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= Σj,k(xj,k-`x)2=(7-7)2+(11-7)2. . . +(8-7)2=72

vb=njΣj(`xj,-`x)2=3(9-7)2+5(7-5)2+4(8-7)2=36

vw=v-vb=72-36=36

We can also obtain vw directly by observing that it is equal to

(7-9)2+(11-9)2+(9-9)2+(4-5)2+(6-5)2+(8-5)2+(5-5)2+(2-5)2+(10-8)2+(8-8)2+(6-8)2+(8-8)2

The data can be summarized in the analysis of variance table:

 Variation Degrees of freedom Mean Square F Vb=36 a-1=2 ŝb2=36/2=18 Vw=36 n-a=9 ŝw2=36/4=9

For 2 and 9 degrees of freedom we find from F-table that F0.95=4.26, F0.99=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.