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

`6.76-2.01(2.55)/(√(50))=6.03`

`6.76+2.01(2.55)/(√(50))=7.49`

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

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)(s)/(√(n))` (1)

We can calculate the mean and the sample deviation we can use the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\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(2.55)/(√(50))=6.03`

`6.76+2.01(2.55)/(√(50))=7.49`

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