Question

5. The number of defects in 4 different samples of 80 units coming off of a production line are as follows: {1, 2, 4, 5}

If I took samples of size 2 from this list, with replacement, there are 16 different permutations.

List them below:

Find the mean of each sample, then create a table showing the sampling distributions of the sample means.

Find the mean of the sampling distribution.

Find the mean of the 4 data values. What do you notice about the mean of the sampling distribution and the mean of the data values?

Answer 1

The different samples of size 2 and their means are given in the below table which shows the samoling distribution of the sample means

Sample number	Sample	Mean
1	(1,1)	1
2	(1,2)	1.5
3	(1,4)	2.5
4	(1,5)	3
5	(2,1)	1.5
6	(2,2)	2
7	(2,4)	3
8	(2,5)	3.5
9	(4,1)	2.5
10	(4,2)	3
11	(4,4)	4
12	(4,5)	4.5
13	(5,1)	3
14	(5,2)	3.5
15	(5,4)	4.5
16	(5,5)	5
		Mean = 48

Mean of the sampling distribution = Mean / Number of Samples

= 48/16

= 3

Mean of data values = (1 + 2 + 4 + 5) / 4

= 12/4

= 3

So Mean of data values is equal to mean of sampling distribution of means