Question

Suppose we want to determine the minimum sample size required to give us a 95% confidence interval that estimates, to within $500, the average salary of a state employee. Also, suppose that from a previous experiment, we know that the population standard deviation is $6300. What is the minimum sample size required

Answer 1

Answer:

The minimum sample size is 610.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find M as such

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

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Population standard deviation is $6300, so $\sigma = 6300$

Suppose we want to determine the minimum sample size required to give us a 95% confidence interval that estimates, to within $500, the average salary of a state employee.

This is n when M = 500. So

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

$500 = 1.96*\frac{6300}{\sqrt{n}}$

$500\sqrt{n} = 1.96*6300$

$\sqrt{n} = \frac{1.96*6300}{500}$

$(\sqrt{n})^{2} = (\frac{1.96*6300}{500})^{2}$

$n = 609.89$

Rouding up

The minimum sample size is 610.