Question

Case 13-6

As part of the process for improving the quality of their cars, Toyota engineers have identified a potential improvement to the process that makes a washer that is used in the accelerator assembly. The tolerances on the thickness of the washer are fairly large since the fit can be loose, but if it does happen to get too large, it can cause the accelerator to bind and create a potential problem for the driver. (Note: This part of the case has been fabricated for teaching purposes and none of these data were obtained from Toyota.) Let’s assume that as a first step to improving the process, a sample of 40 washers coming from the machine that produces the washers was taken and the thickness measured in millimeters. The following table has the measurements from the sample:

2.0 1.6 2.0 2.1 2.0 1.8 1.7 1.9 1.7 1.8

1.8 2.2 2.1 2.2 1.9 1.8 2.1 1.6 1.8 1.6

2.1 2.4 2.2 2.1 2.1 1.8 1.7 1.9 1.9 2.1

2.0 2.4 2.0 2.1 1.9 2.3 1.7 2.0 1.9 2.2

Set up X¯¯¯ and range control charts for the current process. Assume the operators will take samples of 10 washers at a time, and that the data table above represents four samples of ten washers each. Use Exhibit 13.7. (Round your intermediate calculations to 4 decimal places and final answer to 3 decimal places.) CLX¯¯¯ =

CLR =

LCLX¯¯ =

LCLR =

UCLX¯¯=

UCLR =

Exhibit 13.7

$exhibit_13.7.jpg$

Answer 1

We have four samples of 10 observations each. This means the constants to use is

A2 = 0.31

D3 = 0.22

D4 = 1.78

The value x_bar is calculated by averaging the observations for each sample. Then the value of x_bar_bar is obtained by finding the average of x_bar values.

The value or R is calculated by finding the range of each same. This means the difference between the larges observation and the smallest observation is the R value for each sample. Then the values of R are averaged to find R_bar.

The calculated values are shown below.

											X_bar	R
Sample 1	2	1.6	2	2.1	2	1.8	1.7	1.9	1.7	1.8	1.86	0.5
Sample 2	1.8	2.2	2.1	2.2	1.9	1.8	2.1	1.6	1.8	1.6	1.91	0.6
Sample 3	2.1	2.4	2.2	2.1	2.1	1.8	1.7	1.9	1.9	2.1	2.03	0.7
Sample 4	2	2.4	2	2.1	1.9	2.3	1.7	2	1.9	2.2	2.05	0.7
											1.9625	0.625
											x_bar_bar	R_bar

Now, that we have the values of x_bar_bar, R_bar, A2, D3 and D4 we can calculate the required control limits.

CLx = 1.9625

CLr = 0.625

LCLx = 1.9625 – 0.31*0.625 = 1.76875

LCLr = 0.22*0.625 = 0.1375

UCLx = 1.9625 + 0.31*0.625 = 2.15625

UCLr = 1.78*0.625 = 1.1125