q8 im10 final

CHAPTER 12

13 Statistical Thinking and ApplicationsChapter 10 - Statistical Thinking and Applications 2

CHAPTER 10Statistical Thinking and ApplicationsTeaching NotesThis chapter describes concepts of statistics, statistical thinking, statistical methodology, sampling, experimental design, and process capability. Students should be encouraged to take a big picture perspective on this framework, rather than the approach of: How do I get the right answer?

There is a Bonus Materials file on the Premium website that reviews the basic concepts and techniques of statistics that are relevant to the technical areas of statistical process control (SPC). These topics are typically covered in business or engineering statistics course that students should have had prior to taking a course using this text. Key objectives for this chapter include:

To establish the importance of statistics as the "bridge" between quality of design and quality of conformance. The proper use of statistics is highlighted as a quality improvement tool.

To help students appreciate the importance of statistical thinking in order to understand inter-related processes, process variation, and the need to reduce it in order to assure quality in operations.

To review definitions and concepts of statistics, and relate them to quality control applications. Spreadsheet techniques for statistical analysis with Excel( software are also emphasized.

To introduce the use of DOE as a tool for drawing conclusions regarding controllable process factors and/or comparing methods for process development or improvement.

To help students to understand the concept of process capability and its effects on quality and conformance to specifications.

To introduce the concept of statistical measurement of service quality

The Instructors Resource folder on the course website has a number of Baldrige video clips which give an inside view of organizations that have received the Baldrige award. A couple of those, that are especially appropriate for this chapter, have scenes that show how statistical thinking and concepts can enhance an organizations quest for world-class quality.ANSWERS TO QUALITY IN PRACTICE QUESTIONSImproving Quality of a Wave Soldering Process Through the Design of Experiments1. The first experimental design at the HP plant did not achieve the true optimum combination of factors, because not all combinations were tested. It is theoretically possible that a better combination of factors exists among those that were not tested. Thus, the ones that were tested could be considered a random sampling of all of the possibilities. It is also likely that some interaction effects were at work, so some of the combinations that produced a higher number of defects had to be eliminated.

2. Experimental design allows the experimenter to systematically evaluate two or more methods to determine which is better, or to determine the levels of controllable factors to optimize process yields or minimize variation of a response variable. Therefore, it is generally faster and more efficient than using one at a time, trial-and-error methods.

Applying Statistical Analysis in a Six Sigma Project at GE-Fanuc1. This case showed how wave soldering technology was applied to electronic circuit boards in a process that is very similar to the Improving Quality of a Wave Soldering Process Through the Design of Experiments case, above. There are many variables that must be taken into account in order to improve the process, some of which are the true root causes of the problems in the process, but many are not. Thus, the critical Xs (independent variables) must be separated from the insignificant many, in order to solve the process problems. Many individuals were involved because this was a cross-functional process problem. No one team member had all of the knowledge, but together, they had all that was needed.

2. The Ni-Au boards could have originally been selected for any number of reasons. Reports on the advantages of using Ni-Au boards could have been read in technical journals, a persuasive salesperson for the vendor may have sold it as a solution to other problems, the process may have been changed to correct problems that existed at that time, or it might simply have been a matter of not having the methodology and expertise to do the comparative analysis needed to find out that the Ni-Au boards were no better than less expensive ones.

3. The F-test shows that the vendors, finishes, and interaction effects are all highly significant for this particular independent variable, Wave Solder Skips. Thus, the team must further analyze the defects to find out why the vendors products vary from their competitors, how the finishes differ, and what the interaction effects are.

ANSWERS TO REVIEW QUESTIONS

1.Statistical thinking is a philosophy of learning and taking action based on the principles of: a) all work occurs in a system of interconnected processes; b) variation exists in these processes; and c) understanding and reducing variation are the keys to success in improving processes. It is important to both managers and workers because without it, consistent, predictable processes cannot be established or improved upon.

2.Common causes of variation occur as a natural part of the process and are difficult to change without making a major change in the system of which they are a part. Special causes of variation arise from sources outside the system and can generally be traced back to a specific change that has occurred and needs correction. For example, a process may be stable and running well until the supplier of a critical material is changed. The new vendor's material causes the process to go out of control (becomes unstable), so the "solution" to the special cause is to have the vendor correct the deficiency, or return to the previous supplier for materials.

3.The two fundamental mistakes which managers can make in attempting to improve a process are: (1) To treat as a special (or outside) cause any fault, complaint, mistake, breakdown, accident, or shortage which is actually due to common causes, and (2) to attribute to common causes any fault, complaint, mistake, breakdown, accident, or shortage which is actually due to a special cause. In the first case, tampering with a stable system can increase the variation in the system. In the second case, the opportunity to reduce variation is missed because the amount of variation is mistakenly assumed to be uncontrollable.

