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WCU ECO 251 - ECO 251 Third Exam

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251y9872 11/27/98 ECO251 QBA1 Name __________________ THIRD EXAM Section Enrolled: MWF TR 10 11 12:30 NOVEMBER 24, 1998 Part I. Do all the Following (10+ Points) Penalty for not doing question 1. Show your work! You are following a stock and wish to compare its return to the Dow-Jones index. The return for 9 periods on the stock (y) is compared below to the return on “buying the Dow”(x).Period x(Dow) y(Stock) 2yxy 1 12 12 144 144 2 6 15 225 90 3 2 -4 16 -8 4 4 1 1 4 5 4 2 4 8 6 5 -1 1 -5 7 6 -8 64 -48 8 -6 -2 4 12 9 3 3 9 9Total 36 18 468 206Note that 7170.4,4 xsxCompute the following:1. The sample variance for the stock (4).2. The sample covariance between the stock and the Dow (2)3. The sample correlation between the stock and the Dow (2)4. Interpret the correlation (1)Answers to questions 5) and 6) must be based on the results in questions 1-4. Do not recompute the answers after changing y!5. If all the numbers in the y column were higher by 3 (i.e. 15, 18, -1, 4 etc.), what would the variance, covariance and correlation computed above be? (1.5)6. If all the numbers in the y column were twice as high (i.e. 24, 30, -8, 2 etc.), what would the variance, covariance and correlation computed above be? (1.5)Solution:1) 2918nyy,  5482946812222nynysy (34847.754 ys) 2)196xy,   75.16192492061nyxnxysxy.3)4832.0547170.475.16yxxyxysssr4) The positive sign of rxy, the sample correlation, indicates that x and y tend to move together. If we square rxy, we get approximately .23, which on a zero to one scale indicates a relatively weak relationship.251y9872 11/27/985) We are leaving x alone, but replacing yby 3y. From the syllabus supplement and the outline if 3v  ydcyso that 1c and 3da)      541222222yyvcyVarcvVar b) If dcybaxw  vand , ),(),( yxacCovdcybaxCovwv and 1a and0b    75.1675.1611),(  yxacCovwv.c)            4832.4832.14471.11,  signsignracsignrvwCorrxywv6) We are leaving x alone, but replacing yby y2. From the syllabus supplement and the outline ifydcy 2v so that 2c and 0da)        2165442222222yyvcyVarcvVar b) If dcybaxw  vand , ),(),( yxacCovdcybaxCovwv and 1a and0b    50.3375.1621),(  yxacCovwv.c)             4832.4832.24471.21,  signsignracsignrvwCorrxywv2251y9872 11/27/98Part II. Do the following problems ( do at least 40 points ). Show your work! Note: You only need 40 points out of the 72 below to get an A+ -. Do the parts that look easiest!1. The following table represents the joint probability of x and y.Eventx 0 1 2  yP yyP yPy2 2 .25 .10 .05 .40 .80 1.60 y 3 .05 .25 .05 .35 1.05 3.15 4 .15 .10 ? .25 1.00 4.00  xP .45 .45 .10 1.00 2.85 8.75 So  85.2yyEand 75.82yE  xxP 0+ .45 + .20 = 0.65 xxE  xPx2 0+ .45 + .40 = 0.85 2xEa. Fill in the missing number. (1)b. Are x and yindependent? Why?(2)c. Compute xy, the covariance of x and y, and interpret it. (3)d. Compute xy, the correlation of x and y, and interpret it. (3)e. Find the probability that yx is less than 6. (1)f. (i) Find the distribution of yx . (2)(ii) Using only the results of a)-d), find the mean and variance of yx .(3)Solution: a) Since the table must total 1.00, the missing number is 0.b)x and yare independent if      yPxPyxP ,. In the lower right corner  25.100 , so x and yare not independent .c)    yxxyPyxE ,,   yxxyxyEyxCov , = 0(2)(.25) + 1(2)(.10) + 2(2)(.05)  0025.085.265.085.1  + 0(3)(.05) + 1(3)(.25) + 2(3)(.05) Negative, so x and y move in opposite + 0(4)(.15) + 1(4)(.10) + 2(4)(0) = 1.85 directions.d)     0048.00000232.06275.04275.00025.6275.04275.00025.2yxxyxyWe measure the strength of a correlation by squaring it. If we square -.0048, we get .000023. On a zero to one scale, this is tiny, so correlation is very weak.    6538.04275.04275.065.085.02222xxxxExVar    79215.06275.06275.085.275.82222yyyyEyVare) See below. Since the sums of x and y are all below 6,   16  yxP3251y9872 11/27/98f-i) 010.15.05.25.05.05.10.25.432210iesprobabilit654543432432210sumsyxyx The tables above show the sums of x and yand their probabilities . If we add the probabilities we find the distribution below. yx  yxP    yxPyx    yxPyx 2 2 02P.25 = .25 .50 1.00 3     2130P.05 +.10 = .15 .45 1.35 4       223140P.15 +.25+.05 = .45 1.80 7.20 5     3241P.10 +.05 = .15 .75 3.75 6  42P0 = .00 .00 0.00 1.00 3.50 13.30 Only the yx  and  yxP  columns are needed. From this,  50.3 yxE and     22yxyxEyxVar  05.350.330.132. But this is not how I asked you to compute  yxE  and  yxVar . f-ii)      50.385.265.0 yExEyxEyxyx       05.1)0025.0(26275.04275.0,2  yxCovyVarxVaryxVar4251y9872 11/27/982. I sell both coal and oil. Although the prices of oil and coal fluctuate, I maintain a constant markup of $16.00 per ton for coal and $0.06 per gallon on oil. My fixed costs for coal storage are $600 a month. My fixed costs for oil storage are $300/month.My mean sales of


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