Question 1172781: What should the control limits be placed in the design of the process sets alpha = 0.01 with the following parameters (assume that the sample size condition has been verified)?
(a) mu.png= 100,sigma.png= 20, and n = 25 cases per batch
(b) mu.png= 2000,sigma.png= 2000, and n = 100 cases per batch
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! To determine the control limits for an X-bar chart, we need to calculate the upper control limit (UCL) and the lower control limit (LCL). We'll use the following formula:
* UCL = μ + Z(α/2) * (σ / √n)
* LCL = μ - Z(α/2) * (σ / √n)
Where:
* μ = population mean
* σ = population standard deviation
* n = sample size
* Z(α/2) = the Z-score corresponding to the desired significance level (alpha).
Given that alpha = 0.01, we need to find the Z-score for alpha/2 = 0.005. This means we are looking for the Z-score that leaves 0.005 in the upper tail of the standard normal distribution. Using a Z-table or calculator, we find that Z(0.005) ≈ 2.576.
**a. μ = 100, σ = 20, n = 25**
1. **Calculate the standard error of the mean (SEM):**
* SEM = σ / √n = 20 / √25 = 20 / 5 = 4
2. **Calculate the control limits:**
* UCL = 100 + 2.576 * 4 = 100 + 10.304 ≈ 110.30
* LCL = 100 - 2.576 * 4 = 100 - 10.304 ≈ 89.70
Therefore, the control limits for (a) are approximately 89.70 and 110.30.
**b. μ = 2000, σ = 2000, n = 100**
1. **Calculate the standard error of the mean (SEM):**
* SEM = σ / √n = 2000 / √100 = 2000 / 10 = 200
2. **Calculate the control limits:**
* UCL = 2000 + 2.576 * 200 = 2000 + 515.2 ≈ 2515.2
* LCL = 2000 - 2.576 * 200 = 2000 - 515.2 ≈ 1484.8
Therefore, the control limits for (b) are approximately 1484.8 and 2515.2.
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