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SOLUTIONS TO APPENDIX I - SUPPLEMENTARY PROBLEMS USING MINITAB CHAPTER 3 – QUALITY IMPROVEMENT AND COST REDUCTION Q1. A bank investigating problems in account transfers collected data on 100 orders and stratified errors by Cost Center and Category. What conclusions can be drawn as to the biggest problem areas? (filename: Gryna & Chua.MPJ, columns C22, C23, C24) SOLUTION: Pareto analysis indicates that Cost Centers 266 and 263 account for a cumulative 88% of the documented errors. Similar analysis shows that Documentation errors are the major leading defect category, with 56% of errors falling into this category. 100 100 80 80 60 60 40 40 20 20 0 Cost Center Count Percent Cum % 266 60 60.0 60.0 263 28 28.0 88.0 264 8 8.0 96.0 364 3 3.0 99.0 Other 1 1.0 100.0 Percent Count Pareto Chart of Cost Center 0 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Error Category 100 100 80 80 60 60 40 40 20 20 0 m cu o D Count Percent Cum % ta en n tio tR pu n I 56 56.0 56.0 e rd co 18 18.0 74.0 ng di o C Pr 11 11.0 85.0 g si n s e oc er sf n a Tr O 6 6.0 91.0 tR pu t u 6 6.0 97.0 d or ec Percent Count Pareto Chart of Error Category 0 3 3.0 100.0 Q2. A company seeking to improve its Work Order Fulfillment process collected data on consecutive work orders to determine what departments and work order types contributed most to overall work load. What can you conclude from the collected data? What departments and work order types should be investigated further? (filename: Gryna & Chua.MPJ, columns C26, C27, C28) SOLUTION: Pareto analysis shows that the department with the greatest number of work orders is the Filling Department, followed by the Manufacturing and Shipping & Receiving Departments. Unplanned Maintenance is the most frequent type of work order, far ahead of Planned Maintenance Work Orders that fall into second place. Further analysis of Work Order Type stratified by Department shows that Unplanned Maintenance Work Orders originating in the Filling Department is the logical group to investigate first for improvement opportunities. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Pareto Chart of Department 600 100 500 80 400 60 300 40 200 20 100 0 Department Percent Count 700 g llin Fi Count Percent Cum % 314 50.7 50.7 uf an M g in ur t ac g in pp i Sh & ce Re 110 17.8 68.5 g in iv e at or p r Co 84 13.6 82.1 g in ag k c Pa 65 10.5 92.6 0 46 7.4 100.0 Pareto Chart of Work Type 600 100 500 80 400 60 300 40 200 20 100 Work Type 0 e e nc nc a a en en nt nt ai ai M M ed ed n n an an pl Pl n U Count 379 85 Percent 61.2 13.7 Cum % 61.2 75.0 Percent Count 700 e ov M 77 12.4 87.4 lla ta s In n tio 56 9.0 96.4 m Re al ov 0 22 3.6 100.0 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Pareto Chart of Work Type by Department ce an ce en nan t n te ai M ain on l M ed ti n la va n ed e al m o l a nn t v p s a o Un Pl M In Re Department = Corporate Department = Filling Department = Manufacturing 300 200 Count 100 Department = Packaging 300 200 100 Department = Shipping & Receiving l e e e on va nc anc ov ti o M ll a na n m a e tn e nte R st ai ai In M M d d e e n n an lan pl P Un 0 Work Ty pe Unplanned Maintenance Planned Maintenance Mov e Installation Remov al 0 e e e n al nc nc Mov atio ov na ena ll m a e t R n st a ai In M M d d e e nn nn l a Pla np te in U Work Type PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. CHAPTER 15 – INSPECTION, TEST AND MEASUREMENT Q1. A pharmaceutical company conducts manual 100% inspection of incoming glass vials prior to filling. This is done because defects discovered after filling a vial with a drug are quite costly, as the product must be discarded. The company would like to determine if a sampling procedure would be more cost effective. Major types of defects scored were cracks, spikes and birdswings (thin strands extending from one side to another). Parameters are: Number of items in lot Number of items in sample Proportion defective in lot Damage cost incurred if defective slips past inspection Inspection cost per item Probability lot will be accepted by sampling plan 10,000 1000 Calculate from data $3,250 $1.20 0.96 Calculate the total cost of no inspection, sampling, and 100% inspection. What is the break-even point? What do you recommend? Would your conclusion change if an automated inspection system were installed that dropped costs of inspection by half (to 0.60 per vial; all other parameters unchanged)? (Gryna 022805.MPJ; C6-T Vial Status) SOLUTION: Calculate the total cost of no inspection, sampling, and 100% inspection. Proportion defective in the lot is needed to complete the calculations. Using the data provided, the proportion defective in the lot is estimated to be 0.0045. No inspection: NpA = (10,000) (0.0045) ($3,250) = $146,250 Sampling: nI + (N-n)pAPa + (N-n)(1-Pa)I = 1000 ($1.20) + (10,000 – 1000) (0.0045) ($3250) (0.96) + (10,000- 1000) (1-0.96) ($1.20) = $127,997 100% inspection: NI = 10,000 ($1.20) = $12,000 What is the break-even point? Pb = I/A = $1.20 / $3,250 = 0.000369 What do you recommend? The lowest cost option of the three under consideration is 100% inspection. Given this, recommend continuing with 100% inspection. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Would your conclusion change if an automated inspection system were installed that dropped costs of inspection by half (to 0.60 per vial; all other parameters unchanged)? No. The cost of inspection is negligible compared to the costs of a defective vial making its way through the filling process. A 50% reduction in inspection costs would be insufficient to warrant a change. Q2. A pipette is an instrument used to transfer small volumes of liquid from one vessel to another with high precision and accuracy. A molecular biologist suspects that poor experimental results may be the result of a faulty pipette, but does not know which of three pipettes might be at fault. The three pipettes (1000 μl, 200 μl and 20 μl units; designations indicate maximum intended volume) are tested at settings to dispense half the maximum rated volume (that is, 500, 100 and 10 μl, respectively). She takes thirty measures for each pipette. Which pipette has the greatest bias on a percentage basis? What is the precision of each unit (reported as one standard deviation)? Gryna 022805.MPJ; Columns C8-C10; 1000ul, 200ul, 20ul) SOLUTION: Descriptive Statistics: 1000 ul, 200 ul, 20 ul Variable 1000 ul 200 ul 20 ul N 30 30 30 N* 0 0 0 Variable 1000 ul 200 ul 20 ul Maximum 503.89 99.700 10.310 Mean 499.77 98.203 10.042 SE Mean 0.446 0.145 0.0313 StDev 2.44 0.796 0.171 Minimum 494.12 96.220 9.630 Q1 498.13 97.868 9.923 Median 499.53 98.020 10.065 Q3 501.72 98.830 10.195 Which pipette has the greatest bias on a percentage basis? The 200 μl pipette has the greatest absolute bias (100-98.203) and percentage bias ((100-98.203)/100). What is the precision of each unit (reported as one standard deviation)? 1000 μl: 2.44 200 μl: 0.796 20 μl: 0.171 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Q3. The scientist also suspects that the 1000 μl pipette may not be as accurate at lower volumes, and that she should use the 250 μl pipette for volumes at or under 250 μl. Prepare a scatterplot of data comparing actual dispensed volumes against pipette volume setting. Is the suspicion correct? (Gryna 022805.MPJ; Columns C11-12; Setting Volume, Dispensed Volume) SOLUTION: The scientist’s suspicion is correct. At volumes below 200 μl, there is a bias towards dispensing less than the set volume. A scatter plot comparing dispensed volumes and setting volumes is shown below. Given the data range, the bias is difficult to see in this scatterplot. Two additional scatterplots help visualize the problem; these are (a) the deviation in dispensed volume from setting volume plotted against setting volume, and (b) percentage deviation in dispensed volume plotted against setting volume. Both of these scatterplots show a sharp increase in bias at and below the 250 μl setting. Scatterplot of Dispensed Volume vs Setting Volume 1000 Dispensed Volume 800 600 400 200 0 0 200 400 600 Setting Volume 800 1000 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Scatterplot of Set - Dispensed vs Setting Volume 25 Set - Dispensed 20 15 10 5 0 0 200 400 600 Setting Volume 800 1000 Scatterplot of % Deviation vs Setting Volume 40 % Deviation 30 20 10 0 0 200 400 600 Setting Volume 800 1000 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Q4. A project team was investigating the sanding process in a wood finishing operation. Laminated (hard wood) components were sent through a rough sanding