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WCU ECO 251 - ECO 251 Final Exam

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251y0143 12/29/01 ECO251 QBA1 Name KEY FINAL EXAM Class ________________ DECEMBER 14, 2001For problem references, see 251y0141.Part I. Do all the Following (20 Points) Make Diagrams! Show your work! A. z N~ ( , )0 11.  90.190.1  zP   9426.4713.290.102  zP2.  50.114.3  zP   0660.4332.4992.050.1014.3  zPzP3.  04.343.1 zP   0752.4236.4988.43.1004.30  zPzP4.  75.2zP   9970.4970.5.75.200  zPzPNote that probabilities cannot be above 1 or below 0. A large fraction of the errors made on this or any other exam are exactly what you were reminded of in "Things you should Never do on an exam…." Look at it before every exam!5. 23.z 23.z is the 77th percentile of the distribution, the point with a probability of .23 above itor .77 below it. Since 77% is below this point and 50% is below zero, from the diagram .2700.023. zzP the closest we can come to this probability using the Normal table is .2704.74.00 zP So 74.023.z. 1251y0143 12/03/01B.  3,5.1~ Nx. 1.  90.190.1  xP 13.013.135.190.135.190.1 zPzP   4225.0517.3708.13.00013.1  zPzP2.  50.114.3  xP35.150.135.114.3zP 00.155.1  zP    0981.3413.4394.000.1055.1  zPzP3.  04.343.1 xP 51.002.035.104.335.143.1 zPzP   2030.1950.0090.51.00002.0  zPzP4.  75.2xP 42.035.175.2 zPzP   6628.1628.5.42.000  zPzP. 5. 23.x - 23.x is the 77th percentile of the distribution, the point with a probability of .23 above it or .77below it. Since 1.5 is the mean and median of this distribution, 23.x must be above it. On the previous page we found 74.023.z. So 17.23.zx  72.3374.05.1 .To check to see if this is correct:  72.3xP 35.172.3zP      .23.2296.2704.5.74.00074.0  zPzPzP2251y0143 12/03/01II. (4 points-2 point penalty for not trying .) Show your work! We are investigating the cost of remodeling a kitchen. Contractors are asked for estimates, with the following results: contractor Estimate($thousands) 1 20.4 2 22.1 3 25.3 4 28.7 5 19.9 Compute the sample standard deviation, s, for the contractors' estimates (not the number of thecontractor!). Show your work. (4) Solution: x 2x1 20.4 416.162 22.1 488.41 3 25.3 640.09 4 28.7 823.69 6 19.9 396.01 Total 116.4 2764.364.116x, 36.27642x and 5n. So28.2354.116nxx 642.134568.54428.23536.276412222nxnxsx6935.3642.13 xs.3251y0143 12/03/01III. Do at least 4 of the following 6 Problems (at least 12 each) (or do sections adding to at least 48 points - Anything extra you do helps, and grades wrap around) . Show your work! Please indicate clearly what sections of the problem you are answering! If you are following a rule like    xaEaxE  please state it! If you are using a formula, state it! If you answer a 'yes' or 'no' question, explain why! If you are using the Poisson or Binomial table, state things like n, p or the mean. Avoid crossing out answers that you think are inappropriate - you might get partial credit.1. Using the sample mean and sample standard deviation from the sample of 6 on the previous page (and assuming that a confidence interval is a valid thing to do under the circumstances):a. Compute a 90% confidence interval for the average estimate. (4)b. Compute a 90% confidence interval for the average estimate assuming that the sample of 6 was taken ina town where there are only 31 kitchen remodelers. (4)c. Compute a 90% confidence interval for the average estimate using the sample mean that you found from the sample of 6 on the previous page but assuming that there are only 31 kitchen remodelers in town and that you know that the population standard deviation is 11.1 ($thousands) (4)d. What must be true about the distribution from which the data on the previous page comes for a confidence interval to be valid for such a small sample? (1) Solution: From the previous page 28.23nxx, 642.131222nxnxsx and6935.3642.13 xs.For all these sections the confidence level is 90.1  and the significance level is10.90.1 .5n.Repeat after me! "z goes with  (sigma - population variance); t goes with s (sample variance)!"a) The degrees of freedom are 4151 n. 05.210.2. From the t-table,   132.2405.12ttn, .6518.156935.3nssxxPutting this all together  xnstx12 522.328.236518.1132.228.23  or 19.758 to 26.602. More formally,  90.602.26758.19 P.b) The population size, 31N, is less than 20 times the sample size 6n, so 1NnNnssxx 5377.19309.06518.113153156835.3. It is still true that    132.2405.12ttn, so xnstx12 278.328.235377.1132.228.23  or 20.002 to 26.558. More formally,  90.558.26002.20 P.c) 1NnNnxx 621.49309.0964.413153151.11. From the t-table645.105.2zz. xzx2 602.725.23621.4645.128.23  or 15.648 to 30.852. More formally %90852.30648.15 P.d) For a confidence interval based on z or t to be valid for a sample with ,30n the underlying distribution must be Normal.45251y0143 12/03/012. A manufacturer claims that less than 1% of the jorcillators it sells you are defective.. To be charitable, inthe problem assume that the probability is exactly 1% (Remember, 'at least 2' means 2 or more.)a. Assume that you buy 100 jorcillators and that 5 are defective.(i) Without using a table find the probability that exactly 5 are defective.(3)(ii) To check your answer, find the same probability using a binomial table


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