In: Math
The below data shows the amount of time (in seconds) that animated Disney movies showed the use of tobacco and alcohol. Test the claim that the mean difference in time of tobacco use vs. alcohol use is equal to zero at the 0.1 significance level. You may want to use a spreadsheet to help you solve this problem
Tobacco Alcohol
56.1 | 110.9 |
40.5 | 70.2 |
37.2 | 90.6 |
49.2 | 112.8 |
63.4 | 202.4 |
48 | 93.5 |
52.3 | 92.8 |
52 | 78.1 |
40.5 | 109 |
39.2 | 45.5 |
36.6 | 44.2 |
59.2 | 82.9 |
46.1 | 123.8 |
54.1 | 102.2 |
54.7 | 69.9 |
41.6 | 53.1 |
45 | 79.5 |
50.5 | 69.4 |
45.6 | 137.8 |
63.4 | 135.4 |
48 | 97.6 |
33.6 | 145.3 |
51 | 65 |
35.9 | 81.4 |
42.9 | 43.7 |
Claim: Select an answer p > 0 u ≠ 0 u ≤ 0 p < 0 u > 0 p ≥ 0 p ≤ 0 p ≠ 0 u < 0 p = 0 u ≥ 0 u = 0
which corresponds to Select an answer H1: u > 0 H0: u ≠ 0 H1: u < 0 H0: u ≤ 0 H1: u ≠ 0 H0: u = 0 H0: p ≥ 0
Opposite: Select an answer u < 0 p > 0 p ≤ 0 u = 0 p ≠ 0 u ≠ 0 u ≥ 0 p ≥ 0 u > 0 p = 0 p < 0 u ≤ 0
which corresponds to Select an answer H1: u < 0 H0: u ≠ 0 H1: u ≠ 0 H0: u ≤ 0 H0: p ≥ 0 H0: u = 0 H1: u > 0
The test is: Select an answer / left-tailed / two-tailed / right-tailed
The test statistic is: Select an answer / -6.969 / -7.119 / -6.289 / -6.729 / -6.609 (to 3 decimals)
The Critical Value is: Select an answer / ± 1.711 / ± 1.47 / ± 1.316 / ± 1.36 / ± 1.867 (to 3 decimals)
Based on this we: Select an answer / Fail to reject the null hypothesis/ Reject the null hypothesis
Conclusion There Select an answer ( does / not does ) appear to be enough evidence to support the claim that the mean difference in time of tobacco use vs. alcohol use is equal to zero.
Solution:
Here, we have to use two sample t test for the difference between two population means. The excel outputs for this test is given as below:
t-Test: Two-Sample Assuming Equal Variances |
||
Tobaco |
Alcohol |
|
Mean |
47.464 |
93.48 |
Variance |
69.46323 |
1342.794 |
Observations |
25 |
25 |
Pooled Variance |
706.1287 |
|
Hypothesized Mean Difference |
0 |
|
df |
48 |
|
t Stat |
-6.1224 |
|
P(T<=t) one-tail |
8.15E-08 |
|
t Critical one-tail |
1.299439 |
|
P(T<=t) two-tail |
1.63E-07 |
|
t Critical two-tail |
1.677224 |
t-Test: Two-Sample Assuming Unequal Variances |
||
Tobaco |
Alcohol |
|
Mean |
47.464 |
93.48 |
Variance |
69.46323 |
1342.794 |
Observations |
25 |
25 |
Hypothesized Mean Difference |
0 |
|
df |
26 |
|
t Stat |
-6.1224 |
|
P(T<=t) one-tail |
8.98E-07 |
|
t Critical one-tail |
1.314972 |
|
P(T<=t) two-tail |
1.8E-06 |
|
t Critical two-tail |
1.705618 |
Answers based on above outputs are given as below:
Claim: u = 0 which corresponds to H0: u = 0
Opposite: u ≠ 0 which corresponds to H1: u ≠ 0
The test is: two-tailed
The test statistic is: -6.289
The Critical Value is: ± 1.711
Based on this we: Reject the null hypothesis
(Because the test statistic value does not lies between the critical values -1.711 and +1.711.)
Conclusion
There not does appear to be enough evidence to support the claim that the mean difference in time of tobacco use vs. alcohol use is equal to zero.