The question asks for the condition(s) under which the variance of the sample mean from simple random sampling with replacement ($\bar{Y}$) is equal to the variance of the sample mean from stratified random sampling with replacement using proportional allocation ($\bar{Y}_{st}$).
We need to find when $Var(\bar{Y}) = Var(\bar{Y}_{st})$. Equating the formulas:
$ \frac{S^2}{n} = \frac{1}{n} \sum_{h=1}^{L} W_h S_h^2 $This simplifies to:
$ S^2 = \sum_{h=1}^{L} W_h S_h^2 $The overall population variance ($S^2$) can be expressed in terms of within-stratum variances and between-stratum variance:
$ S^2 = \sum_{h=1}^{L} W_h S_h^2 + \sum_{h=1}^{L} W_h (\mu_h - \mu)^2 $where $\mu_h$ is the mean of stratum $h$ and $\mu$ is the overall population mean.
For the condition $S^2 = \sum_{h=1}^{L} W_h S_h^2$ to hold, the between-stratum variance term must be zero:
$ \sum_{h=1}^{L} W_h (\mu_h - \mu)^2 = 0 $Since $W_h > 0$ (assuming all strata exist) and $(\mu_h - \mu)^2 \ge 0$, this equality holds if and only if $(\mu_h - \mu)^2 = 0$ for all strata $h$. This implies:
$ \mu_h = \mu \quad \text{for all } h = 1, 2, ..., L $This means the mean of each stratum must be equal to the overall population mean. Consequently, all stratum means must be equal to each other ($\mu_1 = \mu_2 = ... = \mu_L$).
The sufficient condition for $Var(\bar{Y}) = Var(\bar{Y}_{st})$ is that all the stratum means are equal.
Correct Option Analysis:
Therefore, the correct answer is that all the stratum means are equal.
Suppose there are $k$ groups each consisting of $N$ boys. We want to estimate the mean age $\mu$ of these $kN$ boys. Fix $1 < n < N$ and consider the following two sampling schemes.
I. Draw a simple random sample without replacement of size $kn$ out of all $kN$ boys.
II. From each of the $k$ groups draw a simple random sample with replacement of size $n$.
Let $\bar{Y}$ and $\bar{Y}_G$ be the respective sample mean ages for the two schemes. Which of the following are true?