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A company manufacturing oil seals wants to establish and R control charts on the process. There are 25 preliminary samples of size 4 on the internal diameter of the seal. The summary data (in mm) are as follows:

∑ =1,253.75, ∑=14.08.
(1) Find the control limits that should be used on the and R control charts.
(2) Assume that the 25 preliminary samples plot in control on both charts. Estimate the process mean and standard deviation.

User Sigex
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(1) To find the control limits for the x-bar and R charts, we first need to calculate the center line and the standard deviation of the sample means and ranges.

The center line for the x-bar chart is:
CLx-bar = (∑x) / (n * k) = 1,253.75 / (4 * 25) = 7.92

The center line for the R chart is:
CL(R) = (∑R) / (n * k) = 14.08 / (4 * 25) = 0.14

The standard deviation of the sample means is:
σx-bar = R / d2 = 0.14 / 1.023 = 0.1366

The standard deviation of the sample ranges is:
σR = D4 * R-bar = 2.282 * 0.14 = 0.319

The upper and lower control limits for the x-bar chart are:
UCLx-bar = CLx-bar + A2 * σx-bar = 7.92 + 0.577 * 0.1366 = 7.999
LCLx-bar = CLx-bar - A2 * σx-bar = 7.92 - 0.577 * 0.1366 = 7.840

The upper and lower control limits for the R chart are:
UCLR = D3 * R = 0 * 0.14 = 0
LCLR = D3 * R = 0 * 0.14 = 0

(2) Assuming that the 25 preliminary samples plot in control on both charts, we can estimate the process mean and standard deviation using the x-bar chart. The center line of the x-bar chart is the estimate of the process mean, which is 7.92. The standard deviation of the sample means is the estimate of the process standard deviation, which is 0.1366.
User Kanishk Panwar
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