confidence Interval} = \left( \bar{x} - \text{ME}, \bar{x} + \text{ME} \right)$

Let's perform these calculations.The sample mean rating for Miami International Airport is $6.00$.

The 95% confidence interval for the population mean rating is approximately:

$(5.19, 6.81)$

This means we are 95% confident that the true mean rating of Miami International Airport by business travelers lies between 5.19 and 6.81.

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

Related Questions:

1. What is the interpretation of a confidence interval in statistics?
2. How does the sample size affect the width of the confidence interval?
3. Why do we use the t-distribution instead of the normal distribution in this case?
4. What would happen to the confidence interval if the confidence level was increased to 99%?
5. How can we determine if the sample mean is a good estimate of the population mean?

Tip: To improve the accuracy of confidence intervals, consider increasing the sample size.