MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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Question
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.
expand button
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.
(a) Show that μz = 0 and express s½ in terms of the quantities in (3).
Express in terms by x, y and R.
(b) Using the formulas in Lecture 3, find an expression for Ŵ(z) in terms
of s, n and N, that is unbiased for V(z).
(c) Expand
-
-
με
r − R = 1 (r − R) + (1 − ) (r − R).
-
-
(5)
με
Show that
(r - R) =
(6)
με
με
It can be argued that the second term in (5) is small compared to
the first term. This is because ñ 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).
expand button
Transcribed Image Text:(a) Show that μz = 0 and express s½ in terms of the quantities in (3). Express in terms by x, y and R. (b) Using the formulas in Lecture 3, find an expression for Ŵ(z) in terms of s, n and N, that is unbiased for V(z). (c) Expand - - με r − R = 1 (r − R) + (1 − ) (r − R). - - (5) με Show that (r - R) = (6) με με It can be argued that the second term in (5) is small compared to the first term. This is because ñ 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).
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