Question

Question: Refer to Exercise 5.3.

a. Find the sampling distribution of ${\mathbf{s}}^2\mathbf{.}$

b. Find the population variance ${\mathbf{\sigma }}^2\mathbf{.}$

c. Show that ${\mathbf{s}}^2$is an unbiased estimator of ${\mathbf{\sigma }}^2$.

d. Find the sampling distribution of the sample standard deviation $\mathbf{s}$.

e. Show that s is a biased estimator of $\mathbf{\sigma }\mathbf{.}$

Short answer

A sampling distribution is the chance distribution of a pattern in data from a greater sample size gathered from a certain population. It depicts the spectrum of probable outcomes of statistics, including the mean as well as the mode of a parameter, in the same way as it characterizes a population.

Step-by-step solution

 Step 1: (a) The data is given below

Given data,

${\text{s}}^{\text{2}}=\frac{\Sigma {(\text{χ}-{\displaystyle \stackrel{-}{\chi }})}^2}{\text{n}-1}$

The calculation is given below:

The sample (1, 1) $\stackrel{-}{\chi }=1$

$\begin{array}{c}{\text{s}}^{\text{2}}=[{(1-1)}^2+{(1-1)}^2]/1\\ =0\end{array}$

The sample (1, 3) $\stackrel{-}{\chi }=1$

$$\begin{gathered} {\text{s}}^{\text{2}}=[{(1-1)}^2+{(1-1)}^2]/1 \\ =0 \end{gathered}$$

The sample (1, 2) $\stackrel{-}{\chi }=1.5$

$$\begin{gathered} {\text{s}}^{\text{2}}=[{(1-15)}^2+{(2-1.5)}^2]/1 \\ =0 \end{gathered}$$

To calculate the ${\mathbf{\text{s}}}^{\mathbf{\text{2}}}$values for every sample remaining

Sample Distribution of ${\text{s}}^{\text{2}}$is

(b) Population variance

The calculation is given below:

$$\begin{gathered} \text{Populationvariance}{\sigma }^{\text{2}} \\ ={(1-2.7)}^2\left(0.2\right)+{(2-2.7)}^2\left(0.3\right)+{(3-2.7)}^2\left(0.2\right)+{(4-2.7)}^2\left(0.2\right)+{(5-2.7)}^2\left(0.1\right) \\ =0.578+.147+.018+.338+.529 \\ =1.61 \end{gathered}$$

(c) Unbiased estimator

The calculation is given below:

To demonstrate the ${\text{s}}^{\text{2}}$is an unbiased estimator ${\sigma }^{\text{2}}$

Using the sampling distribution of ${\text{s}}^{\text{2}}$

$$\begin{gathered} =\left(0\right)\left(0.22\right)+\left(0.5\right)\left(0.36\right)+\left(2\right)\left(0.24\right)+\left(4.5\right)+\left(.14\right)+\left(8\right)\left(.04\right) \\ =0+.18+.48+.63+.32 \\ =1.61 \end{gathered}$$

(d) Sampling distribution

The calculation is given below:

(e) Sampling distribution

The calculation is given below:

S from the sampling distribution

$$\begin{gathered} =\left(0\right)\left(0.22\right)+\left(0.707\right)\left(0.36\right)+\left(1.414\right)\left(0.24\right)+\left(2.121\right)+\left(0.14\right)+\left(2.828\right)\left(.04\right) \\ =0+.255+.339+.297+.113 \\ =1.004 \end{gathered}$$

But,

$$\begin{gathered} \sigma =\sqrt{{\sigma }^2} \\ =\sqrt{1.61} \\ =1.269 \\ \ne 1.004 \end{gathered}$$

So, s is a biased estimator of localid="1662357666676" $\sigma$.