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dangina [55]
3 years ago
6

Please help me with this I'll give brainliest to best answer!

Mathematics
1 answer:
Vesnalui [34]3 years ago
7 0

Answer:

Liner

Step-by-step explanation:

17,21,18,20,24

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A survey conducted in July 2015 asked a random sample of American adults whether they had ever used online dating (either an onl
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Answer:

The 99% confidence interval is (0.0787, 0.1613).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval 1-\alpha, we have the following confidence interval of proportions.

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

In which

Z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

The survey included 411 adults between the ages of 55 and 64, and 50 of them said that they had used online dating. This means that n = 411, \pi = \frac{50}{411} = 0.12

The 99% confidence interval is

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{411}} = 0.12 - 2.575\sqrt{\frac{0.12*0.88}{411}} = 0.0787

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{411}} = 0.12 + 2.575\sqrt{\frac{0.12*0.88}{411}} = 0.1613

The 99% confidence interval is (0.0787, 0.1613).

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Step-by-step explanation:

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