Question

Sampling Distributions

This exercise is to be completed in a group.

For the population: 1 3 5 7 9

Find the mean and standard deviation.

Mean Calculation:

1 + 3 + 5 + 7 + 9 = 25

25 / 5 = 5

The mean is 5

Standard Deviation Calculation:

( 1 - 5 ) ² = ( -4 ) ² = 16

( 3 - 5 ) ² = ( -2 ) ² = 4

( 5 - 5 ) ² = ( 0 ) ² = 0

( 7 - 5 ) ² = ( 2 ) ² = 4

( 9 - 5 ) ² = ( 4 ) ² = 16

16 + 4 + 0 + 4 + 16 = 40

( 1 / 5 ) × 40 = 8

√( 8 ) = 2.8284

The standard deviation is ≈ 2.83

Now determine how many samples of size 2 can be formed from this population of size 5, disregarding order. Use the appropriate formula.

List all the samples of size 2 that can be formed from this population. Be sure the list is consistent with the answer obtained from the formula. Otherwise, check your work.

Calculate the mean of each of the samples of size 2.

Create a probability distribution of all the sample means from the sample of size 2. (HINT: If you have a mean of 7 in the list of sample means and it occurs 3 times out of 10, then the probability is 0.3 for that mean value.)

Calculate the mean and standard deviation of that probability distribution using the appropriate formulas.

Answer 1

The population: 1 3 5 7 9

To find,

the mean and standard deviation

Mean Calculation:

1 + 3 + 5 + 7 + 9 = 25

25 / 5 = 5. The mean is 5

Standard Deviation Calculation:

( 1 - 5 ) ² = ( -4 ) ² = 16

( 3 - 5 ) ² = ( -2 ) ² = 4

( 5 - 5 ) ² = ( 0 ) ² = 0

( 7 - 5 ) ² = ( 2 ) ² = 4

( 9 - 5 ) ² = ( 4 ) ² = 16

16 + 4 + 0 + 4 + 16 = 40

( 1 / 5 ) × 40 = 8

√( 8 ) = 2.8284

The standard deviation is ≈ 2.83

We can form 52 = 25 samples of size 2 can be formed from this population of size 5, disregarding order.

i.e. using simple random sampling with replacement.

List of all the samples of size 2 that can be formed from this population.

{ (1,1) ,(1,3), (1,5) ,(1,7),(1,9) ,(3,1) ,(3,3) ,(3,5), (3,7) ,(3,9) ,(5,1) ,(5,3) ,(5,5) ,(5,7) ,(5,9) ,(7,1) ,(7,3) ,(7,5) ,(7,7) , (7,9), (9,1) ,(9,3) ,(9,5) ,(9,7) ,(9,9)}

Sample means from the sample of size 2

Sample	1,1	1,3	1,5	1,7	1,9	3,1	3,3	3,5	3,7	3,9	5,1	5,3	5,5	5,7	5,9	7,1	7,3	7,5	7,7	7,9	9,1	9,3	9,5	9,7	9,9
Mean	1	2	3	4	5	2	3	4	5	6	3	4	5	6	7	4	5	6	7	8	5	6	7	8	9

A probability distribution of all the sample means

Mean	Probabilty
1	0.04
2	0.08
3	0.12
4	0.16
5	0.20
6	0.16
7	0.12
8	0.08
9	0.04
Total	1

The mean of that probability distribution:

= 1*0.04 + 2*0.08+ 3*0.12 + 4*0.16+ 5*0.20 +6*0.16 +7*0.12 +8*0.08 +9*0.04

= 5

The Standard deviation of that probability distribution:

  

= 1*0.04 + 4*0.08+ 9*0.12 + 16*0.16+ 25*0.20 + 36*0.16 + 49*0.12 + 64*0.08 + 81*0.04

= 29

= 29-25 = 4

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