Q237 · Practice questionTechnology-freeComplex unfamiliar9 marks
QUESTION 237 (9 marks)
Independent samples from a normal population with mean and standard deviation 10 have sizes 25 and 100, with means . Consider and pooled mean . For independent A,B, use .
a)[3 marks]
Determine the mean and variance of U.
b)[3 marks]
Determine the mean and variance of P, and identify the more precise estimator.
c)[3 marks]
For , define the unbiased estimator . Use calculus to find the value of a giving least variance and relate the result to P.
WORKED SOLUTION
9 marksPractice marking scheme
ANSWER
(a) ; . (b) ; ; P is more precise. (c) , minimum variance ; the resulting estimator is P.
Worked solution
(a) Component variances are 100/25=4 and 100/100=1. Both weights equal 1/2, so the mean is mu and variance is .
(b) Weights are 1/5 and 4/5, summing to 1, so the mean is mu. Variance is , also 100/125 for all 125 observations. This is smaller than 5/4, giving P a smaller standard error.
(c) The variances of the two means are 4 and 1. Independence gives . Thus at a=1/5, inside the domain; . The minimum is 4/5 and the weights 1/5,4/5 give exactly the pooled mean P. The endpoint variances 1 and 4 are larger.
Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: Find both component variances.
Part a: Find the mean.
Part a: Calculate variance 5/4.
Part b: Find the mean.
Part b: Calculate variance 4/5.
Part b: Compare precision by variance.
Part c: Form the variance as a function of the weight.
Part c: Find and justify the global minimising weight.
Part c: Give the minimum and identify the pooled estimator.
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