Question 10.4: A process has mean μ = 3 and standard deviation σ = 1, Sampl...
A process has mean μ = 3 and standard deviation σ = 1, Samples of size n = 4 are taken. If a special cause shifts the process mean to a value of 3.5, find the ARL.
We first compute the probability p that a given point plots outside the control limits. Then ARL = 1 / p. The control limits are plotted on the basis of a process that is in control. Therefore they are at \mu \pm 3 \sigma_{\bar{X}}, where μ = 3 and \sigma_{\bar{X}}=\sigma / \sqrt{n}=1 / \sqrt{4}=0.5. The lower control limit is thus at I .5, and the upper control limit is at4.5. If \bar{X} is the mean of a sample taken after the process mean has shifted, then \bar{X}\sim N\left(3.5,0.5^2\right). The probability that \bar{X} plots outside the control limits is equal to P(\bar{X}<1.5)+P(\bar{X}>4.5). This probability is 0.0228 (see Figure 10.6). The ARL is therefore equal to 110.0228 = 43.9. We will have to observe about 44 samples, on the average, before detecting this shift.
