252y0212 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 8 6 28 2 8 15 8 z 1 P 15 x 28 2 P P 1 17 z 3 37 6 6 P 0 z 3 37 P 0 z 1 17 4996 3790 1206 8 8 9 8 z P 2 83 z 0 4977 2 P 9 x 8 P 6 6 14 8 14 8 z P 3 67 z 1 00 3 P 14 x 14 P 6 6 P 3 67 z 0 P 0 z 1 00 4999 3413 8412 0 8 14 8 z P 3 67 z 1 33 4 P 14 x 0 P 6 6 P 3 67 z 0 P 1 33 z 0 4999 4082 0917 18 8 5 F 18 The cumulative probability up to 18 P x 18 P z 6 P z 1 67 P z 0 P 0 z 1 67 5 4525 9525 6 A symmetrical interval about the mean with 68 probability We want two points x 84 and x 16 so that P x 84 x x16 6800 From the diagram if we replace x by z P 0 z z 16 3400 The closest we can come is P 0 z 0 99 3389 or P 0 z 1 00 3413 So z 16 0 995 and x z 16 8 0 995 6 8 5 97 13 97 8 2 03 8 z or 2 03 to 13 97 To check this note that P 2 03 x 13 97 P 6 6 P 0 995 z 0 995 2 P 0 z 0 995 2 2910 5820 58 7 x 085 P 0 z z 085 4150 We want a point x 085 so that P x x 085 085 This is the 91 5 percentile From the diagram if we replace x by z The closest we can come is P 0 z 1 37 4147 So z 085 1 37 and x z 085 8 1 37 6 8 8 22 or 16 22 16 22 8 P z 1 37 To check this note that P x 16 22 P z 6 P z 0 P 0 z 1 37 5 4147 0853 085 252y0212 2 20 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 2840 2380 2800 2860 2300 2850 2650 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 2840 8065600 1 2 84 8 0656 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 2850 8122500 6 2 85 8 1225 7 2650 7022500 7 2 65 7 0225 18680 50184600 18 68 50 1846 n 7 x 18680 n 7 x 18 68 x x 2 x 501284600 x 18680 2668 6 n x 7 2 2 2 50 1846 x 18 68 2 6686 thousands n 7 x 2 nx 2 50184600 7 2668 6 50 1846 7 2 6686 2 s s2 n 1 6 n 1 6 55769 71 or s 236 16 0 0557697 or s 0 2362 s s 55769 71 236 16 0 0557697 0 2362 sx 89 2586 sx 0 08927 7 7 n 7 n 7 b From the problem statement 10 From Table 3 of the syllabus supplement if the population 2 x nx variance is unknown 2 x t 2 s x and t n2 1 t 605 1 943 So So 2668 6 1 943 89 2586 2668 6 173 4 2 6686 1 943 0 08927 2 6686 0 1735 or 2495 2 to 2842 0 or 2 495 to 2 842 thousands 2 252y0212 2 20 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 2630 For your convenience the data are repeated below Patient capacity 1 2840 2 2380 3 2800 4 2860 5 2300 6 2850 7 2650 Test to see if the mean breathing capacity is now above 2630 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 2630 or 2 63 thousand This gives us 2630 Since this does not contain an equality it must be an alternative hypothesis so we have H 0 2630 and H 1 2630 b Our facts from the previous page are n 7 x 2668 6 s 236 16 or better s x 89 2586 or 0 0847 thousand In addition 10 and 0 2630 Since our alternate hypothesis is 2630 and is one sided our one sided critical value must be above 2630 Our value of 6 t is t n 1 t 10 1 440 xcv 0 t s x 2630 1 440 89 2586 2758 6 or 2 759 …
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