(12) Suppose N – 60,000, n =15,000, and s =50
(a) compute the standard error of mean (x) using the finite population correction factor
(b) Repeat part a assuming n =30,000
(c)Repeat part a assuming n = 60,000
(d)Compare parts (a), (b), and (c), and describe what happens to the standard error of the mean (x) as n is increased.
a. SE = 50/sqrt(15000) * sqrt((60000-15000)/60000-1))
To compute the standard error of the mean (x) using the finite population correction factor, we can follow these steps:
(a) Given N = 60,000, n = 15,000, and s = 50, let's calculate the standard error of the mean:
1. Calculate the finite population correction factor, FPC, using the formula:
FPC = sqrt((N - n) / (N - 1))
FPC = sqrt((60,000 - 15,000) / (60,000 - 1))
= sqrt(45,000 / 59,999)
≈ 0.94870 (rounded to five decimal places)
2. Calculate the standard error of the mean (x) using the formula:
x = s / sqrt(n) * FPC
x = 50 / sqrt(15,000) * 0.94870
≈ 0.38536 (rounded to five decimal places)
Therefore, the standard error of the mean (x) is approximately 0.38536.
(b) Now, let's repeat part (a) assuming n = 30,000. We'll use the same steps as before:
1. Calculate the finite population correction factor (FPC):
FPC = sqrt((N - n) / (N - 1))
= sqrt((60,000 - 30,000) / (60,000 - 1))
≈ 0.70711 (rounded to five decimal places)
2. Calculate the standard error of the mean (x):
x = s / sqrt(n) * FPC
= 50 / sqrt(30,000) * 0.70711
≈ 0.28868 (rounded to five decimal places)
The standard error of the mean (x) with n = 30,000 is approximately 0.28868.
(c) Let's repeat part (a) assuming n = 60,000:
1. Calculate the finite population correction factor (FPC):
FPC = sqrt((N - n) / (N - 1))
= sqrt((60,000 - 60,000) / (60,000 - 1))
= sqrt(0 / 59,999)
= 0
2. Calculate the standard error of the mean (x):
x = s / sqrt(n) * FPC
= 50 / sqrt(60,000) * 0
= 0
The standard error of the mean (x) with n = 60,000 is 0.
(d) When comparing parts (a), (b), and (c), we observe the following:
- As n increases from 15,000 to 30,000, the standard error of the mean (x) decreases from approximately 0.38536 to 0.28868. This indicates that the estimate of the mean becomes more precise.
- However, when n reaches the population size (60,000) in part (c), the standard error of the mean becomes 0. This occurs because the sample size equals the population size, resulting in perfect estimation. In this case, there is no variability between the samples.
In summary, as the sample size (n) increases, the standard error of the mean decreases, indicating a more accurate estimate. However, when the sample size equals the population size, the standard error becomes 0, representing perfect estimation.