Question 10.6: Refer to Example 10.4. If the sample size remains at n = 4, ......
Refer to Example 10.4. If the sample size remains at n = 4, what must the value of the process standard deviation 𝜎 be to produce an ARL of 10 when the process mean shifts to 3.5?
Let 𝜎 denote the new process standard deviation. The new control limits are 3±3\sigma ∕ \sqrt{n}, or 3 ± 3𝜎 ∕ 2. If the process mean shifts to 3.5, then \overline{X} ∼ N(3.5, \sigma² ∕ 4). The probability that \overline{X} plots outside the control limits is equal to P(\overline{X} < 3\ − \ 3\sigma ∕ 2) + P(\overline{X} > 3 + 3\sigma ∕ 2). This probability is equal to 1 ∕ ARL = 1 ∕ 10 = 0.10 (see Figure 10.8). The process mean, 3.5, is closer to 3 + 3𝜎 ∕ 2 than to 3 − 3𝜎 ∕ 2. We will assume that the area to the left of 3 − 3𝜎 ∕ 2 is negligible and that the area to the right of 3 + 3𝜎 ∕ 2 is equal to 0.10. The z-score for 3 + 3𝜎 ∕ 2 is then 1.28, so
\frac{(3\ +\ 3\sigma ∕ 2)\ −\ 3.5}{\sigma ∕ 2} = 1.28
Solving for 𝜎, we obtain 𝜎 = 0.58. We finish by checking that the area to the left of 3 − 3𝜎 ∕ 2 is negligible. Substituting 𝜎 = 0.58, we obtain 3 − 3𝜎 ∕ 2 = 2.13. The z-score is (2.13−3.5) ∕ (0.58 ∕ 2) = −4.72. The area to the left of 3− 3𝜎 ∕ 2 is indeed negligible.
