Chapter 7, Survey Sampling
Solution 1
Population mean:
Population varaince:
${\sigma}^2 = \frac 1 N \sum_{i=1}^{i=N} {(x_i - \mu)}^2$
$\quad = \frac { {(1-3.4)}^2 + {(2-3.4)}^2 + {(2-3.4)}^2 + {(4-3.4)}^2 + {(8-3.4)}^2 } 5$
$\quad = \frac {31.2} 5 = 6.24$
Samples of size 2:
Note that every sample containing $2$ appears twice except for $(2,2)$. Since in a sample order does not matter and both $2$’s are taken from the population, so there is only one possible sample for $(2,2)$.
Thus we have total ${ 5 \choose 2 } = 10$ samples. The probability of occurance of a sample is the fraction of times it appears in the total(10) samples. For eg: Prob. of choosing sample $(1,2)$ is $\frac 2 {10}$.
Sampling Distribution of the mean of samples of size 2:
| Sample | Mean( ${\mu}_i = {\bar X}$ ) | Probability( $p_i$ ) |
|---|---|---|
| $(1,2)$ | $\frac {1+2} 2 = 1.5$ | $\frac 2 {10}$ |
| $(1,4)$ | $\frac {1+4} 2 = 2.5$ | $\frac 1 {10}$ |
| $(1,8)$ | $\frac {1+8} 2 = 4.5$ | $\frac 1 {10}$ |
| $(2,2)$ | $\frac {2+2} 2 = 2.0$ | $\frac 1 {10}$ |
| $(2,4)$ | $\frac {2+4} 2 = 3.0$ | $\frac 2 {10}$ |
| $(2,8)$ | $\frac {2+8} 2 = 5.0$ | $\frac 2 {10}$ |
| $(4,8)$ | $\frac {4+8} 2 = 6.0$ | $\frac 1 {10}$ |
Mean of all the sample means, $\mathrm{E} {\bar X} = \sum_{i=1}^{i=N} p_i {\mu}_i$
$\quad = 3.4$.
Variance of all the sample means, $\mathrm{Var} ({\bar X}) = \mathrm{E} { ( {\mu}_i - \mathrm{E} {\bar X} ) }^2$.
For better visibility, using the following table for computation:
| Sample | Mean( ${\mu}_i = {\bar X}$ ) | ${ ( {\mu}_i - \mathrm{E} {\bar X} ) }^2$ | Probability( $p_i$ ) |
|---|---|---|---|
| $(1,2)$ | $1.5$ | ${ ( 1.5 - 3.4 ) }^2 = 3.61$ | $\frac 2 {10}$ |
| $(1,4)$ | $2.5$ | ${ ( 2.5 - 3.4 ) }^2 = 0.81$ | $\frac 1 {10}$ |
| $(1,8)$ | $4.5$ | ${ ( 4.5 - 3.4 ) }^2 = 1.21$ | $\frac 1 {10}$ |
| $(2,2)$ | $2.0$ | ${ ( 2.0 - 3.4 ) }^2 = 1.96$ | $\frac 1 {10}$ |
| $(2,4)$ | $3.0$ | ${ ( 3.0 - 3.4 ) }^2 = 0.16$ | $\frac 2 {10}$ |
| $(2,8)$ | $5.0$ | ${ ( 5.0 - 3.4 ) }^2 = 2.56$ | $\frac 2 {10}$ |
| $(4,8)$ | $6.0$ | ${ ( 5.0 - 3.4 ) }^2 = 6.76$ | $\frac 1 {10}$ |
Clearly, Variance is the dot product of last two columns, $\mathrm{Var} ({\bar X}) = 2.34$.
Comparing the values of $\mathrm{E} {\bar X}$ and population mean, $\mu$, both are equal to $3.4$, in agreement with the Theorem mentioned in problem.
Comparing the values of $\mathrm{Var} ({\bar X})$ and population variance, $\sigma$, as per theorem $\mathrm{Var} ({\bar X}) = \frac { {\sigma}^2 } n ( 1 - \frac {n-1} {N-1} )$. Here, $n = 2$ is the sample size and $N = 5$, total population.
Putting these values ion RHS, $\frac {6.24} 2 ( 1 - \frac 1 4 ) = 2.34$, which is same as the $\mathrm{Var} ({\bar X})$.
$$\tag*{$\blacksquare$} $$