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xenn [34]
2 years ago
14

Find the margin of error if p = 71% and n = 100. round to four decimal places.

Mathematics
1 answer:
aivan3 [116]2 years ago
7 0

Using the z-distribution, it is found that the margin of error for the 95% confidence interval is of 0.0889.

<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}}

In which:

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

The margin of error is given by:

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

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

The estimate and the sample size are given as follows:

\pi = 0.71, n = 100

Hence, the margin of error is:

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

M = 1.96\sqrt{\frac{0.71(0.29)}{100}}

M = 0.0889.

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

#SPJ1

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