5. Incorrect. The
answer is false not true.
Compute the means and totals across the rows and columns, as well as the grand mean in the following way:
|
|
Product 1 |
Product 2 |
Product 3 |
Product 4 |
Row Totals |
Row Means |
|
Machine A |
4.5 |
6.4 |
7.2 |
6.7 |
24.8 |
6.2 |
|
Machine B |
8.8 |
7.8 |
9.6 |
7.0 |
33.2 |
8.3 |
|
Machine C |
5.9 |
6.8 |
5.7 |
5.2 |
23.6 |
5.9 |
|
Column Totals |
19.2 |
21.0 |
22.5 |
18.9 |
Grand Total=81.6 |
|
|
Column Means |
6.4 |
7.0 |
7.5 |
6.3 |
Grand Mean=6.8 |
|
Let vr=variation of row means from the grand mean= 4[(6.2-6.8)2+(8.3-6.8)2+(5.9-6.8)2]=13.68
Let vc=variation of column means from the grand mean= 3[(6.4-6.8)2+(7.0-6.8)2+(6.3-6.8)2]=2.82
v=total variation = (4.5-6.8)2+(6.4-6.8)2+(7.2-6.8)2+(6.7-6.8)2+(8.8-6.8)2+(7.8-6.8)2+(9.6-6.8)2+(7.0-6.8)2+(5.9-6.8)2+(6.8-6.8)2+(5.7-6.8)2+(5.2-6.8)2=23.08
ve=random variation=vr-vc=6.58
Given 
, 
, 
(a) F=6.84/1.097=6.24 with d.f.=2,6
(b) F=0.94/1.097=0.86with d.f.=3,6
At 5% level of significance with 2, 6 degrees of freedom F0.95=5.14. Since 6.24>5.14, we can reject the hypothesis that the row means are equal and conclude that at the 5% level there is a significant difference in productivity due to the type of machine.
Since the F value corresponding to differences in means is less than 1, we can conclude that there is no significant difference in productivity due the type of product.