Statistical Quality Design & Control Fall 2003 Odette School of Business University of Windsor

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1 Name (print, please) ID Statistical Quality Design & Control 7-5 Fall Odette School of Business University of Windsor Midterm Exam Solution Wednesday, October 15, 5: 6:5 pm Instructor: Mohammed Fazle Baki Aids Permitted: Calculator, straightedge, and a one-sided formula sheet. Time available: 1 hour minutes Instructions: This exam has 17 pages including this cover page and pages of tables. Please be sure to put your name and student ID number on each page. Show your work. Grading: Question Marks: 1 /1 /15 /1 /6 5 /1 6 /1 Total: /65

2 Question 1: (1 points) Circle the most appropriate answer 1.1 Specifications state product or service characteristics in terms of a target value or dimension. a. statistically computed b. randomly generated c. desired d. six sigma 1. The enable problem solvers to answer the question, How will we know the right changes have been made a. pareto chart b. measures of performance c. why why diagram d. continuous improvement 1. The helps identify causes for nonconforming or defective products or services. a. performance measure b. continuous improvement c. control chart d. cause-and-effect diagram 1. The is a graphical technique that is used to analyze the relationship between two different variables a. scatter diagram b. pareto chart c. histogram d. control chart 1.5 If type I error increases, which per period cost will increase? a. Sampling cost b. Searching cost c. Cost of operating in out-of-control condition d. a and b 1.6 If type II error increases, which per period cost will increase? a. Sampling cost b. Searching cost c. Cost of operating in out-of-control condition d. a and b 1.7 If a sample average falls outside the control limits, a. a rework/scrap process is initiated b. a Type I error occurs c. the process is out of control d. a search for an assignable cause is initiated

3 1.8 If an item measurement falls outside the specification limit a. a rework/scrap process is initiated b. a Type I error occurs c. the process is out of control a. a search for an assignable cause is initiated 1.9 Why and R charts are used instead of and s chart? a. For a large subgroup, R chart loses too much information. b. If the process mean changes, standard deviation changes, but not range. c. Excel produces good and R charts, but not and s chart d. a and b 1.1 The purpose of a Chi-square test is to check if the a. process is in control b. item is defective c. process is capable d. sample averages fits the assumed pattern

4 Question : (15 points) Control charts for and R are kept on the manufacture of a drive shaft. After subgroups have been drawn and inspected, and the diameters recorded, the following are computed: 6mm and R mm. The subgroup size is 5. a. ( points) Assuming the process is in a state of control, compute the and Rchart central line and control limits for the next production period. 6 R 17.5 n 5, A.577 (See Appendix ), mm, R.875 mm m m UCL D UCL, D R + A D R (See Appendix ) R.11 (.875).59mm, LCL A R.577(.875) (.875) mm, LCL D R (.875) mm R.951mm b. ( points) Estimate the standard deviation σ of the process assuming that it is in statistical control. d.6 (See Appendix ) R.875 σ.76 mm d.6 c. ( points) If the specified nominal dimension and tolerances were.±.75 mm, what proportion of items are too large? What proportion of items are too small? Proportion of items too large P USL P ( ) ( ).75 µ, σ.75 P z P.76 (See Appendix 1) [ ]..% Proportion of items too small P LSL P.76 ( z 1.996) ( ) (.5 µ, σ.76).5 P z P.76. (SeeAppendix 1).% ( z 1.996)

5 d. ( points) Estimate the standard deviation σ of the sample mean. σ.76 σ.168 mm n 5 e. ( points) Find the probability that a point will fall outside the control limits on the chart when there is no change in the process mean. n 5, µ mm, σ.168 mm, UCL.59 mm, LCL Probabilit y P P ( Type I error), α.951mm ( UCL or LCL ) P( UCL ) + P( LCL ) (.59 µ, σ.168 ) + P(.951 µ,.168 ) P z + P z P (See Appendix 1) [ ] + [.1 ].6 σ ( z ) + P( z ) f. ( points) Suppose that a sudden change in the process occurs that increases µ by. mm but does not change σ after the change in the process mean. ( ) Probabilit y Type II error, β P P. Find the probability that a point will fall outside the control limits on the chart ( LCL UCL µ '., σ.168 ) ( µ '., σ.168 ).951. P.168 P ( z.67 ) P( z.67 ) P( z ).7 (See Appendix 1) z.168 5

