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Answer: Have two independent normal distributions with unknown variances. Wrens: x¯1=21.18 mm., s12=0.6418, n1=10.

Reed warblers: x¯2=22.14 mm., s22=0.4116, n2=10.

Assume σ12=σ22=σ2 (unknown). Estimate σ2 using

s2=(n1-1)s12+(n2-1)s22n1+n2-2=9s12+9s2218=0.5267.

Also x¯1-x¯2=21.18-22.14=-0.96, s2(1n1+1n2)=0.1053, t18(2.5%) =2.101.

If μ1=μ2 then the two groups of eggs have the same mean length.

To test H0:μ1=μ2 vs. H1:μ1μ2 at 5% level, reject H0 if |x¯1-x¯2s2(1/n1+1/n2)| t8(2.5%). Here |x¯1-x¯2s2(1/n1+1/n2)|=|-0.9601052|=2.95 so reject the null hypothesis of equal means at 5% level. The two groups of eggs are significantly different at 5% level.

This does not necessarily imply cuckoos can control their egg size. It has been proposed that a cuckoo lays its egg in the particular nest for which it is best adapted. For further info rmation see: Wyllie, I. (1981) The Cuckoo. Batsfo rd: London.

Davies, N.B. and Brooke, M. Coevolution of the cuckoo and its host, Scientific American, January 1991, p. 6673.

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