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 $${\sigma _{\bar X}} = \frac{\sigma }{{\sqrt n }}\sqrt {\frac{{N – n}}{{N – 1}}} $$
Pearl Lamptey
September 18 @ 6:05 pm
Please I want samples of size 3 N=4 with replacement.
Khizra
August 17 @ 10:27 pm
Take all possible samples of size 3 with replacement from population comprising 10 12 14 16 18 make sampling distribution and verify
Aimen Naveed
September 10 @ 12:23 pm
Draw all possible sample of size n = 3 with replacement from the population 3,6,9 and 12. Form a sampling distribution of sample means. Hence state and verify relation between (a). Mean of the sampling distribution of the mean and the population mean; (b). Variance of the sampling distribution of the mean and the population variance. Please tell me this question as soon as possible
Aimen Naveed
September 10 @ 12:25 pm
Draw all possible sample of size n = 3 with replacement from the population 3,6,9 and 12. Form a sampling distribution of sample means. Hence state and verify relation between (a). Mean of the sampling distribution of the mean and the population mean; (b). Variance of the sampling distribution of the mean and the population variance. Please tell me this question as soon as possible
(a). Discuss the relevance of the concept of the two types of errors in following case. “Let’s say that you want to increase conversions on a banner displayed on your website. For that to work out, you’ve planned on adding an image to see if it increases conversions or not.You start your A/B test running a control version (A) against your variation (B) that contains the image. After 5 days, the variation (B) outperforms the control version by a staggering 25% increase in conversions with an 85% level of confidence.You stop the test and implement the image in your banner. However, after a month, you noticed that your monthtomonth conversions have decreased. (b) what is a biased sample? How bias can be eliminated? tell this question