252y0211 2 27 02 ECO252 QBA2 FIRST HOUR EXAM February 19 2002 Name Key Hour of class registered Class attended if different Show your work Make Diagrams How many of you looked at Things You Should Never Do before this exam I 14 points Do all the following x N 9 6 28 2 9 14 9 z 1 P 14 x 28 2 P P 0 83 z 3 20 6 6 P 0 z 3 20 P 0 z 0 83 4993 2967 2026 9 9 7 9 z P 2 67 z 0 4962 2 P 7 x 9 P 6 6 13 9 13 9 z P 3 67 z 0 67 3 P 13 x 13 P 6 6 P 3 67 z 0 P 0 z 0 67 4999 2486 7485 0 9 13 9 z P 3 67 z 1 50 4 P 13 x 0 P 6 6 P 3 67 z 0 P 1 50 z 0 4999 4332 0667 17 9 5 F 17 The cumulative probability up to 17 P x 17 P z 6 P z 1 33 P z 0 P 0 z 1 33 5 4082 9082 6 A symmetrical interval about the mean with 58 probability We want two points x 79 and x 21 so that P x 79 x x 21 5800 From the diagram if we replace x by z P 0 z z 21 2900 The closest we can come is P 0 z 0 81 2910 So z 21 0 81 and x z 21 9 0 81 6 9 4 86 13 86 9 4 14 9 z or 4 14 to 13 86 To check this note that P 4 14 x 13 86 P 6 6 P 0 81 z 0 81 2 P 0 z 0 81 2 2910 5820 58 7 x 095 We want a point x 095 so that P x x 095 095 This is the 90 5 percentile From the diagram if we replace x by z P 0 z z 095 4050 The closest we can come is P 0 z 1 31 4049 So z 095 1 31 and x z 095 9 1 31 6 9 7 86 or 16 86 16 86 9 P z 1 31 To check this note that P x 16 86 P z 6 P z 0 P 0 z 1 31 5 4049 0951 095 252y0211 2 21 02 II 6 points 2 point penalty for not trying part a A new product is tried on seven patients Their breathing capacity after using the product is shown below Note You may want to move the decimal point to the left and work in thousands Patient capacity 1 2 3 4 5 6 7 2850 2380 2800 2860 2300 2650 2640 a Compute the sample standard deviation s of the breathing capacity Show your work 3 b Compute a 90 confidence interval for the mean breathing capacity 3 Solution a Original data Data divided by 1000 2 Row Row x x x x2 1 2850 8122500 1 2 85 8 1225 2 2380 5664400 2 2 38 5 6644 3 2800 7840000 3 2 80 7 8400 4 2860 8179600 4 2 86 8 1796 5 2300 5290000 5 2 30 5 2900 6 2650 7022500 6 2 65 7 0225 7 2640 6969600 7 2 64 6 9696 18480 49088600 18 48 49 0886 n 7 x 18480 n 7 x 18 48 x x 2 x 49088600 x 18480 2640 n x x 7 49 0886 x 18 48 2 640 thousands n 7 x nx 2 49088600 7 2640 49 0886 7 2 640 2 s2 n 1 6 n 1 6 50233 33 or s 224 13 0 05023333 or s 0 22413 s s 50233 33 224 13 0 05023333 0 22413 sx 84 712 sx 0 084712 7 7 n 7 n 7 b From the problem statement 10 From Table 3 of the syllabus supplement if the population s2 2 nx 2 2 variance is unknown 2 x t 2 s x and 2 t n2 1 t 605 1 943 So 2640 1 943 84 712 2640 164 6 or 2475 4 to 2804 6 So 2 640 1 943 0 084712 2 640 0 1646 or 2 4754 to 2 8046 thousands 2 252y0211 2 21 02 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 You have not done a hypothesis test unless you have stated your hypotheses run the numbers and stated your conclusion Use a 95 confidence level unless another level is specified 1 The population mean for similar patients to those mentioned on the previous page who had not used the new medicine was 2628 For your convenience the data are repeated below Patient 1 2 3 4 5 6 7 capacity 2850 2380 2800 2860 2300 2650 2640 Test to see if the mean breathing capacity is now above 2628 using the sample mean and standard deviation you found in part II a State the null and alternate hypothesis 2 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 Each time state clearly whether you reject the null hypothesis and why e Do a 90 two sided confidence interval for the variance 3 f Extra credit Assume that the data does not come from a normal distribution i State a confidence interval for the median using the highest and lowest values and give the confidence level 4 ii Do the same using the second highest and second lowest values and give the confidence level 3 Solution From Table 3 of the Syllabus Supplement Interval for Confidence Hypotheses Interval Mean H 0 0 x t 2 s x Unknown DF n 1 H1 0 Test Ratio t x 0 sx Critical Value x cv 0 t 2 s x a You were asked to see if mean breathing capacity is now above 2628 or 2 628 thousand This gives us 2628 Since this does not contain an equality it must be an alternative hypothesis so we have H 0 2628 and H 1 2628 b Our facts from the previous page are n 7 x 2640 s 224 13 or better s x 84 712 or 0 0847 thousand In addition 10 and 0 2628 Since our alternate hypothesis is 2628 and is one sided our one sided critical value must be above 2628 Our value of t is 6 t n 1 t 10 1 440 xcv 0 t s x 2628 1 440 84 712 2750 0 or 2 750 thousand c Make a diagram Show an almost Normal curve with a …
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