Question

Very important

Answer 1

Using the z-distribution, we have that:

• For a 99% confidence level, a sample size of 127 is needed.

• For a 95% confidence level, a sample size of 74 is needed, meaning that a decrease in the confidence level decreases the needed sample size, as M and n are inverse proportional.

What is a z-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

The margin of error is:

$M = z\frac{\sigma}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• z is the critical value.

• n is the sample size.

• $\sigma$ is the standard deviation for the population.

For a 99% confidence interval, $\alpha = 0.99$, hence z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$, so the critical value is z = 2.575.

The margin of error and population standard deviation are:

$M = 3, \sigma = 13.1$

Hence we have to solve for n to find the needed sample size, as follows:

$M = z\frac{\sigma}{\sqrt{n}}$

$3 = 2.575\frac{13.1}{\sqrt{n}}$

$3\sqrt{n} = 2.575 \times 13.1$

$\sqrt{n} = \frac{2.575 \times 13.1}{3}$

$(\sqrt{n})^2 = \left(\frac{2.575 \times 13.1}{3}\right)^2$

n = 126.4.

Rounding up, for a 99% confidence level, a sample size of 127 is needed.

For the 95% confidence interval, we have that z = 1.96, hence:

$M = z\frac{\sigma}{\sqrt{n}}$

$3 = 1.96\frac{13.1}{\sqrt{n}}$

$3\sqrt{n} = 1.96 \times 13.1$

$\sqrt{n} = \frac{1.96 \times 13.1}{3}$

$(\sqrt{n})^2 = \left(\frac{1.96 \times 13.1}{3}\right)^2$

n = 73.3.

Rounding up, for a 95% confidence level, a sample size of 74 is needed, meaning that a decrease in the confidence level decreases the needed sample size, as M and n are inverse proportional.

More can be learned about the z-distribution at brainly.com/question/25890103

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