Determine which of the following four population size and sample size combinations would not require the use of the the finite population correction factor in calculating the standard error.
a. N = 2500; n = 75
b. N = 150; n = 25
c. N = 15,000; n = 1,000
d. N = 1500; n = 300
If your sample is less than 5% of the population, then the population correction factor would not be required.
I'll let you determine which one fits.
To determine which of the given population size and sample size combinations would not require the use of the finite population correction factor in calculating the standard error, we need to compare the sample size (n) to the population size (N) using a specific criterion.
The finite population correction factor is used when the sample size (n) is a large proportion of the population size (N). This is typically the case when the sample size exceeds 5-10% of the population size.
Let's apply this criterion to each combination:
a. N = 2500; n = 75
In this case, n/N = 75/2500 = 0.03, which is much less than 0.1 (10%). Thus, we do not need to use the finite population correction factor.
b. N = 150; n = 25
Here, n/N = 25/150 = 0.17, which is greater than 0.1 (10%). Therefore, we need to use the finite population correction factor.
c. N = 15,000; n = 1,000
For this combination, n/N = 1000/15000 = 0.07, which is less than 0.1 (10%). So, we do not need to use the finite population correction factor.
d. N = 1500; n = 300
Similarly, n/N = 300/1500 = 0.2, which is greater than 0.1 (10%). Thus, we need to use the finite population correction factor.
Based on the criterion mentioned earlier, the combinations that would not require the use of the finite population correction factor are:
a. N = 2500; n = 75
c. N = 15,000; n = 1,000
These combinations have sample sizes that are smaller proportions of their respective population sizes.