Six Sigma Green Belt Practice Quesitons (ASQ)

b. Gage Technique

a. Creating rational subgroups

Perform a measurement system capability re-analysis.

Beta (probability of making Type II error)

Prototype, Pre-Launch, Production

Kaoru Ishikawa

Creating new products

Measles chart

Walter A. Shewhart

Gantt Chart

Precision

Process Capability Study

ANOVA

Center the process

Accommodating

82.47%

• 1. To find the cumulative frequency for the top three causes in the Pareto chart provided, perform the following calculations:
• Total Count = Add the numerical values for all causes
• Total Count = 44 + 22 + 14 + 10 + 7 = 97
• 2. Next, calculate the percentage for all five causes by dividing the count for each individual cause by the Total Count and multiplying by 100%:
• Operator errors = (44 / 97) * 100% = 45.36%
• Equipment failures = (22 / 97) * 100% = 22.68%
• Design errors = (14 / 97) * 100% = 14.43%
• Material failures = (10 / 97) * 100% = 10.31%
• Environmental issues = (7 / 97) * 100% = 7.22%

• 3. Finally, add the percentages for the three highest causes to determine the cumulative percentage:
• Cumulative Percentage (Top 3) = 45.36% + 22.68% + 14.43%
• Cumulative Percentage (Top 3) = 82.47%
• 

Process

Hold a brainstorming session.

Analysis of Variance method.

A detailed listing of activities required to complete a project.

• 3
• (Total degree of freedom, treatment degree of freedom, and error degrees of freedom)

35. 

• Z = 1.96
• Sigma = 36,000
• E = 12,000

Main effect

b. Roof

• 

c. A

d. Stratified sampling

d. 5s

• 

d. A3 report

b. Relational Matrix

• 

c. 7.81

• In this example, the confidence interval is 95% (alpha = 0.05) in the one tail, right side of the distribution. The degrees of freedom (DF) is found using the following formula:
• DF = (number of rows - 1) * (number of columns - 1)
• In this case, there are two rows and four columns. The DF is as follows:
• DF = (2 - 1) * (4 -1) = 3
• From a chi-squared distribution chart, the critical value for a right-sided, one-tail test with DF of 3 at a 95% (0.05) confidence level is 7.81.

b. Review opportunities

d. Encourage everyone to participate equally

b. The sign for the alternate hypothesis is <

a. Cyclical

c. Non-parallel lines

c. It is a statistically significant factor contributing to the customer complaints.

d. Visual factory

a. State the problem and solution

d. Sensitivity

c. Incorrectly calculated control limits

d. Collect and analyze data

d. The total number of measurements minus two

d. An ongoing monitoring plan is developed.

c. Process decision program charts

c. Positional

a. Replication

b. The activity's earliest and latest times of completion are the same

a. To compare the observed and expected frequencies of test outcomes

d. Special cause

• 

a. Foundation

Control limits are closer together for short-term process evaluation than long-term process evaluation.

a. Manual project management methods are easy to learn

a. Master Black Belt

c. Measurements made by workers and machinery to ensure normal operation

a. Rejecting the null hypothesis when it is true

c. Perform a failure modes and effects analysis

b. B

d. 22.94%


The binomial probability distribution function is stated as follows:

P(r) = n!/(r! * (n-r)!) * p^r*(1 - p)^(n-r)

• n = Size of the given sample
• r = Occurrences of the number of defective units
• p = Probability of defective units

• n = 20
• r = 2
• p = 0.15

The calculation using this formula is as follows:

• P(2) = 20!/(2! * (20-2)!) * (0.15)^2*(1 - 0.15)^(20 - 2)
• P(2) = 2432902008176640000 / 12804747411456000 * 0.0225 * 0.05365
• P(2) = 190 * 0.0225 * 0.05365
• P(2) = 0.22935 = 22.94%

a. The sign for the alternate hypothesis is !=

c. np Chart

d. By finding the sum of the squared deviation of each treatment average form the grand average

b. Benchmarking is a structured method of adopting the best features of successful processes

c. Process A is capable, process B is incapable

d. 54.7%

• 
• 

b. c charts

a. Continuous flow manufacturing

c. Management

b. UCL = 13.410, LCL = 12.350

• 
• 
• 

• 
• 
• 

d. 50%

• P(1 or 3 or 6) = P(1) + P(3) + P(6)
• P(1 or 3 or 6) = 1/6 + 1/6 + 1/6
• P(1 or 3 or 6) = 3/6 = 1/2 or 50$

a. I (Catastrophic category/Class I)

d. Pareto chart

d. Range method 

b. Significant

d. 13.36%

• 
• n = Size of the given sample
• r = Occurrences of the number of defective computer chips
• p = Probability of defective computer chips

• 

c. 5S

a. Median

c. Square root (sum (X - X-bar^2)^2 / (n-1))

c. Activity network diagram

a. Kaizen blitz

c. Confirm the results

a. B, C, A (Training and launch, Implementation, DFSS)

b. Kaizen projects can last weeks or months, whereas Kaizen blitz lasts days

c. Results from focus groups typically have high face validity

a. Employees and managers

d. Risk analysis

b. TPM

c. Residuals

d. Critical quality free

a. 3.5%

Given that there are 20 blue poker chips, 35 green poker chips, and 24 red poker chips, the sample space is 79. To calculate the probability of pulling out a blue chip, followed by a green chip, followed by a red chip, begin with the probability of pulling out a blue chip.