operation. Occasionally there were places on the components where all the laminated material was removed exposing the underlying fiber board. This defect was referred to as a sand-through. The team wanted to explore the ability of the operators to consistently and accurately characterize a sand-through. Using the data below, determine if the team can have confidence in the operators’ results. (Filename: Sand through Attribute Gage Study.mpj) SOLUTION: Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Attribute (Expert) Go No No No No No No No No No No No No No Go Go Go No Go No go no no no no no no no no no no no no no go go no no go no Inspector 1 go no no no no no no no no no no no no no go go no no go no go no no no no no no no no no no no no no go go no no go no Inspector 2 go no no no no no no no no no no no no no go no go no go no Answer: (Minitab Stat>Quality Tools>Gage Study>Attribute Gage Study) As the printout shows below, one inspector has good results and one inspector has marginal results. The team decided the data was reasonably reliable but that the training for sand-through inspection should be reviewed and additional attribute studies should be planned to evaluate the other inspectors who make these evaluations. Within Appraiser Assessment Agreement Appraiser # Inspected # Matched Percent (%) 1 20 20 100.0 ( 86.1, 100.0) 2 20 18 90.0 ( 68.3, 98.8) 95.0% CI PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. # Matched: Appraiser agrees with him/herself across trials. Each Appraiser vs Standard Assessment Agreement Appraiser # Inspected # Matched Percent (%) 1 20 19 95.0 ( 75.1, 99.9) 2 20 18 90.0 ( 68.3, 98.8) 95.0% CI # Matched: Appraiser's assessment across trials agrees with standard. Assessment Disagreement Appraiser # no/go Percent (%) # go/no Percent (%) # Mixed Percent (%) 1 1 20.0 0 0.0 0 0.0 2 0 0.0 0 0.0 2 10.0 Between Appraisers Assessment Agreement # Inspected # Matched Percent (%) 95.0% CI 20 18 90.0 ( 68.3, 98.8) # Matched: All appraisers' assessments agree with each other. All Appraisers vs Standard Assessment Agreement # Inspected # Matched Percent (%) 95.0% CI 20 18 90.0 ( 68.3, 98.8) Q5. Sanding wood components is often done on a machine called a Timesaver Sander. The reference surface is the flat belt that takes the product into the sanding area. For a uniform sanding of the wood surface the part must lie reasonably flat on the Timesaver belt. At regular intervals flatness is checked at an inspection station prior to sanding. The project team investigating whether the product flatness could be leading to defects decided to verify the ability to measure flatness. The data given below is from their study. Decide whether the team has a measurement system problem and what should be done about it. (filename: Flatness Gage.mpj) SOLUTION: Part Number 1 Curt 1.10 Michelle 1.10 Sherrie 1.15 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 1.10 1.35 1.35 1.20 1.10 1.00 1.00 0.75 0.70 0.70 0.50 1.50 1.45 0.20 0.10 1.10 1.10 1.35 1.40 1.10 1.30 1.40 1.20 1.20 1.00 1.00 0.85 0.85 0.65 0.75 1.55 1.50 0.20 0.20 1.20 1.25 1.35 1.40 1.15 1.30 1.30 1.15 1.15 1.00 1.00 0.85 0.85 0.80 0.70 1.45 1.45 0.00 0.20 1.10 1.15 1.30 1.29 Answer: (Minitab: Stat>Quality Tools>Gage Study>Crossed A portion of the printout from Minitab is shown here. From the printout it seems the MSA result is in the “dependent” zone. % Study Var is a bit high but not totally unacceptable. Since the flatness measurement is a reasonableness check the team could proceed with the project while looking into ways to improve flatness measurements. Also, from the graphical output, it appears a couple of parts gave the most variation in measurements. The team could look at these again to identify the source of the measurement deviation. Source StdDev (SD) Study Var (5.15*SD) %Study Var (%SV) Total Gage R&R Repeatability Reproducibility Operator Operator*Part Part-To-Part Total Variation 0.060256 0.049598 0.034217 0.021645 0.026501 0.391281 0.395893 0.31032 0.25543 0.17622 0.11147 0.13648 2.01509 2.03885 15.22 12.53 8.64 5.47 6.69 98.83 100.00 Number of Distinct Categories = 9 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Gage R&R (ANOVA) for Response Reported by : Tolerance: M isc: G age name: D ate of study : Components of Variation Response by Part Percent 100 % Contribution % Study Var 0.8 50 0 1.6 0.0 Gage R&R Repeat Reprod 1 Part-to-Part 2 3 Sample Range 0.2 2 _ R=0.042 LCL=0 2 1 3 10 2 Operator-Code 3 Operator-Code * Part Interaction _ _ UCL=1.103 X=1.024 LCL=0.945 1.5 Average Sample Mean 0.5 9 0.0 1.5 1.0 8 0.8 Xbar Chart by Operator-Code 1 7 1.6 UCL=0.1372 0.0 6 Response by Operator-Code 3 0.1 5 Part R Chart by Operator-Code 1 4 Operator-Code 1 1.0 2 3 0.5 1 2 3 4 5 6 Part 7 8 9 10 Project: FLATNESS GAGE.MPJ; 11/12/2004; Six Sigma Project 002-04 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. CHAPTER 17 – BASIC CONCEPTS OF STATISTICS AND PROBABILITY Q1. Glass slides used in microscopy must be of uniform thickness to maintain optical properties. A company manufacturing slides suspected that the slides it was producing were not as uniform as desired. Given 100 measures of slide thickness, a) Summarize the data in tabular form b) Summarize the data in graphical form c) State your conclusions regarding the manufacturing process (Gryna & Chua.MPJ, Column C1) SOLUTION: a) Descriptive Statistics: Thickness Variable Sum Thickness 305.8971 Variable Kurtosis Thickness -1.15 N Mean SE Mean TrMean StDev Variance CoefVar 100 3.0590 0.00643 3.0585 0.0643 0.00414 2.10 Minimum Q1 Median Q3 Maximum Range Skewness 2.9308 2.9997 3.0556 3.1190 3.2005 0.2698 0.07 b) Graphical Summary for Thickness PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Summary for Thickness A nderson-Darling N ormality Test 2.96 3.00 3.04 3.08 3.12 3.16 3.20 A -S quared P -V alue < 1.88 0.005 M ean S tDev V ariance S kew ness Kurtosis N 3.0590 0.0643 0.0041 0.06874 -1.14782 100 M inimum 1st Q uartile M edian 3rd Q uartile M aximum 2.9308 2.9997 3.0556 3.1190 3.2005 95% C onfidence Interv al for M ean 3.0462 3.0717 95% C onfidence Interv al for M edian 3.0294 3.0879 95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals 0.0565 0.0747 Mean Median 3.03 3.04 3.05 3.06 3.07 3.08 3.09 c) The analysis shows evidence of bimodality in the sample. The double peaks suggest there are two processes being used to manufacture the glass slides, e.g., two different machines, types of machines, or shifts. Q2. An outdoor advertiser interested in decreasing billboard print costs while maintaining color quality tested three new types of ink, along with its current ink. The number of hours to reach a quality set point under accelerated photo-bleaching conditions was recorded for each ink type. a) Generate boxplots for the data b) What conclusions can be drawn based on the graphical display? (Gryna & Chua.MPJ, Columns C6, C7) SOLUTION: a) Boxplots PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Boxplot of Hours vs Ink 300 290 Hours 280 270 260 250 240 230 Current Type 1 Type 2 Type 3 Ink b) The boxplots indicate that the Type 3 ink takes substantially longer to reach the quality set point than does the current ink, and therefore is more resistant to photo-bleaching. Furthermore, the variation is less than that of the other inks, suggesting that it may have more consistent performance. Ink Types 1 and 2 are similar in performance, with marginally higher median hours to reach the quality set point. Assuming all other factors are equal, Type 3 ink is the best. Q3. A company manufacturing pass-through autoclaves for heat sterilization of hospital equipment received complaints that pressure within the chamber is not being maintained for one of its models. Design engineers suspected the problem originated from the new type of gasket used seal the exit door. To test this hypothesis, engineers tested the two types of gasket and obtained the pressure (kgf/cm2) at which gaskets failed under constant, peak temperature. a) Prepare boxplots for the data b) Do the data support the hypothesis? Why or why not? (Gryna & Chua.MPJ, Columns C9, C10) SOLUTION: a) Boxplots PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Boxplot of Pressure 3.108 New Original 3.106 3.104 Pressure 3.102 3.100 3.098 3.096 3.094 3.092 3.090 Panel variable: Gasket b) No, the data do not support the hypothesis. The median pressure at which failure occurs is higher for the new gaskets, rather than lower. