Statistics problem

Question

An 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 50 business travelers is selected and each traveler is asked to provide a rating for a certain airport. The ratings obtained from the sample of 50 business travelers follow.

6	4	6	8	7	7	6	3	3	8
10	4	8	7	8	7	5	9	4	8
4	3	8	5	4	4	4	4	8	3
5	6	2	5	9	9	8	4	8	9
9	4	9	7	8	3	10	8	9	6

Develop a 95% confidence interval estimate of the population mean rating for this airport. (Round your answers to two decimal places.)

Answer 1

Hi Steele,

All confidence intervals require, at minimum, a mean(x-bar), standard deviation(s), critical value (CV), and ultimately, standard error (SE). Formula is:

CI=x-bar +/- CV*SE

So, first and foremost, we need to compute the mean. This is (6 + 4 + 6 + 8 + 7...)/50. We also need a standard deviation. The formula for that is too tedious to perform for 50 ratings, so use statistical software, an online calculator, or a graphics calculator. I came up with:

x-bar=6.26

s=2.24

Now, we were not told anything about this population, so we cannot assume normality. That means we have to use t as our critical value, not z. So, we need to talk about degrees of freedom. Degrees of freedom (df)=n-1

n=sample size

n=50

df=50-1

df=49

So, go to the t-table, look for the column with 95% confidence and the row with 49. On my t-table, the closest I get to this is 40:

t₄₀=2.021

That is our critical value (CV).

Now, last but not least, we need the standard error, which has its own formula:

SE=s/sqrt(n)

s=sample standard deviation

n=sample size

Here,

s=2.24

n=50

SE=2.24/sqrt(50)

SE=0.32

Thus, we have all we need to compute:

CI= x-bar +/- CV*SE

x-bar=6.26

CV=2.021

SE=0.32

CI=6.26 +/- (2.021*0.32)

CI=(5.61, 6.91)

I hope this helps.