4.The Red Bead experiment emphasizes that little, if anything, can improve quality in a poorly-managed production system. In the experiment, managers control incoming material (white and red beads) and work procedures so rigidly that there is little room for change. It is their mistake that there is an input of red bead production material the workers cannot stop the red beads from coming. Management inspects the beads only after they (and the mistakes involved) have been made. No amount of encouragement, threats, or promises of rewards will improve quality production when it is inevitable, by the nature of the process, that red beads will be produced. Furthermore, the managers have mistakenly believed that the variables in the process are controllable, and therefore that the workers are simply not trying hard enough in their labors. The final point of the Red Bead experiment is that all factors of a process must be examined to locate and correct negative variations.

The Funnel experiment is designed to show how people can and do affect the outcome of a process and create unwanted variation by "tampering" with the process, or indiscriminately trying to remove common causes of variation. The system of dropping the ball through the funnel towards the target is damaged by the variation of each participant moving the funnel around to "get a better aim" at the target. The lesson is that once a plan or process is determined to be correct and is set in motion, no components of the process should be tampered with. The process should be adjusted only if the entire process has been thoroughly examined and found to be in need of change in some way.

5.The methods for the efficient collection, organization, and description of data are called descriptive statistics. Statistical inference is the process of drawing conclusions about unknown characteristics of a population from which the data were taken. Predictive statistics is used to develop predictions of future values based on historical data. The three differ in approach, purpose, and outcomes. Descriptive statistics simply summarize and report on existing conditions, inference helps to make decisions about population characteristics based on sample data. Predictive statistics attempt to look into the future and state what will be the results, if certain assumptions hold. All three of these can be important to a manager who is trying to describe the current characteristics of a process, or make inferences about whether a process is in control, or predict future values of instrument readings in order to determine whether it is properly calibrated.

6. Discrete variables are used to measure whether tangible or intangible output from a process is acceptable or not acceptable (good or bad; defective, or not defective). Discrete variables are often used to classify the quality level of customer service. Was the patient in the hospital satisfied or dissatisfied; customer compliments versus complaints for a tour firm; did the marketing research firm accurately or inaccurately prepare the report? Continuous variables are used to measure quantifiable characteristics, such as time, temperature, weight, dimensions (in inches or millimeters). They are only limited by the accuracy of the measuring instrument being used (for example, atomic clocks can measure time to millionth of a second or finer accuracy). Thus, in quality control applications, continuous variables are used to measure dimensions of parts in an automotive supplier, temperature in drying processes, or times required to service each customer in a bank.

7. A population is a complete set or collection of objects of interest. A sample is a subset of objects taken from a population.

8. The standard error of the mean is the (estimated) standard deviation of the population ( divided by

(( /

). The standard deviation is, of course, a measure of variability within a population, where the standard error of the mean is a measure of the standard deviation within the sampling distribution.

9. The central limit theorem is extremely useful in that it states (approximately) that a sampling distribution can be defined as the distribution obtained by taking a large number of samples of size n from any population with a mean and a standard deviation, (, and calculating their means. The mean of the sample means for this probability distribution will approach , and the standard deviation of the distribution will be ( /

, as larger and larger sample sizes are taken. The CLT is extremely important in any SQC techniques that require sampling.

10.The two factors that influence sampling procedure are the method of selecting the sample and the sample size. Methods include simple random sampling, stratified sampling, systematic sampling and cluster sampling. Sample size is dependent on the amount of variation in the population (measured by () and the amount of error that the decision maker can tolerate at a specified confidence level.

11. The basic questions that must be addressed in a sampling study are:

a. What is the objective of the study?

b. What type of sample should be used?

c. What possible error might result from sampling?

d. What will the study cost?

All of these questions are inter-related, and only the last two are quantifiable. Anyone making a decision regarding these questions needs to have some understanding of the options, relative accuracy, and cost tradeoffs of each potential alternative in sampling studies. The long-run consequences of making a wrong choice in the design of a sampling study can be quite serious. For example, if a pharmaceutical firm makes a substantial error in the design or execution of a sampling study on a new drug, the result could be felt in the direct costs of running a correct study, the likelihood that customers would sue the company for injury suffered in taking the drug, or for the lack of any substantial benefit coming from having taken an ineffective drug. In addition, the firm's reputation could be permanently damaged.

12.Methods of sample selection, listed previously, include: simple random sampling, stratified sampling, systematic sampling and cluster sampling. Simple random sampling is useful where one needs to gather information from a moderately large, homogeneous population of items. For example, if a MBA director wished to find out the attitudes of 300 MBA students toward various policies, procedures, and services provided to the students, s(he) might use a simple random sample to determine whom the survey should be sent to. An automobile insurance company could use a stratified sample to determine accident rates of customers, stratified according to their ages. An auditor might use a systematic sampling to sample accounts receivable records by choosing every 50th record out of a file cabinet. Cluster sampling could be used by management analysts within city government to determine satisfaction levels of residents on a neighborhood by neighborhood (cluster) basis. Judgment sampling should be avoided, except as a way to gather preliminary, impressionistic data before beginning a true sampling study.

