In: Statistics and Probability
| 
 Compact cars  | 
 Midsize cars  | 
 Full-size cars  | 
| 
 64  | 
 46  | 
 47  | 
| 
 65  | 
 43  | 
 45  | 
| 
 69  | 
 52  | 
 40  | 
| 
 Average: 66  | 
 Average: 47  | 
 Average: 44  | 
The average for all can be found as 52.3.
a). Find SSG.
b). Find SSE.
c). Find SST
d). Find F-value
e). Find F-critical value from the Table.
f). Make your decision.
(a)
From the given data, the following Table is calculated:
| Compact cars | Midsize cars | Full - size cars | |
| Sum = | 198 | 141 | 132 | 
| n = | 3 | 3 | 3 | 
| Mean = | 66 | 47 | 44 | 
![]()  | 
13082 | 6669 | 5834 | 
| Std. Dev. | 2.646 | 4.583 | 3.606 | 
| SS | 14 | 42 | 26 | 
Total Sample Size =N = 3 +3 + 3 = 9
Total Sum of values is given by:

Total Sum of Squated values is given by:

Total Sum of Squares = SST is given by:

Error Sum of Squated values= SSE is given by:
SSE = 14 + 42 + 26 = 82
Group Sum of Squares = SSG is given by:
SSG = SST-SSE = 936 - 82 = 854
So,
SSG = 854
(b)
Error Sum of Squated values= SSE is given by:
SSE = 14 + 42 + 26 = 82
So,
SSE = 82
(c)
Total Sum of Squares = SST is given by:

So,
SST = 936
(d)
MSG = SSG/dfG = 854/2 = 427
MSE = SSE/dfE = 82/6 = 13.667
So,
F Value = 427/13.667 = 31.244
So,
F Value = 31.244
(e)
F-critical value from the Table. = 5.143
(f)
Since calculated value of F = 31.244 is greater than critical value of F = 5.143, the difference is significant. Reject null hypothesis.
Conclusion:
The data do not upport the claim that the mean pressure applied to
the driver’s head during a crash test is equal for each types of
car.