Practice questions from Chapter 3, 7, 10

QUALITY MANAGEMENT AND SIX SIGMA (Chapter 10) A manufacturing company has been inspecting units of output from a process. Each product inspected is evaluated on five criteria. If the unit does not meet standards for the criteria it counts as a defect for the unit. Each unit could have as few as zero defects, and as many as five. After inspecting 2,000 units, they discovered 33 defects. What is the DPMO measure for this process? Defects per Million Opportunities (D P M O) DPMO = 33 5 ∗ 2,000 ∗ 1,000,000 = 3,300 21.Design specifications require that a key dimension on a product measure 100 ± 10 units. A process being considered for producing this product has a standard deviation of four units. a. What can you say (quantitatively) regarding the process capability? C pk = min { UTL − X 3 σ , X − LTL 3 σ } = min { 110 − 100 3 ( 4 ) , 100 − 90 3 ( 4 ) } = min { .8333 , .8333 } = 0.8333 b. Suppose the process average shifts to 92. Calculate the new process capability. C pk = min { UTL − X 3 σ , X − LTL 3 σ } = min { 110 − 92 3 ( 4 ) , 92 − 90 3 ( 4 ) } = min { 1.5000 , .1667 } = 0.1667 c. What can you say about the process after the shift? Approximately what percentage of the items produced will be defective? Many defects will be produced. Assuming a normal distribution, the left tail (Prob{x<90}) is z = (90 - 92)/4 = -0.50, which corresponds to a probability of . 3085. The right tail (Prob{x>110} = 1-Prob{x<110})is z = (110-92)/4 = 4.5,

which approximately corresponds to a probability of .1-1 = 0. Therefore 0.3085+0=0.3085 or approximately 31% are outside the specification limits. 22.C-Spec, Inc., is attempting to determine whether an existing machine is capable of milling an engine part that has a key specification of 4 ± .003 inches. After a trial run on this machine, C-Spec has determined that the machine has a sample mean of 4.001 inches with a standard deviation of . 002 inch. a. Calculate the C pk for this machine. Process Capability Index( C pk larger than one indicates process is capable) C pk = min { UTL − X 3 σ , X − LTL 3 σ } = min { 4.003 − 4.001 3 ( .002 ) , 4.001 − 3.997 3 ( .002 ) } = min { .333 , .667 } = 0.333 b. Should C-Spec use this machine to produce this part? Why? No, the machine is not capable of producing the part at the desired quality level.

26.You are the newly appointed assistant administrator at a local hospital, and your first project is to investigate the quality of the patient meals put out by the food-service department. You conducted a 10-day survey by submitting a simple questionnaire to the 400 patients with each meal, asking that they simply check off that the meal was either satisfactory or unsatisfactory. For simplicity in this problem, assume that the response was 1,000 returned questionnaires from the 1,200 meals each day. The results are as follows: c. Construct a p-chart based on the questionnaire results, using a confidence interval of 95.5 percent, which is two standard deviations. p = 600 10 ( 1000 ) = .06 S p = √ p ( 1 − p ) n = √ .06 ( 1 − .06 ) 1000 = .0075 UCL =´ p +( 2 ) s p = .06 + 2 ( .0075 )= .075 LCL =´ p − ( 2 ) s p = .06 − 2 ( .0075 ) = .045

b. The chart indicates that the process is out of control. The administrator should investigate the quality of the patient meals.

29.The following table contains the measurements of the key length dimension from a fuel injector. These samples of size five were taken at one-hour intervals. Construct a three-sigma ´ X -chart and R -chart (use Exhibit 10.13) for the length of the fuel injector. What can you say about this process? ´ X = .499, ´ R = .037 n = 5 → A 2 = .58, D 3 = 0, D 4 = 2.11 Control limits for X-bar chart: UCL ´ X , LCL ´ X = ´ X ± A 2 ´ R = .499 ± .58 ( .037 ) .520,.478 Control limits for R chart: UCL = D 4 ´ R = 2.11 ( .037 ) .078 LCL = D 3 ´ R = 0 ( .037 )= 0