Refer to Example 10.4. An upward shift to what value can be detected with an ARL of 20?

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Let m be the new mean to which the process has shifted. Since we have specified an upward shift, m > 3. In Example 10.4 we computed the control limits to be 1.5 and 4.5. lf [latex]\bar{X}[/latex] is the mean of a sample taken after the process mean has shifted, then [latex]\bar{X}\sim N\left(m, 0.5^2\right)[/latex]. The probability that [latex]\bar{X}[/latex] plots outside the control limits is equal to [latex]P(\bar{X}<1.5)+P(\bar{X}>4.5)[/latex] (see Figure 10.7). This probability is equal to 1/ ARL = 1/20 = 0.05. Since m > 3,m is closer to 4.5 than to 1.5.We will begin by assuming that the area to the left of 1.5 is negligible and that the area to the right of 4.5 is equal to 0.05. The z-score of 4.5 is then 1.645, so (4.5- m)/0.5 = 1.645. Solving form, we have m = 3.68. We finish by checking our assumption that the area to the left of 1.5 is negligible. With m = 3.68, the z-score for 1.5 is (1.5-3.68) /0.5 = - 4.36. The area to the left of 1.5 is indeed negligible.

Question 10.5: Refer to Example 10.4. An upward shift to what value can be ...

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