The Six Sigma Handbook, Fifth Edition

Black Belt Overview Questions

1) The average and standard deviation of a key product performance indicator is calculated from the sixteen piece sample as 83.2 and 18.7, respectively. The two-sided 95% upper confidence limit on the mean is approximately:
Feedback:

The t-distribution is used to calculate a confidence interval on the sample mean. A 95% two-sided confidence interval splits the 5% confidence to the two sides (2.5% to each side), so we use the t-value corresponding to the degrees of freedom (n-1) = 15 and an alpha of 0.025, which equals 2.131. The upper confidence is calculated as 83.2+(2.131*18.7/SQRT(16)) = 93.16.

2) The average and standard deviation of a key product performance indicator is calculated from the sixteen piece sample as 83.2 and 18.7, respectively. The one-sided 95% upper confidence limit on the mean is approximately:
Feedback:

The one-sided confidence limit for the mean is: 83.2 + (1.753)*18.7/SQRT(16) = 91.40 Where 1.753 is the critical value of the t-statistic for alpha = 0.05 and 15 degrees of freedom.

3) For the data below, determine the critical value of the test statistic, at a 10% significance level, for a test to determine if the mean equals 60.

45             70

52             78

61             82
Feedback:

A two-sided test is used to test whether a computed sample average equals a given value. The t-statistic for n-1 = 5 degrees of freedom and alpha = 0.10 / 2 = 0.05 (two-sided) is 2.015.

4) What distribution is used when testing to determine if the variance of two samples are equal?
5) A sample of 15 items has a mean of 26.4 and a sample standard deviation of 5.1. A 95% confidence interval on the mean is:
Feedback:

A t-value of 2.145 is used for the two-sided 95% confidence interval with 14 degrees of freedom.

6) A sample of 100 items has a mean of 26.4 and a sample standard deviation of 5.1. A 95% confidence interval on the mean is:
Feedback:

A t-value of 1.98, or a z-value of 1.96 given the relatively large sample size, is used for the two-sided 95% confidence interval.

7) The following statistics were calculated from two samples.

Sample #1 Sample #2
Average 26.4 21.2
Sample Sigma 5.1 4.7
Sample Size 15 9

The computed value of the test statistic for comparing if the two samples are likely to be from populations with the same mean, at a 5% significance assuming normality and equal variances, is:
Feedback:

The t-statistic for testing the equality of the means is calculated as: = (26.4-21.2)/(4.96*SQRT((1/15)+(1/9))), where 4.96 is the pooled standard deviation.

8) The following statistics were calculated from two samples.

Sample #1 Sample #2
Average 26.4 21.2
Sample Sigma 5.1 4.7
Sample Size 15 9

The critical value of the test statistic for comparing if the two samples are likely to be from populations with the same mean, at a 5% significance assuming normality and equal variances, is:
Feedback:

The value of the t-statistic for 22 (=15+9-2) degrees of freedom at alpha/2 = .025 is 2.074

9) The following statistics were calculated from two samples.

Sample #1 Sample #2
Average 26.4 21.2
Sample Sigma 5.1 4.7
Sample Size 15 9

The approximate probability that the two samples are from populations with the same mean, assuming normality and equal variances, is:
Feedback:

The t-statistic for testing the equality of the means is calculated as: (26.4-21.2)/(4.96*SQRT((1/15)+(1/9))) = 2.49, where 4.96 is the pooled standard deviation. This value of t (2.49) at 22 degrees of freedom lies between the alpha values of .01 and .025 in a t-table. Excel's TDIST function [=TDIST(2.49, 22, 2)] calculates the alpha value as 2% for a two-sided test.

10) The following statistics were calculated from two samples.

Sample #1 Sample #2
Average 26.4 21.2
Sample Sigma 5.1 4.7
Sample Size 15 9

The calculated value of the test statistic for comparing if the two samples are likely to be from populations with equal variance, at a 5% significance, is:
Feedback:

The F-statistic for testing the equality of the variances is calculated as: (5.1^2)/(4.7^2) = 1.17

11) If a statistical test has a 3% significance, then:

12) When testing the equality of means between two samples, the probability of rejecting a false null hypothesis is improved when:

13) Given the data shown below for two inspectors measuring five pieces three times each using the same measurement instrument, what is the approximate total R&R as a percent of process variation if the process sigma value is estimated as 0.05?

