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.