Question

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.

Answer 1

(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.