Question

In 2010, an online security firm estimated that 64% of computer users don't change their passwords very often. Because this estimate may be outdated, suppose that you want to carry out a new survey to estimate the proportion of students at your school who do not change their password. You would like to determine the sample size required to estimate this proportion with a margin of error of 0.05.

Using 0.65 as a preliminary estimate, what is the required sample size if you want to estimate this proportion with a margin of error of 0.05?

Answer 1

Answer:

The required sample size is 350 to estimate the proportion with a margin of error of 0.05          

Step-by-step explanation:

We are given the following in the question:

$\hat{p} = 0.65$

Confidence level = 95%

Margin of error = 0.05

Confidence interval:

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

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,

$0.05 = 1.96\sqrt{\dfrac{(0.65)(1-0.65)}{n}}\\\\n = \dfrac{(1.96)^2(0.65)(1-0.65)}{(0.05)^2}\\\\n = 349.5856\\n \approx 350$

Thus, the required sample size is 350 to estimate the proportion with a margin of error of 0.05