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
Show your answers with practical examples
(e) Prepare a two –way analysis of variance table
(f) Test at 5% level of significance the null hypothesis that the population mean yields are
identical for all four varieties of corn
(g) Test at 5% level of significance the null hypothesis that the population mean yields are
the same for all three brands of fertilizer
(e)
Source | Sum of Squares | Degrees of Freedom | Mean Square | F
-----------------------------------------------------------------------------------------------------------------------------------
Variety | 168.33 | 3 | 56.11 | 5.16
Fertilizer | 692.33 | 2 | 346.17 | 31.93
Error | 78.00 | 6 | 13.00 |
Total | 938.67
(f) To test if the population mean yields are identical for all four varieties of corn, we use the F-test. The critical value at a 5% level of significance with degrees of freedom (3, 6) is 5.99. Since the calculated F-value of 5.16 is less than the critical value, we fail to reject the null hypothesis. This suggests that there is no significant difference in yields between the four varieties of corn.
(g) To test if the population mean yields are the same for all three brands of fertilizer, we use the F-test. The critical value at a 5% level of significance with degrees of freedom (2, 6) is 5.14. Since the calculated F-value of 31.93 is higher than the critical value, we reject the null hypothesis. This indicates that there is a significant difference in yields between the three brands of fertilizer.