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Q4. A dry cereal manufacturer has a machine for filling boxes that subsequently are sold by weight. The machine is reset depending on the type of cereal (flake, puffed), but boxes are all filled with a weight specification of 16 ounces ± 0.05 ounces. Data were collected across several runs of the two types of cereal. a) Are the weights normally distributed? b) What are the means and standard deviations of weight? c) What percentage of each type will not meet the weight specification? d) Would you recommend any changes to the filling process? If so, what? (Gryna & Chua.MPJ, Columns C3, C4) SOLUTION: a) Probability plots and the Anderson-Darling test for normality indicate the weights are normally distributed for both the flake and puffed types of cereal. Probability Plot of Flake Normal 99.9 Mean StDev N AD P-Value 99 Percent 95 90 16.00 0.02058 100 0.361 0.440 80 70 60 50 40 30 20 10 5 1 0.1 15.950 15.975 16.000 Flake 16.025 16.050 16.075 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Probability Plot of Puff Normal 99.9 Mean StDev N AD P-Value 99 Percent 95 90 16.00 0.004853 100 0.305 0.562 80 70 60 50 40 30 20 10 5 1 0.1 15.980 15.985 15.990 15.995 16.000 16.005 16.010 16.015 Puff b) The mean weights and standard deviations are: Descriptive Statistics: Flake, Puff Variable Flake Puff N 100 100 Mean 16.005 15.998 StDev 0.0206 0.00485 While the means are nearly identical and on target, the puffed cereal has a much more narrow distribution of weight than does the flake cereal. c) All of the samples taken are within the 16 ± 0.05 ounce specification. However, note that the filling process for the flake cereal is much more likely to produce out of spec product because of the greater spread in weights. (While not required as part of the solution, a capability analysis shows Cpk values of <1 and >2 for the flake and puffed cereal filling processes, respectively. This indicates poor capability for the flake process). d) Descriptive Statistics: Flake, Puff Variable Flake Puff N 100 100 Mean 16.005 15.998 StDev 0.0206 0.00485 Minimum 15.952 15.979 Maximum 16.045 16.009 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Histogram of Flake, Puff Normal 15.96 15.98 16.00 16.02 16.04 90 Flake Puff 80 Puff 70 Frequency Flak e Mean 16.00 StDev 0.02058 N 100 Mean StDev N 60 16.00 0.004853 100 50 40 30 20 10 0 15.96 15.98 16.00 16.02 16.04 d) Recommend reducing the variation in the flake cereal filling process. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. CHAPTER 18 – STATISTICAL TOOLS FOR ANALYZING DATA Q1. The United States Golf Association (USGA) sets restrictions on the mass and size of golf balls. The ball mass must not exceed 1.620 ounces (0.04593 kg), and should have a minimum diameter of 1.680 inches. A golf ball manufacturer took a random sample of 1000 balls from its production line and measured the weight and diameter. a) What proportion of balls in the sample fail to meet the weight specification? b) Diameter specification? c) Both specifications? (filename: Gryna & Chua.MPJ, columns C30, C31) SOLUTION: a) Four balls have a mass that exceeds 0.04593 kg (these are: 0.045931, 0.045932, 0.045935, 0,045941, 0.045951). This yields a proportion of (4/1000) = 0.004, or 0.4%. b) Two balls have a diameter that is less than the minimum of 1.680 inches (these are: 1.679520, 1.679912). this yields a proportion of (2/1000) = 0.002, or 0.2%. c) There are no balls that fail to meet both specifications. Scatterplot of Weight vs Diameter 1.68 0.04595 0.04593 Weight 0.04590 0.04585 0.04580 0.04575 1.680 1.682 1.684 1.686 1.688 Diameter 1.690 1.692 1.694 1.696 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Q2. The USGA also regulates the coefficient of restitution (COR) of the ball. A COR of 0.8 means that if a golf ball is dropped from a height of 10 feet, it will rebound to a height of 8 feet. Through the process of impedance matching, clubface and ball hardness can be adjusted to attain a specific COR. A golf ball manufacturer recently changed its ball design with the goal of reducing variation in the COR. Given the original and new ball designs, a) Was the company successful in reducing COR variation (assume identical clubface test conditions)? b) Does the new design meet the target standard deviation of 0.00230? c) What proportion of the sampled balls from the original and new designs exceed the COR limit of 0.83? (filename: Gryna & Chua.MPJ, columns C32, C33) SOLUTION: a) Yes, the company was successful in reducing COR variation; this is evident by the standard deviation and other measures of variation). Descriptive Statistics: COR (Original), COR (New) Variable COR (Original) COR (New) Mean 0.81989 0.82008 SE Mean 0.000108 0.0000743 Variable COR (Original) COR (New) Median 0.81990 0.82010 Maximum 0.83190 0.82820 StDev 0.00341 0.00235 Variance 0.0000117 0.00000552 CoefVar 0.42 0.29 Minimum 0.81000 0.81190 Range 0.02190 0.01630 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Histogram of COR (Original), COR (New) 0 3 6 9 2 5 8 1 81 .81 .81 .81 .82 .82 .82 .83 0. 0 0 0 0 0 0 0 COR (Original) 200 COR (New) Frequency 150 100 50 0 8 0. 10 3 6 9 2 5 8 1 81 .81 .81 .82 .82 .82 .83 0. 0 0 0 0 0 0 b) Yes, the target of standard deviation of 0.00230 has been met. Although the new standard deviation is 0.00235, the 95% confidence interval includes the target value. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Summary for COR (New) A nderson-Darling N ormality Test 0.8125 0.8150 0.8175 0.8200 0.8225 0.8250 0.8275 A -S quared P -V alue 0.37 0.429 M ean S tDev V ariance S kew ness Kurtosis N 0.82008 0.00235 0.00001 -0.043637 0.164743 1000 M inimum 1st Q uartile M edian 3rd Q uartile M aximum 0.81190 0.81850 0.82010 0.82160 0.82820 95% C onfidence Interv al for M ean 0.81994 0.82023 95% C onfidence Interv al for M edian 0.81990 0.82030 95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals 0.00225 0.00246 Mean Median 0.8199 0.8200 0.8201 0.8202 0.8203 c) Three balls in the original sample exceed the specification of 0.83 (these are: 0.8319, 0.8308, 0.8306), which yields a proportion of 3/1000 = 0.003. None of the balls sampled from the new ball design exceed the specification. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Dotplot of COR (Original) 0.810 0.813 0.816 0.819 0.822 COR (Original) 0.825 0.828 0.831 Each symbol represents up to 2 observations. Dotplot of COR (New) 0.8125 0.8150 0.8175 0.8200 COR (New) 0.8225 0.8250 0.8275 Each symbol represents up to 3 observations. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Q3. A penetrometer is a device used to measure the firmness of fruit. A horticultural company has developed a new variety of tomato that it believes is firmer and therefore more suitable for shipping in lower-cost, less protective containers. Firmness measurements (kg force needed to penetrate the skin with a 0.8 cm tip) were collected on a random sample of 75 tomatoes at the breaker stage of fruit development from each of an older variety, and the new variety (one measure per fruit). a) State a hypothesis to evaluate the claim b) Evaluate the claim using a standard hypothesis testing approach. c) Evaluate the claim using the confidence limit approach. (filename: Gryna & Chua.MPJ, columns C36, C37) SOLUTION: a) The hypothesis to be tested is that the new variety is firmer than the old variety. More specifically, the null (H0) and alternative (HA) hypotheses are: H0: There is no difference in firmness between the old and new varieties HA: The new variety is firmer than the old variety. b) The data for both the old and new varieties are normally distributed, and have approximately equal variances. Given this, a two-sample t-test with pooled variances is appropriate. The test result indicates that the null hypothesis can be rejected, and that the new variety is significantly more firm than the old variety. Probability Plot of Old Variety Normal 99.9 Mean StDev N AD P-Value 99 Percent 95 90 5.916 0.4726 75 0.370 0.417 80 70 60 50 40 30 20 10 5 1 0.1 4.5 5.0 5.5 6.0 Old Variety 6.5 7.0 7.5 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Probability Plot of New Variety Normal 99.9 Mean StDev N AD P-Value 99 Percent 95 90 6.105 0.3892 75 0.281 0.633 80 70 60 50 40 30 20 10 5 1 0.1 5.0 5.5 6.0 6.5 New Variety 7.0 7.5 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Test for Equal Variances for Old Variety, New Variety F-Test Test Statistic P-Value Old Variety 1.47 0.097 Lev ene's Test Test Statistic P-Value New Variety 0.30 0.35 0.40 0.45 0.50 0.55 95% Bonferroni Confidence Intervals for StDevs 3.08 0.081 0.60 Old Variety New Variety 5.0 5.5 6.0 6.5 7.0 7.5 Data Two-sample T for New Variety vs Old Variety New Variety Old Variety N 75 75 Mean 6.105 5.916 StDev 0.389 0.473 SE Mean 0.045 0.055 Difference = mu (New Variety) - mu (Old Variety) Estimate for difference: 0.188440 95% lower bound for difference: 0.071428 T-Test of difference = 0 (vs >): T-Value = 2.67 P-Value = 0.004 148 Both use Pooled StDev = 0.4329 DF = c) Minitab automatically generates a confidence interval for the difference between the means (for a two-sided test), or a lower or upper bound for a one-sided test. Because we are interested in whether the firmness of the new variety is greater than that of the old variety, a one-sided, 95% lower bound for the difference is appropriate. Minitab provides this value with the two-sample t-test discussed in (b), above. Because the value of 0.0714 is greater than zero, we can conclude that the new variety is significantly firmer than the old variety at an alpha level of 0.05. If a single-sided test were done, the resulting 95% CI would be reported by Minitab as follows: 95% CI for difference: (0.048742, 0.328138). Note that this interval also does not include zero. