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.
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.
Thank you so much for the help! I completely overlooked using the population correction factor.
You are welcome!
I'm glad the explanation helped.
To find the value of the standard error of the mean in this case, you correctly used the formula for the standard error:
Standard Error (SE) = σ / sqrt(n)
where σ is the population standard deviation and n is the sample size. In this case, σ = 10 and n = 50.
However, it seems that you made an error in calculating the finite population correction factor (when N is significantly smaller than infinite population). The formula for the finite population correction factor is:
Finite Population Correction Factor = sqrt((N - n) / (N - 1))
where N is the population size and n is the sample size.
In your case, N = 500 and n = 50. Therefore,
Finite Population Correction Factor = sqrt((500 - 50) / (500 - 1)) = sqrt(450 / 499) = 0.953
Now, let's correct your calculation:
Standard Error = σ / sqrt(n) * Finite Population Correction Factor
= 10 / sqrt(50) * 0.953
≈ 1.343
Hence, the correct value of the standard error of the mean in this case (rounded to two decimal places) is 1.34, not 1.49.
Make sure to double-check your calculations and apply the finite population correction factor correctly.