P(blue) = 20 / 79 = 0.253 = 25.3%

Next, calculate the probability of pulling out a green chip given that the entire sample space has been reduced by one.

P(green) = 35 / 78 = 0.449 = 44.9%

Next, calculate the probability of pulling out a red chip given that the entire sample space has now been reduced by two.

P(red) = 24 / 77 = 0.312 = 31.2%

Finally, multiply the previously calculated probabilities to find the probability of pulling out a blue chip, followed by a green chip, followed by a red chip.

• P(blue then green then red) = P(blue) x P(green) x P(red)
• = 0.253 x 0.449 x 0.312
• = 0.035 = 3.5%

a. High and low levels

c. Mode

b. DMAIC

d. Waiting

b. A

a. Machine deterioration

a. 6*standard deviation / (USL - LSL)

a. B

b. Measure

a. 0.0096%

Answer: D. Multivoting

Answer: B. CQFA

The cost, quality, features, and availability (CQFA) model describes the four factors that make up value in the eyes of the customer.

Answer: D. 0.20

The ratio of the precision distribution to the precision total, also known as the precision/tolerance ratio, is 0.20 in this example. This value was calculated by suing the following formula: 

• 6*standard deviation / (USL - LSL)
• 6*2 / (130 - 70)
• 12/60 = 0.20

Answer: C. Cross Functional Teams

Answer: D. As a sample size increases, the distribution of averages calculated from the individual samples will approach a normal distribution.

Answer: B. A high school student is an external customer

Answer: D. The difference in means is calculated from data collected in tandem from 2 samples. 

The paired t test is used to test the difference between 2 sample means. Data collected in pairs from the 2 sample. The difference (d) between the collected data is then calculated for each pair. after the difference has been calculated for each pair, the mean of the differences(d-bar) and the standard deviation of the differences (s) are calculated. The critical value t is then calculated using the following formula:

t = d-bar / (s / square root of number of samples)

This critical value is then compared to a value from a t distribution table to determine if the null hypothesis can be rejected.

Answer: B. -1 to 1

The range of the correlation coefficient r is from -1 to 1. the correlation coefficient is a vital statistic in Six Sigma analysis. A comparison of correlation coefficient for various data pairs can give an indication of which are the stronger relationships and may help point the way for future analysis of causes.

Answer: C.Pp

Pp is a process performance index that indicates a process's ability to produce products within specification over a long period of time. Pp does not take process centering into account. Pp is calculated using the  following formula:

Pp = (USL - LSL) / 6 sigma

Answer: Left wall

Answer: C. A data value from a sample that is the form of a number, which can be used to make an inference about a population

Answer: D. Metric should not focus on a single individual measure.

Answer: A. Poka-yoke

Answer: C. Recurring cycles

Recurring cycles on an average control chart (X-bar chart) can be caused by changes in the environment's temp and humidity. They may also be caused by a tired operator or by the regular rotation of a machine or operator. Recurring cycles can be corrected by ensuring that temp and humidity of the physical environment remains constant. it can also  be corrected by regularly servicing the equipment, rotating the operators and evaluating machine maintenance.

• Answer: C. Scatter diagram 
• Scatter diagrams, also known as correlation charts, display XY-coordinate data points to show a relationship between tow variables. Once data points are plotted, a correlation between the variables can be determine is a center line can easily be lifted between the data points. If the fitting center line is straight, it may indicate a positive or negative relationship between the variables, while a curved lines indicate non-linear relationship. If a center line cannot be easily drawn between the data points, it typically indicates no correlation.

Answer: B. F distribution

Named for its inventor Ronald Fisher, the F distribution tests the null hypothesis that the variance of all population means under study are equal. The F Distribution is similar to the student's t Test, except that it examines the variance of multiple sample means,, rather than the means themselves.

Answer: C. To consistently develop near perfect products and service

Answer: B. BPM

Business process improvement (BPM) focuses on creating value for stakeholders by focusing on overall business processes. Instead of focusing processes within individual vertical units in the organization, BPM looks at the processes overall, and looks at the process flows through each unit.

Answer: C. u charts and c charts

Answer: C. The facilitator creates a drafts of the balance scorecard.

There are typically eight steps performed when developing a balance scorecard: 

• Step1: The scope of the card is defined
• Step2: A facilitator gathers information from senior management for the scorecard.
• Step3: The facilitator presents the gather information at an executive workshop to develop a draft of the scorecard's measures. 
• Step4: The facilitator creates a rough draft of the scorecard and a new report.
• Step5: A second executive workshop is held to refine the drafted scorecard and objectives for the measure are defined.
• Step6: A third workshop is conducted to finalized vision, objectives and measures.
• Step7: A task team develops an implements plan.
• Step8: The balance scorecard is periodically reviewed.

Answer: A. the null hypothesis can be rejected or failed to be rejected.

Answer: A. Theory of Constraints

The theory of Constraints (TOC), also known as constraint management, focuses on improving a process by eliminating bottlenecks. TOC allows management to determine on which part of a process a Six Sigma project should be focused. A good first step is constraints management is to develop a process chart.