Sample Appraiser 1 Appraiser 2
1 2.165 2.16
2.16 2.185
2.155 2.175
2 2.16 2.175
2.155 2.165
2.155 2.175
3 2.165 2.16
2.145 2.165
2.155 2.16
4 2.125 2.135
2.115 2.125
2.115 2.13
5 2.18 2.215
2.185 2.195
2.185 2.2

Feedback:

The R&R can be conveniently calculated using the Green Belt XL software, such as shown below.

14) Given the data shown below for two inspectors measuring five pieces three times each using the same measurement instrument, a control chart of the Repeatability data demonstrates which of the following?

Sample Appraiser 1 Appraiser 2
1 2.165 2.16
2.16 2.185
2.155 2.175
2 2.16 2.175
2.155 2.165
2.155 2.175
3 2.165 2.16
2.145 2.165
2.155 2.16
4 2.125 2.135
2.115 2.125
2.115 2.13
5 2.18 2.215
2.185 2.195
2.185 2.2

Feedback:

The control chart for Repeatability can be conveniently calculated using the Green Belt XL software, such as shown below. The chart is out of control, with four out of ten of the points beyond the control limits. This does not indicate anything about the statistical stability of the repeatability statistic, but rather demonstrates that the repeatability is nearly small enough relative to the process variation.

15) Pass / Fail data uses which scale of measurement?
Feedback:

A Nominal measurement scale indicates only the presence or absence of an attribute.

16) When attendees are asked to rate the effectiveness of training using the scale 1 to 5, with 5 being the highest level of effectiveness, they are using which scale of measurement?
Feedback:

In the Ordinal measurement scale, the items can be rank ordered so that an item can be said to have more or less of the attribute than the other items.

17) Elapsed Time data uses which scale of measurement?
Feedback:

In the Ratio measurement scale, zero represents the absence of the attribute. The data can be added, subtracted, multiplied and divided with meaning.

18) Elapsed Time data uses which scale of measurement?
Feedback:

In the Interval measurement scale, the graduations of the scale are equal, but zero does not indicate the absence of the attribute. The data can also be added, subtracted, multiplied and divided with meaning.

19) Gage Linearity Analysis detects:
20) Gage R&R Studies are best done using:
21) Gage R&R Studies may be used to:
22) When using an X-Bar / Range chart for evaluating Gage Repeatability:
23) A criteria for acceptance of Gage R & R is:
24) To demonstrate compliance to a requirement that the Cpk index be at least 1.33 based on a ±3s spread, the quality engineer computed Cpk from fifty units selected at random from the production lot before it was delivered to the customer. Which of the following statements describes this approach to capability analysis?
Feedback:

Few methods have suffered from more abuse in recent years than capability analysis. Many large companies have mandated SPC, complete with capability analysis, of all key suppliers. The contractual requirements have been enforced at times by auditors who apparently don’t have a good grasp of the underlying assumptions involved. The subtleties of capability analysis should be understood. More recently, the Cpm index, sometimes called the Taguchi capability index, has been enjoying popularity. Like Cpk, the Cpm index accounts for both the process central tendency and its spread. However, Cpm works with a target value, while Cpk is based on the nearest tolerance limit. Cpm also has certain mathematical advantages over Cpk, such as a known probability distribution.

25) Which of the following are NOT distribution properties.
Feedback:

There are three basic properties of a distribution: location, spread, and shape. The location refers to the typical value of the distribution, such as the mean. The spread of the distribution is the amount by which smaller values differ from larger ones. The standard deviation and variance are measures of distribution spread. The shape of a distribution is its pattern — peakedness, symmetry, etc. A given phenomenon may have any one of a number of distribution shapes, e.g., the distribution may be bell-shaped, rectangular-shaped, etc.

26) The following measurements for a sample with Dimension X are representative of a process known to be in statistical control.

