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Answer:

E[T]=E[a1X1+a2X2]=a1E[X1]+a2E[X2]=a1μ+a2μ=(a1+a2)μ.

If we require E[T]=μ, then a1+a2=1, so that a2=1-a1.

Since E[T]=μ, then T is said to be an unbiased estimator of the mean μ.

Var [T]= Var [a1X1+a2X2]=a12Var[X1]+a22Var[X2]=a12σ2+a22σ2=(a12+a22)σ2.

Since a2=1-a1, Var[T]={a12+(1-a1)2}σ2=(2a12-2a1+1)σ2. Differentiate this with respect to a1 to find the minimum.

dda1 Var [T]=(4a1-2)σ2,

which is zero when a1=12. Hence Var[T] is a minimum when a1=a2=12 so T=12(X1+X2) . Alternative derivation: write a1=12+ε, a2=12-ε. Then

Var [T]=(a12+a22)σ2={(12+ε)2+(12-ε)2}σ2=(12+2ε2)σ2,

and is a minimum if ε=0.

What does this question show '? In part (a) you chose a2 to restrict attention to linear combinations of the Xi which were unbiased estimators of the mean μ, so E[T]=μ. In part (b) you then showed that of all such unbiased estimators, the sample mean X¯ is the one with smallest variance, so giving values closest to the true mean μ.

Worked Example: Lecture 15.

The following data give the noise level (in decibels) generated by fourteen different chain saws powered in one of two different ways.

Petrol‐powered chain saws

Electric‐powered chain saws

103103105106108105106

97959493919594

At the 5% level of significance, test whether the average noise level of petrol‐powered chain saws is higher than for electric‐powered chain saws.

Answer: Testing H0 : μ1=μ2 vs. H1 : μ1>μ2, i.e. H0 : μ1-μ2=0 vs. H1 : μ1-μ2>0. Have two independent samples with unknown variance. Need to assume variances are equal.

Worked Example: Lecture 15.

The following data give the length (in mm.) of cuckoo (cuculus canorus) eggs found in nests belonging to wrens (A) and reed warblers (B).

A:19.822.121.520.922.021.022.321.020.320.9¯

B: 23. 2 22. 0 22. 2 21. 2 21. 6 21. 6 21. 9 22. 0 22. 9 22.8

Assuming the variances for each group are the same, is there any evidence at the 5% level to suggest that the egg size differs between the two host species?

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