91Ó°ÊÓ

The sample of size \(n = 100\) \(\overline X = \hat p\) is the sample proportion of persons with particular attributes of interest.

The standard deviation of population proportion is given by:

\(\sigma = \sqrt {\frac{{p\left( {1 - p} \right)}}{n}} \sqrt {\frac{{N - n}}{{N - 1}}} \)..........(1)

We know that\(0 < p < 1\). Hence\(p\left( {1 - p} \right)\)is maximized when\(p = 0.5\).

Hence,

\(p\left( {1 - p} \right) \le 0.5 \times p\left( {1 - 0.5} \right)\)

\(p\left( {1 - p} \right) \le 0.25\).............(2)

Consider the finite population correction factor.

The sample size may range between\(1 \le n < N\), hence\(\frac{{N - n}}{{N - 1}} \le 1\)........(3)

From equation (1), it is clear that for a fixed value of n, the standard deviation of p

maximizes when both\(p\left( {1 - p} \right)\)\(\frac{{N - n}}{{N - 1}}\)are maximum. Using equations (2) and (3)

\(\begin{aligned}{}\sigma &= \sqrt {\frac{{p\left( {1 - p} \right)}}{n}} \sqrt {\frac{{N - n}}{{N - 1}}} \\ &\le \sqrt {\frac{{0.25}}{{100}}} \times \sqrt 1 \\ &= 0.05\end{aligned}\)

Hence, we have proved that \(\sigma \le 0.05\)

Therefore. No matter how large the population is the standard deviation of \(\overline X \) is at most 0.05