       42, 52, 64, 45, 53, 56, 70, 57, 49, 62

       Which of the following best approximates the upper and lower control limits of the process capability? (Use generally accepted sigma limits for the United States.)
Feedback:

When n = 10 the process capability of 6 is sometimes approximated as 2R. This set of data yields the following
X-bar = 55
R = 70 – 42 = 28
Thus the limits are
Lower Limit = 55 – 28 = 27
Upper Limit = 55 + 28 = 83
The sample standard deviation for this data set is 8.678. Using the mean plus and minus 3 standard deviations gives
Lower Limit = 55 - 3(8.678) = 28.97
Upper Limit = 55 + 3(8.768)= 81.04
Using the range to estimate standard deviation produces considerable error when the sample size is greater than 5, as evidenced by the above calculations. The simple range estimation technique is no longer needed inasmuch as inexpensive calculators are able to compute the sample standard deviation.

27) Which one of the following best describes machine capability?
Feedback:

Statistically, analysis of the capability of a new machine is as simple as using a slightly modified process capability index. However, when new machinery is involved, a number of other issues arise. When looking at new machinery it is not possible to examine all of the potential sources of variation, such as different operators, tool changes, fixtures, materials, etc. Other factors tend to work to increase the variability of new equipment. Kane (1989) lists several factors that work to affect the variability of new machines.

FACTORS INCREASING FACTORS REDUCING
VARIABLITY VARABILITY
AT MACHINE TRYOUT AT MACHINE TRYOUT
Machine break-in Low production volume
Special gauging Specially trained operators
Nonstandard machine hookup Strict environmental controls
(e.g., power, coolants)  


Usually, but not always, the combined affect of these factors is to decrease variability during the machine capability analysis. Statistically, the usual approach is to use stricter capability requirements for new equipment than for standard process capability analysis. In addition, the purchaser must use a modified approach to estimate the capability of the new machine prior to the purchase. Kane recommends the following approach to a capability analysis, which he calls a machine tryout.

Site— It is important to conduct the tryout at the machine builder’s site. This gives the opportunity to correct any deficiencies quickly. In many cases any lack of capability can be corrected by “fine-tuning” machine parameters. This process is time consuming and often requires a variety of engineering expertise. The required personnel and facilities are most accessible at the machine builder’s site.

Parts and material— As the process flow diagrams indicate, machines are often part of a sequential process of manufacturing a finished product. Machines that operate on various stages of semifinished products require parts or materials in the required semifinished stage of processing. For a new process all machinery is purchased at the same time, so it can be difficult to obtain the required parts. In many cases, these parts must be specially manufactured at a high cost. The quality of these special parts can be much better or worse than the normal production parts. To often, too few parts are available at a tryout to adequately evaluate a new machine.

Cycle time— The rate at which a machine produces parts is often related to the variability of the output. Thus, it is important to perform the tryout at the same cycle time used in normal production. This is a common oversight.

Tooling— Machine output is directly influenced by the type of tools used. It is important to use the tooling expected to be used in normal production. If temporary nonstandard tooling must be used, the machine cycle time and capability criteria must not change.

Machine adjustments— During normal production, machines are sometimes adjusted to attempt to change the mean of the process output. A machine tryout should make as few adjustments as possible. For some machines, a few adjustments are a normal part of the process. However, there is a tendency to want to make many adjustments during a tryout. This should be avoided if at all possible. When unavoidable adjustments and changes are made, they should be recorded in a detailed chronological log. The log should also include information on cycle times, feeds and speeds, tools, coolants, and setup notes. All modifications, even those believed to be of no consequence, should be recorded.

Time— The time required to perform a tryout to obtain the required capability is generally underestimated. In many cases the fine-tuning process requires several months. Sufficient time must be included in the delivery schedule.

Rerun of tryout— Once the machine is installed at the purchaser’s plant, the tryout should be repeated. This will detect any damage incurred in shipping, handling, and installation, as well as establish a baseline for future comparisons. Analysis of the data should include the following steps:

    1. Measurement error analysis. The variability of the measurement system should be removed from the data to determine the true machine variability.

    2. Run chart. A run chart in the production order should be prepared to evaluate obvious machine problems. It is good practice to number the parts in production order and to save the parts used in the final machine tryout so that they can be reinspected later if necessary.

    3. X-bar and R chart. An X-bar and R chart should be prepared to evaluate the stability of the machine. The detailed part-by-part log should be kept to allow the correlation of the X-bar and R data with individual sources of variation.

