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The variable X under study have rectangular distribution with interval (a, a+ d ) the interval is divided into k equal subintervals which form k equal strata of equals sizes . From each stratum simple random sample of n/ k units is drawn . Let V1 and V2 be the variance in estimator of population mean based on stratified and simple random sampling of size n .

prove that V1/V2 = 1/K²

My attempt

V2 = Var($\ y_R $ ) = $\ ( \frac{1}{n} - \frac{1}{\ k^2} ) \ S^2 $

As it is given in the question that total k stratum are of k subsamples in each of total population should be $\ k^2 $

V1 = Var($\ y_st $ ) = $\ ( \frac{1}{n} - \frac{1}{\ k^2} ) \sum_{i= 0 }^{k} \ p_i \ S_i^2 $

So v1/v2 = $\frac{\sum_{i=0}^{\ k} \ S_i^2 }{k \ S^2} $

Is this correct approach ??

Please help if you know how to solve

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    $\begingroup$ The sum in the definition of $V_1$ should be from 1 to $k$, not $0$ to $k^2$ $\endgroup$ Jul 4, 2021 at 3:32

1 Answer 1

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The ratio of the variances depends on the ratio of the within-stratum variances of $X$ to the overall variance of $X$. In this case, the stratum-specific variance for an interval of length $k$ is proportional to $1/k^2$ (specifically, it's $(b-a)/12k^2$)

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  • $\begingroup$ hey , I have tried to use your hint ans solve please review my answer . $\endgroup$
    – simran
    Jul 3, 2021 at 8:03

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