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Liula [17]
3 years ago
7

PLS HELP 10 POINTS ASAP RN PLSSSSS OR I WILL FAIL CLASS

Mathematics
1 answer:
Ivahew [28]3 years ago
7 0
A is .898 B is 1 and C is 1.009
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Using the z-distribution, it is found that the lower bound of the 99% confidence interval is given by:

d. 68.39%.

<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 sample size and estimate are given by:

n = 2002, \pi = 0.71

Hence, the lower bound is given by:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.71 - 2.575\sqrt{\frac{0.71(0.29)}{2002}} = 0.6839

Hence the lower bound is of 68.39%, which means that option D is correct.

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

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