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.