In: Math
To test whether the enactment of seatbelt legislation had any association with injuries to vehicle users, a random sample of 7676 individuals taken over a two-year period (one year prior legislation and one year post legislation) was used. The results are detailed below.
Post Legislation |
Prior Legislation |
Total |
|
No Injury |
1320 |
5430 |
6750 |
Minimal Injury |
70 |
552 |
622 |
Major Injury |
32 |
224 |
256 |
Fatal Injury |
10 |
38 |
48 |
Total |
1432 |
6244 |
7676 |
Please write a summary statement for this hypothesis test.
Hypotheses are:
H0: The enactment of seat-belt legislation had no association with injuries to vehicle users.
Ha: The enactment of seat-belt legislation had an association with injuries to vehicle users.
Expected frequencies will be calculated as follows:
Following table shows the expected frequencies;
Post Legislation | Prior legislation | Total | |
No Injury | 1259.25 | 5490.75 | 6750 |
Minimal Injury | 116.038 | 505.962 | 622 |
Major Injury | 47.758 | 208.242 | 256 |
Fatal Injury | 8.955 | 39.045 | 48 |
Total | 1432 | 6244 | 7676 |
Following table shows the calculations for chi square test statistics:
O | E | (O-E)^2/E |
1320 | 1259.25 | 2.930762359 |
70 | 116.038 | 18.26554615 |
32 | 47.758 | 5.199433896 |
10 | 8.955 | 0.12194584 |
5430 | 5490.75 | 0.672141784 |
552 | 505.962 | 4.189044719 |
224 | 208.242 | 1.19243267 |
38 | 39.045 | 0.02796837 |
Total | 32.59927579 |
The test statistics is:
Degree of freedom: df =( number of rows -1)*(number of columns-1) = (4-1)*(2-1)=3
The p-value using excel function "=Chidist(32.599,3)" is: 0.0000
Since p-value is less than 0.05 so we reject the null hypothesis. That is there is evidence to conclude that the enactment of seatbelt legislation had any association with injuries to vehicle users.