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ANEK [815]
2 years ago
5

Help please T-T i give brainliest *dramatic-c o u g h s* Im DyInG pLeAsE

Mathematics
2 answers:
harkovskaia [24]2 years ago
7 0

Answer:

ill give a guess because I cant read it

allsm [11]2 years ago
6 0

hello since you go the answer im just gonna write this

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Please help me brainliest if correct, please no bots and links
Novay_Z [31]

Answer:

The other one.

Step-by-step explanation:

The other person that answered your question is correct, just confirming it for youuu! :)

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3 years ago
A survey asked whether respondents favored or opposed the death penalty for people convicted of murder. Software shows the resul
scoray [572]

Answer:

The 95% confidence interval for those opposed is: (0.298, 0.334).

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 level of 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:

1786 of the 2611 were in favor, so 2611 - 1786 = 825 were opposed. Then

n = 2611, \pi = \frac{825}{2611} = 0.316

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.316 - 1.96\sqrt{\frac{0.316*0.684}{2611}} = 0.298

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.316 + 1.96\sqrt{\frac{0.316*0.684}{2611}} = 0.334

The 95% confidence interval for those opposed is: (0.298, 0.334).

4 0
3 years ago
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