Mid term

1. The unit costs ($) for manufacturing a component for a washing machine are as follows: Direct labor: 20; direct materials: 5; indirect labor and materials: 20% of direct labor; fixed and administrative costs: 30; selling costs: 10. a. Assuming a 90% first-pass yield, what should the unit selling price be if the manufacturer desires a 20% profit margin for conforming product? (4) b. The manufacturer has identified a secondary market for nonconforming product, that they can sell at $75/unit. Conforming product they wish to sell at 30% above costs. What is the expected profit per unit sold if the company has the some first-pass yield as in part a)? (4) (1) Direct labor = $20 Direct Materials = $5 Indirect labor and materials = $20 x 20% = $4 Fixed and admin cost = $30 Selling costs = $10 Total cost = Direct labor + Direct Material + Indirect labor and materials + Fixed admin costs + Selling cost = $20 + $5 + $4 + $30 + $10 = $69 Profit Margin = (Selling Price x Yield - Total Costs) / Total Costs 20% = (Selling Price x 90% - 69) / 69 Thus, Selling Price = $92 (2) Revenue per unit = (non confirming % x Selling Price) + (Yield x Selling Price for confirming product) = [(100-90) % x 75] + [90% x (69 + 30%)] = 7.50 + 80.73 = $88.23 Expected Net profit = $88.23 - $69.00 = $19.23 (3) Revenue per unit = (non confirming % x Selling Price) + (Yield x Selling Price for confirming product) = [(100-98) % x 75] + [98% x (69 + 30%)]

= 1.50 + 87.91 = $89.41 Expected Net profit = $89.41 - $69.00 = $20.41 2. p1 = 0.90 p2 = 0.95 p3 = 0.80 u1 = $20 u2 = $50 u3 = $40 i1 = $0.70 i2 = $1.50 i3 = $3 First-pass yields (pi), unit processing cost (ui), and unit inspection cost (ii) for each operation are shown above. a. If no inspection is performed, what is the unit cost for a conforming part? (3) b. What should be the unit selling price of a conforming product if a 15% profit margin is desirable? (2) c. Suppose the inspection process correctly identifies each part. Assume that inspection is conducted after Operations 2 and 3. Nonconforming parts are not forwarded to the next operation. What is the unit cost per conforming part? (5) d. A process improvement team improved the first pass yield of Operation 3 to 0.95, and also reduced the unit processing cost to $35. If no inspection is performed in any operation, what is the unit cost for a conforming part? By what percentage has capacity increased now compared to that in part a? (4) (1) (Of an object or action) require the payment of (a specified sum of money) before it can be acquired or done. (2) estimate the price of. We now consider the case where inspection is conducted only after the third operation: (3) As first-pass yields, processing costs per 1000 parts is =1000(20) +1000(50) +1000(40) +1000(.95)(.90)(.80) (20) = $123680 (4) Inspection costs per 1000 parts = 1000 *3 = $3000 3 2 1

(b) Df=9, n2=10, x2’=10.3, s2=2.263 At 99%, ta/2=3.25 The confidence level is 10.3+/- 3.25*(2.263/sqrt (10)) = (7.974,2.626) We are 99% confident that the true mean loading time is between 7.974 to 12.626. (c) sp=sqrt((n-1) s1^2+(n2-1) s2^2)/(n1+n2-2)) =3.15 Se= sqrt (3.15*3.14*(1/8+1/10)) =1.494 Df=8+10-2=16, At 98%, ta/2=2.584, The confidence level is (15.88-10.3) +/- 2.584*(1.494) = (1.72,9.44) Assumption Randomly selected. Independent Normally distributed. (d) H0: u1=u2 H1: u1>u2 Df=16 P=0.001 A=0.05, p<a So, reject h0. (e) The probability of conducting that the mean loading time reduced after periods when actually it is not reduced is 0.001. 4. To determine customer acceptance of a product, from Company A, in a random sample of 200 people, 80 said that they prefer the product. a. Find a 95% confidence interval for the proportion of people that prefer the product. Interpret it. (3) b. What assumptions are made in part a)? (2)

c. A similar product from Company B was chosen. Based on a sample of 300 people, 160 people preferred the product. Find a 98% confidence interval for the difference in the proportion of people, that prefer the product, between that of Company A and Company B. Interpret it. (4) d. Test hypothesis to determine if the product preference is higher for Company B than that of Company A. Use a level of significance of 0.05. Clearly state the null and alternative hypothesis, test statistic, p-value, and your decision. (4) (a) Na=200, xa=200 Pa=80/200=0.4 E=z*sqrt(p*(1-p)/n) Area=(1+c)/2=0.9750 According to z table, Area = 0.9750 corresponds to 1.9 and 0.06, thus z critical value = 1.96 Zc=1.96 E=1.96*sqrt (0,0012) =0.0679 Thus, the confidence interval is (0.3320,0.4679) (b) Sample is selected at random and independently Number of successes = 80 > 10 and Number of failures = 200-80= 120 > 10 (c) Nb=300, Xb=160 Pa=80/200=0.4, Pb=160/300=0.5333 Area= (1+0.98)/2=0.9900 Area 0.9901 is closest to 0.9900, thus corresponding z value is 2.3 and 0.03 Zc=2.33 Thus, e=2.33*sqrt (0.0012+0.0008296) =0.1050 Thus, the confidence interval is (-0.2383, -0.0283) We are 98% confident that true difference in the population proportion of people, that