    4. Histograms. Histograms should be prepared to examine the distribution of the data for each of the various sources of variation.
28) When an initial study is made of a repetitive industrial process for the purpose of setting up a Shewhart control chart, information on the following process characteristic is sought:
Feedback:

Shewhart control chart are used to control process capability.

29) To produce adequate quality, the process capability target should be:
Feedback:

To make a quality product, the target capability should always be tighter than the specifications.

Statistically, analysis of the capability of a new machine is as simple as using a slightly modified process capability index. However, when new machinery is involved, a number of other issues arise. When looking at new machinery it is not possible to examine all of the potential sources of variation, such as different operators, tool changes, fixtures, materials, etc. Other factors tend to work to increase the variability of new equipment. Kane (1989) lists several factors that work to affect the variability of new machines.

FACTORS INCREASING FACTORS REDUCING
VARIABLITY VARABILITY
AT MACHINE TRYOUT AT MACHINE TRYOUT
Machine break-in Low production volume
Special gauging Specially trained operators
Nonstandard machine hookup Strict environmental controls
(e.g., power, coolants)  


Usually, but not always, the combined affect of these factors is to decrease variability during the machine capability analysis. Statistically, the usual approach is to use stricter capability requirements for new equipment than for standard process capability analysis. In addition, the purchaser must use a modified approach to estimate the capability of the new machine prior to the purchase. Kane recommends the following approach to a capability analysis, which he calls a machine tryout.

Site — It is important to conduct the tryout at the machine builder’s site. This gives the opportunity to correct any deficiencies quickly. In many cases any lack of capability can be corrected by “fine-tuning” machine parameters. This process is time consuming and often requires a variety of engineering expertise. The required personnel and facilities are most accessible at the machine builder’s site.

Parts and material — As the process flow diagrams indicate, machines are often part of a sequential process of manufacturing a finished product. Machines that operate on various stages of semifinished products require parts or materials in the required semifinished stage of processing. For a new process all machinery is purchased at the same time, so it can be difficult to obtain the required parts. In many cases, these parts must be specially manufactured at a high cost. The quality of these special parts can be much better or worse than the normal production parts. To often, too few parts are available at a tryout to adequately evaluate a new machine.

Cycle time — The rate at which a machine produces parts is often related to the variability of the output. Thus, it is important to perform the tryout at the same cycle time used in normal production. This is a common oversight.

Tooling — Machine output is directly influenced by the type of tools used. It is important to use the tooling expected to be used in normal production. If temporary nonstandard tooling must be used, the machine cycle time and capability criteria must not change.

Machine adjustments — During normal production, machines are sometimes adjusted to attempt to change the mean of the process output. A machine tryout should make as few adjustments as possible. For some machines, a few adjustments are a normal part of the process. However, there is a tendency to want to make many adjustments during a tryout. This should be avoided if at all possible. When unavoidable adjustments and changes are made, they should be recorded in a detailed chronological log. The log should also include information on cycle times, feeds and speeds, tools, coolants, and setup notes. All modifications, even those believed to be of no consequence, should be recorded.

Time — The time required to perform a tryout to obtain the required capability is generally underestimated. In many cases the fine-tuning process requires several months. Sufficient time must be included in the delivery schedule.

Rerun of tryout — Once the machine is installed at the purchaser’s plant, the tryout should be repeated. This will detect any damage incurred in shipping, handling, and installation, as well as establish a baseline for future comparisons. Analysis of the data should include the following steps:

    1. Measurement error analysis. The variability of the measurement system should be removed from the data to determine the true machine variability.

    2. Run chart. A run chart in the production order should be prepared to evaluate obvious machine problems. It is good practice to number the parts in production order and to save the parts used in the final machine tryout so that they can be reinspected later if necessary.

    3. X-bar and R chart. An X-bar and R chart should be prepared to evaluate the stability of the machine. The detailed part-by-part log should be kept to allow the correlation of the X-bar and R data with individual sources of variation.

