3. Incorrect. The
answer is true not false.
We have 
With a-1=3-1=2 and a(b-1)=3(4-1)=9 degrees of freedom.
a) Referring to F-table with n1 and n2, we get F0.95=4.96. Since F=6>F0.95 we can reject the null hypothesis of equal mean at the 0.05 level.
b) Referring to the F-table with n1 and n2, we get F0.99=8.02. Since F=6<F0.95, we cannot reject the null hypothesis of equal mean at the 0.05 level.
The analysis of variance table for Problems 1-3 is shown the following table:
|
Variation |
Degrees of freedom |
Mean Square |
F |
|
vb=8 |
a-1=2 |
ŝb2=8/2=4 |
|
|
vw=6 |
a(b-1)=9 |
ŝw2=6/9=2/3 |
|
|
Total v=14 |
ab-1=3*4-1=11 |
|
|
