Question

For the following​ situation, find the mean and standard deviation of the population. List all samples​ (with replacement) of the given size from that population. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population. The number of DVDs rented by each of three families in the past month is 2​, 14​, and 3. Use a sample size of 2

Answer 1

Given Data: 2, 14 and 3

Mean of Population, M = (2 + 14 + 3) / 3 = 19 / 3 = 6.333

Number of DVD(j)	(j - m)^2
2	18.77489
14	58.78289
3	11.10889

Variance = (18.77489 + 58.78289 + 11.10889) / 3 = 29.555

Population Standard deviation, = = 5.436

With replacement, Number of samples possible = 3² = 9

Samples(x)	Mean of Sample(m)
2,2	2
14,14	14
3,3	3
2,14	8
14,2	8
14,3	8.5
3,14	8.5
2,3	2.5
3,2	2.5

Samples are:

Mean of sampling Distribution = 6.333

This is same as Population Mean

Samples(x)	Mean of Sample(m)	(m - 6.333)^2
2,2	2	18.7749
14,14	14	58.7829
3,3	3	11.1089
2,14	8	2.7789
14,2	8	2.7789
14,3	8.5	4.6959
3,14	8.5	4.6959
2,3	2.5	14.6919
3,2	2.5	14.6919
Mean	6.333

Variance of sampling distribution = 1/9 * ( sum of all (m - 6.333)^2)

Variance = 14.7778

Sample Distribution Standard Deviation, s = 3.844

Now sample size, n = 2

Sample Distribution Standard Deviation, s = / = 5.436 / = 3.844 which is same as sample distribution standard deviation

So, s =/