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Answer: Have two independent normal distributions with unknown variances. Wrens: ${\overline{\mathit{x}}}_{\mathrm{1}}\mathrm{=}\mathrm{2}\mathrm{1}\mathrm{.}\mathrm{1}\mathrm{8}$ mm., ${\mathit{s}}_{\mathrm{1}}^{\mathrm{2}}\mathrm{=}\mathrm{0}\mathrm{.}\mathrm{6}\mathrm{4}\mathrm{1}\mathrm{8}\mathrm{,}$ ${\mathit{n}}_{\mathrm{1}}\mathrm{=}\mathrm{1}\mathrm{0}\mathrm{.}$

Reed warblers: ${\overline{\mathit{x}}}_{\mathrm{2}}\mathrm{=}\mathrm{2}\mathrm{2}\mathrm{.}\mathrm{1}\mathrm{4}$ mm., ${\mathit{s}}_{\mathrm{2}}^{\mathrm{2}}\mathrm{=}\mathrm{0}\mathrm{.}\mathrm{4}\mathrm{1}\mathrm{1}\mathrm{6}\mathrm{,}$ ${\mathit{n}}_{\mathrm{2}}\mathrm{=}\mathrm{1}\mathrm{0}\mathrm{.}$

Assume ${\mathit{\sigma }}_{\mathrm{1}}^{\mathrm{2}}\mathrm{=}{\mathit{\sigma }}_{\mathrm{2}}^{\mathrm{2}}\mathrm{=}{\mathit{\sigma }}^{\mathrm{2}}$ (unknown). Estimate ${\mathit{\sigma }}^{\mathrm{2}}$ using

${\mathit{s}}^{\mathrm{2}}\mathrm{=}\frac{\mathrm{\left(}{\mathit{n}}_{\mathrm{1}}\mathrm{-}\mathrm{1}\mathrm{\right)}{\mathit{s}}_{\mathrm{1}}^{\mathrm{2}}\mathrm{+}\mathrm{\left(}{\mathit{n}}_{\mathrm{2}}\mathrm{-}\mathrm{1}\mathrm{\right)}{\mathit{s}}_{\mathrm{2}}^{\mathrm{2}}}{{\mathit{n}}_{\mathrm{1}}\mathrm{+}{\mathit{n}}_{\mathrm{2}}\mathrm{-}\mathrm{2}}\mathrm{=}\frac{\mathrm{9}{\mathit{s}}_{\mathrm{1}}^{\mathrm{2}}\mathrm{+}\mathrm{9}{\mathit{s}}_{\mathrm{2}}^{\mathrm{2}}}{\mathrm{1}\mathrm{8}}\mathrm{=}\mathrm{0}\mathrm{.}\mathrm{5}\mathrm{2}\mathrm{6}\mathrm{7}\mathrm{.}$

Also ${\overline{\mathit{x}}}_{\mathrm{1}}\mathrm{-}{\overline{\mathit{x}}}_{\mathrm{2}}\mathrm{=}\mathrm{2}\mathrm{1}\mathrm{.}\mathrm{1}\mathrm{8}\mathrm{-}\mathrm{2}\mathrm{2}\mathrm{.}\mathrm{1}\mathrm{4}\mathrm{=}\mathrm{-}\mathrm{0}\mathrm{.}\mathrm{9}\mathrm{6}\mathrm{,}$ $\sqrt{{\mathit{s}}^{\mathrm{2}}\mathrm{\left(}\frac{\mathrm{1}}{{\mathit{n}}_{\mathrm{1}}}\mathrm{+}\frac{\mathrm{1}}{{\mathit{n}}_{\mathrm{2}}}\mathrm{\right)}}\mathrm{=}\mathrm{0}\mathrm{.}\mathrm{1}\mathrm{0}\mathrm{5}\mathrm{3}\mathrm{,}$ ${\mathit{t}}_{\mathrm{1}\mathrm{8}}$(2.5%) $\mathrm{=}\mathrm{2}\mathrm{.}\mathrm{1}\mathrm{0}\mathrm{1}\mathrm{.}$

If ${\mathit{\mu }}_{\mathrm{1}}\mathrm{=}{\mathit{\mu }}_{\mathrm{2}}$ then the two groups of eggs have the same mean length.

To test ${\mathrm{H}}_{\mathrm{0}}\mathrm{:}{\mathit{\mu }}_{\mathrm{1}}\mathrm{=}{\mathit{\mu }}_{\mathrm{2}}$ vs. ${\mathrm{H}}_{\mathrm{1}}\mathrm{:}{\mathit{\mu }}_{\mathrm{1}}\mathrm{\ne }{\mathit{\mu }}_{\mathrm{2}}$ at 5% level, reject ${\mathrm{H}}_{\mathrm{0}}$ if $\mathrm{|}\frac{{\overline{\mathit{x}}}_{\mathrm{1}}\mathrm{-}{\overline{\mathit{x}}}_{\mathrm{2}}}{\sqrt{{\mathit{s}}^{\mathrm{2}}\mathrm{\left(}\mathrm{1}\mathrm{/}{\mathit{n}}_{\mathrm{1}}\mathrm{+}\mathrm{1}\mathrm{/}{\mathit{n}}_{\mathrm{2}}\mathrm{\right)}}}\mathrm{|}$ $\mathrm{\ge }$t8(2.5%). Here $\mathrm{|}\frac{{\overline{\mathit{x}}}_{\mathrm{1}}\mathrm{-}{\overline{\mathit{x}}}_{\mathrm{2}}}{\sqrt{{\mathit{s}}^{\mathrm{2}}\mathrm{\left(}\mathrm{1}\mathrm{/}{\mathit{n}}_{\mathrm{1}}\mathrm{+}\mathrm{1}\mathrm{/}{\mathit{n}}_{\mathrm{2}}\mathrm{\right)}}}\mathrm{|}\mathrm{=}\mathrm{|}\frac{\mathrm{-}\mathrm{0}\mathrm{.}\mathrm{9}\mathrm{6}}{\sqrt{\mathrm{0}\mathrm{1}\mathrm{0}\mathrm{5}\mathrm{2}}}\mathrm{|}\mathrm{=}\mathrm{2}\mathrm{.}\mathrm{9}\mathrm{5}$ 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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