2 25 00 252y0012 ECO252 QBA2 FIRST HOUR EXAM February 22 2000 Name KEY Hour of class registered Class attended if different Show your work Make Diagrams Note that all diagrams are available on a separate sheet I 14 points Do all the following x N 2 7 3 2 1 2 z P 0 43 z 0 14 1 P 1 x 3 P 7 7 P 0 43 z 0 P 0 z 0 14 1664 0557 2221 11 2 11 2 z P 1 86 z 1 29 2 P 11 x 11 P 7 7 P 1 86 z 0 P 0 z 1 29 4686 4015 8701 2 2 24 2 z P 3 71 z 0 4999 3 P 24 x 2 P 7 7 0 2 P z 0 29 P z 0 P 0 29 z 0 4 P x 0 P z 7 5 1141 3859 12 5 2 9 2 z P 1 00 z 1 50 5 P 9 x 12 5 P 7 7 P 0 z 1 50 P 0 z 1 00 4332 3413 0919 6 A symmetrical interval about the mean with 83 probability We want two points x 915 and x 085 so that P x 915 x x 085 8300 From the diagram if we replace x by z P 0 z z 085 4150 The closest we can come is P 0 z 1 37 4147 So z 085 1 37 and x z 085 2 1 37 7 2 9 59 or 7 59 to 11 59 To check this note that P 7 59 x 9 59 11 59 2 7 59 2 P z 7 7 P 1 37 z 1 37 2 P 0 z 1 37 2 4147 8294 83 7 x 13 We want a point x 13 so that P x x 13 13 From the diagram if we replace x by z P 0 z z 13 3700 The closest we can come is P 0 z 1 13 3708 So z 13 1 13 and x z 11 2 1 13 7 2 7 91 or 9 91 To check this note that 9 91 2 P x 9 91 P z P z 1 13 P z 0 P 0 z 1 13 5 3708 7 5 3708 1292 13 2 25 00 252y0012 II 6 points 2 point penalty for not trying part a A truck dealer asserts that a new model of truck has lower gas consumption than your present fleet model A random sample of 7 runs is made with the new model with gas consumption as below Assume that we were sampling from a normal distribution run 1 2 3 4 5 6 7 Consumption 11 56 10 97 12 74 12 08 12 66 12 84 11 15 a Compute the sample standard deviation s of gasoline consumption Show your work 3 b Compute a 90 confidence interval for the mean gas consumption 3 Solution a 3 12 74 162 3076 4 12 08 145 9264 5 12 66 160 2756 2 x Days Home 6 12 84 164 8656 x x 84 00 7 11 15 124 3225 1 x 11 56 12 00 133 6336 n 7 Total 84 00 1011 6722 2 10 97 120 3409 s x 2 2 1011 6722 7 12 00 2 n 1 6 0 612033 or s 0 782326 2 nx b From the problem statement 10 From Table 3 of the syllabus supplement if the population variance is unknown sx s x t 2 s x and t n 1 t 605 1 943 2 0 612033 0 782326 0 29569 So 7 7 n 12 00 1 943 0 29569 12 00 0 5745 or 11 425 to 12 575 2 2 25 00 252y0012 III Do at least 3 of the following 4 Problems at least 10 each or do sections adding to at least 30 points Anything extra you do helps and grades wrap around Show your work State H 0 and H 1 where appropriate Use a 95 confidence level unless another level is specified 1 A truck dealer asserts that a new model of truck has lower gas consumption than your present fleet model The data is on the previous page On the basis of a great deal of experience you know that the mean for your present model is 12 3 miles per gallon For your convenience the data are repeated below run 1 2 3 4 5 6 7 Consumption 11 56 10 97 12 74 12 08 12 66 12 84 11 15 Evaluate the dealer s statement using the sample mean and standard deviation you found in part II a Choose the correct null hypothesis from the list below and write the alternative hypothesis 2 H0 H0 H0 H0 12 3 12 3 12 3 12 3 H0 H0 x H0 x H0 x 12 3 12 3 12 3 12 3 H 0 x 12 3 H 0 x 12 3 b Find a critical value appropriate for this problem using a confidence level of 90 3 c Use your critical value to test the hypothesis State clearly whether you reject the null hypothesis 2 d Repeat the test using i a test ratio 2 and ii a confidence interval 2 e Test the hypothesis that the mean consumption of gas is exactly 12 1 for the dealer s fleet at the 90 confidence level 2 f Assume that the data does not come from a normal distribution and evaluate the statement that the median is 12 8 4 Solution From Table 3 of the Syllabus Supplement Interval for Confidence Hypotheses Test Ratio Critical Value Interval Mean H 0 0 x 0 x z 2 x x cv 0 z 2 x z Known H 1 0 Mean Unknown x t 2 s x x H 0 0 t H 1 0 x 0 sx x cv 0 t 2 s x DF n 1 a The dealer s statement is 12 3 Because this does not contain an equality it cannot be a null hypothesis So we must test its opposite H 0 12 3 and our alternate hypothesis is H 1 12 3 6 1 440 b n 7 12 2 DF n 1 6 10 t n 1 t 0 From the previous page x 12 00 and s x s n 10 0 612033 0 782326 0 29569 7 7 Critical Value Since this is a one sided test xcv 0 t s x 12 3 1 440 0 29569 12 3 0 4258 or 11 87 It might help to remember that x 0 is always in the accept region 3 2 25 00 252y0012 c The critical value in b means that we reject H 0 if the sample mean is below 11 87 Since x 12 00 is not below this critical value do not reject H 0 d i Test Ratio t x 0 12 00 12 3 1 015 This is in the accept region sx 0 29569 …
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