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 sampling 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 are drawn without replacement from the population consisting of four numbers 4, 5, 5, 7. Find the sample mean $$\bar X$$ for each sample and make a sampling distribution of $$\bar X$$. Calculate the mean and standard deviation of this sampling distribution. Compare your calculations with the 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 $${\sigma _{\bar X}} = \frac{\sigma }{{\sqrt n }}\sqrt {\frac{{N – n}}{{N – 1}}} $$