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Case: Quality Management—Toyota
Quality Control Analytics at Toyota
As part of the process for improving the quality of their cars,
Toyota engineers have identifi ed a potential improvement 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:
1.9 2.0 1.9 1.8 2.2 1.7 2.0 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 2.0 1.8 1.7 1.9 1.9
2.1 2.0 2.4 1.7 2.2 2.0 1.6 2.0 2.1 2.2
Questions
1 If the specification is such that no washer should be greater
than 2.4 millimeters, assuming that the thick-nesses are
distributed normally, what fraction of the output is expected to
be greater than this thickness?
The average thickness in the sample is 1.9625 and the standard
deviation is .209624. The probability that the thickness is
greater than 2.4 is Z = (2.4 – 1.9625)/.209624 = 2.087068 1 -
NORMSDIST(2.087068) = .018441 fraction defective, so
1.8441 percent of the washers are expected to have a thickness
greater than 2.4.
2 If there are an upper and lower specification, where the upper
thickness limit is 2.4 and the lower thick-ness limit is 1.4, what
fraction of the output is expected to be out of tolerance?
The upper limit is given in a. The lower limit is 1.4 so Z = (1.4
– 1.9625)/.209624 = -2.68337. NORMSDIST(-2.68337) =
.003644 fraction defective, so .3644 percent of the washers are
expected to have a thickness lower than 1.4. The total expected
fraction defective would be .018441 + .003644 = .022085 or
about 2.2085 percent of the washers would be expected to be
out of tolerance.
3 What is the Cpk for the process?
4 What would be the Cpk for the process if it were centered
between the specification limits (assume the process standard
deviation is the same)?
The center of the specification limits is 1.9, which is used for
X-bar in the following:
5 What percentage of output would be expected to be out of
tolerance if the process were centered?
Z = (2.4 – 1.9)/.209624 = 2.385221
Fraction defective would be 2 x (1-NORMSDIST(2.385221)) =
2 x .008534 = .017069, about 1.7 percent.
6 Set up X - and range control charts for the current process.
Assume the operators will take samples of 10 washers at a time.
Observation
Sample
1
2
3
4
5
6
7
8
9
10
X-bar
R
1
1.9
2
1.9
1.8
2.2
1.7
2
1.9
1.7
1.8
1.89
0.5
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
3
2.1
2.4
2.2
2.1
2.1
2
1.8
1.7
1.9
1.9
2.02
0.7
4
2.1
2
2.4
1.7
2.2
2
1.6
2
2.1
2.2
2.03
0.8
Mean:
1.9625
0.65
From Exhibit 10.13, with sample size of 10, A2 = .31, D3 = .22
and D4 = 1.78
The upper control limit for the X-bar chart = 1.9625 + .31 x .65
= 2.164
The lower control limit for the X-bar chart = 1.9625 - .31 x .65
= 1.761
The upper control limit for the Range chart = 1.78 x .65 = 1.157
The lower control limit for the Range chart = .22 x .65 = .143
7 Plot the data on your control charts. Does the cur-rent process
appear to be in control?
With respect to the control limits, the process appears to be in
control, though it should be noted that it is difficult to have
confidence in that conclusion based on just four samples.
Students should also point out that there appears to be a
positive trend, both in process mean and variability. Again, it
would be difficult to say so with much confidence based on just
four samples.
8 If the process could be improved so that the standard
deviation were only about .10 millimeter, what would be the
best that could be expected with the processes relative to
fraction defective?
The best that could be done is for the process to be centered at
1.9, and given a standard deviation of .10, then Z = (2.4-1.9)/.1
= 5. The fraction defective = 2 x (1 –NORMSDIST(5)) =
5.73303E-07 which would only be about 573 defects per billion
washers.
