When we need information about the parameters of a population (especially a finite population) we collect many samples of a given size n from the population, then set up the probability distribution for the means of all our samples.
We then use this new distribution, The Sampling Distribution of the Sample Means, to find the mean of and an estimate for the standard deviation of the parent population.

The Sampling Distribution of the Sample Means

Definition: The probability distribution of the sample means
derived from all possible samples of a given size from a given population.

To display this distribution, we list the value of each sample mean x-bar and
its corresponding probability P(x-bar).
(an option is to list the frequency of each mean: see part 3 of Practice question)

The Mean & Standard Deviation of the Sampling Distribution of the Means

The mean of a sampling distribution of the means (called mu x bar) is always equal to the mean of the parent population.

The standard deviation of a sampling distribution of the means (called sigma x bar) is always less than the standard deviation of the parent population. This is because the range of the sample means data is smaller than the range of the population sampled. Since the standard deviation measures the spread of the distribution, and the sampling distribution is always packed tighter around the sampling mean, r _x-bar < r .

In the example that follows, the range of the parent population is 13 - 3 = 10.
The range of the sampling distribution of the means is 12 - 4 = 8.

Formulae for mu x bar and sigma x bar

The number of samples size n we will get from a finite population size N is _N C _n .
So, if there are 10 data items in the population and we take samples of 3 items each,
we will get ₁₀ C ₃ or 120 samples and their means.

If a given sample mean x occurs f times in the distribution, then P(x) = f / ₁₀ C ₃.

To find the probability of any particular sample mean, we divide the frequency
by the total number of samples.

If x is the mean of 3 different samples from a total of 12 samples,
then P(x) = 3/12 = ¼ or 0.25

Example:

Random samples of size 2 are selected from the finite population consisting
of the numbers 3, 5, 7, 9, 11, 13.

a) Find the mean and standard deviation of this population.

The mean l = 48/6 = 8

the standard deviation r = 3.42

b) List the 15 possible random samples (n = 2) that can be selected from
this population and calculate their means.

15 random samples (n = 2) and l their means.

(3, 5) l = 4	(3, 13) l = 8	(5, 13) l = 9	(9, 11) l = 10
(3, 7) l = 5	(5, 7) l = 6	(7, 9) l = 8	(9, 13) l = 11
(3, 9) l = 6	(5, 9) l = 7	(7, 11) l = 9	(11, 13) l = 12
(3, 11) l = 7	(5, 11) l = 8	(7, 13) l = 10

c) Use the results of part (b) to construct the sampling distribution
of the means of these samples.

the sampling distribution of the mean.

l	P(l)
4	1/15
5	1/15
6	2/15
7	2/15
8	3/15
9	2/15
10	2/15
11	1/15
12	1/15
Total	15/15 = 1

d) Calculate the mean l and variance r ² of the probability distribution in part (c)
and compare them with the results obtained in part (a).

mean l = S [x $ P(x) ] = 8	variance = r 2 = = 4.67
so the means are the same = 8	r = 2.16 < 3.42

Practice:

We are doing a study with respect to length of service on
6 neighbourhood social workers. (data in table)

Don(D) 4 yrs	Gary(G) 3 yrs	Sue(S) 3 yrs
Bob(B) 2 yrs	Kirk(K) 4 yrs	Alan(A) 3 yrs

We select all the samples of 4 such social workers from the population of 6 to create a sampling distribution of the means.

1) find the mean and standard deviation of this population

2) List the 15 samples size 4 and their means from this population

3) List the sample mean, frequency and probability for each sample mean.

4) Find the mean and standard deviation for this sampling distribution of the means.

5) Compare l-x bar with l and r -x bar with r.

Solution

1) The population mean l = 3.16667, and the standard deviation r = 0.68718

2)15 random samples (n = 4) and l their means.

DGSB (4, 3, 3, 2) l = 3	DGSK (4, 3, 3, 4) l = 3.5	DGSA (4, 3, 3, 3) l = 3.25	DGBK (4, 3, 2, 4) l = 3.25
DGBA (4, 3, 2, 3 l = 3	DGKA (4, 3, 4, 3) l = 3.5	DSBK (4, 3, 2, 4) l = 3.25	DSBA (4, 3, 2, 3) l = 3
DSKA (4, 3, 4, 3) l = 3.5	DBKA (4, 2, 4, 3) l = 3.25	GSBK (3, 3, 2, 4) l = 3	GSBA (3, 3, 2, 3) l = 2.75
GSKA ( 3, 3, 4, 3) l = 3.25	GBKA (3, 2, 4, 3) l = 3	SBKA (3, 2, 4, 3) l = 3

3)

the Sampling Distribution of the Sample Means.

Sample Mean	Frequency	Probability
2.75	1	1/15
3	6	6/15
3.25	5	5/15
3.5	3	3/15
Total	15	15/15 = 1

4)

5) Both means are the same. As expected r -x bar < r .

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