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nikitadnepr [17]
4 years ago
9

You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proporti

on is approximately 32%. You would like to be 90% confident that your estimate is within 4% of the true population proportion. How large of a sample size is required?
Mathematics
1 answer:
Ksenya-84 [330]4 years ago
5 0

Answer:

We need a sample size of at least 369.

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}}

For this problem, we have that:

\pi = 0.32

90% confidence level

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

How large of a sample size is required?

We need a sample size of at least n.

n is found when M = 0.04. So

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

0.04 = 1.645\sqrt{\frac{0.32*0.68}{n}}

0.04\sqrt{n} = 1.645\sqrt{0.32*0.68}

\sqrt{n} = \frac{1.645\sqrt{0.32*0.68}}{0.04}

(\sqrt{n})^{2} = (\frac{1.645\sqrt{0.32*0.68}}{0.04})^{2}

n = 368.02

Rounding up

We need a sample size of at least 369.

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