An agricultural experiment designed to assess differences in yields of corn for four different
varieties, using three different fertilizers, produced the results (in tonnes per acre) shown in the
following table.
Fertilizer Variety
A B C D
1 86 88 77 84
2 92 91 81 93
3 75 80 83 79
Step by step show your answers practically in the questions below
(1) Prepare a two –way analysis of variance table
(2) Test at 5% level of significance the null hypothesis that the population mean yields are
identical for all four varieties of corn
(3) Test at 5% level of significance the null hypothesis that the population mean yields are
the same for all three brands of fertilizer
(1) Two-way analysis of variance table:
Variety Fertilizer Yields Ybar Ytilde SS
A 1 86 Ybar1 Ytilde1 SS11
A 2 92 Ybar2 Ytilde2 SS12
A 3 75 Ybar3 Ytilde3 SS13
B 1 88 Ybar4 Ytilde4 SS14
B 2 91 Ybar5 Ytilde5 SS15
B 3 80 Ybar6 Ytilde6 SS16
C 1 77 Ybar7 Ytilde7 SS17
C 2 81 Ybar8 Ytilde8 SS18
C 3 83 Ybar9 Ytilde9 SS19
D 1 84 Ybar10 Ytilde10 SS110
D 2 93 Ybar11 Ytilde11 SS111
D 3 79 Ybar12 Ytilde12 SS112
Ybar1 = (86+92+75+84)/4 = 84.25
Ybar2 = (88+91+80+93)/4 = 88.00
Ybar3 = (77+81+83+79)/4 = 80.00
Ybar4 = (86+88+77+84)/4 = 83.75
Ybar5 = (92+91+81+93)/4 = 89.25
Ybar6 = (75+80+83+79)/4 = 79.25
SS1 = 4(84.25-86)^2 + 4(88-84.25)^2 + 4(80-84.25)^2 + 4(83.75-86)^2 + 4(89.25-88)^2 + 4(79.25-88)^2 = 57.25
SS2 = 4(84.25-92)^2 + 4(88-91)^2 + 4(80-81)^2 + 4(83.75-92)^2 + 4(89.25-91)^2 + 4(79.25-81)^2 = 70.50
SS3 = 4(84.25-75)^2 + 4(88-80)^2 + 4(80-83)^2 + 4(83.75-75)^2 + 4(89.25-80)^2 + 4(79.25-83)^2 = 117.25
SS4 = 4(84.25-84)^2 + 4(88-93)^2 + 4(80-79)^2 + 4(83.75-84)^2 + 4(89.25-93)^2 + 4(79.25-79)^2 = 64.50
Fertilizer Ytilde Fbar Ftilde SS3 = SS4
1 Ytilde1 Fbar1 Ftilde1 SS3
2 Ytilde2 Fbar2 Ftilde2 SS4
3 Ytilde3 Fbar3 Ftilde3
Fbar1 = (86+88+77+84)/4 = 83.75
Fbar2 = (92+91+81+93)/4 = 89.25
Fbar3 = (75+80+83+79)/4 = 79.25
SS3 = 4(83.75-86)^2 + 4(89.25-88)^2 + 4(79.25-80)^2 = 80.25
SS4 = 4(83.75-84)^2 + 4(89.25-93)^2 + 4(79.25-79)^2 = 68.75
Total SS = SS1 + SS2 + SS3 + SS4 = 57.25 + 70.50 + 80.25 + 68.75 = 276.75
(2) Test at 5% level of significance the null hypothesis that the population mean yields are identical for all four varieties of corn:
Ho: μ1 = μ2 = μ3 = μ4
Ha: At least one mean is different
Using the F-test with 3 and 8 degrees of freedom:
MS1 = SS1/3 = 57.25/3 = 19.08
MS2 = SS2/3 = 70.50/3 = 23.50
F = MS1/MS2 = 19.08/23.50 = 0.81
F(3, 8, 0.05) = 4.07
Since 0.81 < 4.07, we fail to reject the null hypothesis. There is not enough evidence to suggest that the mean yields are different for all four varieties of corn.
(3) Test at 5% level of significance the null hypothesis that the population mean yields are the same for all three brands of fertilizer:
Ho: μ1 = μ2 = μ3
Ha: At least one mean is different
Using the F-test with 2 and 9 degrees of freedom:
MS3 = SS3/2 = 80.25/2 = 40.13
MS4 = SS4/2 = 68.75/2 = 34.38
F = MS3/MS4 = 40.13/34.38 = 1.17
F(2, 9, 0.05) = 4.26
Since 1.17 < 4.26, we fail to reject the null hypothesis. There is not enough evidence to suggest that the mean yields are different for all three brands of fertilizer.