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Black_prince [1.1K]
2 years ago
11

Construct a 99​% confidence interval to estimate the population proportion with a sample proportion equal to 0.36 and a sample s

ize equal to 100 .
Mathematics
1 answer:
vivado [14]2 years ago
8 0

Using the z-distribution, the 99​% confidence interval to estimate the population proportion is: (0.2364, 0.4836).

<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.

In this problem, we have a 99% confidence level, hence\alpha = 0.99, z is the value of Z that has a p-value of \frac{1+0.99}{2} = 0.995, so the critical value is z = 2.575.

The estimate and the sample size are given by:

\pi = 0.36, n = 100.

Then the bounds of the interval are:

  • \pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.36 - 2.575\sqrt{\frac{0.36(0.64)}{100}} = 0.2364
  • \pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.36 + 2.575\sqrt{\frac{0.36(0.64)}{100}} = 0.4836

The 99​% confidence interval to estimate the population proportion is: (0.2364, 0.4836).

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

#SPJ1

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