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Relative Standard Error (RSE) is one of the main measures to assess the quality of survey indicators. If only sample data are available, RSE can be computed using estimated mean $\bar{y}$ and estimated standard deviation $\hat{\sigma}$:

$$ RSE^{sample} = \frac{\hat{\sigma}}{\sqrt{n}}*\frac{1}{\bar{y}} $$

population = c(1,10,3,4,6,12,3)
n_sample = 5
samples = sample(population,n_sample,replace=FALSE)
ybar = mean(samples)
se_sample = sd(samples)/sqrt(n_sample)
RSE_sample = se_sample/ybar

If both mean and variance of the population are available, does it make sense computing RSE using both population parameters ($\mu$ and $\sigma$)?

$$ RSE^{pop} = \frac{\sigma}{\sqrt{n}}*\frac{1}{\mu} $$

mu_pop = mean(population)
se_pop = sd(population)/sqrt(n_sample)
RSE_pop = se_pop/mu_pop
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1 Answer 1

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RSE is calculated by dividing the standard error of the estimate by the estimate itself. So if you have the population variance $\sigma$, the RSE for sample mean will be given by

$RSE=\frac{\sigma}{\sqrt{n}}\frac{1}{\bar{y}}$

There will be no need to include the population mean, as standard error is a term solely related with sample estimates. Relative standard error measures how large a standard error is relative to the size of the estimated value. If you replace the sample estimate with population parameter while calculating RSE, that defeats the whole purpose.

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