    4. Histograms. Histograms should be prepared to examine the distribution of the data for each of the various sources of variation.
30) Machine capability studies on four machines yielded the following information:
Machine Average ( ) Capability (6s)
#1 1.495 0.004”
#2 1.502 0.006”
#3 1.500 0.012”
#4 1.498 0.012”
The tolerance on the particular dimension is 1.500± .005”. If the average value can be readily shifted by adjustment to the machine, then the best machine to use is:
Feedback:

Since we can, by assumption, shift the average value, the Average column can be ignored. Thus the best Choice is the machine with the least dispersion, which is machine #1.

31) Machine capability studies on four machines yielded the following information:
Machine Average ( ) Capability (6s)
#1 1.495 0.004”
#2 1.502 0.006”
#3 1.500 0.012”
#4 1.498 0.012”
The tolerance on the particular dimension is 1.500± .005”. If the average value can be readily shifted by adjustment to the machine, then the best machine to use is:
Feedback:

We seek the machine with the smallest proportion discrepant (highest Process Capability Cpk). Assuming a normal distribution for each machine:

      Cpk = MIN(ZL/3 , ZU/3)

      ZL =(Average-LSL)/sigmax

      ZU =(USL-Average)/sigmax

      For example, Machine 1 Cpk = 0 since:

      zu= (1.505-1.495) / (.004/6) = 15 ; zl = (1.495-1.495) / (.004/6) = 0;

      The z and capability statistics for each machine are:

Machine Z low spec Z high spec Cpk
1 0.0 +15.0 0.0
2 -7.0 +3.0 1.0
3 -2.5 +2.5 0.83
4 -1.5 +3.5 0.5


Machine #2 has the highest Cpk value so is the best choice.
32) Using the range method calculate the process capability standard deviation to nearest 0.0001 of the following:

8 A.M. 9 A.M. 10 A. M. 11 A.M.
0.001 0.003 0.001 0.005
–0.001 0.004 –0.002 0.006
0.003 0.003 –0.003 0.005
0.002 0.004 0.002 0.005
0.001 0.002 0.000 0.006
33) When using Process Capability estimates:
34) If we replace the statistical control limits with limits defined by customer requirements:
Feedback:

Using customer specifications in place of statistical control limits results in process tampering, which increases process variation.

35) All of the following statements regarding process histograms are true EXCEPT:
36) If a probability plot shows that a set of data does NOT fit a Normal distribution, then:
Feedback:

The probability plot will NOT indicate if the process is in control. The Normal distribution is not required for a process to be in control.

37) A customer requires that inventory replenishments be received within 10 days of due date, but no earlier than the due date. The number of days from due date to shipment receipt is in statistical control, with an average of 7 days and a process standard deviation of 0.75 days. Assuming a Normal distribution, the process capability Cp is approximately:
Feedback:

Cp = (USL-LSL)/(6*sigmax)

Cp = (10-0)/(6*.75) = 2.22

38) A customer requires that inventory replenishments be received within 10 days of due date, but no earlier than the due date. The number of days from due date to shipment receipt is in statistical control, with an average of 7 days and a process standard deviation of 0.75 days. Assuming a Normal distribution, the process capability Cpk is approximately:
Feedback:

Cpk = MIN(ZL/3 , ZU/3)

ZL =(Average-LSL)/sigmax = (7-0)/.75 = 9.33

ZU =(USL-Average)/sigmax = (10-7)/.75 = 4

Cpk = 4/3 = 1.33

39) The cycle time of a process (the time taken from beginning to completing the process) is in statistical control, with an average of 7 days and a process standard deviation of 0.75 days. Assuming a Normal distribution, if the customer requires the process be completed in ten days, then process capability Cp is approximately:
Feedback:

In this problem, there is no lower specification (LSL), so Cp = (USL-LSL)/(6*sigmax) cannot be calculated.

40) If Cp is larger than Cpk, then
41) When calculating the process capability indices Cp and Cpk, use
42) Before calculating the process capability indices Cp and Cpk:
43) Based on subgroups of n=5 the grand average is 105 and the average range is 10. The lower control limit for the X-bar chart is
Feedback:

The control limits for an X bar chart are

UCL = X-doublebar + A2R-bar

LCL = X-doublebar - A2R-bar

A2 is a factor based on the sample size. For a sample size of 5, A2 is 0.577. The lower control limit is

UCL = 105 - 0.577(10) = 99.23

44) Based on subgroups of n = 5 the grand average is 105 and the average range is 10. The lower control limit for the range chart is:
Feedback:

Control limits for both the averages and the ranges charts are computed such that it is highly unlikely that a subgroup average or range from a stable process would fall outside of the limits. All control limits are set at plus and minus three standard deviations from the center line of the chart. Thus, the control limits for subgroup averages are plus and minus three standard deviations of the mean from the grand average; the control limits for the subgroup ranges are plus and minus three standard deviations of the range from the average range. These control limits are quite robust with respect to non-normality in the process distribution. See the Appendix Table of control chart constants for subgroups of 25 or less.

