Question

The International Air Transport Association surveyed business travellers to determine the assessment of international airports. The maximum possible rating was 10. Suppose a simple random sample of 50 travelers rated Miami Airport, and another simple random sample of 50 travelers rated Los Angeles airport. The answers were as follows.

Miami:

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

Los Angeles:

10     9     6     7     8     7     9      8     10     7     6     5     7

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

5      3     1     8     9      6      8     5     4      6    10    9     8

3     2     7      9     5     3     10     3     5     10    8

With α = 0.025, perform all the steps discussed in the course with the respective appropriate null and alternative hypotheses to determine that the two airports are highly competitive.

Answer 1

There are two populations here. We need to compare the assesments of both the airports. We assume that the population variances are same. Also we will conduct an independent t-test for the difference of the population means since n > 30. There is nowhere mentioned that test is to see if any one is better so we will conduct two tailed test.

	Miami (1)	X12	LA (2)	X22
1	6	36	10	100
2	4	16	9	81
3	6	36	6	36
4	8	64	7	49
5	7	49	8	64
6	7	49	7	49
7	6	36	9	81
8	3	9	8	64
9	3	9	10	100
10	8	64	7	49
11	10	100	6	36
12	4	16	5	25
13	8	64	7	49
14	7	49	3	9
15	8	64	5	25
16	7	49	6	36
17	5	25	8	64
18	9	81	7	49
19	5	25	10	100
20	8	64	8	64
21	4	16	4	16
22	3	9	7	49
23	8	64	8	64
24	5	25	6	36
25	5	25	9	81
26	4	16	9	81
27	4	16	5	25
28	4	16	3	9
29	8	64	1	1
30	4	16	8	64
31	5	25	9	81
32	6	36	6	36
33	2	4	8	64
34	5	25	5	25
35	9	81	4	16
36	9	81	6	36
37	8	64	10	100
38	4	16	9	81
39	8	64	8	64
40	9	81	3	9
41	9	81	2	4
42	5	25	7	49
43	9	81	9	81
44	7	49	5	25
45	8	64	3	9
46	3	9	10	100
47	10	100	3	9
48	8	64	5	25
49	9	81	10	100
50	6	36	8	64
Total	317	2239	336	2534

Mean =

SD =

	Miami (1)	LA (2)
Mean	6.34	6.72
Variance	4.6780	5.6343
n	50	50
Pooled Variance	5.15612

Pooled variance =t

(There is no difference between the assesment rates)

(There is a difference between the assesment rates)

Test Stat =

Test Stat = -0.8367

p-value = 2P ( > T.S. )

=2P(> 0.84)

= 2 * 0.2024 .............found using t-dist tables

p-value = 0.4048

Since p-value > 0.025

We do not reject the null hypothesis at 2.5%.

There is insufficient evidence to suppport that there is difference between assesment. Both are equally competitive.