Statistical Quality Control Questions and Answers

Q: Suppose two control charts have the same in-control average run length (ARL). Then which of the following is true ?

The true statement is The charts can be compared using out-of-control ARL.Matching the in-control ARL makes two charts equally prone to false alarms. The fair comparison is then how quickly each detects a real shift, that is, which has the smaller out-of-control ARL for the shifts of interest. This is how Shewhart, CUSUM and EWMA charts are usually compared.Equal in-control ARLs do not mean equal performance: a CUSUM chart detects small shifts much faster than a Shewhart chart with the same in-control ARL. Charts can of course be designed to share an in-control ARL, and ARL comparisons apply to charts for variability as well as for the mean.

Q: For a production process, the sample ranges are found to be 1.2, 1.5, 1.1, 1.4 and 1.5. Suppose that the subgroup size is 5. Then, what will be the process standard deviation, given that \(A_{2} = 0.577\) and \(d_{2} = 2.326\) ?

The process standard deviation is 0.576.The process standard deviation is estimated from the average range as \(\hat{\sigma} = \bar{R}/d_{2}\):\[ \begin{aligned} \bar{R} &= \frac{1.2 + 1.5 + 1.1 + 1.4 + 1.5}{5} = 1.34 \\ \hat{\sigma} &= \frac{1.34}{2.326} \approx 0.576 \end{aligned} \]\(A_{2}\) is used for the control limits of the \(\bar{X}\) chart, not for \(\sigma\); for example \(A_{2}\bar{R} = 0.577 \times 1.34 \approx 0.773\). None of 0.511, 2.463 or 2.322 equals \(\bar{R}/d_{2}\).

Suppose two control charts have the same in-control average run length (ARL). Then which of the following is true ?

A.Both the charts will perform in similar way
B.The charts can be compared using out-of-control ARL
C.Two charts cannot have the same in-control ARL
D.The charts must be only to monitor population mean

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The true statement is The charts can be compared using out-of-control ARL.

Matching the in-control ARL makes two charts equally prone to false alarms. The fair comparison is then how quickly each detects a real shift, that is, which has the smaller out-of-control ARL for the shifts of interest. This is how Shewhart, CUSUM and EWMA charts are usually compared.

Equal in-control ARLs do not mean equal performance: a CUSUM chart detects small shifts much faster than a Shewhart chart with the same in-control ARL. Charts can of course be designed to share an in-control ARL, and ARL comparisons apply to charts for variability as well as for the mean.

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For a production process, the sample ranges are found to be 1.2, 1.5, 1.1, 1.4 and 1.5. Suppose that the subgroup size is 5. Then, what will be the process standard deviation, given that \(A_{2} = 0.577\) and \(d_{2} = 2.326\) ?

A.0.511
B.2.463
C.2.322
D.0.576

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The process standard deviation is 0.576.

The process standard deviation is estimated from the average range as \(\hat{\sigma} = \bar{R}/d_{2}\):

\[ \begin{aligned} \bar{R} &= \frac{1.2 + 1.5 + 1.1 + 1.4 + 1.5}{5} = 1.34 \\ \hat{\sigma} &= \frac{1.34}{2.326} \approx 0.576 \end{aligned} \]

\(A_{2}\) is used for the control limits of the \(\bar{X}\) chart, not for \(\sigma\); for example \(A_{2}\bar{R} = 0.577 \times 1.34 \approx 0.773\). None of 0.511, 2.463 or 2.322 equals \(\bar{R}/d_{2}\).

Discuss in Forum

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