The control limits for a range charts are

LCL = D3R-bar

UCL = D4R-bar

D3 and D4 are constants based on the sample size. For a sample of size 5, D3 is 0. The lower control limit is

LCL = 0(10) = 0

45) The inspector samples 100 can ends per subgroup and records the number of can ends with coating defects. The average number of defective ends is 25. Compute the lower control limit for the np chart.
Feedback:

np charts are statistical tools used to evaluate the count of defectives, or count of items non-conforming, produced by a process. Np charts can be applied to any variable where the appropriate performance measure is a unit count and the subgroup size is held constant. Note that wherever an np chart can be used, a p chart can be used too.

Like all control charts, np charts consist of three guidelines: center line, a lower control limit, and an upper control limit. The center line is the average count of defectives-per-subgroup and the two control limits are set at plus and minus three standard deviations. If the process is in statistical control, then virtually all subgroup counts will be between the control limits, and they will fluctuate randomly about the center line.

The control charts for an np chart are:



For this example:

n = 100

p = 25/100 = 0.25

46) The inspector samples 100 can ends per subgroup and records the number of can ends with coating defects. The average number of defective ends is 25. Compute the lower control limit for the p chart.
Feedback:

P charts are statistical tools used to evaluate the proportion defective, or proportion non-conforming, produced by a process. P charts can be applied to any variable where the appropriate performance measure is a unit count. P charts answer the question “Has a special cause of variation caused the central tendency of this process to produce an abnormal-ly large or small number of defective units over the time period observed?”

Like all control charts, p charts consist of three guidelines: center line, a lower control limit, and an upper control limit. The center line is the average proportion defective and the two control limits are set at plus and minus three standard deviations. If the process is in statistical control, then virtually all proportions should be between the control limits and they should fluctuate randomly about the center line.

47) In a typical week a large factory has 16 accidents. What is the upper control limit for the number of accidents per week?
Feedback:

C charts are statistical tools used to evaluate the number of occurrences-per-unit produced by a process. C charts can be applied to any variable where the appropriate performance measure is a count of how often a particular event occurs and samples of constant size are used. C charts answer the question “Has a special cause of variation caused the central tendency of this process to produce an abnormally large or small number of occurrences over the time period observed?” Note that, unlike p or np charts, c charts do not involve counting physical items. Rather, they involve counting of events. For example, when using an np chart one would count bruised peaches. When using a c chart they would count the bruises. Control limit equations for c charts Like all control charts, c charts consist of three guidelines: center line, a lower control limit, and an upper control limit. The center line is the average number of occurrences-per-unit and the two control limits are set at plus and minus three standard deviations. If the process is in statistical control then virtually all subgroup occurrences-per-unit should be between the control limits and they should fluctuate randomly about the center line.

48) A stable process has a mean of 53 and an average range of 4, using subgroups of n = 4. The requirements are 50±8. Compute the process standard deviation.
Feedback:

The process standard deviation is



where d2 is a factor dependent on the subgroup size. For n = 4, d2 = 2.059.

49) An X-bar and R chart was prepared for an operation using twenty samples with five pieces in each sample. X-bar was found to be 33.6 and R-bar was 6.20. During production a sample of five was taken and the pieces measured 36, 43, 37, 25, and 38. At the time this sample was taken:
Feedback:

The control limits for the range chart are:

LCLR = D3Rbar = (0)(6.2) = 0

UCLR = D4Rbar = (2.114)(6.2) = 13.1

The control limits for the X-bar chart are:

Lower Limit:

Xdoublebar - A2Rbar = 33.6 - .577 x 6.2 = 30.0

Upper Limit:

Xdoublebar + A2Rbar = 33.6 + .577 x 6.2 = 37.2

For the 5 piece sample:

Xbar = 35.8

R = 18

Thus X-bar is in control, the Range is not.

