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Question

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?

The correct answer is
Strata means are same

The question asks for the conditions under which the variance of a sample mean from Simple Random Sampling With Replacement (SRSWR) is equal to the variance of a sample mean from stratified sampling with proportional allocation.

Variance of SRSWR Sample Mean (T1)

For a sample mean $T_1$ obtained through SRSWR from a population with variance $\sigma^2$, the variance is given by:

where $n$ is the sample size.

Variance of Stratified Sample Mean (T2)

For a stratified sample mean $T_2$ drawn using proportional allocation from $L=4$ strata, the variance is:

where $W_h = N_h/N$ is the stratum weight, $N_h$ is the size of stratum $h$, $N$ is the total population size, $\sigma_h^2$ is the variance within stratum $h$, and $n_h$ is the sample size from stratum $h$.

With proportional allocation, $n_h = n W_h$. Substituting this into the variance formula gives:

Condition for Variance Equality

We need to find conditions such that $\text{Var}(T_1) = \text{Var}(T_2)$. Equating the two expressions:

This simplifies to:

Decomposition of Population Variance

The overall population variance $\sigma^2$ can be expressed in terms of within-stratum variances and between-stratum variance:

Here, $\mu$ is the population mean, $\mu_h$ is the mean of stratum $h$, and $\sum_{h=1}^4 W_h (\mu_h - \mu)^2$ is the between-stratum variance component (often denoted as $B^2$).

Identifying the Sufficient Condition

For the equality $\sigma^2 = \sum_{h=1}^4 W_h \sigma_h^2$ to hold, the between-stratum variance component must be zero:

Since $W_h > 0$ (strata have positive weights), this condition implies that $(\mu_h - \mu)^2 = 0$ for all $h=1, 2, 3, 4$. This means:

This result indicates that the mean of each stratum must be equal to the overall population mean. This is precisely the condition stated in Option 3.

Conclusion

Therefore, the condition that strata means are same ($\mu_h = \mu$ for all $h$) is sufficient for $\text{Var}(T_1) = \text{Var}(T_2)$.

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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. 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 $Var(\bar{Y}) = Var(\bar{Y}_{st})$?
  4. 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
  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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