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Q4. A seed company testing germination conditions for new cotton varieties obtained data for the number of days required for 50% germination under standard, controlled conditions. Data were collected for 40 samples of seed for each of four varieties. a) Plot the data. What conclusions can be made? b) Conduct an ANOVA for the data. c) Is there a significant difference in germination time among the varieties tested? (Gryna & Chua.MPJ, Columns C12-15 SOLUTION: a) Histograms of the data show the data from each variety to be approximately normally distributed; this is confirmed by the Anderson-Darling tests for normality (all p>0.05). Further, the data for each variety have approximately equal variances, as shown by the non-significant Bartlett’s and Levene’s test statistics. Varieties A and B have the shortest mean germination time, while C and D have mean germination times of at least one day longer. Summary for Variety A A nderson-Darling N ormality Test 10.5 11.0 11.5 12.0 12.5 A -S quared P -V alue 0.37 0.420 M ean S tDev V ariance S kew ness Kurtosis N 11.268 0.539 0.291 0.002541 -0.406569 40 M inimum 1st Q uartile M edian 3rd Q uartile M aximum 10.300 10.925 11.300 11.675 12.500 95% C onfidence Interv al for M ean 11.095 11.440 95% C onfidence Interv al for M edian 11.100 11.400 95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals 0.442 0.693 Mean Median 11.1 11.2 11.3 11.4 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Summary for Variety B A nderson-Darling N ormality Test 10.0 10.5 11.0 11.5 A -S quared P -V alue 0.49 0.214 M ean S tDev V ariance S kew ness Kurtosis N 11.068 0.499 0.249 -0.266262 0.240108 40 M inimum 1st Q uartile M edian 3rd Q uartile M aximum 12.0 9.800 10.700 11.000 11.475 11.900 95% C onfidence Interv al for M ean 10.908 11.227 95% C onfidence Interv al for M edian 10.900 11.200 95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals 0.409 0.641 Mean Median 10.90 10.95 11.00 11.05 11.10 11.15 11.20 Summary for Variety C A nderson-Darling N ormality Test 11.5 12.0 12.5 13.0 13.5 14.0 A -S quared P -V alue 0.63 0.096 M ean S tDev V ariance S kew ness Kurtosis N 12.925 0.685 0.469 0.572018 0.478406 40 M inimum 1st Q uartile M edian 3rd Q uartile M aximum 14.5 11.400 12.425 12.900 13.275 14.700 95% C onfidence Interv al for M ean 12.706 13.144 95% C onfidence Interv al for M edian 12.600 13.000 95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals 0.561 0.879 Mean Median 12.6 12.7 12.8 12.9 13.0 13.1 13.2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Summary for Variety D A nderson-Darling N ormality Test 11.2 12.0 12.8 A -S quared P -V alue 0.46 0.254 M ean S tDev V ariance S kew ness Kurtosis N 12.488 0.626 0.392 0.137107 -0.320390 40 M inimum 1st Q uartile M edian 3rd Q uartile M aximum 13.6 11.200 12.100 12.400 12.975 14.000 95% C onfidence Interv al for M ean 12.287 12.688 95% C onfidence Interv al for M edian 12.100 12.859 95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals 0.513 0.804 Mean Median 12.0 12.2 12.4 12.6 12.8 Test for Equal Variances for Var Stacked Bartlett's Test Test Statistic P-Value Variety A 4.71 0.194 Lev ene's Test Test Statistic P-Value Variety Variety B 1.54 0.205 Variety C Variety D 0.4 0.5 0.6 0.7 0.8 0.9 1.0 95% Bonferroni Confidence Intervals for StDevs b) ANOVA results and residual plots are shown below. The residual plots indicate no problems with regard to interpretation of the ANOVA table. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. One-way ANOVA: Var Stacked versus Variety Source Variety Error Total DF 3 156 159 S = 0.5918 SS 99.338 54.634 153.972 Level Variety Variety Variety Variety A B C D MS 33.113 0.350 R-Sq = 64.52% N 40 40 40 40 Mean 11.268 11.068 12.925 12.488 StDev 0.539 0.499 0.685 0.626 F 94.55 P 0.000 R-Sq(adj) = 63.83% Individual 95% CIs For Mean Based on Pooled StDev ---------+---------+---------+---------+ (--*--) (--*---) (--*--) (--*--) ---------+---------+---------+---------+ 11.40 12.00 12.60 13.20 Pooled StDev = 0.592 Residual Plots for Var Stacked Normal Probability Plot of the Residuals Residuals Versus the Fitted Values 99.9 2 99 1 Residual Percent 90 50 10 0 -1 1 0.1 -2 -1 0 Residual 1 -2 11.0 2 Histogram of the Residuals 12.5 13.0 2 20 Residual Frequency 12.0 Fitted Value Residuals Versus the Order of the Data 30 10 0 11.5 -1.2 -0.6 0.0 0.6 Residual 1.2 1.8 1 0 -1 -2 1 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 Observation Order c) The ANOVA shows there is a statistically significant difference in germination time among the four varieties. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Q5. Marbles are classified according to size, with the standard size having a diameter of approximately 5/8 inches at the widest point. Larger marbles are called shooters, while those measuring 1/2 inches and less in diameter commonly are referred to as “peewees.” Historically, pewee marbles sometimes were the result of errors during production runs of standard and shooter size marbles. Assume fifty marbles were randomly sampled from a production batch. a) What proportion of marbles is ½ inch or less in diameter? b) What are the 95% Confidence Limits for population fraction of pewees (defective) based on this sample? (Gryna & Chua.MPJ, Column C17) SOLUTION: a) Two marbles have a diameter that is ½ inch or less. This represents a proportion of 2/50, or 4% (0.04) of the sample. Dotplot of Marble Diameter 0.500 0.525 0.550 0.575 Marble Diameter 0.600 0.625 b) A one proportion analysis yields a 95% Confidence Interval of 0.49% to 13.7%. Minitab also provides a one proportion test, with 0.5 as the default proportion to test against. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Test and CI for One Proportion Test of p = 0.5 vs p not = 0.5 Sample 1 X 2 N 50 Sample p 0.040000 95% CI (0.004881, 0.137138) Exact P-Value 0.000 Q6. The “rainfast” period of an herbicide is the length of time between application and the point at which rainfall or irrigation does not reduce product performance. Various adjuvants (additives) are available that assist in spreading and sticking product to a leaf surface. A pesticide company claims that its new non-ionic surfactant reduced the rainfast period from six to four hours for a major commercial herbicide. To test this claim, a competitor conducted a controlled study and measured the number of hours required for rainfastness for the product with and without the adjuvant. a) Plot the data b) Do the data support the company’s claim of a reduction in the rainfast period by use of the adjuvant? c) Do the data support the company’s claim of a four hour rainfast period? (Gryna & Chua.MPJ, Columns C19, C20) SOLUTION: a) Boxplots of the data are below (boxplots are appropriate alternatives to histograms when sample sizes are less than 40 points). Note that the data for “No Adjuvant” are negatively skewed, with a numerically higher median rainfast period. