1 IOE 265 – Winter 10 Note: Please make sure you type in your answers in the spaces provided and submit this file electronically via the CTools website under Assignments. Save this file as ‘uniqname_lab_9’ with your own uniqname. Microtronic, Inc produces scan engines for applications in retail and manufacturing. Microtronic sells these scan engines to various providers of bar code systems. One of Microtronic’s customers (a manufacturer of complete bar code systems) recently reported failures of the scan engines in both incoming inspection tests and field failures. Microtronic has formed a project team to identify the root cause(s) of the problem, and to implement corrective actions that will prevent recurrence of this problem in the future. The team began by analyzing several failed units. The analysis team found that the failures of the motor were related to a failed weld joint between a bracket and one of the clamps in the motor. Based on historical analyses, Microtronic knows that two potential sources for failed weld joint defects are (1) insufficient joint strength between the clamp and bracket, and (2) impurities in the clamp material. Microtronic measures joint strength by utilizing a destructive weld test whereby the clamp is pulled until it separates from the bracket (the weld joint breaks). The force required to break the weld joint is referred to as the “shear force” and represents a critical performance output variable of the process. The design specification for shear force is a minimum of 13 lbf. If a unit requires less than 13 lbf to break the weld joint, the Mylar Motor is considered defective. If greater than 13 lbf, it is considered not defective. To evaluate shear force performance, Microtronic randomly selected 60 samples from the marketplace, and shipped them back to the company for analysis. The data file, “Lab 9 Data.MTW,” contains the results of the shear force analysis of these 60 returned samples. One column in the data file also identifies the production batch for each of the returned scan engines. Microtronic tracks this variable to determine whether product failures are related to a special cause in one of the production batch setups. Based on this information, carry out an analysis of the data to determine if the parts are meeting the shear force specification and if there are any differences between the production batches. Using methods you’ve learned so far in IOE 265, with the aid of Minitab, answer the questions on the next page. Lab 10: Hypothesis Testing CASE STUDY: Microtronic, Inc. 2 Note: When deciding which hypotheses to test, assume you are on the side of the consumers, and that what’s in the best interest of the consumers is in the best interest of the manufacturer as well. 1. Look at the data and obtain descriptive statistics for the shear force of all the 60 observations. What is the mean, median and standard deviation of the 60 observations? Now do the same for the two individual batches (P1, P2). (1.5 points) Batch Mean Median Standard Deviation All 13.970 13.961 1.350 P1 14.833 14.750 0.947 P2 13.107 12.786 1.127 2. Create a comparative box plot for shear force by production batch (P1, P2). You should have two box plots side by side. (0.5 points) 3. Are there any apparent differences between the two production batches? (0.5 points) Section 1: Descriptive Statistics 17 16 15 14 13 12 11 P1 S t r e n g t h P2 Boxplot of Strength Panel variable: Batch 3 4. Let µ 0 represent the true population (both batches) mean shear force. Perform a hypothesis test to investigate if there is conclusive evidence that the mean shear force (based on the 60 observations) meets the specification. Give the p-value of the test as well as a 95% confidence interval (or confidence bound, depending upon the test you choose) of µ 0 . (Assume all the data follows a normal distribution and use α=0.05; you should state the null and alternate hypotheses and justify the type of test you use; e.g. z-test vs. t-test, etc.) (2 points) 5. Let µ 1 and µ 2 represent the true mean shear force for P1 and P2, respectively. Perform hypothesis tests (same hypotheses as above) to check if the mean of each batch meets the specification, select the appropriate test based on the normality assumption and reduced sample size for each batch (use α=0.05). (2.5 points) 6. Do the test results support what was observed in Section 1? (0.5 point) Section 2: Inferential Analysis This is consistent with section one. It seems more likely that sample one will be above the specified requirement compared to sample two Yes, p1 is evenly distributed while p2 is not. The deviation in p1 is smaller than the deviation in p2. Also, p1 overall has more strength than p2. P1 has a higher mean and median. Null hypothesis: µ=13 Alternative hypothesis: µ>13 Type of test for P1: T-test Type of test for P2:T-test P-value for P1: 0.00 P-value for P2:0.303 95% CI for P1: 14.540 (lower bound) 95% CI for P2:12.758 (lower bound) Type of test: z test because the sample size is above 40 P-value :0.000 95% CI:13.684 (lower bound) 4 7. What is the power of your test above? if you wanted to detect departures from the shear force mean of 0.5 lbf, 1.0 lbf and 1.5 lbf? Use the same estimate for the standard deviation as in Question 1. (1.5 points) 8. What sample size would you need to detect a departure of the true shear force mean of +0.5 lbf? Assume the same sigma as in Question 1 and a power of 0.95. (0.5 point) 9. Is there a problem with either one of the batches? If so, which one is not meeting the specification? (0.5 point) Section 3: Conclusions Power to detect a difference of +0.5 lbf: 0. 0.889528 Power to detect a difference of +1.0 lbf: 0.999979 Power to detect a difference of +1.5 lbf: 1.00000 Sample size: 79 There is a problem with Batch two the true mean and median are below specs. The p- value also indicates that it lies below specifications as it is higher than .05. There is a problem with batch 2. The mean is 13 and the customer wants a mean of greater than 13 because they want to know which ones are good. Batch 2 had a lower mean and median, and a higher standard deviation and a lower confidence interval than batch 1Batch 2 continually showed less strength than batch 1. .Also, Batch 2 was not as consistent as batch 1, because it has a higher deviation. There is a problem with batch P2. It’s median value falls below 13, which means that some items in the batch are defective. Ryan Rindler Lab 0: Introduction to Minitab Statistical Software