Downloaded from https://www.everyspec.com AMCP OM196 Example No, 12 Tablo 2-15(B) shows some simulated experimental data. There is I factor i with 3 levels. The question Is, do those 3 levels actually have a real effect? For example, level 1 appears to have a much higher mean thz'n do the other 2 levels. 1. Subtract the overall mean from 1. This has already b~en done in Table each datum. 2-15(B), 2. Find the mean of each column. 2. See Table 2-15(B). 3. The column means have a disper- 3. The mean of the "column means" is zerol slon. Estimate the variance of the so the sum of squares Is (0.588), column means. + (- 0.330)3 + (- 0.258)2 - 0.5212.
There are 3 columns with known meani so there are Yj - 2 (2 - 3 - 1) degrees of freedom (for the numerator in step 6): 0.5212/2 0. 2606, This is the estimated variance of the column means. 4, Estimate the population variance 4. To convert it to the estimated population from the column-mean variance, variance, multiply by the number of ele- ments in each column (4). Therefore aj4 x 0. 2606 - 1. 0424. B. Estimate the variance from the 5. Column I sum of squares Is replications. First, get the sam- (2.98 - 0.588)2 + (- 0.39 - 0.588)2 pIe (column) sums of squares. + (- 0.01 - 0.588)2 + (- 0.23 - 0.588)2 Then calculate the degrees of - 7.7049. Column 2 sum of squares is freedom, Calculate the es.
This 4.3390. Column 3 sum of squares is Is an estimate of the population 0.3455. The total replication sum of variance because It was derived squares is 7. 7049 + 4.3390 + P. 3455 from individual data, not the - 12.3894. Each column has 3 (3 a4 - 1) means* degrees of freedom and there are 3 col- umns. There are 3 X 3 * 9 degrees of freedom for the sum of squares; there- fore Wr - 9 for step 6. 4w 12.3894/4 -1.3766. 2-82