251y0244 12/10/02 ECO251 QBA1 Name KEY FINAL EXAM Class ________________ DECEMBER 10, 2002Part I. Do all the Following (16 Points) Make Diagrams! Show your work! 6,3~ Nx1. 50.250.2 xP 08.091.06350.26350.2 zPzP 2867.0319.3186.008.0091.0 zPzP2. 255 xP 67.333.06325635 zPzP Change 19 to 25 3706.1293.4999.33.0067.30 zPzP3. 52 xP 33.017.0635632 zPzP 1968.1293.0675.33.00017.0 zPzP4. 84.1F (Cumulative Probability) 84.1xP 19.06384.1 zPzP 4247.0753.5.019.00 zPzP5. 11.6xP 52.06311.6 zPzP 3015.1985.5.52.000 zPzP6. 31.x (Find 31.z first). (3 points) We want a point 31.x, so that 31.31.xxP. Make a diagram for ,z showing zero in the middle, 50% below zero, and the area above zero divided into 19% between 31.z and zero and 31% above 31.z. From the diagram, .1900.031. zzP. The closest we can come is 1915.50.00 zP. So 50.031.z, and 00.33650.0333.zx, or 6.00 . To check this note that 00.6xP 6300.6zP 50.0 zP 1915.5.50.000 zPzP 31.3085. 7. A symmetrical region around the mean with a probability of 31%. (3 points) We want two points655.x and 345.x, so that 3100.345655. xxxP. Make a diagram for ,z showing zero in the middle, an area in the middle of 31%, split in two by zero so that 15.5% is above zero and 15.5% is above zero. Since .5 - .155 = .345, the points we want are 345..z and 345.z. From the diagram, 1550.0345. zzP. The closest we can come is 1554.40.00 zP. So use40.0345.z, and 40.23640.03285.zx, or 0.60 to 5.40. To check this note that 40.560.0 xP6340.56360.0zP .31.3108.1554.240.00240.040.0 zPzP1251y0244 12/10/02II. (10 points-2 point penalty for not trying part a .) Show your work! We are investigating the reliability of a machine that fills 16ounce bottles. A sample of six is taken with the following results (Assume that it is correct to do a confidence interval from such a small sample): Be careful! There is a lot of chance for rounding error in this problem! 15.23 15.26 15.19 15.30 16.04 16.00 a. Compute the sample standard deviation, s, for the number of ounces. Show your work. (4)b. Compute a 90% Confidence interval for the mean. (6)c. (Extra credit) Can you say that the mean that you got is significantly different from 16? Youmust give a reason for your answer to be read. (2)Solution: Since the confidence level is ,90.1 the significance level is 10.a. x 2x 1 15.23 231.953 2 15.26 232.868 3 15.19 230.736 4 15.30 234.090 5 16.04 257.282 6 16.00 256.000 93.02 1442.92902.93x, 929.14422x and 6n. So5033.15602.93nxx 55033.156929.144212222nxnxsx1630.058151.04038.01630.0 xs. b. .1648.061630.0nssxx The degrees of freedom are 5161 n.05.210.2. From the t-table, 015.2505.12ttn. But what compelled some of you to decide that 0.4038 was s in a) but in b)? Putting this all together xnstx12 332.0503.151648.0015.25033.15 or 15.17 to 15.83. More formally, 90.83.1517.15 P.c. Since 16 is not included in this interval, we can say that the mean is significantly different from 16. Note that Minitab gets a similar solution.MTB > TInterval 90.0 c1.Confidence IntervalsVariable N Mean StDev SE Mean 90.0 % C.I.C1 6 15.503 0.402 0.164 ( 15.173, 15.834)2251y0244 12/10/02III. 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. Choose the problems that you do carefully – most of us are unlikely to be able to do more than half of the entire possible credit in this section!)1. 36 employees are randomly drawn from a large corporation and their sick days checked. Note that the sample size is 36 throughout this problem!a. The sample mean is 5.3. Assume that the population standard deviation is known to be 2.2 andcreate a 90% confidence interval for the average number of sick days. (4)b. Assume that the facts in part a are correct but that the corporation only has 360 employees. Create a 90% confidence interval for the average number of sick days again, highlighting what has changed. (4)c. Assume that the sample mean is 5.3 and the corporation has 360 employees, but that 2.2 is a sample standard deviation. Create a 90% confidence interval for the average number of sick days again, highlighting what has changed. (4)d. The sample mean is 5.3. Assume that the population standard deviation is known to be 2.2 andcreate a 93% confidence interval for the average number of sick days. (3)e. Explain the effect of the following on the size of a confidence interval (keep the reasons brief) (3):(i) A smaller sample size.(ii) A smaller population size.(iii) A lower confidence level.Solution: Note that the sample size is 36 throughout this problem. Since the confidence level is,90.1 the significance level is 10..Repeat after me! "z goes with (sigma - population variance); t goes with s (sample variance)!"Most of you seem to have been so hypnotized by the old exams
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