Question

Suppose you wanted to estimate the time in minutes the average person spends on e-mail a day. It’s estimated that the standard deviation is 16 minutes and the average is 82 minutes. You want the margin of error to be only one minute. What sample size do you need if the confidence level is 99%?

Answer 1

Answer:

We need a sample size of at least 1698.

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:

$\alpha = \frac{1-0.99}{2} = 0.005$

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 $1-0.005 = 0.995$, so $z = 2.575$

Now, find the margin of error 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.

You want the margin of error to be only one minute. What sample size do you need if the confidence level is 99%?

This is at least n when $M = 1, \sigma = 16$. So

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

$1 = 2.575*\frac{16}{\sqrt{n}}$

$\sqrt{n} = 2.575*16$

$(\sqrt{n})^{2} = (2.575*16)^{2}$

$n = 1697.4$

Rouding up

We need a sample size of at least 1698.