5. Correct. The
answer is false
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 v_{r}=variation of row means from the grand mean= 4[(6.26.8)^{2}+(8.36.8)^{2}+(5.96.8)^{2}]=13.68
Let v_{c}=variation of column means from the grand mean= 3[(6.46.8)^{2}+(7.06.8)^{2}+(6.36.8)^{2}]=2.82
v=total variation = (4.56.8)^{2}+(6.46.8)^{2}+(7.26.8)^{2}+(6.76.8)^{2}+(8.86.8)^{2}+(7.86.8)^{2}+(9.66.8)^{2}+(7.06.8)^{2}+(5.96.8)^{2}+(6.86.8)^{2}+(5.76.8)^{2}+(5.26.8)^{2}=23.08
v_{e}=random variation=v_{r}v_{c}=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 F_{0.95}=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 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.