# Mathematical Statistics and Data Analysis - Solutions

### 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 $\mathrm{Var} ({\bar X})$.

$$\tag*{\blacksquare}$$