Question

In an attempt to asses total daily travel taxes in various cities, the Global Business Travel Association conducted a study of daily travel taxes on lodging, rental cars, and meals (GBTA Foundation website, October 30, 2012). The data in the file named "TravelTax" are consistent with the findings of that study for business travel to Chicago. Assume the population standard deviation is known to be $8.50 and develop a 95% confidence interval of the population mean total daily travel taxes for Chicago.

Lodging, Car Rental, and Meal Taxes ($)
38.02
41.89
47.44
40.14
47.69
44.02
32.93
33.32
33.90
36.53
43.01
33.22
35.65
39.47
38.86
42.42
47.71
45.03
53.73
36.07
49.95
31.81
63.43
31.50
64.13
42.14
46.78
30.45
47.35
39.42
39.79
43.63
34.03
35.74
48.43
30.92
44.45
41.63
27.23
41.03
43.87
46.07
34.34
39.57
36.40
48.18
42.92
35.57
45.68
43.14
55.21
34.73
26.40
40.66
37.66
28.63
38.17
45.16
46.19
49.58
48.14
45.61
34.22
44.34
55.14
37.00
48.78
52.77
49.07
38.14
45.35
42.59
41.27
20.93
30.70
43.00
50.91
38.56
41.31
39.18
33.57
25.09
30.05
19.28
32.47
53.97
47.32
35.61
44.11
38.97
37.48
40.55
52.03
46.86
46.07
34.89
42.93
49.64
44.35
41.66
39.80
50.31
36.59
33.40
45.51
44.43
47.65
36.24
32.03
32.81
31.64
40.18
34.13
33.22
48.60
40.66
43.37
41.08
34.11
52.91
31.76
56.03
33.61
40.11
53.77
34.73
35.58
48.28
31.89
34.71
37.44
34.77
31.82
39.98
44.66
31.78
21.69
43.23
54.00
46.55
44.85
42.15
25.67
50.25
44.62
44.30
45.09
35.12
34.06
31.77
36.55
36.79
29.65
47.66
25.91
47.27
32.31
42.52
46.86
48.75
38.81
33.66
27.24
46.40
50.95
39.74
35.47
35.95
29.51
62.77
37.72
45.80
39.91
33.28
39.63
28.57
38.14
36.04
40.13
35.51
31.49
39.49
37.69
50.34
44.10
45.61
46.51
53.00
35.40
25.64
47.65
47.49
32.76
45.41
28.77
37.32
28.64
50.67
33.77
40.88

Answer 1

($39.132, $41.488)

A step-by-step explanation is mentioned in the attached image

Answer 2

The 95% confidence interval of the population mean total daily travel taxes for Chicago is ($39.13, $41.49).

Explanation:

The first step, before building the confidence interval, is finding the mean of the data set.

We are given 200 values, and the with the help of a calculator, the mean of this values is of 40.31.

Confidence interval:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population(8.50, as given in the problem) and n is the size of the sample(200).

`M = 1.96(8.50)/(√(200)) = 1.18`

The lower end of the interval is the sample mean subtracted by M. So it is 40.31 - 1.18 = $39.13.

The upper end of the interval is the sample mean added to M. So it is 40.31 + 1.18 = $41.49.

The 95% confidence interval of the population mean total daily travel taxes for Chicago is ($39.13, $41.49).