Question

We intend to estimate the average break time of Dunder Mifflin employees. From a previous study, we believe that the average time is 42 minutes with a standard deviation of 6 minutes. We want our 99 percent confidence interval to have a margin of error of no more than plus or minus 2 minutes. What is the smallest sample size that we should consider?

Answer 1

Answer:

The minimum sample size that we should consider is of 60 employees.

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.

We want our 99 percent confidence interval to have a margin of error of no more than plus or minus 2 minutes. What is the smallest sample size that we should consider?

We need to find n for which $M = 2, \sigma = 6$

So

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

$2 = 2.575*\frac{6}{\sqrt{n}}$

$2\sqrt{n} = 2.575*6$

Simplifying by 2

$\sqrt{n} = 2.575*3$

$(\sqrt{n})^{2} = (2.575*3)^{2}$

$n = 59.67$

Rounding up

The minimum sample size that we should consider is of 60 employees.

Answer 2

Answer:

$n=(\frac{2.58(6)}{2})^2 =59.907 \approx 60$

So we need a sample of at least 60 in order to satisfy the condition.

Step-by-step explanation:

Notation

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=6$ represent the population standard deviation  assumed

n represent the sample size  

Solution

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (2)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (3)

The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.005;0;1)", and we got $z_{\alpha/2}=2.58$, replacing into formula (3) we got:

$n=(\frac{2.58(6)}{2})^2 =59.907 \approx 60$

So we need a sample of at least 60 in order to satisfy the condition.