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Question

Suppose $\bar{Y}$ is the sample mean of the study variables corresponding to a sample of size n using simple random sampling with replacement scheme and $\bar{Y}_{st}$ is the sample mean of the study variables corresponding to a sample of size n using stratified random sampling with replacement scheme under proportional allocation. Which of the following is/are sufficient condition/conditions for $Var(\bar{Y}) = Var(\bar{Y}_{st})$?

The correct answer is
All the stratum means are equal

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 ($\bar{Y}_{st}$).

Variance Formulas

  • For Simple Random Sampling (SRS) with replacement, the variance of the sample mean is: $Var(\bar{Y}) = \frac{S^2}{n}$ where $S^2$ is the population variance and $n$ is the sample size.
  • For Stratified Random Sampling (StRS) with replacement and proportional allocation ($n_h = n W_h$, where $W_h = N_h/N$), the variance of the sample mean is: $Var(\bar{Y}_{st}) = \sum_{h=1}^{L} W_h^2 \frac{S_h^2}{n_h} = \sum_{h=1}^{L} W_h^2 \frac{S_h^2}{n W_h} = \frac{1}{n} \sum_{h=1}^{L} W_h S_h^2$ where $L$ is the number of strata, $W_h$ is the stratum weight, and $S_h^2$ is the variance within stratum $h$.

Condition for Variance Equality

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 $

Population Variance Decomposition

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.

Deriving the Sufficient Condition

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 ($\mu_1 = \mu_2 = ... = \mu_L$).

Conclusion

The sufficient condition for $Var(\bar{Y}) = Var(\bar{Y}_{st})$ is that all the stratum means are equal.

Correct Option Analysis:

  • Option 1: Equal stratum sizes ($N_h$ equal) does not guarantee equal means.
  • Option 2: Equal stratum totals ($N_h \mu_h$ equal) does not guarantee equal means.
  • Option 3: Equal stratum means ($\mu_h$ equal) is the derived condition.
  • Option 4: Equal stratum variances ($S_h^2$ equal) does not guarantee equal means.

Therefore, the correct answer is that all the stratum means are equal.

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Important Questions from Ratio And Regression

  1. A simple random sample (without replacement) of size $n$ is drawn from a finite population of size $N (\ge 7)$. What is the probability that the $4^{\text{th}}$ population unit is included in the sample but the $6^{\text{th}}$ population unit is not included in the sample?
  2. For a data set $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$ the following two models were fitted using least square method.
    Model 1: $y_i = \beta_0 + \beta_1 x_i \quad i = 1, 2, \dots n$
    Model 2: $y_i = \beta_0^* + \beta_1^* x_i + \beta_2^* x_i^2 \quad i = 1, 2, \dots n$
    Let $\hat{\beta}_0, \hat{\beta}_1$ be least square estimates of $\beta_0, \beta_1$ from model 1 and $\hat{\beta}_0^*, \hat{\beta}_1^*, \hat{\beta}_2^*$ be the least square estimates from model 2.
    Let $A = \sum_1^n \left(y_i - (\hat{\beta}_0 + \hat{\beta}_1 x_i)\right)^2$,
    $B = \sum_1^n \left(y_i - (\hat{\beta}_0^* + \hat{\beta}_1^* x_i + \hat{\beta}_2^* x_i^2)\right)^2$
    Then
  3. Consider the problem of drawing a sample of size 2 from a finite population of size 20. The sampling is done with replacement using probability proportional to size sampling scheme. The normed size measures $p_1, \cdots, p_{20}$ are given by $p_i = \frac{1}{40}$, $i = 1, \cdots, 10, \; p_i = \frac{3}{40}$, $i = 11, \cdots, 20$. The expected number of distinct units drawn is
  4. Consider a finite population of size $N$. Let $T_1$ be the sample mean based on a sample of size $n$ under simple random sampling with replacement (SRSWR) scheme. Let $T_2$ be the sample mean based on a stratified random sample of size $n$ where the samples are drawn from each of 4 strata using SRSWR scheme under proportional allocation. Then which of the following are sufficient conditions for $\text{Var}(T_1) = \text{Var}(T_2)$ to hold?
  5. 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?

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