# In Gelman's 8 school example, why is the standard error of the individual estimate assumed known?

Context:

In Gelman's 8-school example (Bayesian Data Analysis, 3rd edition, Ch 5.5) there are eight parallel experiments in 8 schools testing the effect of coaching. Each experiment yields an estimate for the effectiveness of coaching and the associated standard error.

The authors then build a hierarchical model for the 8 data points of coaching effect as follows:

$$y_i \sim N(\theta_i, se_i) \\ \theta_i \sim N(\mu, \tau)$$

Question In this model, they assume that $$se_i$$ is known. I do not understand this assumption -- if we feel that we have to model $$\theta_i$$, why don't we do the same for $$se_i$$?

I've checked the Rubin's original paper introducing the 8 school example, and there too the author says that (p 382):

the assumption of normality and known standard error is made routinely when we summarize a study by an estimated effect and its standard error, and we will not question its use here.

To summarize, why don't we model $$se_i$$? Why do we treat it as known?

• I assume because they know the total number of schools in the area, so the SE is a function of the sample size and the estimate? – Learning stats by example Jan 8 at 22:08
• The sample size is known and fixed, but the standard error depends on the standard deviation of the data as well, and I'm not sure why we're treating that as fixed. – Heisenberg Feb 5 at 23:45
• If you are happy making your results fully conditional on the assumption of fixed standard errors, then there is nothing wrong with making (and stating) that condition. Still, why? Absence of a defensible prior? Or perhaps if the standard errors are given a broad, uninformative prior, the rest of the analysis just washes away. I dunno. – Peter Leopold Feb 6 at 0:15

In the school example, they rely on large sample size to assume that the variances are known "for all practical purposes" (p119), and I expect they estimate them using $$\frac{1}{n-1} \sum (x_i - \overline{x})^2$$ and then pretend those are the exact known values.
• Sure. Estimators tend to get more precise with increasing $n$ and the variance is no exception; because $\hat{\sigma}^2$ is a chi-square (times some constant), it has standard deviation $\sqrt{2 \sigma^4 /(n-1)}$. A denominator of about 30 schools seems like it was enough for them. – Drew N Apr 26 at 20:38