# Does it make sense to calculate Relative Standard Error using population parameters?

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


## 1 Answer

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.