Question

Darren wants to estimate the mean age in a population of trees. He'll sample nnn trees and build a 90\%90%90, percent confidence interval for the mean age. He doesn't want the margin of error to exceed 333 years. Preliminary data suggests that the standard deviation for the ages of trees in this population is \sigma=16σ=16sigma, equals, 16 years. Which of these is the smallest approximate sample size required to obtain the desired margin of error?

Answer 1

Using the z-distribution, as we have the standard deviation for the population, it is found that the smallest sample size required to obtain the desired margin of error is of 77.

What is a z-distribution confidence interval?

The confidence interval is:

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

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.

The margin of error is given by:

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

In this problem, we have that the parameters are given as follows:

$M = 3, z = 1.645, \sigma = 16$.

Hence, solving for n, we find the sample size.

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

$3 = 1.645\frac{16}{\sqrt{n}}$

$3\sqrt{n} = 1.645 \times 16$

$\sqrt{n} = \frac{1.645 \times 16}{3}$

$(\sqrt{n})^2 = \left(\frac{1.645 \times 16}{3}\right)^2$

$n = 76.97$

Rounding up, the smallest sample size required to obtain the desired margin of error is of 77.

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