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WCU ECO 252 - ECO 252 First Hour Exam

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252y0111 2/28/01 ECO252 QBA2 Name ____key_______________ FIRST HOUR EXAM Hour of class registered _____ February 20, 2001 Class attended if different ____Show your work! Make Diagrams!I. (14 points) Do all the following. 9,3~ Nx1.  99  xP 67.033.193993`9 zPzP   6568.2486.4082.67.00033.1  zPzP2. 90 xP 67.033.093993`0 zPzP   3779.2486.1293.67.00033.0  zPzP3.  369 xP 67.367.0932893`9 zPzP   2513.2486.4999.67.0067.30  zPzP4. 5xP 22.0935 zPzP   5871.0871.5.22.000  zPzP5. 05  xP 33.089.0930935 zPzP   1840.1293.3133.033.0089.0  zPzP6. A symmetrical interval about the mean with 87% probability. We want two points 935.065. and xx, so that 8700.935.065. xxxP. Make a diagram.If you do a diagram for z, it will show two points, 065.z and 935.z. 935.z (which has 93.5% above it!) is below zero and 065.z is above zero. Since zero is halfwaybetween these two points, the diagram will show 87% split between the two sidesof zero, so that 43.5% is between 935.zand zero, and 43.5% is between zero and 065.z. The probability below 935.z and the probability above 065.z are both 6.5%. From the diagram, if we replace x by z,  4350.0065. zzP. The closest we can come is 4345.51.10 zP. So 51.1065.z, and  951.13065.zx59.133 or -10.59 to 16.59.To check this note that  59.1659.10  xP 51.151.19359.169359.10 zPzP   %.878690.4345.251.102  zP7. 125.x We want a point 125.x, so that 125.125.xxP. Make a diagram. The diagram for z will show one point , 125.z which has 12.5% above it (and 87.5% below it!) and is above zero because zero has only 50% below it. Since zero has 50% above it, the diagram will show 37.5% between zero and 125.z. From the diagram, if we replace x by z,  3750.0125. zzP. The closest we can come is 3749.15.10 zP. So 15.1125.z, and  35.103915.13135.zx, or 13.35.To check this note that  35.13xP 9335.13zP  15.1 zP   3749.5.15.100  zPzP %.5.121251. 2252y0111 2/21/01II. (6 points-2 point penalty for not trying part a.) The Watched Potts Bank of Pottstown wishes to monitor the debt-to-equity ratio of the firms to which it has made commercial loans. A random sample of 8 of the firms is taken. Assume that the bank issampling from a population with a normal distribution. firm Debt/equity ratio 1 1.31 2 1.05 3 1.45 4 1.21 5 1.196 1.787 1.378 1.41 a. Compute the sample standard deviation, s, of the debt-to-equity ratios. Show your work! (3)b. Compute a 99% confidence interval for the mean debt-to-equity ratio, .(3)Solution: a) Firm x (Ratio) x2 1 1.31 1.7161 2 1.05 1.1025 3 1.45 2.1025 4 1.21 1.4641 5 1.19 1.4161 6 1.78 3.1684 7 1.37 1.8769 8 1.41 1.9881 Total 10.77 14.8347 b) From the problem statement 01.. From Table 3 of the syllabus supplement, if the population variance is unknown  x t sx2 and 499.37005.12ttn.07741.082190.08047941.0nssx. So 271.0346.107741.0499.334625.1  or 1.075 to 1.617. 34625.1877.10nxx 734625.188347.1412222nxnxs 047941.0 or 2190.0s.3252y0111 2/21/01III. 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. 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.Error: On this page 1.42 should read 1.32. Remove section f.1. The Watched Potts Bank of Pottstown is a subsidiary of the Umongous Bancorp. The Bancorp is threatening to remove the current officers of the bank if the firms to which they are making loans have an average debt-to-equity ratio that exceeds 1.32. Are the officers in trouble? The data are on the previous page. For your convenience the data are repeated below.firm Debt/equity ratio 1 1.31 2 1.05 3 1.45 4 1.21 5 1.19 6 1.78 7 1.37 8 1.41Test to see if the mean is above 1.32 using the sample mean and standard deviation you found in part II. a. Choose the correct null hypothesis from the list below and write the alternative hypothesis. (2)32.1:0xH32.1:0H32.1:0H32.1:0xH32.1:0H32.1:0xH32.1:0pH32.1:0H32.1:0xH32.1:0pH32.1:0H32.1:0xHb. Find a critical value appropriate for this problem, using a confidence level of 95%.(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)e. Test the hypothesis that the average debt-to-equity ratio is exactly 1.32 at the 95% confidence level. (2)Solution: From Table 3 of the Syllabus Supplement: It's remarkable how many of you don't know that 'the mean is above 1.32' means '32.1!'Interval for Confidence IntervalHypotheses Test Ratio Critical ValueMean ( Known)xzx20100:H:Hxxz0xcvzx20Mean ( Unknown)xstx21nDF0100:H:Hxsxt0xcvstx20a) We are afraid that 32.1. Because this does not contain an equality, it cannot be a null hypothesis.So we must test its opposite 32.1:0H and our alternate hypothesis is 32.1:1Hb)   895.1,05.,71,32.1,8705.10ttnDFnn (If you chose a 2-sided test, 365.27025.t)From the previous page: 34625.1x and


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