Some notes on finite population correction factor:
If the population is small and the sample is large (more than 5% of the small population), use the finite population correction factor.
For standard error of the mean, use:
sd/√n
If you need to adjust for the finite population correction factor, use:
sd/√n * √[(N-n)/(N-1)]
N = number in population
n = number in sample
Is the sample size of 50 more than 5% of the population size of 500? Yes, it is, so use the finite population correction factor.
With your data:
10/√50 * √[(500-50)/(500-1)] =
1.4142 * 0.9496 = 1.34
Check these calculations.
I hope this helps.
Hi can someone check my answer. I'm not sure why the computer keeps marking it wrong.
**Suppose a random sample of size 50 is selected from a population with σ = 10. Find the value of the standard error of the mean in each of the following cases (use the finite population correction factor if appropriate).
Question: The population size is N = 500 (to 2 decimals).
My work:
1.4142 sqrt ( 499/450) = 1.49
I keep getting 1.49 but the computer is marking it wrong. I'm not sure what I am doing incorrectly. Thanks for the help.
3 answers
Thank you so much for the help! I completely overlooked using the population correction factor.
You are welcome!
I'm glad the explanation helped.
I'm glad the explanation helped.