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In univariate statistical process control, for the $\bar{X}$ chart with $n$ subgroups each of $m$ samples, we calculate the center line, upper and lower control limits using this equation (1):

enter image description here

Here, $Z_{1-\alpha/2}$ is a quantile of the normal distribution, $\bar{\bar{X}}$ is the mean of subgroup sample means, $\bar{s}$ is the mean of sample standard deviations of subgroups, and $d_3(m)=\frac{\Gamma(\frac{m}{2})}{\Gamma(\frac{m-1}{2})} \sqrt{\frac{2}{m-1}}$ is a normalising constant such that $\frac{\bar{s}}{d_3(m)} = \hat{\sigma}$ (here $\Gamma$ is the gamma function).

What I'm confused by is the fact that we're using the quantiles of the normal distribution at all. Since we're estimating the standard deviation from data, $\hat{\sigma}$ should follow a Chi squared distribution, meaning that $X$ (and therefore $\bar{X}$ and $\bar{\bar{X}}$) should follow a student $t$ distribution. I get that the $t$ distribution converges to $N(0, 1)$ with infinite sample size, but why make that assumption when we could calculate more accurate control limits using the $t$-distribution?

(1) Qiu, P. (2014). Introduction to Statistical Process Control.

Migwell
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  • Sure, you could frame the problem like that: however, the use of the $t$ distribution would merely change the values of $Z_{1-\alpha/2}$ and a compensating change in $d_3(m)$ would produce a completely equivalent chart. – whuber Mar 31 '21 at 16:59
  • You could do that, but I've added the definition of $d_3$ from the same source and I don't believe that it does attempt to transform the variance from normal to $t$. Otherwise it would surely involve $n$ as well. – Migwell Apr 01 '21 at 00:08
  • Why should it involve $n$ at all? – whuber Apr 01 '21 at 01:34

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