Refer to Example 10.4. If the san1ple size remains at n = 4, what must the value of the process standard deviation a be to produce an ARL of 10 when the process mean shifts to 3.5?

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Let σ denote the new process standard deviation. The new control limits are [latex]3 \pm 3 \sigma / \sqrt{n}[/latex], or [latex]3 \pm 3 \sigma / 2[/latex]. If the process mean shifts to 3.5, then [latex]\bar{X}\sim N\left(3.5, \sigma^2 / 4\right)[/latex]. The probability that [latex]\bar{X}[/latex] plots outside the control limits is equal to [latex]P(\bar{X}<3-3 \sigma / 2)+P(\bar{X}>3+3 \sigma / 2)[/latex].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 0.1 0. The z-score for 3 + 3σ /2 is then 1.28, so

[latex]\frac{(3+3 \sigma / 2)-3.5}{\sigma / 2}=1.28[/latex]

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 = 2.1 3. The z-score is (2 .1 3 - 3.5)/(0.58/2) = - 4.72. The area to the left of 3-3σ / 2 is indeed negligible.

Question 10.6: Refer to Example 10.4. If the san1ple size remains at n = 4,...

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