4. Correct. The
answer is false,
Sample 1 
507 
511 
509 


Sample 2 
504 
506 
508 
505 
502 
Sample 3 
510 
508 
506 
508 

It is convenient to subtract some number, e.g. 500, obtaining table 9.4.

Total 
Mean 

Sample 1 
7 
11 
9 


27 
9 
Sample 1 
4 
6 
8 
5 
2 
25 
5 
Sample 1 
10 
8 
6 
8 

32 
8 
`x=Grand mean=84/12=7 
In this table we have indicated the row totals, the sample or group means and the grand mean. We then have
v= Σ_{j,k}(x_{j,k}`x)^{2}=(77)^{2}+(117)^{2}_{. . . }+(87)^{2}=72
v_{b}=n_{j}Σ_{j}(`x_{j}_{,}`x)^{2}=3(97)^{2}+5(75)^{2}+4(87)^{2}=36
v_{w}=vv_{b}=7236=36
We can also obtain v_{w} directly by observing that it is equal to
(79)^{2}+(119)^{2}+(99)^{2}+(45)^{2}+(65)^{2}+(85)^{2}+(55)^{2}+(25)^{2}+(108)^{2}+(88)^{2}+(68)^{2}+(88)^{2}
The data can be summarized in the analysis of variance table:
Variation 
Degrees of freedom 
Mean Square 
F 
V_{b}=36 
a1=2 
ŝ_{b}^{2}=36/2=18 
_{} 
V_{w}=36 
na=9 
ŝ_{w}^{2}=36/4=9 
For 2 and 9 degrees of freedom we find from Ftable that F_{0.95}=4.26, F_{0.99}=8.02. Thus we can reject the hypothesis of equal means (i.e. no difference in the three types of bulbs) at the 0.05 but not the 0.01 level.