13.Systematic errors in sampling can come from bias, non-comparable data, uncritical projection of trends, causation, and improper sampling. They may be avoided by approaches discussed in the chapter. Basically, careful planning of the sampling study, awareness of possible systematic error causes, and careful execution of the study can help to avoid most of the common errors listed above.

14.The purpose of design of experiments is to set up a test or series of tests to enable the analyst to compare two or more methods to determine which is better, or to determine levels of controllable factors to optimize the yield of a process, or minimize variability of a response variable.

15.A factorial experiment is a specific type of experimental design that considers all combinations of levels of each factor. For example, a factorial experiment might be set up to determine the profitability of an ice cream store based on external and broad quality related factors. External factors, which could influence profits, would be days/times of the week and external temperature. Quality factors might be quality of flavors available and perceived quality characteristics (time, timeliness, empathy, etc.) of the workers.

16.Simple factorial experiments often require many iterations of the experiment to be run before reliable results are obtained. ANOVA permits testing multiple combinations of factors, simultaneously. Therefore it is often faster and cheaper to perform ANOVA than using simple factorial experiments. Conclusions regarding interaction effects can also be tested more rigorously using analysis of variance.

17.ANOVA is a methodology for drawing conclusions about equality of means of multiple populations. In its simplest form one-way ANOVA it compares means of observed responses of several different levels of a single factor. ANOVA tests the hypothesis that the means of all populations are equal against the alternative hypothesis that at least one mean differs from the others.

SOLUTIONS TO PROBLEMS CHAPTER 10

Note: Data sets for several problems in this chapter are available in the Excel workbook C10Data.xls on the Premier website accompanying this chapter. Click on the appropriate worksheet tab as noted in the problem (e.g., Prob. 10-1) to access the data.

1.Use the data for Twenty First Century Laundry for the weights of loads of clothes processed through their washing department in a week. (See Prob. 10-1 in C10Data.xls on the Premier website). Apply the Descriptive Statistics and Histogram tools in Excel to compute the mean, standard deviation, and other relevant statistics, as well as a frequency distribution and histogram for the following data. From what type of distribution might you suspect the data are drawn?Answer

1. The following results were obtained from the Twenty First Century Laundry data

Column1

Mean32.920

Standard Error2.590

Median25.500

Mode14.000

Standard Deviation25.899

Sample Variance670.741

Kurtosis0.233

Skewness0.994

Range106.000

Minimum1.000

Maximum107.000

Sum3292.000

Count100.000

Largest(1)107.000

Smallest(1)1.000

Confidence Level(95.0%)5.139

The conclusion that can be reached from looking at the summary statistics and the histogram is that these data are exponentially distributed, with descending frequencies. These data may show that small, low-weight batches are most frequently processed, as represented by the histogram. This has implications for the number and size of washers and dryers that should be installed in the laundry. More detailed data can be found in the Excel( solution spreadsheet coded as P10-1&2.xls.

2.The times for carrying out a blood test at Rivervalley Labs were studied in order to learn about process characteristics. Apply the Descriptive Statistics and Histogram analysis tools in Excel to compute the mean, standard deviation, and other relevant statistics, as well as a frequency distribution and histogram, for the data taken from 100 tests and found in the Prob. 10-2, Excel data set. From what type of distribution might you suspect the data are drawn?Answer

2. One of the advantages of using Excel( spreadsheets is that a great deal of analysis can be done easily. The summary statistics follow. Also shown is the histogram constructed by using Excels( Data Analysis tools (found under the Tools heading on the spreadsheet). For best results in constructing the histogram, it is suggested that students set up their own bins so as to provide 7 to 10 approximately equal sized class intervals for the data. Note that if the program finds that the classes shown in the bins do not extend over the upper or lower range of the data, it will automatically compensate by adding a Less or More category for the outliers.

Descriptive Statistics

Column1

Mean3.578

Standard Error0.081

Median3.600

Mode3.600



Kurtosis-0.267

Skewness-0.223

Range3.600

Minimum1.700

Maximum5.300

Sum357.800

Count100.000

Largest(1)5.300

Smallest(1)1.700

Confidence Level(95.0%)0.161

The conclusion that can be reached from looking at the summary statistics and the histogram is that these data are fairly normally distributed, with some slight skewing to the left. More detailed data can be found in the Excel( solution spreadsheet P10-1&2.xls.

3.The data (Prob. 10-3 in C10Data.xls found on the Premier website for this textbook) represent the weight of castings (in kilograms) from a production line in the Fillmore Metalwork foundry. Based on this sample of 100 castings, compute the mean, standard deviation, and other relevant statistics, as well as a frequency distribution and histogram. What do you conclude from your analysis?Answer

3.Descriptive statistics for the Fillmore Metalwork foundry are shown below.


