Question

List all possible samples of size n=3, with replacement, from the population (1,3,5). Calculate the mean of each sample. Construct a probability distribution of the sample means and compute the mean, variance, and standard deviation of the sample means and compare to the mean, variance, and standard deviation of the population.

Answer 1

for population:

x	P(x)	xP(x)	x^2P(x)
1	1/3	0.33	0.33
3	1/3	1.00	3.00
5	1/3	1.67	8.33
	total	3.00	11.67

mean =E(X)=3.00

Variance =E(X²)-(E(X))² =11.67-3² =2.67

and standard deviation =sqrt(2.67)=1.633

for probability distribution of the sample means :

x1	x2	x3	P(x1,x2)	xbar		xbar*P(xbar)	xbar^2*P(xbar)
1	1	1	1/27	1.000		0.037	0.037
1	1	3	1/27	1.667		0.062	0.103
1	1	5	1/27	2.333		0.086	0.202
1	3	1	1/27	1.667		0.062	0.103
1	3	3	1/27	2.333		0.086	0.202
1	3	5	1/27	3.000		0.111	0.333
1	5	1	1/27	2.333		0.086	0.202
1	5	3	1/27	3.000		0.111	0.333
1	5	5	1/27	3.667		0.136	0.498
3	1	1	1/27	1.667		0.062	0.103
3	1	3	1/27	2.333		0.086	0.202
3	1	5	1/27	3.000		0.111	0.333
3	3	1	1/27	2.333		0.086	0.202
3	3	3	1/27	3.000		0.111	0.333
3	3	5	1/27	3.667		0.136	0.498
3	5	1	1/27	3.000		0.111	0.333
3	5	3	1/27	3.667		0.136	0.498
3	5	5	1/27	4.333		0.160	0.695
5	1	1	1/27	2.333		0.086	0.202
5	1	3	1/27	3.000		0.111	0.333
5	1	5	1/27	3.667		0.136	0.498
5	3	1	1/27	3.000		0.111	0.333
5	3	3	1/27	3.667		0.136	0.498
5	3	5	1/27	4.333		0.160	0.695
5	5	1	1/27	3.667		0.136	0.498
5	5	3	1/27	4.333		0.160	0.695
5	5	5	1/27	5.000		0.185	0.926
			total			3.000	9.889

mean of sampling distribution of sample means =E(Xbar)=3.00

and variance of sampling distribution of sample means =9.889-3² =0.889

and standrd deviation of sampling distribution of sample means =sqrt(0.889)=0.943

from above population mean =sampling distribution of sample means

standrd deviation of sample means =population standard deviation/sqrt(3)

variance of sample means =population variance/3