Determine which of the following four population size and sample size combinations would not require

the use of the finite population correction factor in calculating the standard error.
A. N = 1500; n = 300
B. N = 2500; n = 75
C. N = 15,000; n = 1,000
D. N = 150; n = 25

I think that it's A, could you please revise this question.

calculate the percent of sample of each

A. 300/1500=20 percent
B. 75/2500= 3 percent
C. 1000/15000=6 percent
D. 25/150=16 percent
The Finite Population Correction Factor (FPC) is used when you sample without replacement from more than 5% of a finite population. It’s needed because under these circumstances, the Central Limit Theorem doesn’t hold and the standard error of the estimate (e.g. the mean or proportion) will be too big. In basic terms, the FPC captures the difference between sampling with replacement and sampling without replacement.

thanks for the explanation. is there any way you can review my other question?

To determine which of the population size and sample size combinations would not require the use of the finite population correction factor, we need to compare the sample size to the population size.

The general guideline is that if the sample size (n) is less than or equal to 5% of the population size (N), then the finite population correction factor is not necessary.

Let's go through each option:

A. N = 1500; n = 300
In this case, n/N = 300/1500 = 0.2, which is greater than 0.05. Therefore, the finite population correction factor would be needed. Option A is not the correct answer.

B. N = 2500; n = 75
Here, n/N = 75/2500 = 0.03, which is less than 0.05. So, the finite population correction factor is not necessary. Option B could be the correct answer.

C. N = 15,000; n = 1,000
The ratio n/N = 1000/15000 = 0.067, which is greater than 0.05. Hence, the finite population correction factor would be required. Option C is not the correct answer.

D. N = 150; n = 25
In this case, n/N = 25/150 = 0.167, which is greater than 0.05. Therefore, the finite population correction factor would also be needed. Option D is not the correct answer.

Based on the analysis, the combination that would not require the use of the finite population correction factor is:

B. N = 2500; n = 75

To determine which population size and sample size combination would not require the use of the finite population correction factor, we need to compare the sample size (n) to the population size (N).

The finite population correction factor is used when the sample size is a large proportion of the population size (typically more than 5-10% of the population). It adjusts the standard error calculation to account for the finite population, ensuring more accurate estimates.

Let's examine each combination:

A. N = 1500; n = 300
In this case, n/N = 300/1500 = 0.2, which is less than 10%. The sample size is not a large proportion of the population size, so the finite population correction factor would not be required.

B. N = 2500; n = 75
Here, n/N = 75/2500 = 0.03, which is significantly less than 10%. The sample size is also not a large proportion of the population size, so the finite population correction factor would not be necessary.

C. N = 15,000; n = 1,000
In this combination, n/N = 1000/15000 = 0.0667, which is still less than 10%. Therefore, the finite population correction factor would not be required.

D. N = 150; n = 25
Here, n/N = 25/150 = 0.1667, which is greater than 10%. The sample size is a large proportion of the population size, so the finite population correction factor would be needed.

Based on the analysis, options A, B, and C (N = 1500; n = 300, N = 2500; n = 75, and N = 15000; n = 1000) would not require the use of the finite population correction factor. Option D (N = 150; n = 25) would require using the correction factor.

Therefore, the correct answer is A.