Column1

Mean

38.654

Standard Error

0.031

Median

38.600

Mode

38.600

Standard Deviation

0.306

Sample Variance

0.094

Kurtosis

4.112

Skewness

-0.324

Range

2.300

Minimum

37.300

Maximum

39.600

Sum

3865.400

Count

100.000

Largest(1)

39.600

Smallest(1)

37.300

Confidence Level(95.0%)

0.061

The conclusion that can be reached from looking at the summary statistics and the histogram is that these data are not strongly normally distributed, due to significant skewing to the right. More detailed data can be found in the Excel( solution spreadsheet P10-3&4.xls.

4.The data (Prob. 10-4 in C10Data.xls found on the Premier website for this textbook) show the weight of castings (in kilograms) being made in the Fillmore Metalwork foundry and were taken from another production line. Compute the mean, standard deviation, and other relevant statistics, as well as a frequency distribution and histogram. Based on this sample of 100 castings, what do you conclude from your analysis?

Answer

4.Data from the Fillmore Metalwork foundry are shown below.


Column1

Mean38.6270

Standard Error0.0474

Median38.6000

Mode38.7000



Range2.6000

Minimum37.3000

Maximum39.9000

Sum3862.7000

Count100.0000

The conclusion that can be reached from looking at the summary statistics and the histogram is that these data are not normally distributed, due to pronounced skewing to the right. More detailed data can be found in the Excel( solution spreadsheet P10-3&4.xls.

5.San Juan Green Tea is sold in 1/2 liter (500 milliliter) bottles. The standard deviation for the filling process is 10 milliliters. If the process requires a 1 percent, or smaller, probability of over-filling, defined as over 495 milliliters, what must the target mean for the process be?

Answer

5.For San Juan Green Teas bottling process, the values for the 1% cutoff and the standard deviation are:

x = 495 ml; ( = 10 ml

For a total probability of 1% for overfilling:

P(x > upper fill limit) = 0.5000 - P = 0.5000 - .4900 = 0.01

Using the Normal Table, Appendix A, z = 2.33

x - 495 -

z = --------- = --------------- = 2.33

( 10

= 471.7 ml

( The process mean should be 471.7 ml., so that there is only a 1% probability of overfilling.

6.Texas Punch was made by Frutayuda, Inc. and sold in 12-ounce cans to benefit victims of Hurricane Ike. The mean number of ounces placed in a can by an automatic fill pump is 11.8 with a standard deviation of 0.12 ounce. Assuming a normal distribution, what is the probability that the filling pump will cause an overflow in a can, that is, the probability that more than 12 ounces will be released by the pump and overflow the can?Answer

6.For cans of Texas Punch the mean, = 11.8; the standard deviation, ( = 0.12

x - 12 11.8

z = --------- = --------------- = 1.667 = 1.67

( 0.12

P(x > 12) = 0.5000 P ( 0 < z < 1.67 )

P(z > 12) = 0.5000 - 0.4525 = 0.0475

Thus, there is a 4.75% probability of an overflow.

(Results are based on the Standard Normal Table, Appendix A)

7.Kiwi Blend is sold in 900 milliliter (ml) cans. The mean volume of juice placed in a can is 876 ml with a standard deviation of 12 ml. Assuming a normal distribution, what is the probability that the filling machine will cause an overflow in a can, that is, the probability that more than 900 ml will be placed in the can?

Answer

7.The mean, for the Kiwi Blend product is = 876; the standard deviation, ( = 12, x = 900.

x - 900 876

z = --------- = ---------------- = 2.0

( 12

P(x > 900) = 0.5000 - P ( 0 < z < 2.0 )

P(z > 900) = 0.5000 - 0.4772 = 0.0228

Thus, there is a 2.28 % probability of an overflow.


8.Wayback Beer bottles have been found to have a standard deviation of 5 ml. If 95 percent of the bottles contain more than 250 ml, what is the average filling volume of the bottles? Answer

8. Given that the standard deviation for Wayback Beer, as ( = 5 ml., x = 250, and P (z > x) = 0.95

Since z = - 1.645 for x > 0.95

250 -

-1.645 = ----------- ; ( = 241.775 ml.

5

(Results are based on the Standard Normal Distribution Table, Appendix A)

9.The mean filling weight of salt containers processed by the Piedra Salt Co. is 15.5 ounces. If 2 percent of the containers contain more than 16 ounces, what is the standard deviation of the filling weight of the containers?

Answer

9.Given that the process mean filling weight is = 15.5 oz. for the Piedra salt containers,

By looking up 0.5000 - 0.0200 = 0.48, we find z = 2.05

16 15.5

z = 2.05 = -------------

(( ( = 0.2439 oz.