50) A process is checked by inspection of random samples of four shafts after a polishing operation, and X-bar and R charts are maintained. A person making a spot check picks out two shafts, measures them accurately, and plots the value of each on the X-bar chart. Both points fall just outside the control limits. He advises the department foreman to stop the process. This decision indicates that
Feedback:

The person plotted individual readings on a chart for averages. He is not using the chart correctly.

51) In which of the following examples would it is useful to establish statistical correlation?
52) Linear Regression uses which of the following statistical methods to estimate correlation?
53) Weak correlation implies:
54) In estimating correlation and the parameters of a linear regression model, we should vary the independent variable over a sufficient range so that:
55) If the cycle time of a process is predicted by Cycle Time = 5.25 * (Number of items) + 4.3, with a correlation of 0.8, then it is fair to say:
Feedback:

Statistical correlation, as evidenced by the reasonable correlation index of 0.8, does not prove an association between the independent variable (number of items) and the dependent variable (cycle time). It is possible the independent variable is correlated with another variable (under the conditions of this data sampling) that is the true source of influence to the variation in cycle time. Furthermore, the absence of other independent variables in the expression should not imply that no other variables influence cycle time.

56) Larger correlation implies:
57) The flaring of the Confidence intervals about the Regression Line is evidence of:
58) Multiple Regression techniques may be used to:
59) Assuming the coefficients in the expression Y = ß0+ ß1X1 + ß2X2 + ß3X3 are non-zero:
60) In the expression Cycle Time (in minutes) = 3 + 1.4*(Number of Orders) - 2.1*(Number of clerks) - .034*(Process Distance):
Feedback:

A given coefficient indicates the change in the response for each unit change in its factor.

61) In the Interaction Plot below, it is evident that:

62) In the Interaction Plot below, it is evident that:

Feedback:

Each of the statements is true.

63) Considering only the Interaction Plot below, it is evident that:

Feedback:

When Furnace Temperature is low (the 1223 line), then the variation in Part Density is smallest. Choice c is incorrect since we have not defined "best conditions".

64) A model of the form Y = ß0+ ß1X1 + ß2X2+ ß12X12:
65) A model of the form Y = ß0+ ß1X1 + ß11X12+ ß2X2+ ß12X12:
Feedback:

Note the X12 term.

66) To estimate the regression model Y = ß0+ ß1X1 + ß2X2+ ß12X12:
Feedback:

Choice b is incorrect as there are no higher order terms. Choice c may not be correct if there are no extra runs (degrees of freedom) to calculate error, as when only four trials are used.

67) If an ANOVA analysis does not provide results of the F-test, then:
Feedback:

F-tests cannot be performed on the regression when there are insufficient degrees of freedom (unique trials or experimental conditions) to estimate error. Often, factors can be removed from the regression to free up trials for error estimation.

68) Some useful purposes for model transformations include:
69) To detect non-constant variance in the response:
70) When transforming data to stabilize the variance:
71) Given the following data, construct a 95% confidence interval for the mean.

45 70
52 78
61 82
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The sample average of the 6 values is 64.67, and the sample standard deviation is 14.61. Assuming the variance is unknown, a 95% confidence interval for the mean is calculated using a t-value of 2.57 (from the t-table with alpha/2=.025 and 5 degrees of freedom) as:

         Upper Limit = 64.67 + (2.57*14.61/SQRT(6)) = 80.0

        Lower Limit = 64.67 - (2.57*14.61/SQRT(6)) = 49.34

72) A random sample of 50 items yielded 40 successes. Determine a 95% confidence interval for the proportion of successes.
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n = 50, p = 40/50 = 0.8, and z (.025) = 1.96

The confidence limits are calculated as:

        Upper = 0.8 + 1.96 * SQRT[ 0.8 * (1 - 0.8) / 50 ] = 0.91

        Lower = 0.8 - 1.96 * SQRT[ 0.8 * (1 - 0.8) / 50 ] = 0.69

73) For the data below, determine the critical value of the test statistic, at a 10% significance level, for a test to determine if the mean is greater than 60.
45             70
52             78
61             82
Feedback:

The mean of the sample is 64.67, and the sample standard deviation is 14.61. A one-sided test is used to test whether a computed sample average is either greater than or less than a given value. The t-statistic for n-1 = 5 degrees of freedom and alpha = 0.10 (one-sided) is 1.476.

