The use of the finite population correction factor when sampling without replacement from finite populations will
A) increase the standard error of the mean.
B) not affect the standard error of the mean.
C) reduce the standard error of the mean.
D) only affect the proportion, not the mean.
The correct answer and explanation is :
The correct answer is:
C) reduce the standard error of the mean.
Explanation:
When sampling from a finite population without replacement, the finite population correction (FPC) factor is used to adjust the standard error of the mean (SEM). The formula for SEM with FPC is:
[
\text{Adjusted SEM} = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N – n}{N – 1}}
]
Where:
- ( \sigma ) is the population standard deviation,
- ( n ) is the sample size,
- ( N ) is the total population size,
- ( \sqrt{\frac{N – n}{N – 1}} ) is the finite population correction factor.
The FPC factor accounts for the reduced variability when we sample without replacement. Since each selected unit is not returned to the population, the remaining elements are less variable, leading to a lower standard error.
Why Does the FPC Reduce Standard Error?
- In infinite or very large populations, the FPC factor is close to 1, meaning it has little effect.
- In smaller populations, as ( n ) approaches ( N ), the FPC factor gets smaller, reducing the SEM significantly.
- This reduction happens because, as we take more samples from a finite population, there is less variability left, making the estimates more precise.
Thus, using the FPC reduces the standard error of the mean when sampling without replacement, making the sample mean a more accurate estimate of the population mean.
