The following data represent the number of units produced by three workers, each
ID: 3228793 • Letter: T
Question
The following data represent the number of units produced by three workers, each working on the same machine for four different days: Use alpha = .05 in all your tests below. a) Write down the appropriate two-way ANOVA model with all relevant assumptions. b) Test if differences among the Machines are significant? c) Test if differences among the Workers are significant? d) Test if interactions between the Machines and workers are significant? e) Draw interaction plot. If the interactions are significant, describe how the means of one factor behave at various levels of the other factor. f) Are the tests in b) and c) meaningful given your conclusions in d) above? Comment.Explanation / Answer
(a) The two null hypothesis are -
H0w : The number of units produced does not depend on the worker.
H0T : The number of units produced does not depend on the type of machine.
The experiment has two factors (Factor Machine, Factor Worker) at a = 3(M1, M2, M3) and b = 3(A,B,C) levels.
Thus there are ab = 3 × 3 = 9 different combinations of machines and workers. With each combination there are 4 number of replicates. This sums up to n = abr = 36 data in total.
The averages of the ANOVA Table is
Assumptions of ANOVA:
(i) All populations involved follow a normal distribution.
(ii) All populations have the same variance (or standard deviation).
(iii) The samples are randomly selected and independent of one another.
(b) SS(within) = (15-15.75)^2 + (16-15.75)^2 + ... + (18-18.5)^2 + (19-18.5)^2
= 48.25
df(within) = (r-1)*a*b = (4-1)*3*3 = 27
MS(within) = 48.25/27 = 1.787
SS(Machines) = 4 * 3 * [ (17.5-17.7)^2 + (18.25-17.7)^2 + (17.33-17.7)^2] = 5.75
df(Machines) = 3-1 = 2
MS(machines) = 5.75/2 = 2.875
F = MS(machines)/ MS(within) = 2.875/1.787 = 1.6
Critical value of F for df = 1, 27 and siginifcance level of 0.05 is 4.21
As, observed F is less than critical value of F, we fail to reject the null hypothesis and conclude that Machines are not significant.
(c) SS(workers) = 4 * 3 * [ (16.67-17.7)^2 + (16.75-17.7)^2 + (19.67-17.7)^2] = 70.13
df(workers) = 3-1 = 2
MS(workers) = 70.13/2 = 35.06
F = MS(workers)/ MS(within) = 35.06/1.787 = 19.62
Critical value of F for df = 1, 27 and siginifcance level of 0.05 is 4.21
As, observed F is greater than critical value of F, we reject the null hypothesis and conclude that Workers are significant.
(d) SS(interaction) = 4 * [(15.75-17.5-16.67+17.7)^2 + (17.75-17.5-16.75+17.7)^2 + (19-17.5-19.67+17.7)^2 + (17.75-18.25-16.67+17.7)^2 + (15.5-18.25-16.75+17.7)^2 + (21.5-18.25-19.67+17.7)^2 + (16.5-17.33-16.67+17.7)^2 + (17-17.33-16.75+17.7)^2 + (18.5-17.33-19.67+17.7)^2 + (15.75-16.67-17.5+17.7)^2 + (17.75-16.67-18.25+17.7)^2 + (16.5-16.67-17.33+17.7)^2 + (17.75-16.75-17.5+17.7)^2 + (15.5-16.75-18.25+17.7)^2 + (17-16.75-17.33+17.7)^2 + (19-19.67-17.5+17.7)^2 + (21.5-19.67-18.25+17.7)^2 + (18.5-19.67-17.33+17.7)^2]
= 67.224
df(interaction) = (3-1)*(3-1) = 4
MS(interaction) = 67.224/4 = 16.81
F = MS(workers)/ MS(within) = 16.81/1.787 = 9.41
Critical value of F for df = 4, 27 and siginifcance level of 0.05 is 2.73
As, observed F is greater than critical value of F, we reject the null hypothesis and conclude that interactions between workers and machines are significant.
A B C Mean M1 15,16,15,17 (15.75) 18,19,16,18 (17.75) 19,16,20,21 (19) 17.5 M2 18,18,17,18 (17.75) 16,16,15,15 (15.5) 18,23,22,23 (21.5) 18.25 M3 16,17,18,15 (16.5) 18,16,17,17 (17) 19,18,18,19 (18.5) 17.33 Mean 16.67 16.75 19.67 17.7Related Questions
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