Supplementary Exercise 12.27 of IPS7e
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A total of N=209 observations, and 3 groups.

Model: X_ij are normally distributed N(mu_i,sigma) and independent.

(a) The pooled variance s^2 and stand.dev. s are calculated as
s^2 = (45*2.5^2 + 110*1.8^2 + 51*1.8^2) / (209-3) = 802.89/206 = 3.8975
  s = sqrt(3.8975) = 1.97
In the ANOVA table, MSE = s^2.

(b)
Source of	Degrees of	Sum of		Mean	  F	  P	
variation	freedom		squares		square
-----------------------------------------------------------------------
country		2		17.22		8.61	  2.21	  >0.10
error		206		802.89		3.8975
total		208		820.11

Table E in IPS (or Table F in PSLS) gives the 90%-percentile of F(2,200)
as 2.33. Using statistical software, we can determine the exact value in F(2,206) 
as P=0.112.

(c) 
H0: mu1=m2=mu3 (same mean weight gains in all 3 countries),
Ha: between some countries the mean weight gains differ.

(d) The F-statistic follows an F(2,206)-distribution under H0. The P-value 
is given above.
Conclusion: There is no evidence against H0, the data are compatible with the
hypothesis that the weight gains are the same in the 3 countries. The "problem"
with these data is that the within-group standard deviations are very large,
compared to the group means.

(e) Skip this question.
(R^2=17.22/820.11=0.021)