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OLEGan [10]
3 years ago
8

2/3 = 20/b solve for b.

Mathematics
1 answer:
liberstina [14]3 years ago
3 0
2b = 3 * 20
2b = 60
b = 30
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At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro
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Answer:

A sample of 1068 is needed.

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}}

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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.

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of \pi, so we use the worst case scenario, which is \pi = 0.5

Then

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

0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.03\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.03}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}

n = 1067.11

Rounding up

A sample of 1068 is needed.

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