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Boxplot of No Adjuvant, With Adjuvant 9 No Adjuvant With Adjuvant 8 7 6 5 4 3 2 1 b) A two-sample, one-sided t-test of the data supports the company’s claim that the adjuvant decreases the rainfast period. Given the data skew and departure from normality, a non-parametric test may be advisable; a Mann-Whitney test for a difference in the medians also supports the claim. Two-Sample T-Test and CI: No Adjuvant, With Adjuvant Two-sample T for No Adjuvant vs With Adjuvant No Adjuvant With Adjuvant N 35 35 Mean 5.77 4.66 StDev 1.52 1.14 SE Mean 0.26 0.19 Difference = mu (No Adjuvant) - mu (With Adjuvant) Estimate for difference: 1.11571 95% lower bound for difference: 0.57965 T-Test of difference = 0 (vs >): T-Value = 3.47 P-Value = 0.000 = 63 DF Mann-Whitney Test and CI: No Adjuvant, With Adjuvant No Adjuvant With Adjuvant N 35 35 Median 6.370 4.570 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Point estimate for ETA1-ETA2 is 1.310 95.0 Percent CI for ETA1-ETA2 is (0.650,1.940) W = 1536.0 Test of ETA1 = ETA2 vs ETA1 > ETA2 is significant at 0.0003 The test is significant at 0.0003 (adjusted for ties) c) Although the claim of a reduced rainfast period is supported (above), the data do not support the claim of a four hour rainfast period. A one-sample t-test against a test mean of four hours indicates a significant difference (rainfast period is not equal to four hours), and the 95% CI does not include four. One-Sample T: With Adjuvant Test of mu = 4 vs not = 4 Variable With Adjuvant N 35 Mean 4.65714 StDev 1.13929 SE Mean 0.19258 95% CI (4.26578, 5.04850) T 3.41 P 0.002 Q7. Multiple Drug Resistant (MDR) bacteria are a growing health problem. Hospitals are a significant point source of drug resistance, and may facilitate the spread of drug resistant bacteria through waste disposal systems. A hypothetical study examined the MDR bacterial count (Colony Forming Units per ml) from the effluent water of hospitals and private residences in three different states. What conclusions may be drawn regarding the incidence of MDR bacteria among the hospitals? Among the residences? (filename: Gryna 022805.MPJ, Columns C22-24, Source, State, CFU/ml) SOLUTION: Boxplots of the data suggest that the CT hospitals have the lowest incidence of MDR bacteria, followed by NY and NJ. For residences, CT had the lowest incidence, followed by NJ and NY. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Boxplot of CFU/ml vs Source, State 9 8 7 CFU/ml 6 5 4 3 2 1 0 State Source CT NJ Hospital NY CT NJ Residence NY The data for the hospitals are approximately normally distributed, and the variances may be assumed to be equal among the states. An ANOVA of the hospital data indicates a significant difference in MDR bacteria incidence among the states. Further, Tukey’s multiple pairwise comparisons show that each state is significantly different from the other. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Probability Plot of CFU/ml Normal 99 Mean StDev N AD P-Value 95 90 4.093 1.715 30 0.245 0.740 Percent 80 70 60 50 40 30 20 10 5 1 0 1 2 3 4 5 CFU/ml 6 7 8 9 Test for Equal Variances for CFU/ml Bartlett's Test Test Statistic P-Value CT 1.31 0.520 Lev ene's Test State Test Statistic P-Value 1.06 0.361 NJ NY 0.5 1.0 1.5 2.0 2.5 95% Bonferroni Confidence Intervals for StDevs One-way ANOVA: CFU/ml versus State Source DF SS MS F P PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. State Error Total 2 27 29 56.49 28.85 85.34 S = 1.034 Level CT NJ NY N 10 10 10 28.25 1.07 26.44 R-Sq = 66.20% Mean 2.440 5.800 4.040 StDev 0.913 1.254 0.893 0.000 R-Sq(adj) = 63.69% Individual 95% CIs For Mean Based on Pooled StDev -----+---------+---------+---------+---(----*-----) (----*-----) (-----*----) -----+---------+---------+---------+---2.4 3.6 4.8 6.0 Pooled StDev = 1.034 Tukey 95% Simultaneous Confidence Intervals All Pairwise Comparisons among Levels of State Individual confidence level = 98.04% State = CT subtracted from: State NJ NY Lower 2.213 0.453 Center 3.360 1.600 Upper 4.507 2.747 -----+---------+---------+---------+---(-----*-----) (-----*-----) -----+---------+---------+---------+----2.0 0.0 2.0 4.0 State = NJ subtracted from: State NY Lower -2.907 Center -1.760 Upper -0.613 -----+---------+---------+---------+---(-----*-----) -----+---------+---------+---------+----2.0 0.0 2.0 4.0 For the residences, the data also may be assumed to be normally distributed, with equal variances among the states. An ANOVA indicates a significant difference in MDR bacteria incidence among the states. A Tukey’s multiple pairwise comparison indicates that CT is significantly different from both NJ and NY, but that NJ and NY are not significantly different at a family-wide error rate of 0.05 (mean difference includes zero). PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Probability Plot of CFU/ml Normal 99 Mean StDev N AD P-Value 95 90 0.9543 0.4045 30 0.475 0.224 Percent 80 70 60 50 40 30 20 10 5 1 0.0 0.5 1.0 CFU/ml 1.5 2.0 Test for Equal Variances for CFU/ml Bartlett's Test Test Statistic P-Value CT 1.53 0.466 Lev ene's Test State Test Statistic P-Value 0.58 0.565 NJ NY 0.1 0.2 0.3 0.4 0.5 0.6 0.7 95% Bonferroni Confidence Intervals for StDevs 0.8 One-way ANOVA: CFU/ml versus State PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Source State Error Total DF 2 27 29 SS 2.3091 2.4365 4.7455 S = 0.3004 Level CT NJ NY N 10 10 10 MS 1.1545 0.0902 R-Sq = 48.66% F 12.79 Mean 0.6020 0.9810 1.2800 P 0.000 R-Sq(adj) = 44.85% StDev 0.2316 0.3015 0.3552 Individual 95% CIs For Mean Based on Pooled StDev ------+---------+---------+---------+--(-----*------) (------*-----) (------*-----) ------+---------+---------+---------+--0.60 0.90 1.20 1.50 Pooled StDev = 0.3004 Tukey 95% Simultaneous Confidence Intervals All Pairwise Comparisons among Levels of State Individual confidence level = 98.04% State = CT subtracted from: State NJ NY Lower 0.0456 0.3446 Center 0.3790 0.6780 Upper 0.7124 1.0114 ---+---------+---------+---------+-----(------*-----) (------*-----) ---+---------+---------+---------+------0.50 0.00 0.50 1.00 State = NJ subtracted from: State NY Lower -0.0344 Center 0.2990 Upper 0.6324 ---+---------+---------+---------+-----(------*------) ---+---------+---------+---------+------0.50 0.00 0.50 1.00 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Q8. For the coming year the mailroom manager wanted to determine his staffing level. He had been receiving some complaints about late mail and wanted to assure had the right resources. His first analysis was to monitor his workload to see if how it compared to his forecast. His resource plan used 3500 pieces per day as a maximum. He collected daily data for three weeks. What can be said about the projected workload? (filename: Transactional.mtw, column C1) SOLUTION: One-Sample T: Pieces of Mail Test of mu = 3500 vs > 3500 Variable Pieces of Mail N 15 Mean 3603.73 StDev 142.77 SE Mean 36.86 95% Lower Bound 3538.81 T 2.81 P 0.007 Reject the Null hypothesis since p is less than alpha of 0.05. The projected workload exceeds his resource plan maximum of 3500. Q9. The packaging department of a book warehouse was being evaluated for performance to goal. The average number of orders processed per person per day was expected to be 28. 20 people were selected at random and the number of orders processed recorded. Can it be concluded that the packaging department is meeting its goal? (filename: Transactional.mtw, column C3) SOLUTION: One-Sample T: Book Orders Test of mu = 28 vs not = 28 Variable Book Orders N 20 Mean 28.5000 StDev 1.6059 SE Mean 0.3591 95% CI (27.7484, 29.2516) T 1.39 P 0.180 p-value is greater than alpha of 0.05. Therefore fail to reject the null hypothesis. Also, note the confidence interval. It includes the goal of 28. The packaging department is meeting its goal. Q10. During a major league baseball game many balls are lost to the public or removed from play for other reasons. To prepare for a major league baseball game PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. umpires need to prepare a number of baseballs prior to the game. The league has set the number of balls to be prepared to 40 to assure a 95% confidence that they would have enough for any game. This was based on an average number of lost balls of 32 and a standard deviation of 4. Based on more recent data from 40 games, is 40 enough balls to complete 95% of games without running out? How many should they prepare to be 99% confident they will not run out? (filename: Transactional.mtw, column C5) SOLUTION: One-Sample T: Baseballs Test of mu = 32 vs not = 32 Variable Baseballs N 40 Mean 38.4500 StDev 4.5118 SE Mean 0.7134 95% CI (37.0071, 39.8929) T 9.04 P 0.000 40 is not enough because the mean value is not at 32 although the standard deviation is about the same. To assure enough balls to complete 95% of games, the umpires should prepare the mean value plus 2X the standard deviation of the original data, since the number of lost balls is normally distributed. Therefore prepare 38.5 + 2*4.5 = 47. (round up to 48, four dozen) 99% confidence requires the mean plus 2.6X the standard deviation. This would be 38.5 + 2.6*4.5 = 49. (1 or 2 more – I think it is worth it) Q11. The post office charges 37 cents to deliver a letter. Each letter