3. Correct. The
answer is true
We have
With a1=31=2 and a(b1)=3(41)=9 degrees of freedom.
a) Referring to Ftable with n_{1} and n_{2}, we get F_{0.95}=4.96. Since F=6>F_{0.95 }we can reject the null hypothesis of equal mean at the 0.05 level.
b) Referring to the Ftable with n_{1} and n_{2}, we get F_{0.99}=8.02. Since F=6<F_{0.95 }we cannot reject the null hypothesis of equal mean at the 0.05 level.
The analysis of variance table for Problems 9.19.3 is shown the following table:
Variation 
Degrees of freedom 
Mean Square 
F 
v_{b}=8 
a1=2 
`s_{b}^{2}=8/2=4 
_{} 
v_{w}=6 
a(b1)=9 
s_{w}^{2}=6/9=2/3 

Total v=14 
ab1=3*41=11 

