At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.05 for the estimation of a population proportion

Answer:

A sample of 385 is needed.

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]

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].

How large a sample:

We need a sample of n.

n is found when M = 0.05.

We dont know the true proportion, so we work with the worst case scenario, which is [tex]\pi = 0.5[/tex]

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.05 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]

[tex]0.05\sqrt{n} = 1.96*0.5[/tex]

[tex]\sqrt{n} = \frac{1.96*0.5}{0.05}[/tex]

[tex](\sqrt{n})^{2} = (\frac{1.96*0.5}{0.05})^{2}[/tex]

[tex]n = 384.16[/tex]

Rounding up

A sample of 385 is needed.

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