(Results are based on the Standard Normal Distribution Table, Appendix A)

10.In filling bottles of E&L Cola, the average amount of over- and under-filling has not yet been precisely defined, but should be kept as low as possible. If the mean fill volume is 11.9 ounces and the standard deviation is 0.05 ounce:

a) What percentage of bottles will have more than 12 ounces (overflow)?

b) Between 11.9 and 11.95?

c) Less than 11.83 ounces?Answer

10.The mean value for E&L Cola, in ounces is: = 11.9; the standard deviation, ( = 0.05

a) P(x > 12.0) = 0.5000 - P ( 11.9 > x > 12.0)

x - (12.0 11.9)

z = --------- = ----------------- = 2.0

( 0.05

P(z > 2.0) = 0.5000 - 0.4772 = 0.0228

Thus, 2.28 % will be filled to more than 12 ounces (overflowing).

b) P (11.9 < x < 11.95)

x - (11.95 11.9)

z = --------- = ----------------- = 1.0

( 0.05

P(0 < z < 1.0) = z = 0.3413

Thus, 34.13 % will have a fill value between 11.9 and 11.95 ounces.

c) For P(x < 11.83) = 0.5000 - P (11.83 < x < 11.9)

x - (11.83 11.9)

z = --------- = ------------------ = -1.4

( 0.05

P(x < 11.83) = 0.5000 - P ( -1.4 < z < 0) = 0.5000 0.4192 = 0.0808

Thus, 8.08 % will be filled to less than 11.83 ounces.


11.The frequency table for Prob. 10-11 found in C10Data.xls on the Premier website shows the weight of castings (in kilograms) being made in the Fillmore Metalwork foundry (also see Prob 10-3, in C10Data.xls, for raw data).

a. Based on this sample of 100 castings, find the mean and standard deviation of the sample. (Note: If only the given data are used, it will be necessary to research formulae for calculating the mean and standard deviations using grouped data from a statistics text.)

b. Prepare and use an Excel spreadsheet, if not already done for Problem 10-3, to plot the histogram for the data.

c. Plot the data on normal probability paper to determine whether the distribution of the data is approximately normal. (Note: The Regression tool in Excel has a normal probability plot that may be used here.) Answer

11. An older statistics book from the library may have to be consulted in order to find background information on the formulae used for grouped statistics in solving this problem (see below). The following, use class midpoints estimated from the given cells. (Note that the data have been re-sorted in ascending order to conform to the format in the Excel statistical package.)

Answer

From the Excel statistical package, we get:

Upper CellCumulative

BoundariesFrequencies%

Cell 137.566.0%

Cell 237.81016.0%

Cell 338.13854.0%

Cell 438.42781.0%

Cell 538.71091.0%

Cell 639.0697.0%

Cell 739.3198.0%

Cell 839.62100.0%

For castings made at the Fillmore Metalwork foundry, when calculated using a hand calculator, the following grouped data results can be obtained:Adj. Cell

MidpointsFrequenciesfxfx^2

Cell 138.16228.608709.66

Cell 238.34153.205867.56

Cell 338.520770.0029645.00

Cell 438.7411586.7061405.29

Cell 538.916622.4024211.36

Cell 639.17273.7010701.67

Cell 739.34157.206177.96

Cell 839.5279.003120.50

1003870.80149839.00

a) fx3870.80

= n = 100 = 38.708 (vs. 38.654 from the data in spreadsheet 10-3)

s = = = 0.2856 (versus

0.306 from the spreadsheet data)

Note that the calculated value from the frequency distribution and from the raw data would probably have been much closer had the data been more uniformly distributed and less skewed around the mean.

b) See answer and spreadsheet printouts in Problem 10-3, above.

c) A normal probability plot shows that the data are approximately normally distributed, with an R square value of 0.825. (See spreadsheet Prob10-3&4 for more details). This value is good, but indicates only a moderately good fit because of the skewed distribution, as noted in 10-3.

The conclusions that can be reached are:

The formula for grouped data gives a close approximation of the statistics from Excel, and from the actual data.

The normal probability plot and the histogram show that these data are approximately normally distributed, with an R-squared value of 0.825.

12.The frequency table for Prob. 10-12 found in C10Data.xls on the Premier website shows the weight of another set of castings (in kilograms) being made in the Fillmore Metalwork foundry (also see Prob. 10-4, in C10Data.xls, for raw data).

a. Based on this sample of 100 castings, find the mean and standard deviation of the sample. (Note: If only the given data are used, it will be necessary to research formulae for calculating the mean and standard deviations using grouped data from a statistics text.)

b. Prepare and use an Excel spreadsheet, if not already done for Problem 10-4, to plot the histogram for the data.

c. Plot the data on normal probability paper to determine whether the distribution of the data is approximately normal. (Note: The Regression tool in Excel has a normal probability plot that may be used here.)

Answer

12.An older statistics book from the library may have to be consulted in order to find background information on the formulae used for grouped statistics in solving this problem (see below). The following, use class midpoints estimated from the given cells. (Note that the data have been re-sorted in ascending order to conform to the format in the Excel statistical package.)

