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1 Many manufacturing problems involve the accurate matching of machine parts such as shafts that fit into a valve hole. A particular design requires a shaft with a diameter of 22.000 mm, but shafts with diameters between 21.900 mm and 22.010 mm are acceptable. Suppose that the manufacturing process yields shafts with diameters normally distributed with a mean of 22.002 mm and a standard deviation of .005 mm. What is the probability of producing an acceptable shaft? (22.01 – 22.002)/0.005 = 1.6 = z-value 22.01 is 1.6 standard deviations above the mean (21.9 – 22.002)/0.005 = -20.4 = z-value 21.9 is 20.4 standard deviations below the mean Probability of being less than 1.6 σ above the mean is 0.9452 Probability of being less than 20.4 σ below the mean is 0.0000 Probability of being between the two points is 0.9452 – 0.0000 = 0.9452 What diameter will be exceeded by 95% of the shafts? If 95% of the shafts are greater than this value, then 5% are less. So, look up probability 0.05 in normal table and get z-value of -1.645. So, the diameter is 1.645 standard deviations below the mean. 22.002 – 1.645 * 0.005 = 21.99378 If a random sample of 36 shafts is chosen from the above process, what is the probability of the sample mean being above 20.003 mm? (22.003 – 22.002)/(0.005/√36) = 1.2 = z-value 22.003 is 1.2 standard deviations above the mean Probability of being less than 1.2 σ above the mean is 0.8849 Probability of being greater than 1.2 σ above the mean is 1.0000 - 0.8849 = 0.1151 2 McDonalds claims that their quarter-pounders have an average weight of at least .25 lbs. The weight standard deviation is known to be .05 lbs. State the null and alternative hypotheses. Null: μ ≥ 0.25 pounds Alternative: μ < 0.25 pounds What test statistic will be used to test the null hypothesis? Sample mean A random sample of 36 quarter-pounders indicates a sample weight of .2475 lbs. At the 0.01 level of significance, is there evidence that the mean weight is less than .25 lbs? This is a one-tail test with a lower critical value, look up 0.01 in the normal table and find a z-value of -2.33. The lower critical value is 2.33 standard deviations below the mean. Lower critical value = 0.25 – 2.33 * (0.05/√36) = 0.230583 Since the sample mean of 0.2475 is greater than the lower critical value, the sample statistic is in the region of nonrejection. Therefore, we accept the null and conclude the average weight is at least a quarter pound. There is not evidence that the mean weight is less than 0.25 pounds. 3 The quality-control manager at a light bulb factory needs to determine whether the average life of a large shipment of light bulbs from a new process is equal to the specified value of 375 hours. The process standard deviation is 95 hours. State the null and alternative hypothesis. Null: μ = 375 hours Alternative: μ ≠ 375 hours What test statistic will be used to test the null hypothesis? Sample mean A random sample of 64 light bulbs indicates a sample mean life of 360 hours. At the 0.05 level of significance is there evidence that the mean life is different from 375 hours? This is a two-tail test with lower and upper critical values. Divide 0.05 by 2 and get 0.025, look up 0.025 in the normal table and find a z-value of -1.96. The lower critical value is 1.96 standard deviations below the mean and the upper critical value is 1.96 standard deviations above the mean Lower critical value = 375 – 1.96 * (95/√64) = 351.725 Upper critical value = 375 + 1.96 * (95/√64) = 398.275 Since the sample mean of 360 is between the two critical values, the sample statistic is in the region of nonrejection. Therefore, we accept the null and conclude the average life of a light bulb is equal to 375 hours. There is not evidence that the mean life is different from 375 hours. 4 A local farmer asked me for help with his feed operation. The manufacturer set up the feed dispenser to dispense 500 lbs of chop. The farmer fears that his young son has been playing with the feed dispensing dial and the setting has been changed. Thirty chop weights are measured and the data is contained in the Feed Weights column of the StatProblems Minitab worksheet. Using Minitab, determine the 99% confidence interval for the true mean weight. 508.27 to 516.40 Interpret the 99% confidence interval. We are 99% confident that the average chop weight is between 508.27 and 516.40. At the 99% confidence level, would I conclude that the young son has changed the setting? Why? Yes, because we are 99% confident that the average weight is greater than 500 pounds. Use Minitab to construct the following three graphs of the chop weights: Histogram Dot Plot Run Chart 5 I charted the points that my favorite Wolfpack player scored over 15 games last year. The data is contained in the Points column. Using Minitab determine the following measures: Mean 22.33 Median 22.00 Mode First Quartile 18.00 Third Quartile 28.00 Range 24.00 Variance 50.38 Standard Deviation 7.10 A company has been having a large amount of late deliveries to customers. In order to improve deliveries, the company has been tracking reasons for late deliveries. The reasons and data are contained in the Late Reason and Quantity columns. Use Minitab to construct the following graphs: Bar Chart Pie Chart Pareto Diagram 6 Pepsi has a bottling plant that is dispensing pepsi in 20 ounce bottles. They are concerned about the variation among two dispensing machines, Machine 1 and Machine 2. They measured fifteen bottles from the two machines and the data is in the Machine 1 and Machine 2 columns. Use Minitab to run an F Test for the Ratio of Two Variances to determine if the amount of variation is equal between the two machines with a level of significance of 0.01. The p-value is 0.000 which is less than 0.01 so we reject the null and conclude that the variation among the machines is not equal. If the probability of a certain employee producing a defective part is 0.07 and the employee produces 20 parts per day, use Minitab to compute the following probabilities (Assume independent parts): Producing two defects 0.252141 Producing less than 3 defects 0.839 Producing at least four defects 1.00000 – 0.95287 = 0.04713 A call center receives an average of 12.35 calls per hour. Assume the calls follow the Poisson distribution. Use Minitab to determine the following probabilities: Receiving exactly 11 calls 0.110572 Receiving no more than 14 calls 0.739476 Receiving more than 12 calls 1.000000 – 0.536004 = 0.463996