2
$\begingroup$

This is going to be a very simple question, but I just need to make sure of a few basic things.

I'm building some Xbar-R charts and have sample size of 4 and am looking over the Control Limits.

I've been told two ways to calculate control limits. Either

$$ \mathrm{UCL = \overline x + 3\sigma } $$ $$ \mathrm{LCL = \overline x - 3\sigma } $$

Or

$$ \mathrm{UCL = \overline x + \overline R *A2} $$ $$ \mathrm{LCL = \overline x - \overline R *A2} $$

I've never really thought much about it, but those values are quite different. I'm probably just missing something, but can someone explain to me the difference and when I should use them? In general, can I just use 3 sigma?

(I apologize if this is a duplicate. I had difficulty finding good keywords to search and couldn't quite find a match.)

$\endgroup$

1 Answer 1

1
$\begingroup$

The terms $\hat{\sigma}$, $s$ and $\overline{R}*A_{2}$ are "equivalent" even though they would result in different values. All terms other than $\sigma$ are simply point estimates of the actual standard deviation. In the case of Shewhart charts, $\overline{x}\pm \overline{R}*A_{2}$ is the preferred term because this method for three standard deviations was what the charts were designed to use.

This paper addresses this issue in great detail with discussion and insight into which form might be the "best": Mahmoud, M. A.; Henderson, G. R.; Epprecht, E. K. & Woodall, W. H. Estimating the Standard Deviation in Quality-Control Applications Journal of Quality Technology, 2010, 42, 348-357

Using the following data, the standard deviation is estimated using a variety of methods and sample sizes. The usual method of sample standard deviation, $s$, is calculated for the complete sample of fifty units. Values $s_{2_a}$ and $s_{2_b}$ use the $\frac{\overline{R}}{d_2}\simeq\hat\sigma$ method with sample sizes of 10 and 5, respectively. Results for $s_{3_a}$ and $s_{3_b}$ use the formula $\frac{\overline{s}}{c_4}=\hat\sigma$ with sample sizes of 10 and 5, respectively.

3.5340, 3.4942, 3.4913, 3.6751, 3.3433, 3.3962, 3.5263, 3.5019, 3.5539, 3.4750, 3.5054, 3.5969, 3.3966, 3.5357, 3.5876, 3.6289, 3.6750, 3.3760, 3.4822, 3.4309, 3.4064, 3.5685, 3.5296, 3.3918, 3.2556, 3.5911, 3.4216, 3.4741, 3.6216, 3.5856, 3.5574, 3.6323, 3.3915, 3.4602, 3.5275, 3.6394, 3.6773, 3.5937, 3.5436, 3.6391, 3.3754, 3.4199, 3.5507, 3.5414, 3.5187, 3.5381, 3.6922, 3.5144, 3.6709, 3.4844

The method $c_4\overline{s} \simeq\hat\sigma$ is used to calculate the values $s_{4_a}$ and $s_{4_b}$ with the noted sample sizes of 10 and 5 while $(\sqrt{(n-1)/n})s=\hat\sigma$ was used to evaluate $s_{5_a}$ and $s_{5_b}$ based on sample sizes of 10 and 5. The pooled standard deviation, $S_{pooled}$, using the method $$S_{pooled}=\sqrt{\frac{\sum^m_{i=1} \left (n_i-1 \right )S^2_i}{\sum^m_{i=1}\left ( n_i-1 \right )}}$$ was used to evaluate $s_{6_a}$, $s_{6_b}$, and $s{6_c}$ based on sample sizes of 10, 5, and then a split between 7 and 3, to reinforce the diversity of the pooled method.

These values are listed for reference below. It is easily demonstrated that $\hat{\sigma}$ has a significant impact on the control limits. The impact can be seen by comparing values , based on the criteria used while using the various estimates of $\hat{\sigma}$. Estimates for $\hat{\mu}$ can also be problematic, with additional impacts on control charts and their limits.

$\hat{\sigma}$, Value, Method, $n$

$s$ 0.0991 $\sqrt{\frac{\sum_{i=1}^{n}\left ( x_i-\bar{x} \right )^2}{n-1}}$ 50

$s_{2_a}$ 0.0923 $\frac{\overline{R}}{d_2}$ 10

$s_{3_a}$ 0.0983 $\frac{\overline{s}}{c_4}$ 10

$s_{4_a}$ 0.0930 $c_4\overline{s}$ 10

$s_{5_a}$ 0.0907 $(\sqrt{(n-1)/n})s$ 10

$s_{6_a}$ 0.0967 $S_{pooled}$ 10

$s_{2_b}$ 0.0983 $\frac{\overline{R}}{d_2}$ 5

$s_{3_b}$ 0.0983 $\frac{\overline{s}}{c_4}$ 5

$s_{4_b}$ 0.0930 $c_4\overline{s}$ 5

$s_{5_b}$ 0.0824 $(\sqrt{(n-1)/n})s$ 5

$s_{6_b}$ 0.0952 $S_{pooled}$ 5

$s_{6_c}$ 0.0966 $S_{pooled}$ 7/3

$\endgroup$
2

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.