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I am flirting with the idea of using econometrics and regression to improve upon the classical SPC control charting techniques. Hopefully with the end of goal of doing trend analysis on mean and variance concurrently while generating control charts, which would add further cues for an out of control process.

While looking into how the control limits are defined for X-bar chart, I understand why we use the empirical values for $d_2$ and $A_2$, but I'm wondering if other alternatives are as robust as this technique.

If we define the X-bar chart with:

$\mu \pm 3 \dfrac{\sigma}{\sqrt{n}} \approx \overline{\overline{X}} \pm 3 \dfrac{\hat{\sigma}}{\sqrt{n}}$

Why can we not estimate:

$\hat{\sigma} = s_x$ instead of using $\hat{\sigma} = \dfrac{\overline{R}}{d_2}$

or alternatively use $\mu \pm 3 \dfrac{\sigma}{\sqrt{n}} \approx \overline{\overline{X}} \pm 3 s_\overline{X}$

I understand that using R is convenient since we have it at hand, but since most control charts are built electronically nowadays I am not worried about convenience. Also, I understand that an Xbar chart without an R chart will not catch changes in subgroup variation.

Please let me know if my question is misleading or framed incorrectly.

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At least one commercial statistical software package (JMP) offers a variant of the $\bar{X}$/$R$ chart that replaces $R$ with $s$ (see https://www.jmp.com/support/help/13/Control_Limits_for_X-bar_and_S-charts.shtml). However, for constant sample sizes, it set the limits for $\bar{X}$ to $\bar{\bar{X}} \pm 3 \frac{\bar{s}}{ c_4\sqrt{n}}$.

The reason you can't just set $\widehat{\sigma} = s_x$ is because $s_X$ is not an unbiased estimator of $\sigma$. See http://www.itl.nist.gov/div898/handbook/pmc/section3/pmc321.htm.

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  • $\begingroup$ Can this be corrected by using Student's t distribution instead of a normal distribution? $\endgroup$ Commented Jun 15, 2017 at 11:44
  • $\begingroup$ Nevermind, looks to me as if the limits are based on the cv of subgroup aligned to 3sigma of entire sample. Therefore testing cv of each subgroup. $\endgroup$ Commented Jun 15, 2017 at 16:20

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