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ss7ja [257]
3 years ago
7

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 pro

portion
Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
6 0

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 \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

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 \pi = 0.5

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.05 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.05\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.05}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.05})^{2}

n = 384.16

Rounding up

A sample of 385 is needed.

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