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WCU ECO 252 - ECO 252 Second Hour Exam

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3/22/00 252y0021 ECO252 QBA2 Name _____key___________ SECOND HOUR EXAM Hour of class registered _____ March 21,2000 Class attended if different ____Show your work! Make Diagrams!I. (14 points) Do all the following. 6,9~ Nx1. 91  xP 4525.067.1699691 zPzP2. 1111  xP 33.033.369116911 zPzP    6289.1293.4996.33.00033.3  zPzP3. 09  xP 50.100.3690699 zPzP    06545.4332.49865.050.1000.3  zPzP4. 15F (The cumulative probability up to 15)  15xP6915zP  00.1 zP   8413.3413.5.00.100  zPzP5. 5.1714 xP 42.183.0695.176914 zPzP    1255.2967.4222.83.0042.10  zPzP6. A symmetrical interval about the mean with 48% probability. We want two points 26.74. and xx, so that 4800.2674. xxxP. From the diagram, if we replace x by z,  2400.026. zzP. The closest we can come is  2389.64.00 zP or  2422.65.00 zP. Since these are about the samedistance from .2400 use 645.026.z, and  87.396645.0926.zx, or 5.13 to 12.87. To check this note that  87.1213.5 xP6987.126913.5zP     4800.24002645.002645.0645.0  zPzP7. 085.x We want a point 085.x, so that 085.085.xxP. From the diagram, if we replace x by z,  4150.0085. zzP. The closest we can come is  4147.37.10 zP. So 37.1085.z, and  22.89637.19085.zx, or 17.22 . To check this note that  22.17xP 6922.17zP  37.1 zP   4147.5.37.100  zPzP 085.0853. 3/22/00 252y0021II. (6 points - 2 point penalty for not trying) A manufacturer of mowers wants to compare the amount of time it takes to mount a mower engine using two different processes. Two independent samples are taken and the time in minutes appears below. ! Note: Most of you didn't bother to tell me what your hypotheses were in any problem but thefirst one. This means that I couldn't really tell whether your critical values etc. were right or wrong! 1x2x 2.1 3.3 4.1 7.3 9.1 5.3 3.1 8.3 2.1 4.3 3.3 Note that 1.41x and 915476.21s  5.821s a. Compute the sample standard deviation for the second process, 2s. Show your work.(3)b. Compute a 95% confidence interval for the difference between the population means 1 and 2 assuming that the variances are equal for the two parent populations. According to your confidence interval, is there a significant difference between the population means (You must tell why!)? (3) a. Solution: 3.52x and 097618.22s  4.422s 2x 22x 3.3 10.89 7.3 53.29 5.3 28.09 8.3 68.89 4.3 18.49 3.3 10.8931.8 190.54b. From page 10 of the Syllabus Supplement: Interval for Confidence IntervalHypotheses Test Ratio Critical ValueDifference between Two Means ( unknown, variances assumed equal) d t sd2s sn ndp 1 11 2DF n n  1 22HH :0:1     001 2tdsd0   211ˆ212222112nnsnsnspd t scvd 022.13.51.4915476.2,50.8,1.4211211xxdssx 92652097618.2,40.4,3.5212222nnDFssx ,05.   211ˆ212222112nnsnsnsp=    262.222222.6900.2200.34940.4550.849025.ts sn ndp  1 11 2    51046.128148.2366667.22222.6615122222.6  30.568.3122nxx 530.5654.19012222222nxnxs 40.4. 097618.22s. 23/22/00 252y0021Confidence Interval: dstd2   417.320.151046.1262.220.1 or -4.62 to 2.22. The interval includes 0, so there is no significant difference between the means.Formally, our hypotheses are 21100:H0:H or 211210:H:H or 0:H0:H211210 We do not rejectH0.33/22/00 252y0021III. 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 0H and 1H where appropriate. Use a 95% confidence level unless another level is specified.1. For your convenience, the data from the previous page, giving samples of the time to mount a mower engine, is copied below. Test the hypothesis that the mean time to mount a mower is greater for method two than method one. 1x2x 2.1 3.3 4.1 7.3 9.1 5.3 3.1 8.3 2.1 4.3 3.3 Note that 1.41x and 915476.21s  5.821sa) Choose the correct null hypothesis from the list below and write the alternative hypothesis. (2)210:H210: xxH 1.4:20H210:H210: xxH 1.4:20H210:H210: xxH 1.4:20H210:H210: xxH 1.4:20H210:H210: xxH 1.4:20Hb) Assume that 21 and 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)find an approximate p-value for your null hypothesis. (1)e) Do a 90% confidence interval for the ratio of the standard deviations of the populations from which 1x and 2x come. On the basis of this comparison, what assumption should we have made about the equality of the variances? (3) Solution: a) The hypothesis says that 21. Since this does not include an equality, it is the alternate hypothesis. The null hypothesis is the opposite, 21 , so that our hypotheses are 211210::HH or0:0:211210HH or 0:0:10HH where 21. Did you make the same mistake on this exam as you did in the first one?43/22/00


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