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Template for 13.2 Process Controlp chartNumber of
samples10Sample size300Data
Elissa Torres: Quality Control: Submodel = 4; Problem size @
10 by 1Results# Defects% DefectsTotal Sample
Size3000Sample 1100.0333333333Total Defects91Sample
280.0266666667Percentage defects0.0303333333Sample
390.03Std dev of p-bar0.0099017208Sample 4130.0433333333z
value3Sample 570.0233333333Sample 670.0233333333Upper
Control Limit0.06004Sample 760.02Center Line0.03033Sample
8110.0366666667Lower Control Limit0.00063Sample
9120.04Sample 1080.0266666667Graph informationSample
10.03333333330.0006281710.03033333330.0600384956Sample
20.02666666670.0006281710.03033333330.0600384956Sample
30.030.0006281710.03033333330.0600384956Sample
40.04333333330.0006281710.03033333330.0600384956Sample
50.02333333330.0006281710.03033333330.0600384956Sample
60.02333333330.0006281710.03033333330.0600384956Sample
70.020.0006281710.03033333330.0600384956Sample
80.03666666670.0006281710.03033333330.0600384956Sample
90.040.0006281710.03033333330.0600384956Sample
100.02666666670.0006281710.03033333330.0600384956
p-chart
3.3333333333333333E-2 2.6666666666666668E-2 0.03
4.3333333333333335E-2 2.3333333333333334E-2
2.3333333333333334E-2 0.02 3.6666666666666667E-2
0.04 2.6666666666666668E-2 6.2817104347761069E-4
6.2817104347761069E-4 6.2817104347761069E-4
6.2817104347761069E-4 6.2817104347761069E-4
6.2817104347761069E-4 6.2817104347761069E-4
6.2817104347761069E-4 6.2817104347761069E-4
6.2817104347761069E-4 3.0333333333333334E-2
3.0333333333333334E-2 3.0333333333333334E-2
3.0333333333333334E-2 3.0333333333333334E-2
3.0333333333333334E-2 3.0333333333333334E-2
3.0333333333333334E-2 3.0333333333333334E-2
3.0333333333333334E-2 6.0038495623189053E-2
6.0038495623189053E-2 6.0038495623189053E-2
6.0038495623189053E-2 6.0038495623189053E-2
6.0038495623189053E-2 6.0038495623189053E-2
6.0038495623189053E-2 6.0038495623189053E-2
6.0038495623189053E-2
Sample
Mean
Enter the sample size then enter the number of defects in each
sample.
Template for Example 18.4ForecastingSimple linear
regressionData
Elissa Torres: Forecasting: Submodel = 15; Problem size @ 8
by 2Forecasts and Error AnalysisTracking SignalPeriodDemand
(y)Period(x)ForecastErrorAbsoluteSquaredAbs Pct ErrCum
errorCum Abs ErrMadTrack Signal (Cum error/MAD)2011
Q1300228.3228.3171.6971.695139.330.242011
Q2200280.6280.61-80.6180.616497.620.40-80.6180.6180.61-
1.002011 Q3220332.9332.90-112.90112.9012747.460.51-
193.51193.5188.40-2.192011
Q4530385.1385.10144.90144.9020995.560.27-
48.61338.41102.52-0.472012
Q1520437.4437.4082.6082.606823.020.1633.99421.0198.540.34
2012 Q2420489.6489.60-69.6069.604843.510.17-
35.61490.6193.72-0.382012 Q3400541.9541.89-
141.89141.8920133.400.35-177.50632.50100.60-1.762012
Q4700594.2594.19105.81105.8111195.950.15-
71.69738.31101.25-0.71Total-
0.00810.0088375.842.26Intercept0.02Average-
0.00101.2511046.980.28Slope1.00BiasMADMSEMAPESE121.3
6Forecast9.029Using Linear Regression
MethodCorrelation0.75Coefficient of
determination0.56QuarterActual AmountTrend from
forecastRation of Actual/TrendSeasonal Factor(Av. Of Same Qtr
for 2011 and
2012)20111300228.311.3111.252200280.610.7120.793220332.9
00.6630.704530385.101.3841.2820121520437.401.192420489.6
00.863400541.890.744700594.191.18Forecast Including
TrendsIntercept=176.1, Slope=52.3FITSt=FIT X SeasonalI-2013
FITS9911.29I-2013 FITS10107.87I-2013 FITS11117.71I-2013
FITS121215.36
Regression
228.3 280.60000000000002 332.9 385.1 437.4
489.6 541.9 594.20000000000005 300 200
220 530 520 420 400 700
If this is trend analysis then simply enter the past demands in
the demand column. If this is causal regression then enter the
y,x pairs with y first and enter a new value of x at the bottom in
order to forecast y.
Complete Example 13.2: Process Control Chart Design, located
in Chapter 13 of the textbook using the Excel spreadsheet,
“Process Control Chart Design.” (HAS BEEN COMPLETED,
SEE ATTACHMENT)
Answer questions 1-8 from Case: Quality Management-Toyota,
located at the end of Chapter 13 in the textbook. (HAS BEEN
COMPLETED, SEE ATTACHMENT)
Refer to the Excel spreadsheet, "Computing Trend and Seasonal
Factor," to complete Example 18.4: Computing Trend and
Seasonal Factor From a Linear Regression Line Obtained With
Excel, located in Chapter 18 of the textbook. (HAS BEEN
COMPLETED, SEE ATTACHMENT)
After working through the examples, write a 150-300-word
paragraph explaining the following:
1. Comparison of the simple moving average, weighted moving
average, exponential smoothing, and linear regression analysis
time series models
2. Description of market research, panel consensus, historical
analogy, and Delphi method qualitative forecasting techniques.
