The general model for the control limits for a control chart for the sample stat
ID: 3182412 • Letter: T
Question
The general model for the control limits for a control chart for the sample statistic w is the following:
UCL = Mu(w) + L * Sigma(w)
CL = Mu(w)
LCL = Mu(w) - L * Sigma(w)
where: L is the distance of the control limit from the center line
Mu(w) is the mean of the sample statistic w
Sigma(w) is the standard deviation of the statistic w
Assume that the sample statistic w is Normally distributed.
a. Find the Probability Limits (value of L) if we want to have a control chart with a
probability of 0.0001 of a type I error in each direction.
Ans: 3.7190
b. What is the in-control Average Run Length (ARL) for this control chart?
Ans: 5000
Final answer provided, need written steps. Thanks
Explanation / Answer
UCL = Mu(w) + L * Sigma(w)
CL = Mu(w)
LCL = Mu(w) - L * Sigma(w)
where: L is the distance of the control limit from the center line
Mu(w) is the mean of the sample statistic w
Sigma(w) is the standard deviation of the statistic w
Assume that the sample statistic w is Normally distributed.
a. Here control chart can have only 0.0001 of a type I error that means alpha = 0.0001 so we will check the probability level in z - table for p- value = 1- 0.0001 = 0.9999
by using calculator we find z - value for p = 0.9999 => z = 3.7190
that means L = 3.7180
b. The average run length (ARL) at a given quality level is the average number of samples (subgroups) taken before an action signal is given. The ARL tells us, for a given situation, how long on the average we will plot successive control charts points before we detect a point beyond the control limits. For an XBar chart, with no change in the process, we wait on the average 1/p points before a false alarm takes place, with pdenoting the probability of an observation plotting outside the control limits. For a normal distribution,here p=0.0001 at one side so p = 0.0001*2 = 0.0002 for two - tailed and the ARL is approximately 1/0.0002 = 5000.
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