Final 3 Flashcards
A Belt gathered the following defect data for a car manufacturing unit and wanted to assemble it into a Pareto chart. Which defects should be focused upon by the Belt as per the following Pareto chart?
a.
Fuel tank, mirror, brakes, and steering
b.
Fuel tank only
c.
Fuel tank and steering
d.
Steering, fuel tank, and mirror
C. A Pareto chart is based on the 80/20 rule stating that 80% of defects are caused by 20% of the reasons. It plots the data in descending order, starting from highest and going to lowest. The first two defects contribute to around 80% of the problem, so they should be focused upon.
A characteristic was inspected in a sample of 1,000 orders, of which 25 orders were found to be incorrect, but 20 of these orders were reprocessed to correct them. What is the throughput yield (TPY) of this process?
a.
0.9
b.
0.95
c.
0.99
d.
0.975
D. TPY = 1 − (rework + reject)/input. TPY = 1 − 25/1000 = 1 − 0.025 = 0.975.
A company produces two products X1 and X2 in Units A and B. Below is the histogram plots on the weight (kg) data for these products from both units. What does the company understand from these plots?
a.
Product X1 from either Unit A or Unit B must have been rejected by the customer.
b.
Product X2 from Unit B has larger variation than from Unit A.
c.
Product X2 from Unit A has larger variation than from Unit B.
d.
Product X1 from either Unit A or Unit B must have been rejected by the customer, and Product X2 from Unit B has larger variation than from Unit A.
e.
Product X1 from either Unit A or Unit B must have been rejected by the customer, and Product X2 from Unit A has larger variation than from Unit B.
C. As depicted in the multiple histogram plots, Product X2 from Unit A has a larger spread in comparison to X2 from Unit B. Therefore, Product X2 from Unit A has larger variation than from Unit B. Though Product X1 has a different mean value for Unit A and Unit B, and the plots are separate, the production of both might meet the customer requirements if specifications are sufficiently large to include both distribution plots within specifications limits. Therefore, option 1 is not an absolute case.
A linear regression model shows an R-squared (adjusted) of 0.90 and a p-value of 0.002. What can you conclude?
a.
Variability is well explained by the regression model.
b.
Only 10% of the variability is explained by the model.
c.
There is a significant impact of the independent variables on the dependent variable.
d.
There is a significant impact of the independent variables on the dependent variable, and variability is well explained by the regression model.
e.
Only 10% of the variability is explained by the model; however, there is a significant impact of the independent variables on the dependent variable.
D. R-squared is a statistical measure of the proportion or percentage of variation in the dependent variable that can be predicted from the set of independent variables in a regression model. It is also known as the coefficient of determination, or the coefficient of multiple determination for multiple regression. A figure of 100% indicates that the model explains all the variability of the response data around its mean.
A(n) ________ is used to determine whether the averages of normally distributed population samples differ from each other with statistical confidence.
a.
t-test
b.
chi-square test
c.
F-test
d.
DOE
A. A 2-sample t-test is used to compare the means of two independent samples that are normally distributed. A chi-square test is used for one normally distributed sample variance against the target. An F-test is used to compare variances of two normally distributed samples.
Box-Cox transformation can be used to:
a.
Convert the data from non-normal to normal.
b.
Convert the data from unstable to stable.
c.
Create the box plot.
d.
All of the above
A. A Box-Cox transformation is a way to transform non-normal data into a normal shape; however, normal conversion is not always guaranteed.
During a Six Sigma project, a Belt has established a linear regression line with an equation of the form Y = 3.41 + 6.28X, where X is the independent variable and Y is the dependent variable. The value of 3.41 in the above equation is called the _____ and the value 6.28 in the given equation is the ______.
a.
intercept; slope
b.
slope; intercept
c.
x-intercept; slope coefficient
d.
slope; slope coefficient
A. In a simple linear regression model, we get output as an equation of a line with the slope and the Y-intercept of the line. In the equation Y = a + bX, b is the slope and a is the y-intercept.
Factory xyz is operating in two shifts. Quality supervisors of the first and second shifts are both claiming to have the highest quality shift. Yesterday, the second shift observed 3 defects out of 450 units produced, and the first shift found 5 defects out of 450 units. With this data, the management got the following output to draw the conclusion. Should management reward the second shift for the highest quality shift?
a.
