Question

The following is a 90% confidence interval for p:(0.26, 0.54). How large was the sample used to construct thisinterval?

Short answer

The required sample size is 34.

Step-by-step solution

Given information

The following is a 90% confidence interval for p is(0.26, 0.54)

Finding the sample size

The formula for the confidence interval for a population proportion is

$(\widehat{p}-z_{\alpha /2}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}},\widehat{p}+z_{\alpha /2}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}})$

Here the confidence interval is (0.26, 0.54)

$\begin{array}{l}Therefore,\\ \widehat{p}-z_{\alpha /2}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=\mathrm{0.26.....}\left(i\right)\\ \widehat{p}+z_{\alpha /2}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=\mathrm{0.54.....}\left(ii\right)\end{array}$

Adding the equation(i) and equation(ii)

$\begin{array}{c}2\widehat{p}=0.26+0.54\\ \widehat{p}=\frac{0.80}{2}\\ \widehat{p}=0.40\end{array}$

The critical value for 90% confidence interval is$z_{\alpha /2}=z_{0.10/2}=z_{0.05}=1.645$

Now putting the value of$\widehat{p}$ and $z_{\alpha /2}$in equation(i) ,

role="math" localid="1658294262963" $\begin{array}{c}0.40-1.645\sqrt{\frac{0.40(1-0.40)}{n}}=0.26\\ 1.645\sqrt{\frac{0.40\times 0.60}{n}}=0.14\\ \frac{0.40\times 0.60}{n}={\left(\frac{0.14}{1.645}\right)}^2\\ n=\frac{0.24}{{\left(\frac{0.14}{1.645}\right)}^2}\\ n=33.135\\ n\simeq 34\end{array}$

The required sample size is 34