(b) Using the formulas in Lecture 3, find an expression for V(z) in terms of s, n and N, that is unbiased for V(z). (c) Expand - - με r − R = (r − R) + (1 − ) (r — R). - I με - (5) Show that - (r− R) = 1. (6) μα με It can be argued that the second term in (5) is small compared to the first term. This is because x is a good estimator of μx and so is close to μx when n is large. Since R is a constant, it follows that - V (r) = V (r − R) ≈ V(2), (7) where the approximation is due to ignoring the second term in (5). (d) A good approximation of V(r) would be 12V(2), where Ŵ(z) is de- rived in 2(b). However the expression 2½ contains the unknown R. Lets be the expression s½ with the unknown R replaced by r. Show that is indeed what is given in (2). 2. This exercise is to provide more details to the arguments leading to V (r) = (N), nN where (1) n r = -1 and s² = (yi – rx;)², Σ - (2) i=1 see page 8 of Lecture 8. - Let zi = yi — Rx; and consider a simple random sample from a population 21,..., Zn Z1, ZN (3) (4) Let μz be the population mean, σ the population variance, z the sample με mean and s² the sample variance.

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(b) Using the formulas in Lecture 3, find an expression for V(z) in terms
of s, n and N, that is unbiased for V(z).
(c) Expand
-
-
με
r − R = (r − R) + (1 − ) (r — R).
-
I
με
-
(5)
Show that
-
(r− R) = 1.
(6)
μα
με
It can be argued that the second term in (5) is small compared to
the first term. This is because x is a good estimator of μx and so is
close to μx when n is large. Since R is a constant, it follows that
-
V (r) = V (r − R) ≈ V(2),
(7)
where the approximation is due to ignoring the second term in (5).
(d) A good approximation of V(r) would be 12V(2), where Ŵ(z) is de-
rived in 2(b). However the expression 2½ contains the unknown R.
Lets be the expression s½ with the unknown R replaced by r. Show
that is indeed what is given in (2).
Transcribed Image Text:(b) Using the formulas in Lecture 3, find an expression for V(z) in terms of s, n and N, that is unbiased for V(z). (c) Expand - - με r − R = (r − R) + (1 − ) (r — R). - I με - (5) Show that - (r− R) = 1. (6) μα με It can be argued that the second term in (5) is small compared to the first term. This is because x is a good estimator of μx and so is close to μx when n is large. Since R is a constant, it follows that - V (r) = V (r − R) ≈ V(2), (7) where the approximation is due to ignoring the second term in (5). (d) A good approximation of V(r) would be 12V(2), where Ŵ(z) is de- rived in 2(b). However the expression 2½ contains the unknown R. Lets be the expression s½ with the unknown R replaced by r. Show that is indeed what is given in (2).
2. This exercise is to provide more details to the arguments leading to
V (r) = (N),
nN
where
(1)
n
r =
-1
and s² = (yi – rx;)²,
Σ
-
(2)
i=1
see page 8 of Lecture 8.
-
Let zi = yi — Rx; and consider a simple random sample
from a population
21,..., Zn
Z1, ZN
(3)
(4)
Let μz be the population mean, σ the population variance, z the sample
με
mean and s² the sample variance.
Transcribed Image Text:2. This exercise is to provide more details to the arguments leading to V (r) = (N), nN where (1) n r = -1 and s² = (yi – rx;)², Σ - (2) i=1 see page 8 of Lecture 8. - Let zi = yi — Rx; and consider a simple random sample from a population 21,..., Zn Z1, ZN (3) (4) Let μz be the population mean, σ the population variance, z the sample με mean and s² the sample variance.
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