Question

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) as shown in the table below: Fertilizer 1 = 86 88 77 84 Fertilizer 2 = 92 91 81 93 Fertilizer 3 = 75 80 83 79 (a) Prepare a two –way analysis of variance table. (b) Test at 5% level of significance the null hypothesis that the population mean yields are identical for all four varieties of corn. (c) Test at 5% level of significance the null hypothesis that the population mean yields are the same for all three brands of fertilizer.

Answer 1

Solution: We can use excel data analysis tool to prepare the two-way analysis of variance table. The excel output is given below:

Anova: Two-Factor Without Replication SUMMARY Count SumAverage Variance 335 83.75022.917 357 89.250 30.917 317 79.25010.917 1 4 Fertilizer 253 84.333 74.333 259 86.333 32.333 241 80.333 9.333 256 85.33350.333 ANOVA Source of Variation MS P-value F crit Blocks Treatments Error 200.67 62.25 132.00 2 100.3333 4.56060.06255.1433 3 20.7500 0.9430.4766 4.7571 Total 394.92

(a) Prepare a two –way analysis of variance table.

Answer: The two-way analysis of variance table is:

(b) Test at 5% level of significance the null hypothesis that the population mean yields are identical for all four varieties of corn.

Answer: Since the p-value is 0.4766, which is greater than the significance level 0.05, we, therefore, fail to reject the null hypothesis and conclude that the population mean yields are identical for all four varieties of corn.

(c) Test at 5% level of significance the null hypothesis that the population mean yields are the same for all three brands of fertilizer.

Answer: Since the p-value is 0.0625, which is greater than the significance level 0.05, we, therefore, fail to reject the null hypothesis and conclude that the population mean yields are the same for all three brands of fertilizer.