Sampling Distribution

2011-11-26 11:16 pm
A population consists of the 4 numbers 2, 4, 7, 9.
a) Find the population variance.
b) If random samples of size 2 are drawn without replacement from this population and order is not counted, there are only 6 possible samples. Find the variance of the sample means of these 6 samples.

Show your steps.

(Ans.: a. Var(X) = 7.25; b. Variance of sample means = 2.4167)
更新1:

To myisland8132: Thanks for your reply. But I wonder how do you come up with the expression in part b)? May you elaborate on that a bit more? thx =)

更新2:

Btw, anyone can help?

回答 (3)

2011-11-26 11:23 pm
✔ 最佳答案
(a) E(X) = (2 + 4 + 7 + 9)/4 = 5.5

E(X^2) = (4 + 16 + 49 + 81)/4 = 37.5

Var(X) = 37.5 - (5.5)^2 = 7.25

(b) Var(X_bar) =(4 - 2)/(4 - 1) * 7.25/2 = 24167

2011-11-27 21:00:50 補充:
It is a formula which can be found out in any textbook of sampling theory
2011-11-27 5:53 am
Part (b)
Sample Sample Mean
2, 4 3
2, 7 4.5
2, 9 5.5
4,7 5.5
4,9 6.5
7,9 8
mean of sample mean, E(X) = 33/6 = 5.5
Sample variance = [(3^2 + 4.5^2 + 5.5^2 + 5.5^2 + 6.5^2 + 8^2) - 6 x 5.5^2]/6
= (196 - 181.5)/6 = 2.416666 = 2.4167.


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