From the Excel statistical package, we get:

Upper CellCumulative

BoundariesFrequencies%

Cell 137.511.00%

Cell 237.834.00%

Cell 338.11014.00%

Cell 438.42438.00%

Cell 538.72765.00%

Cell 639.01681.00%

Cell 739.31394.00%

Cell 839.6498.00%

Cell 939.92100.00%

When calculated using a hand calculator, you may obtain the following, using the grouped data formula:

Adj. Cell

MidpointsFrequenciesfxfx^2

Cell 137.35137.351395.02

Cell 237.65137.651417.52

Cell 337.956227.708641.22

Cell 438.2511420.7516093.69

Cell 538.55261002.3038638.67

Cell 638.8524932.4036223.74

Cell 739.1518704.7027589.01

Cell 839.459355.0514006.72

Cell 939.75279.503160.13

Cell 1040.05280.103208.01

1003877.50150373.71

a)

fx3877.5

= n = 100 = 38.775 (vs. 38.627 from the actual data in

spreadsheet 10-4)

s = = = 0.4887 (versus 0.4737

from the data in

spreadsheet 10-4)

b)See answer and spreadsheet printouts in Problem 10-4, above.

c) A normal probability plot shows these data are approximately normally distributed, with an R square value of 0.947. (See spreadsheet Prob10-3&4 for more details).

The conclusions that can be reached are:

The formula for grouped data gives a close approximation of the statistics from Excel, and from the actual data.

The normal probability plot and the histogram show that these data are approximately normally distributed, with an R-squared value of 0.947.

13.In a filling line at A & C Foods, Ltd., the mean fill volume for rice bubbles is 325 grams and the standard deviation is 20 grams. What percentage of containers will have less than 295 grams? More than 345 grams (assuming no overflow)?Answer

13. For the A & C Foods, Ltd. rice bubble filling process, the mean is = 325 grams and the standard deviation, ( = 20

P(x < 450) = 0.5000 - P ( 450 < x < 475)

x - (295 - 325)

z = --------- = ----------------- = - 1.5

( 20

P(z < -1.5) = 0.5000 - 0.4332 = 0.0668


Thus, 6.68 % will be filled with less than 295 grams

P(x > 345) = 0.5000 P (325 < x < 345)

x - (345 - 325)

z = --------- = ---------------- = 1.0

(

20

P (z > 1.0) = 0.5000 - 0.3413 = 0.1587

Thus, 15.87 % will contain more than 345 grams


14.Tessler Electric utility requires service operators to answer telephone calls from customers in an average time of 0.1 minute or less. A sample of 30 actual operator times was drawn, and the results are given in the following table. In addition, operators are expected to determine customer needs and either respond to them or refer the customer to the proper department within 0.5 minute. Another sample of 30 times was taken for this job component and is also given in the table. If these variables can be considered to be independent, is the average time taken to perform each component statistically different from the standard?

ComponentMean TimeStandard Deviation

Answer0.10230.0183

Service0.52900.0902

Answer

14. Specification for answer time for the Tessler utility is :

H0: Mean response time: ( 1 < 0.10

H1: Mean response time: ( 1 > 0.10

1 = 0.1023, s1 = 0.0183

and the t-test is:

, t29, .05 = 1.699Specification for service time is:

H0: Mean service time: ( 2 < 0.50

H1: Mean service time: ( 2 > 0.50

2 = 0.5290, s2 = 0.0902

and the t-test is:

, t29, .05 = 1.699Because t29, .05 = 1.699, we cannot reject the null hypothesis for t1, but we can reject the hypothesis for t2 . Therefore, there is no statistical evidence that the mean response time exceeds 0.10 for the answer component, but the statistical evidence does support the service component.Note: Problems 1518 address sample size determination and refer to theory covered in the Bonus Material folder for this chapter as contained on the Premier website.

15.You are asked by the owner of the Moonbow Motel to develop a customer satisfaction survey to determine the percentage of customers who are dissatisfied with service. In the past year, 10,000 customers were serviced. She desires a 95 percent level of confidence with an allowable statistical error of 0.01. From past estimates, the manager believes that about 3.5 percent of customers have expressed dissatisfaction. What sample size should you use for this survey?

Answer

15.The size of the population is irrelevant to this customer satisfaction survey, although it is good to know that it is sizable. Therefore, make the following calculations:

n = (z (/2)2 p(1-p) / E2 = (1.96)2 (0.035)(0.965) / (0.01)2 = 1297.5, use 1298

16.Determine the appropriate sample size needed to estimate the proportion of sorting errors at the Puxatawney post office at a 99 percent confidence level. Historically, the sorting error rate is 0.015, and you wish to have an allowable statistical error of 0.02.Answer16.The sample size for the proportion of sorting errors at a post office, using a 99% confidence level is:

n = (z (/2)2 p(1-p) / E2 = (2.33)2 (0.015)(0.985) / (0.02)2 = 200.53, use 201

17.A management engineer at Country Squire Hospital determined that she needs to make a work sampling study to see whether the proportion of idle time in the diagnostic imaging department had changed since being measured in a previous study several years ago. At that time, the percentage of idle time was 8 percent. If the engineer can only take a sample of 850 observations due to cost factors, and can tolerate an allowable error of 0.02, what percent confidence level can be obtained from the study?Answer

