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.

<h3>What is a z-distribution confidence interval?</h3>

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

4 0

2 years ago

Ok please answer this, this is confusing to me

Nezavi [6.7K]

What the question lol

8 0

3 years ago

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