While APA format is not required for the body of this
assignment, solid academic writing is expected, and
documentation of sources should be presented using APA
formatting guidelines, which can be found in the APA Style
Guide.

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Case Quality Management—ToyotaQuality Control Analytics at Toyo.docx

  • 1. Case: Quality Management—Toyota Quality Control Analytics at Toyota As part of the process for improving the quality of their cars, Toyota engineers have identifi ed a potential improvement 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: 1.9 2.0 1.9 1.8 2.2 1.7 2.0 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 2.0 1.8 1.7 1.9 1.9 2.1 2.0 2.4 1.7 2.2 2.0 1.6 2.0 2.1 2.2 Questions 1 If the specification is such that no washer should be greater than 2.4 millimeters, assuming that the thick-nesses are distributed normally, what fraction of the output is expected to be greater than this thickness? The average thickness in the sample is 1.9625 and the standard deviation is .209624. The probability that the thickness is greater than 2.4 is Z = (2.4 – 1.9625)/.209624 = 2.087068 1 - NORMSDIST(2.087068) = .018441 fraction defective, so 1.8441 percent of the washers are expected to have a thickness greater than 2.4. 2 If there are an upper and lower specification, where the upper thickness limit is 2.4 and the lower thick-ness limit is 1.4, what fraction of the output is expected to be out of tolerance? The upper limit is given in a. The lower limit is 1.4 so Z = (1.4 – 1.9625)/.209624 = -2.68337. NORMSDIST(-2.68337) =
  • 2. .003644 fraction defective, so .3644 percent of the washers are expected to have a thickness lower than 1.4. The total expected fraction defective would be .018441 + .003644 = .022085 or about 2.2085 percent of the washers would be expected to be out of tolerance. 3 What is the Cpk for the process? 4 What would be the Cpk for the process if it were centered between the specification limits (assume the process standard deviation is the same)? The center of the specification limits is 1.9, which is used for X-bar in the following: 5 What percentage of output would be expected to be out of tolerance if the process were centered? Z = (2.4 – 1.9)/.209624 = 2.385221 Fraction defective would be 2 x (1-NORMSDIST(2.385221)) = 2 x .008534 = .017069, about 1.7 percent. 6 Set up X - and range control charts for the current process. Assume the operators will take samples of 10 washers at a time. Observation Sample 1
  • 5. Mean: 1.9625 0.65 From Exhibit 10.13, with sample size of 10, A2 = .31, D3 = .22 and D4 = 1.78 The upper control limit for the X-bar chart = 1.9625 + .31 x .65 = 2.164 The lower control limit for the X-bar chart = 1.9625 - .31 x .65 = 1.761 The upper control limit for the Range chart = 1.78 x .65 = 1.157 The lower control limit for the Range chart = .22 x .65 = .143 7 Plot the data on your control charts. Does the cur-rent process appear to be in control? With respect to the control limits, the process appears to be in control, though it should be noted that it is difficult to have confidence in that conclusion based on just four samples. Students should also point out that there appears to be a positive trend, both in process mean and variability. Again, it would be difficult to say so with much confidence based on just four samples. 8 If the process could be improved so that the standard deviation were only about .10 millimeter, what would be the best that could be expected with the processes relative to fraction defective? The best that could be done is for the process to be centered at 1.9, and given a standard deviation of .10, then Z = (2.4-1.9)/.1 = 5. The fraction defective = 2 x (1 –NORMSDIST(5)) = 5.73303E-07 which would only be about 573 defects per billion washers. {
  • 8. , 3 min = = þ ý ü î í ì = þ ý ü î í ì - - = s s LTL X X UTL C pk Template for 13.2 Process Controlp chartNumber of samples10Sample size300Data Elissa Torres: Quality Control: Submodel = 4; Problem size @ 10 by 1Results# Defects% DefectsTotal Sample Size3000Sample 1100.0333333333Total Defects91Sample