17.Using the formula: n = (z (/2)2 p(1-p) / E2 , the engineer at the Country Squire Hospital can solve for z (/2 as follows:

850 = (z (/2 )2 (0.08)(0.92) / (0.02)2

850 = (z (/2 )2 (184)

(z (/2 )2 = 850 / 184

z (/2 = = 2.149, use 2.15

From the Standard Normal Distribution table, Appendix A, we find a probability of 0.4838 for z = 2.15. Because it is only one tail of the distribution, we multiply the area by 2 to get the confidence level of 0.9676. Thus, the management engineer can be almost 97% confident of her results based on this sample size.

18.Localtel, a small telephone company, interviewed 200 customers to determine their satisfaction with service. Nineteen expressed dissatisfaction. Compute the sample size needed to ensure that they can be 90 percent confident of correctly estimating the proportion dissatisfied with an allowable error of 0.04.

Answer

18.First, we must find an estimated p for Localtels sample:

p =__19__ = 0.095

200

Then, using the formula for sample size:

n = (z (/2)2 p (1-p) / E2 , we can solve for n as follows:

n = (1.65)2 (0.095)(0.905) / (0.04)2

n = [(2.723) (0.0860)]/ (0.0016) = 146.36 use 147

Thus, Localtel can be more than 90% confident of their results by using a sample size of 200. Thus, there is no need to take more samples in order to meet the required sample size.

19.Using the Discovery Sampling table found in the Bonus materials on the Premier website, suppose that a population consists of 2,000 units. The critical rate of occurrence is 1 percent, and you wish to be 99 percent confident of finding at least one nonconformity. What sample size should you select?

Answer

19. Using the Discovery Sampling table (in the Sample Size Determination file) in the Bonus materials on the Premier website, for a population of 2,000, and reading the sample size required for a critical 1% rate, with a 99% confidence level yields the solution for the samples size of approximately 400 (use 98.9% confidence, with a critical rate of 1%).

20.The process engineer at Sival Electronics was trying to determine whether three suppliers would be equally capable of supplying the mounting boards for the new gold plated components that she was testing. The table found in Prob. 10-20 on C10Data.xls on the Premier website shows the coded defect levels for the suppliers, according to the finishes that were tested. Lower defect levels are preferable to higher levels. Using one-way ANOVA, analyze these results. What conclusion can be reached, based on these data?

Answer

20. The process engineer at Sival Electronics can develop a one-way ANOVA spreadsheet (see spreadsheet 10-20 for details) that shows:

Supplier 1Supplier 2Supplier 3

Finish 111.96.813.5

Finish 210.35.910.9

Finish 39.58.112.3

Finish 48.77.214.5

Finish 514.27.612.9

SUMMARY

GroupsCountSumAverageVariance

Supplier 1554.610.924.762

Supplier 2535.67.120.697

Supplier 3564.112.821.812

ANOVA

Source of VariationSSdfMSFP-valueF crit

Between Groups84.233242.11717.3770.000293.885

Within Groups29.084122.424

Total113.31714

According to the F-test and probability of P < .05, there is a significant difference between suppliers

21.The process engineer at Sival Electronics is also trying to determine whether a newer, more costly design involving a gold alloy in a computer chip is more effective than the present, less expensive silicon design. She wants to obtain an effective output voltage at both high and low temperatures, when tested with high and low signal strength. She hypothesizes that high signal strength will result in higher voltage output, low temperature will result in higher output, and the gold alloy will result in higher output than the silicon material. She hopes that the main and interaction effects with the expensive gold will be minimal. The data found in Prob. 10-21 on C10Data.xls on the Premier website were gathered in testing of all 2n combinations. What recommendation would you make, based on these data?

Answer

21. Using the following input data:

SignalMaterialTemperatureOutput Voltage

HighGoldLow18

HighGoldHigh12

HighSiliconLow16

HighSiliconHigh10

LowGoldLow8

LowGoldHigh11

LowSiliconLow7

LowSiliconHigh14

the process engineer at Sival Electronics can calculate the main effects as follows:

Signal

High (18 + 12 + 16 + 10)/ 4 = 14

Low ( 8 + 11 + 7 + 14)/ 4 = 10

High - Low = 4

Material

Gold (18 + 12 + 8 + 11)/ 4 = 12.25

Silicon (16 + 10 + 7 + 14)/ 4 = 11.75

Gold - Silicon = 12.25 - 11.75 = 0.5

Temperature

Low (18 + 16 + 8 + 7)/ 4 = 12.25

High (12 + 10 + 11 + 14)/ 4 = 11.75

Low - High = 12.25 - 11.75 = 0.5

The main effects of the signal (high or low) far outweigh the effects of material and temperature, indicating that those factors are insignificant. So, interaction effects will be negligible.

