# Sample Size Determination

#### aweimer

##### New Member
n=8, mean=5096, StDev=1460, SEmean=516.19

I need a 95% confidence interval no wider than \$360. What is the minimum sample size required?

I calculated this as 193, but was told I was incorrect.
The issue may be with my bound, which I calculated to be 0.0706, or 7.06%.

Using the formula n=.25(Z/B)

PLEASE HELP!!

#### aweimer

##### New Member
Could I solve this like so:
(zcrit^2*stdev^2)/me^2??

So, 1.96^2*1460^2 / 180^2 = 253?

#### aweimer

##### New Member
NEVER MIND, I figured this one out. Thank you!

#### Dason

##### Ambassador to the humans
Would you mind posting your solution? Threads like this that end with "don't worry - I figured it out" are pretty worthless to anybody else that stumbles across the thread looking for an answer to the question posed in the original post.

#### aweimer

##### New Member
Of course! The solution to this problem is what I posted in my 1st reply.

Use the equation: (Zcrit^2 * StDev^2) / (Margin of Error^2) = n

In this instance, n = 253.