is supposed to weigh 1 ounce. The officials at the post decided to determine the rate of compliance with the 1 oz specification. To do this they took 100 letters with a 37 cent stamp affixed. Can the officials conclude that the average weight of a letter with a 37 cent stamp is no more than 1.49 ounces? They normally charge for the next ounce if the weight goes over 1.5 ounces. What portion of the letters should be returned for insufficient postage? (filename: Transactional.mtw, column C7) SOLUTION: One-Sample T: Letter Weight (oz) Test of mu = 1.2 vs > 1.2 Variable Letter Weight (o N 100 Mean 1.07530 StDev 0.29691 SE Mean 0.02969 95% Lower Bound 1.02600 T -4.20 P 1.000 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. To determine the % of letters weighing >= 1.5 oz, use CALC>Probability Distributions>Normal, use the above mean and StDev to obtain: Cumulative Distribution Function Normal with mean = 1.0753 and standard deviation = 0.26969 x 1.49 P( X <= x ) 0.937938 1-.937938 = .062062% above 1.49 oz PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Q12. A project team was trying to find the factor that had the most influence on the number of calls answered. Over a period of time they collected and verified data covering 2 shifts, 2 operators, and 2 processes. Which factor has the biggest effect on the number of calls processed? (filename: Transactional.mtw, columns C9-C12) SOLUTION: Two-Sample T-Test and CI: Number, Operator Two-sample T for Number Operator Bill Cathy N 12 12 Mean 27.58 27.92 StDev 3.78 3.06 SE Mean 1.1 0.88 Difference = mu (Bill) - mu (Cathy) Estimate for difference: -0.333333 95% CI for difference: (-3.242996, 2.576329) T-Test of difference = 0 (vs not =): T-Value = -0.24 Both use Pooled StDev = 3.4367 P-Value = 0.814 DF = 22 P-Value = 0.000 DF = 22 Two-Sample T-Test and CI: Number, Process Two-sample T for Number Process Process A Process B N 11 13 Mean 25.09 30.00 StDev 2.84 1.73 SE Mean 0.86 0.48 Difference = mu (Process A) - mu (Process B) Estimate for difference: -4.90909 95% CI for difference: (-6.86763, -2.95055) T-Test of difference = 0 (vs not =): T-Value = -5.20 Both use Pooled StDev = 2.3052 Two-Sample T-Test and CI: Number, Shift Two-sample T for Number Shift Day Evening N 10 14 Mean 28.00 27.57 StDev 2.75 3.84 SE Mean 0.87 1.0 Difference = mu (Day) - mu (Evening) Estimate for difference: 0.428571 95% CI for difference: (-2.520066, 3.377209) T-Test of difference = 0 (vs not =): T-Value = 0.30 Both use Pooled StDev = 3.4340 P-Value = 0.766 DF = 22 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Only the factor Process has a significant effect (all data groups are normal, all subgroup variances are equal). Q13. A team working on inventory accuracy wanted to see if the changes they implemented had had a significant positive effect on the accuracy of inventory as measured by the weekly audit. Each week 100 inventory locations are selected at random and if the item is correct and the quantity is within 1% of the computer count then the location is judged as accurate. The team had 6 weeks of data from early in the project and the 6 most current weeks. Each number is the number of locations judged to be accurate by the auditor. Can the team claim to have made a significant improvement in inventory accuracy? (filename: Transactional.mtw, columns C14-15) SOLUTION: Inv Acc Initial 79 70 82 84 71 81 Inv Acc Now 76 78 75 84 88 86 Two-Sample T-Test and CI: Inv Acc Initial, Inv Acc Now Two-sample T for Inv Acc Initial vs Inv Acc Now Inv Acc Initial Inv Acc Now N 6 6 Mean 77.83 81.17 StDev 5.91 5.53 SE Mean 2.4 2.3 Difference = mu (Inv Acc Initial) - mu (Inv Acc Now) Estimate for difference: -3.33333 95% CI for difference: (-10.69706, 4.03039) T-Test of difference = 0 (vs not =): T-Value = -1.01 Both use Pooled StDev = 5.7242 P-Value = 0.337 DF = 10 There is no statistically significant difference in performance. Q14. A New Product Development team decided to test the effectiveness of two versions of a CAD program that claimed to find design criteria that did not meet recommended industry practice. Twelve designs were selected with varying levels of complexity. Each finding was verified by the design team as a legitimate problem. Is there a difference in the effectiveness of the two applications? PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. (filename: Transactional.mtw, columns C18-C19) SOLUTION: Paired T-Test and CI: CAD App 1, CAD App 2 Paired T for CAD App 1 - CAD App 2 CAD App 1 CAD App 2 Difference N 12 12 12 Mean 11.6667 13.5833 -1.91667 StDev 2.1034 3.1176 3.28795 SE Mean 0.6072 0.9000 0.94915 95% CI for mean difference: (-4.00573, 0.17240) T-Test of mean difference = 0 (vs not = 0): T-Value = -2.02 P-Value = 0.068 There is no statistical difference in performance at the 5% significance level. However, there is a difference at the 10% (alpha = 0.10) significance level. Q15. A project team in one plant found that a department running the same process but in another plant was able to complete the required paperwork in less time than in their plant. To test the theory that the layout of the paper forms was a critical factor the team set up a trial. Five operators from each plant would complete paperwork for a production run using Layout A one time and Layout B the other time. The order in which they used the forms was randomized. Is there a difference in the time required to complete the documentation based on layout? (filename: Transactional.mtw, columns C21-C23) SOLUTION: Operator 1 2 3 4 5 6 7 8 9 10 Layout A 11 43 14 84 52 37 44 57 54 76 Layout B 25 46 43 77 56 45 46 62 58 63 Paired T-Test and CI: Layout A, Layout B PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Paired T for Layout A - Layout B Layout A Layout B Difference N 10 10 10 Mean 47.2000 52.1000 -4.90000 StDev 23.3181 14.2552 11.29848 SE Mean 7.3738 4.5079 3.57289 95% CI for mean difference: (-12.98244, 3.18244) T-Test of mean difference = 0 (vs not = 0): T-Value = -1.37 P-Value = 0.203 There is NO difference in the time required to complete the documentation based on layout. Q16. Two people assigned to review documents for errors were given 12 completed documents. Each reviewer examined the 12 documents and recorded the error count. Is there a difference in ability to identify errors between these two reviewers? (filename: Transactional.mtwm columns C25-C27) SOLUTION: Document 1 2 3 4 5 6 7 8 9 10 11 12 Reviewer 1 5 6 10 6 8 4 10 4 7 11 7 7 Reviewer 2 8 6 7 5 4 4 3 5 3 5 6 4 Paired T-Test and CI: Reviewer 1, Reviewer 2 Paired T for Reviewer 1 - Reviewer 2 Reviewer 1 Reviewer 2 Difference N 12 12 12 Mean 7.08333 5.00000 2.08333 StDev 2.31432 1.53741 2.93748 SE Mean 0.66809 0.44381 0.84798 95% CI for mean difference: (0.21695, 3.94972) T-Test of mean difference = 0 (vs not = 0): T-Value = 2.46 P-Value = 0.032 There is a statistically significant difference. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Q17. A pest control company tested a new insecticide for effectiveness on mosquitoes with the following results: Mosquitoes Killed Surviving New Formula 95 13 Old Formula 81 17 Brand A 73 21 Brand B 72 22 What is the conclusion about the new formula? (filename: Transactional.mtw, columns C29-C31) SOLUTION: Chi-Square Test: Killed, Surviving Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Killed 95 87.99 0.558 Surviving 13 20.01 2.456 Total 108 2 81 79.84 0.017 17 18.16 0.074 98 3 73 76.58 0.168 21 17.42 0.737 94 4 72 76.58 0.274 22 17.42 1.206 94 Total 321 73 394 1 Chi-Sq = 5.491, DF = 3, P-Value = 0.139 Effectiveness is independent of formula used. Q18. A purchasing department was evaluating the possibility of using a new supplier of business cards. Over a six month period orders were sent to the current supplier and two proposed suppliers. Should the purchasing department make the change? Correct Order Problem Orders ACME Cards 33 7 Omega Cards 57 16 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Current Supplier 211 18 (filename: Transactional.mtw columns C33-35) SOLUTION: Chi-Square Test: Correct Orders, Problem Orders Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Correct Orders 33 35.20 0.138 Problem Orders 7 4.80 1.014 2 57 64.25 0.818 16 8.75 6.004 73 3 211 201.55 0.443 18 27.45 3.255 229 Total 301 41 342 1 Total 40 Chi-Sq = 11.672, DF = 2, P-Value = 0.003 1 cells with expected counts less than 5. The outcome of orders is dependent on supplier. Looking more closely, the current supplier’s observed correct orders exceeds the expected correct orders, and the problem orders observed is less than expected. The opposite is true for both ACME and Omega. Don’t change