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alisha [4.7K]
2 years ago
11

You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable preliminary estim

ation for the population proportion. You would like to be 98% confident that you estimate is within 2% of the true population proportion. How large of a sample size is required?
Mathematics
1 answer:
prohojiy [21]2 years ago
5 0

Using the z-distribution, it is found that a sample size of 3,385 is required.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

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

The margin of error is:

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

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 98% confidence level, hence\alpha = 0.98, z is the value of Z that has a p-value of \frac{1+0.98}{2} = 0.99, so the critical value is z = 2.327.

We have no prior estimate, hence \pi = 0.5 is used, which is when the largest sample size is needed. To find the sample size, we solve the margin of error expression for n when M = 0.02, hence:

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

0.02 = 2.327\sqrt{\frac{0.5(0.5)}{n}}

0.02\sqrt{n} = 2.327 \times 0.5

\sqrt{n} = \left(\frac{2.327 \times 0.5}{0.02}\right)

(\sqrt{n})^2 = \left(\frac{2.327 \times 0.5}{0.02}\right)^2

n = 3,385.

A sample size of 3,385 is required.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

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