MS Module 12 E(MSTr) practice exam questions

(The attached PDF file has better formatting.)

●    An experiment has five groups with 7 observations in each group.
●    The five groups have the same population variance σ2 = 2.1
●    An analysis of variance is done on the five groups to test the null hypothesis H0: μ1 = μ2 = μ3 = μ4 = μ5

The true means of the five groups are μ1 = 3.6, μ2 = 3.2, μ3 = 7.9, μ4 = 2.8, μ5 = 0.6, but these values are not known.


Question 12.1: Deviations of group means

What are the deviations of the group means from the overall mean?

Answer 12.1: the overall mean = (3.6 + 3.2 + 7.9 + 2.8 + 0.6) / 5 = 3.62, so the deviations of the group means from the overall mean are

●     α1 = 3.6 – 3.62 = -0.02
●     α2 = 3.2 – 3.62 = -0.42
●     α3 = 7.9 – 3.62 = 4.28
●     α4 = 2.8 – 3.62 = -0.82
●     α4 = 0.6 – 3.62 = -3.02


Question 12.2: Expected value of the treatment mean square

What is the expected value of MSTr, the treatment mean square?

Answer 12.2: E(MSTr) = σ2 + J/(I-1) ×  α2i =

2.1 + (7 / (5 – 1)) × ( (-0.02)2 + (-0.42)2 + (4.28)2 + (-0.82)2 + (-3.02)2 ) = 51.604


Question 12.3: Non-centrality parameter

What is the non-centrality parameter for the analysis of variance?

Answer 12.3: 7 × ( (-0.02)2 + (-0.42)2 + (4.28)2 + (-0.82)2 + (-3.02)2 ) / 2.1 = 94.293