Supplementary Exercise 12.43 of IPS7e
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(a) I = 5, N = 8*5 = 40
MSE = 50, DFE = N-I = 40-5 = 35
MSG = 57, DFG = I-1 = 5-1 = 4
F = MSG/MSE = 57/50 = 1.14 ~ F(4,35) under H0: all population means equal
P = P(F(4,35)>1.14) > 0.1, because 1.14<2.18 = 90% perc. of F(4,25)
                    = 1-0.646159 = 0.354 using Minitab/Stata
Conclusion: there is no evidence at the 5% significance level against H0,
so H0 cannot be rejected. The data do not provide reason to say that there is
any difference between the groups.


(b) I = 3, N = 7*3 = 21
SSE = 90, DFE = N-I = 21-3 = 18, MSE = SSE/DFE = 5
SSG = 40, DFG = I-1 = 2, MSG = SSG/DFG = 20
F = F = MSG/MSE = 20/5 = 4.0 ~ F(2,18) under H0: all population means equal
P = P(F(2,18)>4.0) < 0.05, because 4.0>3.68 = 95% perc. of F(2,15)
                   = 1-0.963466 = 0.037 using Minitab/Stata
Conclusion: there is evidence at the 5% significance level against H0,
so H0 should be rejected - there are some differences between population
means. The evidence is not very strong, however.

Below we show Minitab listings from the Calc-Probability Distribution
menu. The results could also be obtained from the Graph-Probability
Distribution Plot menu.

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 CDF 1.14;
  F 4 35.

F distribution with 4 DF in numerator and 35 DF in denominator

   x  P( X <= x )
1.14     0.646159


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 CDF 4.0;
  F 2 18.

F distribution with 2 DF in numerator and 18 DF in denominator

x  P( X <= x )
4     0.963466