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.

Respuesta :

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 [tex]\pi[/tex], and a confidence level [tex]1-\alpha[/tex], we have the confidence interval with a margin of error of:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

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

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 [tex]\sqrt{\pi(1-\pi)}[/tex]

So

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

[tex]n = 50[/tex]

99% confidence interval

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].

The margin of error is

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

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

[tex]n = 50[/tex]

90% confidence interval

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

The margin of error is

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

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

[tex]n = 300[/tex]

90% confidence interval

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

The margin of error is

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

This produces smallest margin of error.

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

[tex]n = 300[/tex]

99% confidence interval

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].

The margin of error is

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