Question

The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maximum possible rating is 10. Suppose a simple random sample of business travelers is selected and each traveler is asked to provide a rating for the Miami International Airport. The ratings obtained from the sample of business travelers follow. 7 7 3 8 4 4 4 5 5 5 5 4 9 10 9 9 8 10 4 5 4 10 10 10 11 4 9 7 5 4 4 5 5 4 3 10 10 4 4 8 7 7 4 9 5 9 4 4 4 4 Develop a confidence interval estimate of the population mean rating for Miami. Round your answers to two decimal places.

Answer 1

Answer:

$6.26-2.01\frac{2.47}{\sqrt{50}}=5.56$    

$6.26+2.01\frac{2.47}{\sqrt{50}}=6.96$    

So on this case the 95% confidence interval would be given by (5.56;6.96)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)  

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)  

The mean calculated for this case is $\bar X=6.26$

The sample deviation calculated $s=2.47$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=50-1=49$

Assuming a Confidence of 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that $t_{\alpha/2}=2.01$

Now we have everything in order to replace into formula (1):

$6.26-2.01\frac{2.47}{\sqrt{50}}=5.56$    

$6.26+2.01\frac{2.47}{\sqrt{50}}=6.96$    

So on this case the 95% confidence interval would be given by (5.56;6.96)