Question

Consider estimating a population proportion p. What is the most conservative sample size (erring on the large size) required in order to build a 95% confidence interval for a population proportion p with a margin of error of at most 2%?

Answer 1

Answer:

The most conservative sample size is 2401.

Step-by-step explanation:

We are given the following in the question:

We are not given any approximation of s]ample proportion, thus,

$\hat{p} = 0.5$

We have to construct a 95% confidence interval.

Margin of error = 2% = 0.02

Formula for margin of error =

$z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get,

$0.02 = 1.96\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\0.02 = 1.96\times \sqrt{\dfrac{0.5(1-0.5)}{n}}\\\\\sqrt{n} = 1.96\times \dfrac{\sqrt{0.5(1-0.5)}}{0.02}\\\\\sqrt{n}=49\\n = 2401$

Thus, the most conservative sample size is 2401.