Analysis of Sampling Schemes

We need to evaluate two sampling schemes for estimating the mean age ($\mu$) of $kN$ boys across $k$ groups of $N$ boys each.

Scheme I: Simple Random Sample Without Replacement (SRSWOR) of size $kn$ from the total population of $kN$ boys. The sample mean is $\bar{Y}$.

Scheme II: Simple Random Sample With Replacement (SRSWR) of size $n$ from each of the $k$ groups. The overall sample mean is $\bar{Y}_G$.

Scheme I: SRSWOR

• Expected Value: For SRSWOR, the sample mean ($\bar{Y}$) is an unbiased estimator of the population mean ($\mu$).

  $ E(\bar{Y}) = \mu $

  This confirms statement A.
• Variance: The variance of the sample mean in SRSWOR is given by:

  $ \text{Var}(\bar{Y}) = \frac{S^2}{n_I} \left( \frac{N_{pop} - n_I}{N_{pop} - 1} \right) $

  where $S^2$ is the population variance, $n_I = kn$ is the sample size, and $N_{pop} = kN$ is the population size.

  $ \text{Var}(\bar{Y}) = \frac{S^2}{kn} \left( \frac{kN - kn}{kN - 1} \right) $

Scheme II: Stratified SRSWR (by group)

• Expected Value: Let $\mu_i$ be the mean age of the $i$-th group. The sample mean from the $i$-th group using SRSWR is $\bar{Y}_i$, with $E(\bar{Y}_i) = \mu_i$. The overall sample mean $\bar{Y}_G$ is the average of the group sample means:

  $ \bar{Y}_G = \frac{1}{k} \sum_{i=1}^k \bar{Y}_i $

  The expected value is:

  $ E(\bar{Y}_G) = E\left(\frac{1}{k} \sum_{i=1}^k \bar{Y}_i\right) = \frac{1}{k} \sum_{i=1}^k E(\bar{Y}_i) = \frac{1}{k} \sum_{i=1}^k \mu_i = \mu $

  This confirms statement B.
• Variance: The variance of $\bar{Y}_G$ is calculated as:

  $ \text{Var}(\bar{Y}_G) = \text{Var}\left(\frac{1}{k} \sum_{i=1}^k \bar{Y}_i\right) = \sum_{i=1}^k \frac{1}{k^2} \text{Var}(\bar{Y}_i) $

  Since $\text{Var}(\bar{Y}_i) = \frac{\sigma_i^2}{n}$ for SRSWR (where $\sigma_i^2$ is the variance within group $i$):

  $ \text{Var}(\bar{Y}_G) = \sum_{i=1}^k \frac{1}{k^2} \frac{\sigma_i^2}{n} = \frac{1}{kn} \left(\frac{1}{k} \sum_{i=1}^k \sigma_i^2\right) = \frac{\bar{\sigma}^2}{kn} $

  where $\bar{\sigma}^2$ is the average variance within the groups.

Comparison of Variances

• Statement 3: $\text{Var}(\bar{Y})$ may be less than $\text{Var}(\bar{Y}_G)$ We compare $\text{Var}(\bar{Y}) = \frac{S^2}{kn} \left( \frac{kN - kn}{kN - 1} \right)$ and $\text{Var}(\bar{Y}_G) = \frac{\bar{\sigma}^2}{kn}$. The overall population variance $S^2$ can be decomposed as $S^2 = \bar{\sigma}^2 + V_{bg}$, where $V_{bg}$ is the variance between group means. $\text{Var}(\bar{Y}) < \text{Var}(\bar{Y}_G)$ requires:

  $ \frac{S^2}{kn} \left( \frac{kN - kn}{kN - 1} \right) < \frac{\bar{\sigma}^2}{kn} $

  $ S^2 \left( \frac{kN - kn}{kN - 1} \right) < \bar{\sigma}^2 $

  $ (\bar{\sigma}^2 + V_{bg}) \left( \frac{kN - kn}{kN - 1} \right) < \bar{\sigma}^2 $

  This inequality holds if $V_{bg}$ is small relative to $\bar{\sigma}^2$. For instance, if $V_{bg} = 0$ (all group means are equal), then $\text{Var}(\bar{Y}) = \frac{\bar{\sigma}^2}{kn} \left( \frac{kN - kn}{kN - 1} \right)$, which is strictly less than $\text{Var}(\bar{Y}_G) = \frac{\bar{\sigma}^2}{kn}$ because the finite population correction factor $\left( \frac{kN - kn}{kN - 1} \right)$ is less than 1. Thus, $\text{Var}(\bar{Y})$ can be less than $\text{Var}(\bar{Y}_G)$. This confirms statement C.
• Statement 4: $\text{Var}(\bar{Y}) = \text{Var}(\bar{Y}_G)$ if all group means are the same As shown above, if all group means are the same ($V_{bg}=0$), then $S^2=\bar{\sigma}^2$.

  $ \text{Var}(\bar{Y}) = \frac{\bar{\sigma}^2}{kn} \left( \frac{kN - kn}{kN - 1} \right) $

  $ \text{Var}(\bar{Y}_G) = \frac{\bar{\sigma}^2}{kn} $

  Since $\left( \frac{kN - kn}{kN - 1} \right) < 1$, we have $\text{Var}(\bar{Y}) < \text{Var}(\bar{Y}_G)$, not equal. Statement 4 is false.

Conclusion

Based on the analysis, statements A, B, and C are true.