Probability - Solved Final Exam | MATH 3338, Exams of Probability and Statistics

Download Probability - Solved Final Exam | MATH 3338 and more Exams Probability and Statistics in PDF only on Docsity! First Exam Probability MATH 3338-10853 (Fall 2006) September 13, 2006 This exam has 2 questions, for a total of 0 points. Please answer the questions in the spaces provided on the question sheets. If you run out of room for an answer, continue on the back of the page. Name and UH-ID: 1. (a) Let x1, . . ., xn be a sample and c be a constant. Prove that 1. if y1 = x1 + c, . . ., yn = xn + c, then s 2 y = s 2 x, and 2. if y1 = c x1, . . ., yn = c xn, then s 2 y = c 2s2x, where s2x is the sample variance of the x’s and s 2 y is the sample variance of the y’s. (b) The article “A Thin-Film Oxygen Uptake Test for the Evaluation of Automotive Crankcase Lubricants” reported the following data on oxidation-induction time (min) for various commercial oils: 87 103 130 160 180 195 132 145 211 105 145 153 152 138 87 99 93 119 129 1. Calculate the sample variance and standard deviation. 2. If the observations were re-expressed in hours, what would be the resulting values of the sample variance and sample standard deviation? Answer without actually performing the re-expression. Solution: (a) 1. First, ȳ = 1 n n∑ i=1 yi = 1 n n∑ i=1 (xi + c) = 1 n n∑ i=1 xi + c = x̄ + c. Then s2y = 1 n− 1 n∑ i=1 (yi−ȳ)2 = 1 n− 1 n∑ i=1 (xi + c− (x̄ + c))2 = 1 n− 1 n∑ i=1 (xi−x̄)2 = s2x. 2. First, ȳ = 1 n n∑ i=1 yi = 1 n n∑ i=1 (c xi) = c 1 n n∑ i=1 xi = c x̄. Then s2y = 1 n− 1 n∑ i=1 (yi−ȳ)2 = 1 n− 1 n∑ i=1 (c xi − c x̄)2 = c2 1 n− 1 n∑ i=1 (xi−x̄)2 = c2 s2x. Page 1 of 4 Please go to the next page. . . First Exam (continued) MATH 3338-10853 (Fall 2006) September 13, 2006 (b) 1. First compute∑ xi = 87 + 103 + 130 + 160 + 180 + 195 + 132 + 145 + 211 + 105 +145 + 153 + 152 + 138 + 87 + 99 + 93 + 119 + 129 = 2563∑ x2i = 87 2 + 1032 + 1302 + 1602 + 1802 + 1952 + 1322 + 1452 + 2112 + 1052 +1452 + 1532 + 1522 + 1382 + 872 + 992 + 932 + 1192 + 1292 = 368501 Then s2x = 1 n− 1 (∑ x2i − 1 n (∑ xi )2) = (368501− 25632/19)/18 ≈ 1264.8 and sx ≈ 35.564 2. If yi are the observations reexpressed in hours, then yi = cxi with c = 1/60. So s2y = c 2s2x ≈ 1264.8/602 ≈ 0.35133 and sy = csx ≈ 35.564/60 ≈ 0.59273 Page 2 of 4 Please go to the next page. . .