# Mathematical Statistics and Data Analysis - Solutions

### Chapter 7, Survey Sampling

#### (a)

To prove that $s^2$ is unbiased in estimating $\sigma^2$, we need to show that $\mathrm{E}(s^2) = \sigma^2$.

To Prove that we shall first prove the following Lemmas:

Lemma 1

$\mathrm{E}(X_i) = \mu$, where $\mu$ denotes the population mean.

Proof

Since this is a random sampling with replacement and every element having equal chance to be selected:

Thus $\mathrm{E}(X_i) = \sum_{i=1}^{i=n} x_i P(X_i = x_j) = \frac 1 N \sum_{i=1}^{i=n} x_i$

Thus $\mathrm{E}(X_i) = \mu$

Lemma 2

Proof

Now, we will prove the main result $\mathrm{E}(s^2) = \sigma^2$

We have:

Cancelling $n-1$ from both sides, it follows $\mathrm{E}(s^2) = \sigma^2$.

#### (b)

No, it is not an unbiased estimate for sigma. By Jensenâ€™s Inequality:

Thus $\, \mathrm{E}(s) \leq \sigma \,$. It follows that $\, \mathrm{E}(s) \,$ is not always equal to $\sigma$, or $s$ is not an unbiased estimate of $\sigma$.

#### (c)

We need to show $\, \mathrm{E} \left( \frac{s^2}{n} \right) = \sigma^2_{\bar X} \,$

We have:

We have:

#### (e)

To show $\, \mathrm{E} \left( \frac { \hat p (1- \hat p) } {n-1} \right) = \sigma^2_{\hat p} \,$, where $\, \hat p = \bar X = \frac 1 n \sum_{i=1}^{n} X_i \,$, the sample proportion.

We have:

$$\tag*{\blacksquare}$$