Question

Suppose that a sample of size 3 is drawn from a population consisting of the six values 4, 8, 5, 3, 8, and 4, and that the proportion of values that are greater than 4 is recorded. Find the sampling distribution of this statistic by listing all possible such samples of size 3. Find the mean and variance of the sampling distribution.

Answer 1

Solution

We assume that the sampling is without replacement.

Under the above assumption, number of samples of size 3 = ⁶C₃ = 20.

These 20 sample alongwith the value of the statistic, say t = proportion of sample values that are greater than 4, are presented in the following table:

Sample #	Sample values	# values > 4	t
1	4, 8, 5	2	2/3
2	4, 8, 3	1	1/3
3	4, 8, 8	2	2/3
4	4, 8, 4	1	1/3
5	4, 5, 3	1	1/3
6	4, 5, 8	2	2/3
7	4, 5, 4	1	1/3
8	4, 3, 8	1	1/3
9	4, 3, 4	0	0
10	4, 8, 4	1	1/3
11	8, 5, 3	2	2/3
12	8, 5, 8	3	1
13	8, 5, 4	2	2/3
14	8, 3, 8	2	2/3
15	8, 3, 4	1	1/3
16	8, 8, 4	2	2/3
17	5, 3, 8	2	2/3
18	5, 3, 4	1	1/3
19	5, 8, 4	2	2/3
20	3, 8, 4	1	1/3

Frequency Distribution of t

Value (x)	frequency (f)	Relative Frequency {r = f/20}
0	1	0.05
1/3	9	0.45
2/3	9	0.45
1	1	0.05

Sampling Distribution of t

t	p(t)
0	0.05
1/3	0.45
2/3	0.45
1	0.05
Total	1.00

ANSWER 1

Calculation of Mean and Variance

t	p(t)	t.p(t)	t2.p(t)
0	0.05	0	0
1/3	0.45	0.15	0.05
2/3	0.45	0.30	0.20
1	0.05	0.05	0.05
Total	1.00	0.50	0.30

Mean of t = ∑[ t.p(t)]summed over all possible values of t.

                = 0.5 ANSWER 2

Variance of t = ∑[ t².p(t)]summed over all possible values of t – Mean²

                      = 0.30 – 0.25

                      = 0.05 ANSWER 3