# 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 ${\sigma _{\bar X}} = \frac{\sigma }{{\sqrt n }}\sqrt {\frac{{N – n}}{{N – 1}}}$