Examples of Sampling Distribution

Example:

Draw all possible samples of size 2 without replacement from a population consisting of 3, 6, 9, 12, 15. Form the sample distribution of sample means and verify the results.

(i) {\text{E}}\left( {\bar X} \right) = \mu

(ii) {\text{Var}}\left( {\bar X} \right) = \frac{{{\sigma ^2}}}{n}\left( {\frac{{N - n}}{{N - 1}}} \right)

Solution:

We have population values 3, 6, 9, 12, 15, population size N = 5 and sample size n = 2. Thus, the number of possible samples which can be drawn without replacement is

\left( {\begin{array}{*{20}{c}} N \\ n \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right) = 10

Sample No.
Sample Values
Sample Mean
\left( {\bar X} \right)
Sample No.
Sample Values
Sample Mean
\left( {\bar X} \right)
1
3, 6
4.5
6
6, 12
9.0
2
3, 9
6.0
7
6, 15
10.5
3
3, 12
7.5
8
9, 12
10.5
4
3, 15
9.0
9
9, 15
12.0
5
6, 9
7.5
10
12, 15
13.5

The sampling distribution of the sample mean \bar X and its mean and standard deviation are:

\bar X
f
f\left( {\bar X} \right)
\bar Xf\left( {\bar X} \right)
{\bar X^2}f\left( {\bar X} \right)
4.5
1
1/10
4.5/10
20.25/10
6.0
1
1/10
6.0/10
36.00/10
7.5
2
2/10
15.0/10
112.50/10
9.0
2
2/10
18.0/10
162.00/10
10.5
2
2/10
21.0/10
220.50/10
12.0
1
1/10
12.0/10
144.00/10
13.5
1
1/10
13.5/10
182.25/10
Total
10
1
90/10
877.5/10

{\text{E}}\left( {\bar X} \right) = \sum \bar Xf\left( {\bar X} \right) = \frac{{90}}{{10}} = 9
{\text{Var}}\left( {\bar X} \right) = \sum {\bar X^2}f\left( {\bar X} \right) - {\left[ {\sum \bar X\,f\left( {\bar X} \right)} \right]^2} = \frac{{887.5}}{{10}} - {\left( {\frac{{90}}{{10}}} \right)^2} = 87.75 - 81 = 6.75

 

The mean and variance of the population are:

X
3
6
9
12
15
\sum X = 45
{X^2}
9
36
81
144
225
\sum {X^2} = 495

\mu = \frac{{\sum X}}{N} = \frac{{45}}{5} = 9 and {\sigma ^2} = \frac{{\sum {X^2}}}{N} - {\left( {\frac{{\sum X}}{N}} \right)^2} = \frac{{495}}{5} - {\left( {\frac{{45}}{5}} \right)^2} = 99 - 81 = 18

Verification:

(i) E\left( {\bar X} \right) = \mu = 9 (ii) {\text{Var}}\left( {\bar X} \right) = \frac{{{\sigma ^2}}}{n}\left( {\frac{{N - n}}{{N - 1}}} \right) = \frac{{18}}{2}\left( {\frac{{5 - 2}}{{5 - 1}}} \right) = 6.75

Example:

If random samples of size three and drawn without replacement from the population consisting of four numbers 4, 5, 5, 7. Find sample mean \bar X for each sample and make sampling distribution of \bar X. Calculate the mean and standard deviation of this sampling distribution. Compare your calculations with population parameters.

Solution:

We have population values 4, 5, 5, 7, population size N = 4 and sample size n = 3. Thus, the number of possible samples which can be drawn without replacement is \left( {\begin{array}{*{20}{c}} N \\ n \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right) = 4

Sample No.
Sample Values
Sample Mean \left( {\bar X} \right)
1
4, 5, 5
14/3
2
4, 5, 7
16/3
3
4, 5, 7
16/3
4
5, 5, 7
17/3

The sampling distribution of the sample mean \bar X and its mean and standard deviation are:

\bar X
f
f\left( {\bar X} \right)
\bar Xf\left( {\bar X} \right)
{\bar X^2}f\left( {\bar X} \right)
14/3
1
1/4
14/12
196/36
16/3
2
2/4
32/12
512/36
17/3
1
1/4
17/12
289/36
Total
4
1
63/12
997/36

 

{\mu _{\bar X}} = \sum \bar X\,f\left( {\bar X} \right)\,\,\,\, = \,\,\,\frac{{63}}{{12}} = 5.25
{\sigma _{\bar X}} = \sqrt {\sum {{\bar X}^2}\,f\left( {\bar X} \right) - {{\left[ {\sum \bar X\,f\left( {\bar X} \right)} \right]}^2}} \,\,\,\, = \,\,\,\sqrt {\frac{{997}}{{36}} - {{\left( {\frac{{63}}{{12}}} \right)}^2}} = 0.3632

The mean and standard deviation of the population are:

X
4
5
5
7
\sum X = 21
{X^2}
16
25
25
49
\sum {X^2} = 115

\mu = \frac{{\sum X}}{N} = \frac{{21}}{4} = 5.25 and {\sigma ^2} = \sqrt {\frac{{\sum {X^2}}}{N} - {{\left( {\frac{{\sum X}}{N}} \right)}^2}} = \sqrt {\frac{{115}}{4} - {{\left( {\frac{{21}}{4}} \right)}^2}} = 1.0897

\frac{\sigma }{{\sqrt n }}\sqrt {\frac{{N - n}}{{N - 1}}} = \frac{{1.0897}}{{\sqrt 3 }}\sqrt {\frac{{4 - 3}}{{4 - 1}}} = 0.3632

Hence {\mu _{\bar X}} = \mu and {\sigma _{\bar X}} = \frac{\sigma }{{\sqrt n }}\sqrt {\frac{{N - n}}{{N - 1}}}