suppliers! Q19. A project team was evaluating the performance of projects. The team collected results on the relative success of new products and associated variables Product Type, Division, and Region. Is success dependent on any of these association variables? (filename: Transactional.mtw, columns C37-C42) Product Type Low End Middle High End Low End Middle High End Low End Middle High End Division Meters Meters Meters Systems Systems Systems Meters Meters Meters Region US US US US US US ROW ROW ROW Exceeded Goal 0 2 2 1 1 3 0 1 2 Met Goal 5 4 5 4 2 3 3 4 2 Did not meet 2 1 1 1 1 0 3 2 1 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Low End Middle High End Systems Systems Systems ROW ROW ROW 0 1 2 3 2 2 2 1 0 SOLUTION: The format above is typical of a download from a company database. The data is discrete and the desired test is a Contingency Table (chi-square) test. In order to run the test in Minitab the data must be summarized. For this problem, the data must be summarized three times so that Product Type, Division, and Region can be assessed. A tool to assist this is Stat>Basic Stats>Statistics; then check only the sum, uncheck the defaults. Product Type_1 High Low Med Exceeded Goal_1 9 1 5 Met Goal_1 12 15 12 Did Not Meet_1 2 8 5 Division_1 Meters Systems Exceeded Goal_2 7 8 Met Goal_2 23 16 Did not meet_2 10 5 Region_1 ROW US Exceeded Goal_3 6 9 Met Goal_3 16 23 Did not meet_3 9 6 From here, proceed with the contingency table analysis. Product Type 1 Did Exceeded Met Not Goal_1 Goal_1 Meet_1 9 12 2 5.00 13.00 5.00 3.200 0.077 1.800 Total 23 2 1 5.22 3.409 15 13.57 0.152 8 5.22 1.484 24 3 5 4.78 0.010 12 12.43 0.015 5 4.78 0.010 22 Total 15 39 15 69 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chi-Sq = 10.157, DF = 4, P-Value = 0.038 2 cells with expected counts less than 5. {this warning requires us to consider the results with caution. With expected cell counts less than 5 some of the accuracy of the test is diminished. Division Exceeded Goal_2 1 7 8.70 0.331 Met Goal_2 23 22.61 0.007 Did not meet_2 10 8.70 0.196 Total 40 2 8 6.30 0.456 16 16.39 0.009 5 6.30 0.270 29 Total 15 39 15 69 Chi-Sq = 1.268, DF = 2, P-Value = 0.530 Region Exceeded Goal_3 6 6.74 0.081 Met Goal_3 16 17.52 0.132 Did not meet_3 9 6.74 0.758 2 9 8.26 0.066 23 21.48 0.108 6 8.26 0.619 38 Total 15 39 15 69 1 Total 31 Chi-Sq = 1.764, DF = 2, P-Value = 0.414 From this analysis it appears the degree of success in projects is dependent only on the type of product in development. It may suggest that the design group responsible for the particular product type/line may be operating differently than the others. Of course it could be the markets in which the products compete. More investigation would be needed to make those determinations. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Q20. A project team looking into call center effectiveness collected data on the number of calls vs. the quality of the outcome. When the call did not reach a conclusion in an expected amount of time the call was judged to be a defect. Data was collected over 4 weeks. What is the prediction (regression) equation? (filename: Transactional.mtw, columns C44-C46) SOLUTION: Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Number of Calls 139 208 163 296 218 220 238 326 132 135 128 339 336 278 Defective Results 12 28 14 35 16 12 21 28 13 11 11 50 50 39 Day 15 16 17 18 19 20 21 22 23 24 25 26 27 28 Number of Calls 239 129 217 208 175 230 246 257 118 216 287 289 185 227 Defective Results 29 13 22 20 15 35 24 25 11 17 39 46 15 25 Fitted Line Plot Fitted Line Plot C4 = - 11.87 + 0.1632 Number of Calls C4 = 7.44 - 0.0246 Number of Calls + 0.000419 Number of Calls**2 S R-Sq R-Sq(adj) 50 6.04305 76.4% 75.5% S R-Sq R-Sq(adj) 50 5.85184 78.7% 77.0% 40 C4 C4 40 30 30 20 20 10 10 100 150 200 250 Number of Calls 300 350 100 150 200 250 Number of Calls 300 350 Regression Analysis: Defective Results versus Number of Calls The regression equation is Defective Results = - 11.87 + 0.1632 Number of Calls PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission. S = 6.04305 R-Sq = 76.4% R-Sq(adj) = 75.5% Analysis of Variance Source Regression Error Total DF 1 26 27 SS 3077.95 949.48 4027.43 MS 3077.95 36.52 F 84.28 P 0.000 Polynomial Regression Analysis: Defective Results versus Number of Calls The regression equation is Defective Results = 7.44 - 0.0246 Number of Calls + 0.000419 Number of Calls**2 S = 5.85184 R-Sq = 78.7% R-Sq(adj) = 77.0% Analysis of Variance Source Regression Error Total DF 2 25 27 SS 3171.33 856.10 4027.43 MS 1585.66 34.24 F 46.30 P 0.000 Sequential Analysis of Variance Source Linear Quadratic DF 1 1 SS 3077.95 93.38 F 84.28 2.73 P 0.000 0.111 Conclude that there is a statistically significant, positive relationship between number of calls and defective results. A non-linear model has marginally better explanatory power than a linear model. Q21. A project team was assigned to study the environmental system in a large office building. As might be expected, there seemed to be complaints across the entire temperature range. The team took some data to see how the number of complaints related to the room temperature. Based on the data shown, what recommendations could the team propose? (filename: Transactional.mtw, columns C48-C49) SOLUTION: Temperature Complaints Temperature Complaints Temperature Complaints Temperature Complaints 72 9 71 2 69.7 0 70.8 1 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 69.9 71.2 69.3 73.6 71.9 69.7 69.3 72.1 69.1 70.3 72.9 70.2 71.3 70.3 71.1 71 70.2 71.8 70.2 70.4 69.4 68.6 70.2 68.4 0 1 3 16 5 1 2 10 5 0 6 0 1 0 1 2 0 3 0 0 4 5 1 10 71.9 72.6 72.3 69.1 73.4 71.1 71.9 69.8 72.4 69.2 71.5 73.3 70.7 69.7 72.1 70.1 68.7 71.6 70.2 71.6 71.8 70.6 70.7 71.6 7 7 6 7 21 0 5 2 12 2 5 20 1 0 5 0 8 3 0 6 5 0 1 2 69.5 71.9 70.9 70.8 69.9 72.3 73.2 70.7 69.9 70.7 73 72.4 72.2 69.7 69.7 69.9 72.3 70.9 71.5 72 69.7 71.5 70.6 71.4 3 2 0 0 2 8 14 0 1 0 15 6 3 3 3 1 10 0 3 3 3 1 0 2 69 68.7 70.5 71.2 70.2 69.7 69.3 70.2 70 70 70.3 69.7 69.4 69.4 70.8 71.4 69.9 72.2 69.3 70.1 70.8 70 71.2 70.2 6 6 0 3 0 1 2 1 1 0 0 0 4 4 0 5 1 5 1 0 0 2 3 1 Fitted Line Plot Complaints = 9802 - 278.2 Temperature + 1.974 Temperature**2 S R-Sq R-Sq(adj) 20 1.71619 84.2% 83.9% Complaints 15 10 5 0 68 69 70 71 72 Temperature 73 74 Polynomial Regression Analysis: Complaints versus Temperature The regression equation is PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Complaints = 9802 - 278.2 Temperature + 1.974 Temperature**2 S = 1.71619 R-Sq = 84.2% R-Sq(adj) = 83.9% Analysis of Variance Source Regression Error Total DF 2 97 99 SS 1520.67 285.69 1806.36 MS 760.333 2.945 F 258.15 P 0.000 Sequential Analysis of Variance Source Linear Quadratic DF 1 1 SS 513.84 1006.82 F 38.96 341.84 P 0.000 0.000 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission. CHAPTER 19 – STATISTICAL TOOLS FOR DESIGNING FOR QUALITY Q1. Hospital equipment must have a reliable source of power, and ideally is protected from brownouts, surges, unstable (“dirty”) power, in addition to total power loss. Assume an Uninterruptible Power System (UPS) approved by the Food and Drug Administration for a heart-lung ventilator has a MTBF of 100,000 hours. What is the reliability for a period of 10 days? 30 days? 100 days? 365 days? Plot the reliability against time. Assume an exponential distribution. SOLUTION: Using the formula for reliability, R = e –t/µ, we obtain the following: t (Period of failure-free operation, in hours) 10 x 24 = 240 30 x 24 = 720 100 x 24 = 2400 365 x 24 = 8760 µ (MTBF) R (Reliability) 100,000 100,000 100,000 100,000 0.9976 0.9928 0.9763 0.9161 Scatterplot of Reliability vs Days 1.00 0.99 0.98 Reliability 0.97 0.96 0.95 0.94 0.93 0.92 0.91 0 100 200 Days 300 400 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Q2. A hard drive is a critical component of a computer. Many EIDE computer hard drives have a MTBF of approximately 400,000 hours, while MTBF for SCSI drives typically is higher, ranging through 1.2 million hours. Assuming continuous operation, what is the chance of each type of drive running for one year without failure? At what time period will approximately 50% of a population of each drive type have failed? SOLUTION: The reliability needs to found for a period of 1 year, or 365 x 24 = 8760 hours. Using the formula for reliability, R = e –t/µ, we obtain the following: Drive Type µ (MTBF, in hours) R (Reliability) EIDE 400,000 0.9783 SCSI 1,200,000 0.9927 Using the same equation, we can calculate the time period at which R=50%. Because e -0.693 = 0.500, a population of EIDE drives could be expected to reach 50% failed at 