Discussion
If \(\rho_{wsy}\) denotes the intra-class correlation coefficient between pairs of units that are in the same systematic sample and if \(\rho_{wsy} = 0\), then :
Answer: C
Systematic sampling is as efficient as SRSWR but less efficient than SRSWOR.
For a population of size \(N = nk\), the variance of the systematic sample mean in terms of the intra-class correlation is
\[ \begin{aligned} V(\bar{y}_{sy}) &= \frac{\sigma^{2}}{n}\,[1 + (n - 1)\rho_{wsy}] \\ \sigma^{2} &= \frac{N - 1}{N}S^{2} \end{aligned} \]
With \(\rho_{wsy} = 0\) this is \(\sigma^{2}/n\), exactly the variance of the mean under SRSWR. Under SRSWOR the variance is \(\dfrac{N - n}{N} \cdot \dfrac{S^{2}}{n}\), which is smaller than \(\dfrac{N - 1}{N} \cdot \dfrac{S^{2}}{n}\) whenever \(n \gt 1\), because of the finite population correction.
So systematic sampling matches SRSWR and falls just short of SRSWOR. It beats SRSWOR only when \(\rho_{wsy}\) is negative enough, which rules out the options claiming it is as efficient as or more efficient than SRSWOR, or more efficient than SRSWR.
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