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Basile [38]
3 years ago
9

Suppose that the epidemiologist wants to re-estimate the population proportion and wishes for her 95% confidence interval to hav

e a margin of error no larger than 0.04. How large a sample should she take to achieve this? Please carry answers to at least six decimal places in intermediate steps.
Mathematics
1 answer:
Svetllana [295]3 years ago
3 0

Answer:

She needs a sample of at least 385.

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 should she take to achieve this?

She needs a sample of size at least n.

n is found when M = 0.04.

We do not know the true population proportion, so we use \pi = 0.5, which is the case for which we are going to need the largest sample size.

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

Dividing both sides by 0.05

\sqrt{n} = 19.6

(\sqrt{n})^{2} = 19.6^{2}

n = 384.2

Rounding up

She needs a sample of at least 385.

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