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Ipatiy [6.2K]
3 years ago
11

We would like to create a confidence interval.

Mathematics
1 answer:
Vlada [557]3 years ago
6 0

Answer:

c.A 90% confidence level and a sample size of 300 subjects.

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 1-\alpha, we have the confidence interval with a margin of error of:

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

In which

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

In this problem

The proportions are the same for all the options, so we are going to write our margins of error as functions of \sqrt{\pi(1-\pi)}

So

a.A 99% confidence level and a sample size of 50 subjects.

n = 50

99% confidence interval

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 margin of error is

M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.3642\sqrt{\pi(1-\pi)}

b.A 90% confidence level and a sample size of 50 subjects.

n = 50

90% confidence interval

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The margin of error is

M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.2623\sqrt{\pi(1-\pi)}

c.A 90% confidence level and a sample size of 300 subjects.

n = 300

90% confidence interval

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The margin of error is

M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.0950\sqrt{\pi(1-\pi)}

This produces smallest margin of error.

d.A 99% confidence level and a sample size of 300 subjects.

n = 300

99% confidence interval

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 margin of error is

M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.1487\sqrt{\pi(1-\pi)}

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