Question

Using the traditional formula, a \(95\% \) CI for \({p_1} - {p_2}\)is to be constructed based on equal sample sizes from the two populations. For what value of \(n( = m)\)will the resulting interval have a width at most of .1, irrespective of the results of the sampling?

Short answer

The value of \(n = 769\)

Step-by-step solution

To Find the Confidence interval

A confidence interval for \({p_1} - {p_2}\) with confidence level of approximately \(100(1 - \alpha )\% \) is

\({\hat p_1} - {\hat p_2} \pm {z_{\alpha /2}} \times \sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{m} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{n}} \)

For the \(95\% \)confidence interval \(\alpha = 0.05\) and\({z_{\alpha /2}} = {z_{0.025}} = 1.96\).

The confidence interval can be used when \(m \times {\hat p_1},m \times {\hat q_1},n \times {\hat p_1}\), and \(n \times {\hat q_1}\) are at least \(10.\)

Irrespective of the results of the sampling means that

\({\hat p_1} = {\hat p_2} = {\hat q_1} = {\hat q_2} = 0.5\)

Final Proof

Using\(n = m\), the \(95\% \) confidence interval now becomes

\(\begin{array}{c}{{\hat p}_1} - {{\hat p}_2} \pm {z_{\alpha /2}} \times \sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{m} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{n}} \\ = 0.5 - 0.5 \pm 1.96 \times \sqrt {\frac{{0.5 \times 0.5}}{n} + \frac{{0.5 \times 0.5}}{n}} \\ = \pm 1.96 \times \sqrt {\frac{{0.25}}{n} + \frac{{0.25}}{n}} \\ = \pm 1.96 \times \sqrt {\frac{{0.5}}{n}} \\ = \pm \frac{{1.38593}}{{\sqrt n }}\end{array}\)\(\)

\( \Rightarrow \frac{{1.38593}}{{\sqrt n }} - \left( { - \frac{{1.38593}}{{\sqrt n }}} \right) = \frac{{2.7719}}{{\sqrt n }}\)

In order for the resulting interval to have a width at most of\[0.1\], the necessary sample size can be obtained from

\(\frac{{2.7719}}{{\sqrt n }} = 0.1\)

\(n = {\left( {\frac{{2.7719}}{{0.1}}} \right)^2}\)or equally

\[{\rm{n = 769 }}{\rm{.}}\]

The sample size \(n = 769\)