Question

Which of the following CORRECTLY defines the relationship between the variances of sample means for simple random samples drawn with and without replacement from a normal population?

• A. \(\frac{\sigma^2}{n}\gt\frac{\sigma^2}{n}(\frac{N-n}{N-1})\)
• B. \(\frac{\sigma^2}{n}\le\frac{\sigma^2}{n}(\frac{N-n}{N-1})\)
• C. \(\frac{\sigma^2}{n}\lt\frac{\sigma^2}{n}(\frac{N-n}{N-1})\)
• D. \(\frac{\sigma^2}{n}=\frac{\sigma^2}{n}(\frac{N-n}{N-1})\)

Answer 1

The Correct Option is A

To solve this problem, we need to understand the concept of variance in the context of sample means for simple random samples drawn from a finite population. When a sample is drawn from a population, the variance of the sample mean depends on whether the sampling is done with replacement or without replacement.

For a population with variance \( \sigma^2 \) and size \( N \), if a sample of size \( n \) is drawn, the variances of the sample means are defined as follows:

• With Replacement: The variance of the sample mean is given by \(\frac{\sigma^2}{n}\). This is because each sample is drawn independently of others, maintaining the population variance.
• Without Replacement: The variance of the sample mean is adjusted for the finite population correction factor and is given by \(\frac{\sigma^2}{n} \left( \frac{N-n}{N-1} \right)\). This accounts for the reduction in variability because the samples are not independent (each sample affects the remaining population).

Since the factor \(\frac{N-n}{N-1}\) is less than 1, the variance of the sample mean without replacement is less than that with replacement. Therefore, we have:

\(\frac{\sigma^2}{n} > \frac{\sigma^2}{n}\left(\frac{N-n}{N-1}\right)\)

Therefore, the correct answer is the one that states the variance of the sample mean when samples are drawn with replacement is greater than when samples are drawn without replacement.

Hence, the correct option is: \(\frac{\sigma^2}{n} > \frac{\sigma^2}{n} \left( \frac{N-n}{N-1} \right)\)