<p>Homework #13 ME 368 - Engineering Measurements and Instrumentation Due in .pdf format, submitted in ecow2, by noon Thursday, Dec 10, 2009 Fall Semester 2009</p><p>HW Problem 13-1: (Objective: Combined chi-squared and student t analysis)</p><p>A sample of 5 ball bearings are pulled from a production line and their diameters are measured. The standard deviation of the diameters is 0.10 mm. </p><p> a) Determine the range that contains the population standard deviation in [mm] at 95% confidence.</p><p>2 2 [x/10] Using eq. 8.45 with ν = N – 1 = 4, Sx = 0.10 mm, and from table 8.8,  / 2 = 11.1, 1 / 2 = 0.484, obtain 0.060 mm < σ < 0.29 mm (P = 95%). </p><p> b) What is the percent chance that σ is greater than 0.29 mm?</p><p>[x/10] Since the probability that σ is < 0.060 mm should be equal to the probability that σ is > 0.29 mm and the sum of these two probabilities sum to 5%, the percent chance that σ is greater than 0.29 mm is 2.5%.</p><p> c) How many bearings N should be measured so that the population mean diameter can be estimated to within 0.001 mm at 95% confidence? Use Sx = 0.29 mm in your calculation.</p><p> t S t 2  ,P95% x  ,P95% t ,P95%  [x/10] = 0.001 mm. Sx = 0.29 mm. thus =0.00345 N = . N N 0.0000119</p><p>Using the same reasoning as in quiz 10, choose t ,P95% =1.96 from table 8.4. Compute N = 1.9602 = 322 824, or to 2 significant digits, N = 3.2 x 105. 0.0000119</p><p> d) Given the combined probabilities involved in your answer to (c) {there is a probability associated with Sx as well as a probability associated with the student t value you used in (c)}, do you think the N value you computed in (c) will result in i. exactly 95% confidence in an estimate within 0.001 mm, as was the target. ii. better than 95% confidence in an estimate within 0.001 mm OR iii. worse than 95% confidence in an estimate within 0.001 mm and why?</p><p>[x/10] I think better than 95% confidence. It would be exactly 95% confidence if Sx was really 5 0.29 mm for the huge N = 3.2 x 10 sample. Actually, there is only a 2.5% chance that Sx is 0.29 mm or bigger, as determined in (b); there is a 97.5 % chance that Sx is smaller than 0.29 mm and 5 in this (likely) case, N = 3.2 x 10 results in better than 95% confidence. HW Problem 13-2: (Objective: Use student’s t distribution to determine whether 2 samples are from the same population)</p><p>Some students you may know have been measuring the speeds of suction cup darts fired from toy guns. Imagine 10 dart speeds were measured for each of 2 guns, with the resulting data given at http://homepages.cae.wisc.edu/~ssanders/me_368/homework_problem_statements/fall_2009/dart_speeds.xls</p><p>In this problem we will explore the following question: does one of these guns shoot faster than the other? Follow example 8.8 Dunn, and determine:</p><p> a) Statistically, which gun might be faster, gun 1 or gun 2, and why?</p><p>[x/10] By averaging each gun speed, we see that gun 2 is on average 0.87 m/s faster than gun 2.</p><p> b) Statistically, what is the percent confidence that the gun you identified in (a) is actually faster?</p><p>I suggest working the entire problem in Excel and then reporting your answer in such a way that the numbers and formulas you used are visible to the grader in the .pdf you turn in.</p><p>[x/50] Based on the data, we can be 92% confident that gun 2 is actually faster than gun 1. In other words, there is an 8% chance that if we took more data, we could find that the guns are the same speed or that gun 1 is actually faster. An Excel version of the solution is available at http://homepages.cae.wisc.edu/~ssanders/me_368/homework_solutions_with_problem_statemen ts/fall_2009/HW13_darts.xls</p><p>See first the ASSIGNED VERSION at right, answer in cell K23</p><p>It is instructive to play with the RANDOM VERSION at left and the N = 100 version below. It would be nice if someone made a LabVIEW version of this so N can be varied as well…</p>