'O ij ' matrix (observed frequencies) Quality\ Vendor Perfect parts Made-4-U 24 Hours parts Total Rejected 50 50 70 170 Perfect 80 70 90 240 Acceptable 20 30 40 90 Total 150 150 200 500 'E ij ' matrix (Expected frequencies) Quality\ Vendor Shift-1 Shift-2 Shift-3 Rejected 150*170/500 = 51 150*170/500 = 51 2000*170/500 = 68 Perfect 150*240/500 = 72 150*240/500 = 72 200*240/500 = 96 Acceptable 150*90/500 = 27 150*90/500 = 27 200*90/500 = 36 '(O ij - E ij ) 2 / E ij ' matrix Quality\ Vendor Shift-1 Shift-2 Shift-3 Total Rejected (50 - 51) 2 /51 = 0.02 (50 - 51) 2 /51 = 0.02 (70 - 68) 2 /68 = 0.06 0.1 Perfect (80 - 72) 2 /72 = 0.89 (70 - 72) 2 /72 = 0.06 (90 - 96) 2 /96 = 0.38 1.32 Acceptable (20 - 27) 2 /27 = 1.81 (30 - 27) 2 /27 = 0.33 (40 - 36) 2 /36 = 0.44 2.59 Total 2.72 0.41 0.88 4.01

Significance level (α) = 5% The Minitab output is the test statistic = 4.01 Degrees of freedom= (No. of quality classes - 1) *(No. of vendors - 1) = (3 - 1) *(3 - 1) = 4 the P-value of the test = CHISQ.DIST. RT (4.01,4) = 0.405 > α So, the null hypothesis cannot be rejected hence Part quality and Vendor are independent factors 6. The management of a retail outlet is interested in reducing the wait time to check-out of its customers. An estimate of the standard deviation of waiting times is about 6.5 minutes. If management wishes to estimate the mean waiting time to within 1.2 minutes, with a probability of 0.98, how many samples must be selected? (5) Thea=6.5, x’=1.2 Z at o.98=2.33 And at significance level of 0.02, X]+/- z*thea/sqrt(n) <=1.2 Thus, 159.28<=n N should be 160 7. At a call center, the response time (in seconds) to route a call to an appropriate technical person, is shown below for 30 samples taken consecutively. a. Is the distribution of response times normal? Use α = 0.05 to make your decision. Specify null and alternative hypothesis. (3) b. Is the sequence of response times (listed column wise) random? Use α =0.05. Is there any evidence of clustering, mixture, trend, or oscillation? (3) c. Create a box plot and comment on whether the distribution is symmetric. What is the inter-quartile range? What is this a measure of? (4) 49 60 46 58 59 57 50 53 55 62 60 54 52 63 49 64 48 55 47 66 50 66 47 56 58 58 54 68 53 54 (1)

H0: the sequence of random time is random H1: the sequence of random time is not random P>0.05, we do not reject the null hypothesis, The sequence of responce time is random, (3) From the Boxplot, we see that the distribution of responce times is Symmetric. Descriptive Statistics of the Data: 8. Be specific in your response to the discussion questions (3 points each): a. Name four characteristics you would consider in selecting a vendor for a financial institution that is considering outsourcing its information technology related services. Normally, randomly selected, Large sample, independence b. Discuss the concept of Six Sigma quality as a metric. What are other ways to interpret “Six-Sigma quality”? Six Sigma is an approach that provides organizations with tools to improve their business process capabilities. This increased performance and reduced process variation helps reduce defects and improve profits, employee morale and the quality Variabl e N Mean StDev Minimu m Q1 Median Q3 Maximu m IQR Data 30 55.7 6.13 46 50 55 60 68 10

of the product or service. "Six Sigma Quality" is a term commonly used to indicate that a process is well controlled (within the process limits of ±3s from the centerline on the control chart, and within the requirement/tolerance limits of ±6s from the centerline). c. In analyzing survey data from questionnaires, discuss how importance and satisfaction measures may be used for decision making. Measuring customer satisfaction enables companies to identify factors of dissatisfaction. By doing so, they can implement the necessary improvement initiatives before customers abandon the brand. d. What is Quality Function Deployment? Select a product or process of your choice. Name four “Whats” and four “Hows”. How is it used systematically by an organization? In QFD, quality is a measure of customer satisfaction with a product or a service. QFD is a structured method that uses the seven management and planning tools to identify and prioritize customers' expectations quickly and effectively. QFD is a focused methodology for carefully listening to the voice of the customer and then effectively responding to those needs and expectations. The QFD methodology focuses on the most important product or service attributes or qualities. These are composed of customer wows, wants, and musts. e. What is Failure Mode and Effects Criticality Analysis? Select a product or process of your choice. Name three failure modes. With limited resources, how do you decide which ones to address? How is it used by an organization? Failure mode and impact criticality analysis is a method developed to identify all possible failures in the design of an industrial process, product or service. Failure Mode - The possibility of a process failing or the possibility of a defect in the process that could hinder customers or have potential. Effects Analysis - Simply put, understanding and analyzing the results of failure. f. What is a Type I error and a Type II error in hypothesis testing? Explain clearly. If we reject the null hypothesis when it is in fact, true, we have made an error that is called a type I error. On the other hand, if we accept the null hypothesis when it is, in fact, false we have made an error that is called a type II error. In summary: Rejecting a null-hypothesis when it should have been accepted creates a Type I error. Accepting a null-hypothesis when it should have been rejected creates a Type II error. For example, rejecting a student when he is qualified is type I error qualifying a student when he is failed is type I error

3) After this, the analysis function is required. For example, how рrосess might affect customers or рrоduсt. 4) Identify the new рrосess, рrоduсt оr serviсe whiсh is to be сhоsen. 5) For each of them above, find all possible faults. 6) Next, identify each failure mode and determine the cause of each failure mode. 7) Determine the severity of each effect by feeding them their number from 1 (minimum effect) to 10 (which is unavoidable). 8) Аррlying соntrоl method. 9) Set the risk priority and impact severity. 10) Identify recommended actions and make recommendations to minimize all impacts