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Lorico [155]
2 years ago
11

Please help with math, will give brainliest, thanks, and 5 stars

Mathematics
1 answer:
Oksana_A [137]2 years ago
8 0
3. -11
4. 24
5. 2
6. 5
7. 12
8. 2
9. -4
10. 10
11. 6
12. 1 2/3
13. 4
14. -4
15. 14
16. 20
17. C . X<-1
18. G. Y>-6
19. A. a>-9
20. F

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In preparation for an earnings report, a large retailer wants to estimate p= the proportion of annual sales
mr Goodwill [35]

Using the z-distribution, it is found that the 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).

<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 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 sample size and the estimate are given by:

n = 161, \pi = \frac{37}{161} = 0.2298

Hence:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2298 - 1.96\sqrt{\frac{0.2298(0.7702)}{161}} = 0.1648

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2298 + 1.96\sqrt{\frac{0.2298(0.7702)}{161}} = 0.2948

The 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).

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

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2 years ago
Help please, I’m not sure
Lyrx [107]
YOUR ANSWER IS A
HOPE THIS HELPS
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3 years ago
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