In: Statistics and Probability
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.
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) | X1^{2} | LA (2) | X2^{2} | |
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.