Use 6.84 days as a planning value for the population standard deviation. a. Assuming 95% confidence, what sample size would be required to obtain a margin of error of days

The sample size required is , in which M is the desired margin of error. If n is a decimal number, it is rounded up to the next integer.

Step-by-step explanation:

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

Now, we have to find z in the Z-table as such z has a p-value of .

That is z with a pvalue of , so Z = 1.96.

Now, find the margin of error M as such

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

Use 6.84 days as a planning value for the population standard deviation.

This means that .

What sample size would be required to obtain a margin of error of M days?

This is n for the given value of M. So

The sample size required is , in which M is the desired margin of error. If n is a decimal number, it is rounded up to the next integer.

its definatley   this on your answer choice?

"a" determine whether sage found the correct quotient. show your work

step-by-step explanation: