If a population has mean \(\mu\) and variance \(\sigma^2\), the variance of the sample mean \(\bar{X}\) for samples of size \(n\) drawn WITHOUT replacement from a finite population of size \(N\) is:

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CSS Statistics QUESTION #1984

Question 1

\(\dfrac{\sigma^2}{n}\)
\(\dfrac{\sigma^2}{n} \cdot \dfrac{N-n}{N-1}\)✔️
\(\dfrac{\sigma^2}{n} \cdot \dfrac{N-n}{N}\)
\(\dfrac{\sigma^2}{N}\)

Correct Answer Explanation

For sampling without replacement from a finite population, \(Var(\bar{X}) = \dfrac{\sigma^2}{n} \cdot \dfrac{N-n}{N-1}\), where \(\dfrac{N-n}{N-1}\) is the finite population correction (fpc) factor. It reduces variance compared to sampling with replacement.

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