SUGGESTIONS FOR PROJECTS, ETC.1.This project will show that there are different ways to interpret Demings Red Bead experiment, statistically.

2.This experiment is designed to give the student hands on experience in elementary experimental design.

3.This experiment in aerodynamics will give the student some hands on experience in dealing with design problems in which multiple variables can have a significant impact on the operability of the mechanism. ANSWERS TO CASE QUESTIONSCase - The Disciplinary Citation

1.This case study shows Deming's "Red Bead" Experiment in action. Drivers are being blamed for conditions that are not under their control. The problem could be addressed by process measurement, eliminating "special causes" and reducing common causes.

2.A run chart would appear to be a way to begin to understand the process and to determine if it is in control or not. Based on the available data, we have:

Center Line (average) for the chart = 240/40 = 6.0 mistakes

The data show that 12 drivers have exceeded the average. Also, 6 drivers had "no defects". A Pareto chart (see spreadsheet disccase.xls for details) shows that only 8 drivers have more than 10 errors. The characteristics of drivers who are having difficulty should be examined to explain their higher error rates. Are their errors far above normal, or just a little above? Are they well trained? Are they overworked, with more than the average number of difficult orders? Do they have poor equipment? A useful control chart cannot be established, unless "special causes" are dealt with. Analysis should also be done to determine what the good drivers are doing right. Are they more experienced, drive newer cars, have better hearing and vision, etc.?

After corrections, a new chart (called a "c-chart" and discussed in Chapter 12) on the stable process can be set up. Then those who consistently do well can be rewarded and the performance of those who have an unsatisfactory level of errors can be improved.

Case - The Quarterly Sales Report

Obviously, Hagler has got to look beyond the quarterly sales reports to find out what is really going on in his regions. Two simple scatter diagrams (see spreadsheet C10Salecase.xls for details) that plot the entire 5 years of data, divided into large regions and small regions, are very revealing. He might start with those graphs to see what the general trend has been.

The Northeast, Southwest, and Northwest regions show basically flat sales for the 5 year period. The Southwest region had a strange jump in sales in the second quarter of 2001 and again in the third quarter of 2002, possibly due to inventory reduction sales. Otherwise, the three regions showed no consistent trend, while still managing to increase sales in the last quarter of 2003.

Two of the three smaller regions, North Central and Mid-Atlantic, showed upward trends from the third quarter of 2001 to the third quarter of 2003. Unfortunately, Hagler concentrated on the one quarter when sales went down for both of them, while also decreasing for the South Central region, which had been holding its own in a tough market.

To apply concepts of statistical thinking, Hagler must: 1) look at the big picture, graphically, rather than at quarter to quarter variations, 2) find out what each regions unique characteristics are, rather than treating all regions the same, 3) determine what causes variation within and among the regions, 4) determine if variations are due to common causes or special causes (probably the former), and 5) train, help, and support the regional managers if he expects to see different results over the coming year.

See spreadsheet C10SaleCase for further analysis.

Case - The HMO Pharmacy Crisis

It appears that, because of the crisis, top management at Dover is reacting in a typical red Bead manner. The importance of statistical concepts in quality management cannot be overemphasized. Indeed, statistics is essential in implementing a continuous improvement philosophy.

Statistical thinking is a philosophy of learning and action based on the principles that:

1. All work occurs in a system of interconnected processes.

2. Variation exists in all processes.

3. Understanding and reducing variation are keys to success

Understanding processes provides the context for determining the effects of variation and the proper type of managerial action to be taken. By viewing work as a process, we can apply statistical tools to establish consistent, predictable processes, study them, and improve them. While variation exists everywhere, many business decisions do not often account for it, and managers frequently confuse common and special causes of variation. Those inside and outside the pharmacy must understand the nature of variation, before they can focus on reducing it.

The complex interactions of these variations in drugs, equipment, computer systems, professional, clerical, and technical staff, and the environment are not easily understood. Variation due to any of these individual sources could be random; individual sources may not be identified or explainable. However their combined effect in the pharmaceutical system is probably stable and might be predicted statistically. These common causes of variation that are present as a natural part of the process need to be understood before special causes can be separated and eliminated.

To address the problem, Dover should consider using the following steps:

Form a cross-functional group consisting of pharmacists, assistant pharmacists, physicians, nurses, health care insurance experts, and administrative support people who may handle prescriptions or assess pharmaceutical risks.

Develop an understanding of the process by flowcharting all steps, including interconnected processes in various locations.

Do a risk assessment to determine where there are gaps in procedures or potential risk of errors

Develop corrective action teams to work first on the causes with the highest risk if errors in prescription handling occur.

Propose solutions

Pilot test solutions to ensure that problems have been corrected

Disseminate results to all concerned

Track results to determine long-term benefits.

Work on solving problems in priority order

Benchmark best practices for future improvements.

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_1047227205.xlsSheet: Error Data

Sheet: Scatter Diagram

Sheet: Pareto Chart

The Disciplinary Citation Case

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