251y0233 12/02/02 ECO251 QBA1 THIRD EXAM NOVEMBER 20, 2001 Name: _____Key__________(Open this document in 'Print Layout' view!) Section Enrolled: (Circle) MWF 10 MWF 11 TR 11 Since there is no correct answer for the Take-home exam a possible solution appears below. Many errors you made are in ‘Things that you should never do on a Statistics Exam or Anyplace Else.’ Wake up and read it! 251x023a 11/16/01 ECO251 QBA1 THIRD EXAM NOVEMBER 22-23, 2002 TAKE HOME SECTION Name: Seymour Butz Section Enrolled: (Circle) MWF 10 MWF 11 TR 11 TR 12:30 Social Security Number: 234567891Throughout this exam show your work! Please indicate clearly what sections of the problem you are answering! If you are following a rule like xaEaxE , please state it! If you are using a formula, state it! If you answer a 'yes' or 'no' question, explain why!Part I. Do all the Following (10 Points) Show your work! My Social Security Number is 265398248. If I write it as below, so that, for example, ,53x Minitab tells me that the sample mean is 5.222 and the sample standard deviation is 2.682. (Please don’t do these again!) Write your Social Security next to it in the same way and call it .y x 1 2 2 6 3 5 4 3 5 9 6 8 7 2 8 4 9 8 Compute the following, showing the steps clearly as if you had done it by hand. Do not tell me “That is what my calculator said,” though you are welcome to use your calculator, Excel or Minitab to check your work:1. The sample variance of .y (2).2. The sample covariance between xand .y (2)3. The sample correlation between x and .y (2)4. Interpret the correlation (1)Answers to questions 5) and 6) must be based on the mean, standard deviation, covariance and correlation that you found in questions 1-4. Do not recompute the answers after changing x or y!5. If all the numbers in y rise by 20, (so that if y was [2, 3, 4, 5, 6, 7, 8, 9, 1], it is now [22, 23, 24, 25, 26, 27, 28, 29, 21]) what will the new mean and standard deviationof y be? What will be the correlation between x and y be now? (1.5)6. If, instead, all the numbers in y are multiplied by 3.5, (so that if y was [2, 3, 4, 5, 6, 7, 8, 9, 1], it is now [7, 10.5, 14, 17.5, 21, 24.5, 28, 31.5, 3.5]) what will the new mean and standard deviation of y be? What will be the correlation between x andy be now? (1.5)251y0233 11/19/02Worksheetobsx 2x y 2y xy1 2 4 2 4 42 6 36 3 9 183 5 25 4 16 204 3 9 5 25 155 9 81 6 36 546 8 64 7 49 567 2 4 8 64 168 4 16 9 81 369 8 64 1 1 8 47 303 45 265 227 73861.250000.7800000.59285100000.594568225.219444.7822222.59303122222.594722222222yyxxsnynysnyysnxnxsnxx9n,47x , 3032x, 45y, 2852y and 227xyNote that the x and 2x columns and their sums were not needed! 800000.522222.592271nyxnxysxy00000.188 so136135.050000.719444.700000.1yxxyxysssrSolution: From above:1. The sample variance of .y 50000.72ys2. The sample covariance between xand .y 00000.1xys3. The sample correlation between x and .y 136135.0xyr4. Interpretation. The negative sign of the covariance and correlation indicates that x and y tend to move in opposite directions. The correlation squared is about .019. On a zero to one scale, this is extremely weak.5. If all the numbers in y rise by 20, what will the new mean and standard deviation of y be?What will be the correlation between x and y be now? dycEdcyE means that 201201 yEyEworks for sample means too, so that the sample mean rises by 20 to 25.0000.And yVarcdcyVar2, so 50000.750000.7112022 yVaryVar. The standard deviation remains 2.73861. Also, ),(),( yxacCovdcybaxCov , so 00000.100000.111),(11)20,0( yxCovyxCovand if baxw and dcy v, xywvacsign. In this case 01 xw and 201v y 11signacsign, and we started with a 2251y0233 11/19/02correlation of 136135.0xyr, so the correlation between the two is -0.136135. In short, adding 20 to y has no effect.6. If, instead, all the numbers in y are multiplied by 3.5, what will the new mean and standarddeviation of y be? What will be the correlation between x and y be now? (1.5) dycEdcyE means that 05.305.3 yEyEworks for sample means too, so that the sample mean, which was 5.00000 is multiplied by 3.5 to become 17.5.And yVarcdcyVar2, so if our original standard deviation was50000.72ys we find .75.1950000.75.35.305.322 yVaryVar Also, yxacCovdcybaxCov ,, , and 00000.1xysso 5000.300000.15.31),(5.31)05.3,0( yxCovyxCovand if baxw and dcy v, xywvacsign. 5.31signacsign, so the correlation between the two is unchanged at136135.0xyr.Part II: Take your Social Security number again, let 1g be the 2nd and 3rd digits of the number, 2gbe the 2nd, 3rd and 4th digits and 132gg and 143gg . ( For example, my Social Security number is 265398248, so 651g, 6532g, 1306523g, .195653314 ggA jorcillator’s lifespan (failure time) can be represented by a continuous uniform distribution between 1g and 2g years (My jorcillator has a lifespan between 65 and 653 years). 1. What is the probability that it lasts between 3g and 4g years? (1)2. What is the probability that it lasts between 3g and 1000 years? (1)3. What is the mean life of such a jorcillator ? (1)4. What is the standard deviation of the life of such a jorcillator ? (1)5. If I have five such jorcillators, what is the probability that at least one lasts between 3g and 1000 years? (1)Solution: Seymour’s Social Security number was 234567891, so that 341g, 3452g, 683423g, and .102343314 gg Our jorcillator has a lifespan from
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