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2 years ago

PLEASE HELP ME!!

Mathematics

1 answer:

Dmitry_Shevchenko [17]2 years ago

6 0

I think it’s A sorry if it’s wrong!

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6.Suppose the Gallup Organization wants to estimate the population proportion of those who think there should be a law that woul

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Answer:

A sample of $n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$ is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

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

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The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law.

This means that $n = 1012, \pi = \frac{374}{1012} = 0.3696$

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 $Z = 1.96$ .

How large a sample size is needed to be 95% confident with a margin of error of E?

A sample size of n is needed, and n is found when M = E.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$E = 1.96\sqrt{\frac{0.3696*0.6304}{n}}$

$E\sqrt{n} = 1.96\sqrt{0.3696*0.6304}$

$\sqrt{n} = \frac{1.96\sqrt{0.3696*0.6304}}{E}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$

$n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$

A sample of $n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$ is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

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