Question

BridgeRock is a major manufacturer of tires in the U.S.. The company had five manufacturing facilities...


BridgeRock is a major manufacturer of tires in the U.S.. The company had five manufacturing facilities where tires were made and another 20 facilities for various components and materials used in tires. Each manufacturing facility produced 10,000 tires every hour. Quality had always been emphasized at BridgeRock, but lately quality was a bigger issue because of recent fatal accidents involving tires made by other manufacturers due to tread separation. All tire manufacturers were under pressure to ensure problems did not arise in the future. However, even with multiple visual quality inspections in place, adhesion flaws and other internal problems were not visible, so manufacturers randomly pulled tires from the production line and cut them apart to look for defects. Given the large number of steps in building a tire, errors tended to accumulate, which possibly results in larger process quality variability. Another issue was related to the settings on various machines. Over time, these settings tended to vary because of wear and tear on the machines. In such a situation, a machine would produce defective product even if the machine had the correct setting.
To detect process variations, the company implemented a statistical process control (SPC) program. At the extruder, the rubber for the AX-123 tires had thickness specifications of 400±10 thousandth of an inch (thou) (these are the specification limits). The quality team at BridgeRock analyzed many samples of output from the extruder and determined that if the extruder settings were accurate, the output produced by the machine had a thickness that was normally distributed with a mean of 400 thou and a standard deviation of 4 thou.
Answer the following questions:
a. If the extruder setting is accurate, what proportion of the rubber extruder will be with specifications?
b. The quality team asked operators to take a sample of 10 sheets of rubber each hour from the extruder and measure the thickness of each sheet. Based on the average thickness of this sample, operators will decide if the extrusion process is in control or not. Given that z=3 for constructing control limits, what upper and lower control limits should they specify to the operators?
c. If a bearing is worn out, the extruder produces a mean thickness of 403 thou when the setting is 400 thou. The standard deviation is still 4 thou. Under this condition, what proportion of defective sheets will the extruder produce? Assuming the control limits in question b, what is the probability that a sample taken from the extruder with the worn bearings will be out of control? On average, how many hours are likely to go by before the worn bearing is detected?
d. Now consider the case where extrusion is a six sigma process. In this case, the extruder output should have a mean of 400 thou and a standard deviation of 1.667 thou. Still using the same 400±10 thou quality specifications, what proportion of the rubber extruded will be within specifications in this case?
e. Assuming the operators will continue to collect samples of 10 sheets each hour to check if the process is in control, what control limits should they set for the case when extrusion is a six sigma process (keep in mind Z=3)?
f. Return to the case of the worn bearings in question c where extrusion produces a mean thickness of 403 thou when the setting is 400 thou. Now the process standard deviation is 1.667 thou. Under this condition, what proportion of defective sheets will the extruder produce? Assuming the control limits calculated in question e, what is the probability that a sample taken from the extruder with the worn bearings will be out of control? On average, how many hours are likely to go by before the worn bearing is detected?
Mini-Case Hint Information
Before analyzing this case, you need to review the key concepts we discussed in the class. Please pay attention to the following items:
a. USL and LSL are relatively fixed and requested by internal or external customers (We use these for defining defects). At the same time, UCL and LCL are control limits which are usually prepared by quality management professionals (for judging process status). These two sets are quite different (and irrelevant).
b. The sigma quality (say, six sigma quality with 3.4 DPMO) essentially refers to the magnitude of process standard deviation (or variance). On the other hand, in the UCL and LCL formulas, the number of sigma (Z value) relates to the corresponding confidence level or type I error. For example, in most cases, 3 sigma is used, which shows the type I error is about 0.26% or 0.3% (it means if you spot one dot is located above UCL or below LCL, 0.3% of the chance your out of control conclusion is wrong). Z value can be chosen based on your required confidence level (in reality, Z=3 is always used for constructing SPC charts everywhere, including in this class), regardless of the sigma quality of the process (say 4 or 5 or 6 sigma) you want to control. You can see here that the same z value means two different things in SPC analysis.
c. The process (or population) standard deviation is different from the sample standard deviation. The sample standard deviation is usually smaller. (Refer to the formula for making the conversion: ). The fundamental difference is the sample standard deviation measures the sample behavior (including n units), and the process standard deviation measures individual product behavior (single unit).
d. When judging if the process produces defects, you are looking at one unit randomly picked product, to see if the quality dimension is above USL or below LSL. The standard deviation is the overall process (population) standard deviation.
e. When deciding if the process is out of control of not, you are looking at one randomly picked sample (containing n=10 units of product in the case), and judging if the sample mean is above UCL or below LCL. UCL and LCL are computed based on X-double bar ± 3* sample standard deviation. The sample standard deviation is calculated in part c (see the conversion formula).
Please use Excel functions provided below for making the calculations in Excel (It is easier and more accurate). You need to present your answers for all the six questions raised in the case here. If you just provide final answers for these questions, please make sure that you explain what these numbers are, and also attach an appendix for Excel worksheet to show your work.
Hereafter I provide hint information for each of the six case questions:
1. Since the product specifications are given, we can easily derive the USL and LSL values for the rubber sheets extruded. Based on the distribution of the extruder output (mean and sigma are given at the end of the case), we can find the probability of non-defective products by using either Standard Normal Distribution Table or Excel (for accuracy purpose, Excel is preferred, see the following formula).
Proportion of output within specifications
= NORMDIST (USL, mean, sigma, 1) – NORMDIST (LSL, mean, sigma, 1)
Please note that Excel function NORMDIST(X, mean, sigma, 1) is corresponding to the following shaded area:

2. In this case, the sample size is 10 sheets. As we know the process standard deviation is 4 thou, we can easily compute the sample standard deviation (Using the formula on page 1). Consequently, the control charts can be constructed (using the general X-bar chart formulas with known process SD on our lecture PPT slide).
3. If a bearing is worn out, the output distribution becomes a normal distribution with a mean of 403 thou and a standard deviation (sigma) of 4 thou. Use the formula in (1) for estimating the proportion of non-defective sheets (remember to update the mean and sigma values). And then calculate the defective rate.
Now, we want to assess the chance a sample taken from the extruder will be out of control, we need to focus on the sample mean (403 thou) and the sample standard deviation (you calculated in step (2)). The probability can be identified by using the following formula:
Proportion of output within specifications
= NORMDIST (UCL, sample mean, sample sigma, 1) – NORMDIST (LCL, sample mean, sample sigma, 1)
If the probability that a sample taken will be within control limits is X, we know the probability of having an out-of-control sample is (1-X). On average, it will take 1/(1-X) samples before an out-of-control sample appears. This is the number of hours which are likely to go by before the worn bearing is detected assuming one sample will be taken per hour.
4. With the six sigma quality, the process output is normally distributed with a mean of 400 and a standard deviation of 1.667. Use the formula in (1) for estimating the probability. (Please choose at least 8 decimal places for the relevant cells in Excel, otherwise you are going to see 100% instead of 99.999999%)
5. Use the assumption that process output standard deviation is 1.667 thou. Then repeat (2) steps. Calculate the new sample sigma. What Z value do you want to use for constructing the control limits? (The answer is already provided in this teaching note) Does Z=6 make sense? Why? What is going to happen if you choose Z=6?
6. Similar to (3), repeat all steps with different mean and sigma. Compare the results and discuss the power of six sigma quality. The formulas are all the same, but the standard deviations and control limits are different.
5

Homework Answers

Answer #1

tire manufacturing firms

5

component manufacturing firms

20

Specification Limits

400+10 thou

410

400 - 10 thou

390

mean

400

thou

std dev

4

thou

n

10

a) If the extruder setting is accurate, what proportion of the rubber extruder will be within specifications?

Proportion of output within specifications = NORMDIST (USL, mean, sigma, 1) – NORMDIST (LSL, mean, sigma, 1)

Proportion of output within specifications

0.99

b) The quality team asked operators to take a sample of 10 sheets of rubber each hour from the extruder and measure the thickness of each sheet. Based on the average thickness of this sample, operators will decide if the extrusion process is in control or not.

Given that z=3 for constructing control limits, what upper and lower control limits should they specify to the operators?

for population std dev formula is : std dev = sqrt ( (sum of (xi-mean)^2)/n),

for sample std dev formula is : std dev = sqrt ( (sum of (xi-mean)^2)/(n-1))

form the above formula, population variance= population std dev ^2 = ((sum of (xi-mean)^2)/n)

multiplying, the above formula of variance by n/(n-1) we get ( (sum of (xi-mean)^2)/n)* (n/(n-1),i.e. (sum of (xi-mean)^2)/(n-1)

(sum of (xi-mean)^2)/(n-1) is the sample variance or std dev sample ^2

hence, sample std dev = population std dev * sqrt (n/(n-1))

hence,

sample std dev

4.22

z

3

x double bar

400

Using the formulas, UCL = x double bar + z* sample std dev , and LCL = x double bar - z* sample std dev

UCL

412.65

LCL

387.35

c)

mean

403

std dev

4

Specification Limits

400+10 thou

410

400 - 10 thou

390

Proportion of output within specifications = NORMDIST (USL, mean, sigma, 1) – NORMDIST (LSL, mean, sigma, 1)

Proportion of output within specifications

0.96

Proportion of defective sheets = (1-proportion within specs)

Proportion of defective sheets

0.04

what is the probability that a sample taken from the extruder with the worn bearings will be out of control?

UCL

412.65

LCL

387.35

sample mean

403.00

sample std dev

4.22

Proportion of output within specifications= NORMDIST (UCL, sample mean, sample sigma, 1) – NORMDIST (LCL, sample mean, sample sigma, 1)

hence, probability that a sample will be in control

0.99

probability that a sample taken from the extruder with the worn bearings will be out of control

0.01

On average, how many hours are likely to go by before the worn bearing is detected?

probability that a sample taken will be within control limits is X,On average, it will take 1/(1-X) samples before an out-of-control sample appears.

No of samples before an out of control samples appear

89.63

This is the number of hours which are likely to go by before the worn bearing is detected assuming one sample will be taken per hour.

but, 10 samples are taken per hour,

hence,

No. of hours likely to go by before the worn bearing is detected

896

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