74) For the data below compute the appropriate test statistic, for a test to determine if the mean is greater than 60.
45             70
52             78
61             82
Feedback:

The sample mean is calculated as 64.67, and the sample standard deviation is calculated as 14.61. The t-statistic is calculated as

t = (64.67 - 60) / (14.61 / SQRT(6) ) = 0.783

75) The "customer" in a design FMEA is:
76) In development of a Design FMEA, it is important to consider:
77) In defining failure modes in a Design FMEA:
Feedback:

Potential failure modes that occur under limited conditions should be included, as well as those resulting from lower-level subsystem failures.

78) Of the three types of design controls:

1. Those that prevent the cause or failure mode from occurring, or reduce their occurrence.

2. Those that detect the cause and leads to corrective action.

3. Those that detect the failure mode.

Which is the most desirable control?
79) When evaluating severity in a FMEA:
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The Severity relates to the effect of a failure mode. A given function may have several failure modes, each with one or more effects.

80) A high severity in a FMEA is applied to:
Feedback:

The severity is applied to the effect of the process step's failure mode. Any given step may have more than one failure mode and effects, each with different severity ratings.

81) A process characteristic with an acceptable capability index:
Feedback:

An acceptable capability level (1.33 or higher) should be associated with a low occurrence ranking, per the AIAG recommendations.

82) If the severity of a process failure mode and effect is such that the customer is unlikely to notice the error; its process capability marginal so that errors occur from time to time; and there is an extremely low chance that the error will reach the customer, then the RPN is reasonably estimated as:
Feedback:

The Severity can be estimated as 1; the Occurrence estimated as 4; the Detect-ability as a 2, resulting in an RPN of 8.

83) Screening Designs are often used:
84) If the fitted model is a good approximation to the data, then:
85) In a two level experiment, when Factor A is run at its high level, Factor B at its low level, and Factor C at its high level:
Feedback:

The interaction AB will be at its low level, AC at its high level, and ABC at its low level.

86) If we alias a factor F with the ABCD interaction, then:
87) When we fold an entire design:
88) The power of an experiment refers to:
89) Which of the following is the simplest design for generating a response surface:
Feedback:

At least three levels are needed for each factor. Response surfaces are usually considered to have curved surfaces.

90) Objectives of response surface designs include:
91) What is the recommended order of implementing the following activities in a Sequential Response Surface technique?

1. Use a central composite design to define the second order effects.

2. A screening design to exclude insignificant main effects.

3. Generate a first order model including main factor interactions.
92) Consider the following first order model and factor levels:

Y = 125 + 46.6x1 + 12.5x2 - 13.7x3

x1 levels (50,100) ; x2 levels (-25,25); x3 levels (0,200).

If it's desirable to collect data in increments of 5 units of x2, then the preferred increment for collecting data for x3 (in real units) is approximately:
Feedback:

Since the difference between the high and low levels of x2 = 25- (-25) = 50 corresponds to two coded units of x2, data collection in increments of 5 real units of x2 corresponds to 0.2 coded units of x2.

The change in X3 in coded units is calculated as .2*(-13.7)/12.5 = -0.2192, which can be multiplied by 100 (one coded unit of x3) to see that a corresponding change in x3 in real units is approximately -22.
93) Constraint Management’s focusing steps, in order, are:
94) Once we have identified the constraint, and taken actions to make the most of its resources, we should:
Feedback:

The question refers to completing the Identify and Exploit stages. Choice a refers to the Subordination phase. Choice c refers to the step 5 Repeat stage. Choice b is incorrect since increasing the capacity of downstream processes would not have any impact on the constraint or the ability of the system to meet demand.

95) Buffers should:
96) In Goldratt’s Drum-Buffer-Rope, the rope:
97) A key difference between Critical Chain management and PERT/CPM critical path is:
98) According to Constraint Management theory, as Inventory costs increase:
99) We might choose to simulate a process:
100) A necessary pre-requisite for simulation include:
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