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}}}

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