# Thread: Simple Random Sample Question

1. ## Simple Random Sample Question

A Simple Random sample of 25 computer firms was chosen from a total of 80 operating in a region in order to estimate the size of the workforce in that area. The average number of employees in the sample was 250, and the standard deviation of the sample was 100. Obtain

a) an estimate of the total employee size of the 80 firms?

N * std ??? im not sure whats N is in this case

i think the answer is 250 employees * 80 firms?

Anyone care to check if im right or wrong please

b) the standard error of the estimate of the total employees size?

SE (total) = N root 1-f * sd/n

im not sure whats my N , i-f is.

2. Do not post the same question twice: http://www.mathhelpforum.com/math-he...-question.html

Use your sample mean to give an estimate of the total employee size of 80 firms: $80*250 = ?$

Your N is 80. Depending on a confidence level and corresponding Z, your population mean should be in this range: $n\mu - z \sqrt{n\sigma^2} \le x \le n\mu + z\sqrt{n \sigma^2}$

Standard error looks like this: $\frac{\sigma}{\sqrt{n}}$

In these cases:
$\mu$: sample mean
$\sigma$: sample standard deviation
$\sigma^2$: sample variance
n: sample size

3. i was using SE (mean) = N root 1-f * sd/n

but in fact i should have been using the SE(total)..

would that be correct to say that?