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 50 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 50 business travelers follow.

Excel or Minitab users: The data set is available in the file named "Miami." All data sets can be found in your eBook or on your Student CD.
6 4 6 8 7 7 6 3 3 8 10 4 8
7 8 7 5 9 5 8 4 3 8 5 5 4
4 4 8 4 5 6 2 5 9 9 8 4 8
9 9 5 9 7 8 3 10 8 9 6
Develop a 95% confidence interval estimate of the population mean rating for Miami (to 2 decimals).
95% Confidence:
( , )

Answer 1

A confidence interval is a decision criteria used in statistics. There are three decision criteria, the traditional method, the p-value method and the confidence interval. These method are use to decide if the null hypothesis is accepted or rejected.

Using a confidence interval method, we reject the hypothesis if the population parameter has a value outside this interval.

To find the confidence level we have to use this formula:
`(u+z_(value)((o)/(√(n) ))  ; u-z_(value)((o)/(√(n) ))`; where u is the sample mean, o is the standard deviation and n is the sample size.

So, first we have to calculate the mean and standard deviation:

`u=(sum of all values)/(total number of values) = (317)/(50) =6.34\\o = \sqrt{(\sum(x_(i)-u)^(2)  )/(N-1) } = 2.16`.

Using a 95% of confidence level, the z-value is 1.6.

Now, we can find the confidence interval because we already have all needed values.

`(u+z_(value)((o)/(√(n) ))  ; u-z_(value)((o)/(√(n) ))`

`(6.34+1.6((2.16)/(√(50) ))  ; 6.34-1.6((2.16)/(√(50) ))`

`(6.34+0.49) ; 6.34-0.49)`

`(6.83) ; 5.85)`

Hence, the confidence interval is (6.83;5.85)

Answer 2

Confidence interval: (5.74,6.94)

Explanation:

We are given the following data set:

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

Formula:

`\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}`

where
`x_i` are data points,
`\bar{x}` is the mean and n is the number of observations.

`Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}`

`Mean =\displaystyle(317)/(50) = 6.34`

Sum of squares of differences = 0.1156+5.4756+0.1156+2.7556+0.4356+0.4356+0.1156+11.1556+11.1556+2.7556+13.3956+5.4756+2.7556+0.4356+2.7556+0.4356+1.7956+7.0756+1.7956+2.7556+5.4756+11.1556+2.7556+1.7956+1.7956+5.4756+5.4756+5.4756+2.7556+5.4756+1.7956+0.1156+18.8356+1.7956+7.0756+7.0756+2.7556+5.4756+2.7556+7.0756+7.0756+1.7956+7.0756+0.4356+2.7556+11.1556+13.3956+2.7556+7.0756+0.1156 = 229.22

`S.D = \sqrt{(229.22)/(49)} = 2.16`

Confidence interval:

`\mu \pm z_(critical)(\sigma)/(√(n))`

Putting the values, we get,

`z_(critical)\text{ at}~\alpha_(0.05) = 1.96`

`6.34 \pm 1.96((2.16)/(√(50)) ) = 6.34 \pm 0.599 = (5.74,6.94)`