Yes
b.
No
B. Since the p-value of the 2-proportion test here is 0.478, which is more than 0.05, we cannot reject the null hypothesis. Also, 95% CI contains zero within the interval boundaries -0.0078 and 0.0167, so, again, we cannot reject the null hypothesis.
If r-square is more, while adjusted r-square is less for a multiple regression model, what can we infer?
a.
We should recheck the calculations.
b.
We have some useless factors in the model.
c.
Both of the values are always equal.
d.
We should consider the r-square value only.
B. When we add lots of useless variables in the regression model, adding a variable will always increase r-square. However, if we add insignificant variables in the model, the adjusted r-square will be reduced.
In a hypothesis test for means, a sample size of 20 has produced a mean of 9.5 mm with a standard deviation of 0.5 mm. The customer specification for the part is 10 mm. At 5% significance level, what should the customer do?
a.
State that the population mean is greater than 10 mm.
b.
Change their specification to 9.5 mm.
c.
Accept the lot.
d.
Reject the lot.
D. t-value = (9.5 − 10)/(0.5 /sq. rt (20)) = -4.472 t-critical = t(0.025, 19) = - 2.093 (from t-table) t-value = -4.472 < t-critical = -2.093. Reject the null hypothesis, meaning the lot mean is not the same as the target or customer specification. The customer should reject the lot.
In FMEA, RPN value is calculated for each potential failure mode. Which elements does this RPN calculation consist of?
a.
Structure, opportunity, and detection
b.
Failure, probability, and impact
c.
Severity, occurrence, and detection
d.
Failure, effect, and cause
C. In FMEA (failure mode and effects analysis), for each failure mode, severity, occurrence, and detection ratings are determined and RPN (risk priority number) is calculated. RPN is the product of these three ratings.
In the following U chart for an inspection process, if the sample number 6 has 5 defects in total, then how many items are inspected in this sample group?
a.
30
b.
25
c.
40
d.
36
B. For the given U chart, the sample number 6 corresponds to 0.2 defects per unit. Since this sample number has 5 defects in total, the total number of inspected items = 5 defects/0.2 defects per unit = 25 units.
What is the cycle time, in seconds, for a process having a throughput of 8,000 units per hour?
a.
0.45
b.
2.22
c.
266.66
d.
133.33
A. Cycle time = average time per unit = 60 × 60/8,000 = 3,600/8,000 = 0.45 sec per unit.
What is the value of the mean, median, range, and standard deviation for the following data? 3, 2, 3, 4, 6, 2, 8, 11
a.
4.875, 4, 9, 3.526
b.
4.875, 4, 8, 3.226
c.
4.875, 3, 9, 3.526
d.
4.875, 3.5, 9, 3.226
D. Mean = average of the data = (3 + 2 + 3 + 4 + 6 + 2 + 8 + 11)/8 = 4.875. Median = middle value when data is in ascending order. Sorted data: 2, 2, 3, 3, 4, 6, 8, 11. Since we have 3 and 4 in the middle at the 4th and 5th places of the sorted data, the median = average of 3 and 4 = 3.5. Range = maximum − minimum = 11 − 2 = 9. Standard deviation = √(Ʃ(mean − individual value)2 )/ (n − 1)), n = 8 and mean = 4.875, so standard deviation = 3.226.
What should we do to minimize the repeatability error if gage with better precision is not available?
a.
Use the nested MSA design.
b.
Use the signal averaging method.
c.
Use the crossed MSA design.
d.
Calibrate the gage.
B. In the case where an instrument with a smaller least count is not available, the signal averaging method can be used to get precise values. A gage with poor precision adds to measurement errors or noise, so it decreases the signal-to-noise ratio. The signal averaging method helps to increase the signal-to-noise ratio by reducing the measurement variation due to noise. For example, if we get repeated measurements of 12, 12.5, 12, 12.5, 12, and 12, then using signal averaging, the value is 12.16 {(12 + 12.5 + 12 + 12.5 + 12 + 12)/6 = 12.16}, which is better than any of the individual readings, as 12.16 falls between the 12 and 12.5 values.
