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. 1 5 6 7 8 8 8 9 9 9 9 10 3 4 5 5 7 6 8 9 10 5 4 6 5 7 3 1 9 8 8 9 9 10 7 6 4 8 10 2 5 1 8 6 9 6 8 8 10 10 Develop a confidence interval estimate of the population mean rating for Miami. Round your answers to two decimal places.

Answer 1

Answer:

$6.76-2.01\frac{2.55}{\sqrt{50}}=6.03$    

$6.76+2.01\frac{2.55}{\sqrt{50}}=7.49$    

The 95% confidence interval would be given by (6.03;7.49)    

Step-by-step explanation:

Notation

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n=50 represent the sample size  

Solution

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

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

We can 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 $\bar X=6.76$

The sample deviation calculated $s=2.55$

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

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

We assume a standard confidence level 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 tabel 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):

$6.76-2.01\frac{2.55}{\sqrt{50}}=6.03$    

$6.76+2.01\frac{2.55}{\sqrt{50}}=7.49$    

The 95% confidence interval would be given by (6.03;7.49)