400,000 x 0.693 = 277,200 hours, or over 31 years. For SCSI drives, the period is calculated as 1,200,000 x 0.693 = 831600 hours, or over 94 years. Q3. The U.S. National Institute of Justice establishes standards for the ballistic resistance of body armor. The NIJ specifies for all armor types under prescribed test conditions a maximum depth of deformation of 44 mm (1.73 inches). Assume partial specifications as follows: a Type II –A bulletproof vest must be resistant to a 357 Magnum, 158 gr Jacketed Soft Point fired at a minimum velocity of 381 m/s, while a Type II vest is resistant to the same test ammunition with a minimum velocity of 425 m/s. A manufacturer is testing vests intended as Type II-A and Type II armor. a) What is the safety factor for the armor tested under Type II-A test conditions? b) What is the safety factor for testing under Type II test conditions? c) What proportion of manufactured armor meets or exceeds Type II-A specification? Type II specification? (Gryna 022805.MPJ; Columns C1, C2) SOLUTION: a) The safety factor may be calculated after finding the mean and standard deviation in vest deformation, shown below for both types of tests: Descriptive Statistics: 381 m/s, 425 m/s Variable 381 m/s 425 m/s N 100 100 Mean 35.833 41.918 StDev 1.215 1.301 Minimum 33.060 38.450 Maximum 38.621 45.156 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission. The safety factor for armor tested under the Type II-A conditions is (35.8 – 44.0) / 1.215) = -6.7. Taking the absolute value, this becomes 6.7 b) Following the same procedure as above, the safety factor for the armor tested under Type II conditions is abs((41.9 – 44.0)/1.301) = 1.6. c) None of the vests tested under the Type II-A conditions exceeded the 44mm deformation specification. The specification limit is more than 6 standard deviations away from the mean; in practical terms, 100% of the armor will meet or exceed the Type II-A specification. Cumulative Distribution Function Normal with mean = 35.833 and standard deviation = 1.215 x 44 P( X <= x ) 1.00000 For the Type II test, the cumulative distribution function indicates that approximately 94.5% of the armor will meet or exceed the 44 mm specification. Cumulative Distribution Function Normal with mean = 41.918 and standard deviation = 1.301 x 44 P( X <= x ) 0.945235 Q4. The FDA applies Recognized Consensus Standards when evaluating new medical devices. The air burst test is one such standard applied since the late 1970s to supplement the older, tensile test to measure the elasticity and strength of latex male condoms. Assuming a minimum air volume of 16L to pass inspection and a manufacturing process that produces condoms is tested, with results as shown. What is the safety factor relative to the specification? (Gryna 022805.MPJ, Column C4) SOLUTION: After finding the mean and standard deviation as follows: Descriptive Statistics: Air Burst Volume (L) Variable Air Burst Volume N 200 Mean 18.043 StDev 0.715 Minimum 16.202 Maximum 19.853 the safety factor may be calculated as (18.0 – 16.0) / 0.715 = 2.8. That is, the specification limit (assumed to be the worst stress expected) is 2.8 standard deviations away from the mean value. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission. BONUS QUESTION: The FDA applies Recognized Consensus Standards when evaluating new medical devices. The air burst test is one such standard applied since the late 1970s to supplement the older, tensile test to measure the elasticity and strength of latex male condoms. Assuming a minimum air volume of 16L to pass inspection and a manufacturing process that produces condoms is tested, with results as shown. Calculate the average air burst volume (L) and variance. What percentage of product will be expected to fail (assume leaking condoms are eliminated from the test)? (Gryna 022805.MPJ, Column C4) SOLUTION: Calculate the average air burst volume (L) and variance. Descriptive Statistics: Air Burst Volume (L) Variable Air Burst Volume N 200 N* 0 Variable Air Burst Volume Q3 18.586 Mean 18.043 SE Mean 0.0505 StDev 0.715 Minimum 16.202 Q1 17.504 Median 18.043 Maximum 19.853 Mean air burst volume = 18.0 L Variance = (0.0505)2 = 0.00255 What percentage of product will be expected to fail (assume leaking condoms are eliminated from the test)? The data are approximately normally distributed… PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Probability Plot of Air Burst Volume (L) Normal 99.9 Mean StDev N AD P-Value 99 Percent 95 90 18.04 0.7147 200 0.361 0.442 80 70 60 50 40 30 20 10 5 1 0.1 16 17 18 19 Air Burst Volume (L) 20 …so that an approximate Z score may be found as Z = (16.0 – 18.0) / 0.715 = -2.8. From a normal distribution table, the proportion of total area under the curve below 16.0 is 0.0026, meaning that about 0.26% of the products may be expected to fail. More precise estimation can be obtained from Minitab, which yields an estimate of 0.21%. Cumulative Distribution Function Normal with mean = 18.043 and standard deviation = 0.715 x 16 P( X <= x ) 0.0021360 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission. CHAPTER 20 – STATISTICAL PROCESS CONTROL Q1. Municipal public transportation systems typically include multiple bus lines. New bus routes and additional busses periodically are added or removed to accommodate changes in demand. To facilitate service decisions, a city bus route was monitored on consecutive days for one month and total travel time and passenger load recorded for the lunchtime run. Given the data, is there any evidence that travel time is changing? Number of passengers? (Gryna 022805.MPJ; Columns C14-16; Day, Travel Time, Passengers) SOLUTION: There is no clear evidence that either travel time or number of passengers is increasing or decreasing consistently over the 30 day period. Note, however, that there is a strong cycle in number of passengers, reflecting normal patterns of travel due to the work week (fewer passengers on the weekends). Individual value and moving range charts are below. I-MR Chart of Travel Time (minutes) Individual V alue U C L=82.26 80 75 _ X=73.37 70 65 LC L=64.47 1 5 9 13 17 21 O bser vation 25 29 33 37 41 U C L=10.93 M oving Range 10.0 7.5 5.0 __ M R=3.34 2.5 0.0 LC L=0 1 5 9 13 17 21 O bser vation 25 29 33 37 41 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission. I-MR Chart of Passengers 50 Individual V alue U C L=45.97 40 _ X=29.37 30 20 5 5 10 1 5 LC L=12.77 1 9 13 17 21 O bser vation 25 29 33 37 41 U C L=20.39 M oving Range 20 15 10 __ M R=6.24 5 0 LC L=0 1 5 9 13 17 21 O bser vation 25 29 33 37 41 Q2. The incidence of various adverse effects (side effects) from vaccines routinely is monitored via the Vaccine Adverse Event Reporting System. Such surveillance can facilitate trending and decisions regarding administration to populations at risk. Given the hypothetical counts (assume constant number of doses administered) across three adverse event types over time, what can you conclude? (Gryna 022805.MPJ; Columns C27-29, Year, Event Type, Adverse Events) SOLUTION: Control charts indicate no clear trends in adverse event reporting for the three event types over the 30 year period. Note, however, that in year 23 there was an unusually high number of rashes and injection site hypersensitivity incidents reported, well above the upper control limit. Reports of fever also were high in year 23, but just within the control limits. The data suggest that a special cause was present in year 23 that boosted the number of reported adverse events. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chart for adverse event of “Rash.” I-MR Chart of Adverse Events Individual V alue 250 1 200 U C L=166.3 150 _ X=113.0 100 LC L=59.6 50 1 4 7 10 13 16 O bser vation 25 22 19 150 M oving Range 1 28 1 100 U C L=65.6 50 __ M R=20.1 0 2 1 4 7 10 13 16 O bser vation 19 2 LC L=0 22 25 28 Control chart for adverse event “Injection site hypersensitivity.” PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission. I-MR Chart of Adverse Events 1 Individual V alue 180 U C L=164.7 150 _ X=118.8 120 90 LC L=72.9 60 1 4 7 10 13 16 O bser vation 19 22 25 28 1 M oving Range 80 1 60 U C L=56.33 40 __ M R=17.24 20 0 LC L=0 1 4 7 10 13 16 O bser vation 19 22 25 28 Control chart for adverse event “Fever.” PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission. I-MR Chart of Adverse Events Individual V alue 300 U C L=294.6 275 _ X=249.4 250 225 LC L=204.2 200 1 4 7 10 13 16 O bser vation 19 22 25 28 M oving Range 60 U C L=55.54 45 30 __ M R=17 15 0 LC L=0 1 4 7 10 13 16 O bser vation 19 22 25 28 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.