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den301095 [7]
3 years ago
15

help help help\help help help\help help help\help help help\help help help\help help help\help help help\

Mathematics
2 answers:
8_murik_8 [283]3 years ago
8 0

Answer:

i think its D but im not exactly sure

LenaWriter [7]3 years ago
7 0

Answer:

D

Step-by-step explanation:

Subtract 60 from 80 to get 20, and divide 20 by 80 to get 1/4, which can be converted to 25%.

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A random sample of 864 births in a state included 426 boys. Construct a 95% confidence interval estimate of the proportion of bo
oksian1 [2.3K]

Using the z-distribution, it is found that the 95% confidence interval is (0.46, 0.526), and it does not provide strong evidence against that belief.

<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:

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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.

We have that a random sample of 864 births in a state included 426 boys, hence the parameters are given by:

n = 864, \pi = \frac{426}{864} = 0.493

Then, the bounds of the interval are given by:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.493 - 1.96\sqrt{\frac{0.493(0.507)}{864}} = 0.46

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The 95% confidence interval estimate of the proportion of boys in all births is (0.46, 0.526). Since the interval contains 0.506, it does not provide strong evidence against that belief.

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

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