MATH1725 Introduction to Statistics: Worked examples

Answer: Have two independent normal distributions with unknown variances.
Wrens: ¯x 1 = 21.18 mm., s 2 1 = 0.6418, n 1 = 10.
Reed warblers: ¯x 2 = 22.14 mm., s 2 2 = 0.4116, n 2 = 10.
Assume σ 2 1 = σ2 2 = σ2 (unknown). Estimate σ 2 using
s 2 = (n 1 − 1)s 2 1 + (n 2 − 1)s 2 2
= 9s2 1 + 9s2 2
= 0.5267.
n 1 + n 2 − 2 18
( 1
Also ¯x 1 − ¯x 2 = 21.18 − 22.14 = −0.96,
√s 2 + 1 )
= 0.1053, t 18 (2.5%) = 2.101.
n 1 n 2
If µ 1 = µ 2 then the two groups of eggs have the same mean length.
¯x 1 − ¯x 2
To test H 0 : µ 1 = µ 2 vs. H 1 : µ 1 ≠ µ 2 at 5% level, reject H 0 if
√ ∣ s 2 (1/n 1 + 1/n 2 ) ∣ ≥ t 8(2.5%).
¯x 1 − ¯x ∣ ∣
2
∣∣∣
Here
√ ∣ s 2 (1/n 1 + 1/n 2 ) ∣ = −0.96 ∣∣∣
√ = 2.95 so reject the null hypothesis of equal means at 5%
0.1052
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 information see:
Wyllie, I. (1981) The Cuckoo. Batsford: London.
Davies, N.B. and Brooke, M. Coevolution of the cuckoo and its host, Scientific American, January
1991, p.66-73.
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