  • 9. 280.0266666667Percentage defects0.0303333333Sample 390.03Std dev of p-bar0.0099017208Sample 4130.0433333333z value3Sample 570.0233333333Sample 670.0233333333Upper Control Limit0.06004Sample 760.02Center Line0.03033Sample 8110.0366666667Lower Control Limit0.00063Sample 9120.04Sample 1080.0266666667Graph informationSample 10.03333333330.0006281710.03033333330.0600384956Sample 20.02666666670.0006281710.03033333330.0600384956Sample 30.030.0006281710.03033333330.0600384956Sample 40.04333333330.0006281710.03033333330.0600384956Sample 50.02333333330.0006281710.03033333330.0600384956Sample 60.02333333330.0006281710.03033333330.0600384956Sample 70.020.0006281710.03033333330.0600384956Sample 80.03666666670.0006281710.03033333330.0600384956Sample 90.040.0006281710.03033333330.0600384956Sample 100.02666666670.0006281710.03033333330.0600384956 p-chart 3.3333333333333333E-2 2.6666666666666668E-2 0.03 4.3333333333333335E-2 2.3333333333333334E-2 2.3333333333333334E-2 0.02 3.6666666666666667E-2 0.04 2.6666666666666668E-2 6.2817104347761069E-4 6.2817104347761069E-4 6.2817104347761069E-4 6.2817104347761069E-4 6.2817104347761069E-4 6.2817104347761069E-4 6.2817104347761069E-4 6.2817104347761069E-4 6.2817104347761069E-4 6.2817104347761069E-4 3.0333333333333334E-2 3.0333333333333334E-2 3.0333333333333334E-2 3.0333333333333334E-2 3.0333333333333334E-2 3.0333333333333334E-2 3.0333333333333334E-2 3.0333333333333334E-2 3.0333333333333334E-2 3.0333333333333334E-2 6.0038495623189053E-2 6.0038495623189053E-2 6.0038495623189053E-2 6.0038495623189053E-2 6.0038495623189053E-2 6.0038495623189053E-2 6.0038495623189053E-2 6.0038495623189053E-2 6.0038495623189053E-2 6.0038495623189053E-2
  • 10. Sample Mean Enter the sample size then enter the number of defects in each sample. Template for Example 18.4ForecastingSimple linear regressionData Elissa Torres: Forecasting: Submodel = 15; Problem size @ 8 by 2Forecasts and Error AnalysisTracking SignalPeriodDemand (y)Period(x)ForecastErrorAbsoluteSquaredAbs Pct ErrCum errorCum Abs ErrMadTrack Signal (Cum error/MAD)2011 Q1300228.3228.3171.6971.695139.330.242011 Q2200280.6280.61-80.6180.616497.620.40-80.6180.6180.61- 1.002011 Q3220332.9332.90-112.90112.9012747.460.51- 193.51193.5188.40-2.192011 Q4530385.1385.10144.90144.9020995.560.27- 48.61338.41102.52-0.472012 Q1520437.4437.4082.6082.606823.020.1633.99421.0198.540.34 2012 Q2420489.6489.60-69.6069.604843.510.17- 35.61490.6193.72-0.382012 Q3400541.9541.89- 141.89141.8920133.400.35-177.50632.50100.60-1.762012 Q4700594.2594.19105.81105.8111195.950.15- 71.69738.31101.25-0.71Total- 0.00810.0088375.842.26Intercept0.02Average- 0.00101.2511046.980.28Slope1.00BiasMADMSEMAPESE121.3 6Forecast9.029Using Linear Regression MethodCorrelation0.75Coefficient of determination0.56QuarterActual AmountTrend from forecastRation of Actual/TrendSeasonal Factor(Av. Of Same Qtr for 2011 and 2012)20111300228.311.3111.252200280.610.7120.793220332.9 00.6630.704530385.101.3841.2820121520437.401.192420489.6 00.863400541.890.744700594.191.18Forecast Including TrendsIntercept=176.1, Slope=52.3FITSt=FIT X SeasonalI-2013
  • 11. FITS9911.29I-2013 FITS10107.87I-2013 FITS11117.71I-2013 FITS121215.36 Regression 228.3 280.60000000000002 332.9 385.1 437.4 489.6 541.9 594.20000000000005 300 200 220 530 520 420 400 700 If this is trend analysis then simply enter the past demands in the demand column. If this is causal regression then enter the y,x pairs with y first and enter a new value of x at the bottom in order to forecast y. Complete Example 13.2: Process Control Chart Design, located in Chapter 13 of the textbook using the Excel spreadsheet, “Process Control Chart Design.” (HAS BEEN COMPLETED, SEE ATTACHMENT) Answer questions 1-8 from Case: Quality Management-Toyota, located at the end of Chapter 13 in the textbook. (HAS BEEN COMPLETED, SEE ATTACHMENT) Refer to the Excel spreadsheet, "Computing Trend and Seasonal Factor," to complete Example 18.4: Computing Trend and Seasonal Factor From a Linear Regression Line Obtained With Excel, located in Chapter 18 of the textbook. (HAS BEEN COMPLETED, SEE ATTACHMENT) After working through the examples, write a 150-300-word paragraph explaining the following: 1. Comparison of the simple moving average, weighted moving average, exponential smoothing, and linear regression analysis time series models 2. Description of market research, panel consensus, historical analogy, and Delphi method qualitative forecasting techniques. While APA format is not required for the body of this assignment, solid academic writing is expected, and documentation of sources should be presented using APA
  • 12. formatting guidelines, which can be found in the APA Style Guide.