Chapter 7.2, Problem 38E

To determine

To find: the sample size would be required to reduce the standard deviation of the sampling distribution to one-third the value.

Answer to Problem 38E

135

Explanation of Solution

Formula used:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

Calculation:

The standard deviation of the sampling distribution of the sample proportion $\widehat{p}$

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

It can’t to reduce the standard deviation to one- third of the value found in the previous exercise, therefore divide ${\sigma }_{\widehat{p}}$ by 3.

  $\begin{array}{l}\\ \frac{{\sigma }_{\widehat{p}}}{3}=\frac{1}{3}\sqrt{\frac{p\left(1-p\right)}{n}}=\sqrt{\frac{1}{9}\cdot \frac{p\left(1-p\right)}{n}}=\sqrt{\frac{p\left(1-p\right)}{9n}}\end{array}$

It is observed that $n$ was replaced by $9n$ in the expression of the standard deviation, when the standard deviation is reduced to one-half of its value, which implies that the sample size needs to be multiplied by 9.

The sample size in the previous exercise was 15.

  $9n=9(15)=135$

This then implies that if use the sample sizes $135$ then the standard deviation of the sampling distribution of the sample proportions is reduced to one-third of its original value.