Answer:
The required sample size is 329.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
31% of all computer users in a recent year had tried to get on a Wi-Fi network that was not their own in order to save money.
This means that [tex]\pi = 0.31[/tex]
Confidence level is not given, so we assume 95%.
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
What is the required sample size if you want to estimate this proportion with a margin of error of 0.05?
We need a sample size of n. n is found when M = 0.05. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.05 = 1.96\sqrt{\frac{0.31*0.69}{n}}[/tex]
[tex]0.05\sqrt{n} = 1.96\sqrt{0.31*0.69}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.31*0.69}}{0.05}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.31*0.69}}{0.05})^2[/tex]
[tex]n = 328.6[/tex]
Rounding up
The required sample size is 329.