6 Question : (1 points) Control charts for and s are maintained on the resistance in ohms of a certain rheostat coil based on a subgroup size of 9. After subgroups,, and s 969. a. ( points) Determine the central lines and -sigma control limits for the and s charts., s 1,, s m m n 9, A 1.(See Appendix ) UCL B UCL + A s 1, , B s B 1.761(See Appendix ) (.5) 1,5.79, UCL A s 1, 1.(.5) ( ).67, LCL B s.9(.5) s s b. ( points) Estimate the value of σ, assuming that the process is operating in statistical control. σ c s c. ( points) Assuming that the distribution generated by the process is approximately normal, what proportion of the rheostat coils meets specifications of 1, ± 6 ohms? Proportion of coils meeting specification P P ( LSL USL) P( 9 1, 6 µ 1,, σ 5) 9 1, 1, 6 1, P z 5 5 (. z. ) P( z. ) P( z. ) (See Appendix 1) % d. ( points) Among the subgroups, subgroup averages are within 1,±1 ohms. However, the following are subgroup averages outside 1,±1 ohms: 97, 985, 115, 15, 98, 16, 99, 955, 1, 965 Is the data sufficient to compute revised control limits of the control chart? If so, compute the revised control limits of the control chart. Since the LCL 975 and UCL 1, 5, any point with average less than 975 or more than 1,5 will be removed. So, the data is sufficient to compute the new centreline. d, ( ), 6.15,985 new m m 6 d However, to compute the revised control limits, we need to compute the standard deviation of the subgroup averages after removing the subgroups whose means lie outside (975,15). Hence, the data is not sufficient to compute the revised control limits. Or, with the previously computed standard deviation, ( 5) 1,. 56, LCL A ( 5) UCL + Aσ σ 6

7 Question : (6 points) A spherical plastic ball is to be produced against a specification of 1.7±.1 cm (diameter). Suppose that the process is in control with a µ and σ.. a. ( points) Calculate the process capability index C p C p USL - LSL 6σ (. ).8 b. ( points) What is your estimate of the percent defective product? P ( a randomly selecteditemisdefective ) P( USL or LSL) P( USL) + P( LSL) P( 1.71 µ 1.696, σ.) + P( 1.69 µ 1.696, σ.) P z + P z P.. (SeeAppendix 1) [ ] + [.668].67 Hence, 6.7% items are defective. ( z.5) + P( z 1.5) 7

8 Question 5: (1 points) Using the information shown below, compute per period total cost of sampling, searching, and operating in out-of-control condition. Economic design of -bar control chart Sampling cost, a1 $per item Searching cost, a $6per search Cost of operating out of control, a $9per period Prob(out-of-control in one period), π.15 Number of σ by which mean shift, δ 1.5 (when the process is in out-of-control condition) n 5 K.75 α Φ( k).6 βφ(k δ sqrt(n)) Φ(-k δ sqrt(n)).67 E(T)(1 π)/π 7. E(S)1/(1-β). E(C)E(T)+E(S) 1. Sampling cost per cycle a1* n*e(c) Search cost per cycle a*(1+α E(T)) 65. Operating cost per cycle a*e(s) Total cost per cycle, S Total cost per period, S S1/E(C) 8. (. 75) (. ) 6 α Φ. (See Appendix 1) ( ) Φ( ) Φ(. 57) Φ( 5. 6) β Φ E( T ) 7 periods E( S ). periods E C periods ( ) Sampling cost per cycle ()(5)(1.)$1,5.555 Search cost per cycle 6[1+(.6)(7)]$65. Operating cost per cycle 9(.)$,77.7 Total cost per cycle 1, ,77.7 $, Total cost per period,857.18/1. $8. 8

9 Question 6: (1 points) RM Manufacturing makes thermometer for use in the medical field. These thermometers, which read in degrees Celsius, are able to measure temperatures to a level of precision of two decimal places. Each hour, RM manufacturing tests a subgroup of 7 randomly selected thermometers in a solution that is known to be at a temperature of. C. RM manufacturing conducts a test for a total of hours and records average temperature of each of subgroups. The sample data is used to estimate the mean and standard deviation of the subgroup average temperature. The sample data shows that the subgroup average temperature has a mean,. C and a standard deviation, s.5 C. The frequency distribution of the subgroup average temperature is shown below: Lower limit Upper limit Frequency Can we conclude at the.5% level of significance that the subgroup average temperature is normally distributed with a mean, µ. C and a standard deviation, σ.5 C. Use chi-square test. a. ( points) State the null and alternate hypotheses H : The average temperatures of subgroups with size 8 are normally distributed with σ.5 C H A : The average temperatures of subgroups with size 8 are not normally distributed with and σ.5 C b. (7 points) Compute the chi-square test value. µ. C and µ. C Class Interval ( l u) Observed frequency f µ z u σ P ( x u) P( l x u) Φ( z) Expected frequency f e Chi-Square statistic χ Total.877 (Continued ) 9

10 c. ( points) State the conclusion of the test and interpret the conclusion. Degrees of freedom, ν k 1 p 7 1 χ critical χ ν, α χ, (See the χ table) χ test.877 > 11.1 χ critical Conclusion: Reject H Interpretation: There is enough statistical evidence to conclude at a level of significance α. 5 that the average temperatures of subgroups with size 7 are not normally distributed with µ. C and σ.5 C. The test fails. 1