Mathematics: We would like to create a confidence interval. Which of the following would produce the smallest margin of error? Select one: a.A 99% confidence level and a sample size of 50 subjects. b.A 90% confidence level and a sample size of 50 subjects. c.A 90% confidence level and a sample size of 300 subjects. d.A 99% confidence level and a sample size of 300 subjects.

i think the correct answer is f...

We would like to create a confidence interval.

Ответ:

parisw8

18.01.2022

Mathematics

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 $\sqrt{\pi(1-\pi)}$

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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