数模最终版.doc

2、2圣卡路斯滑坡P101 SAN CARLOS MUD SLIDESConsider the decision tree on page 29. According to this tree, the optimal strategy is to use the test as long as its cost is less than $2,217 = EMV(Test) - EMV(Not test) = -7,783 - $10,000. If the result of the test is positive, then it is better to build the retaining wall. If the result of the test is negative, then it is better not to build the wall. If the test is not taken, then do not build the wall.Let N denote the event that the result of the test is negative and S denote the event that the slide occurs. Notice that P(not N | S) =0.9, P(N | not S) = 0.85, and P(S) = 0.01 To compute the probabilities on the branches we do the following: P(N) = P(N | S)P(S) + P(N | not S)P(not S) = (0.1)(0.01) + (0.85)(0.99) = 0.8425 P(S | N) = (0.1)(0.01)/(0.8425) = 0.0012 P(S | not N) = (0.9)(0.01)/(0.1575) = 0.065、1温磬小扁豆P222 THE GENTLE LENTIL RESTAURANTLet N be the number of meals sold per month, P the price of a prix fixe meal, and L the monthly labor costs. Then Sanjays monthly salary without the partnership is:S = PN - 11N - L - $3,995.According to the case, N, P, and L are random variables, and so, S is also a random variable. N obeys a normal distribution with mean 3,000 and standard deviation 1,000. P obeys a discrete distribution given by Table 5.12, and L obeys a uniform distribution ranging from $5,040 to $6,860. To solve the case, we run a Crystal Ball simulation with 1000 trials. The assumption variables are N, P, and L, and the forecast variable is S.Answer to assignment questions:(a) After running a 1000-trial simulation using Crystal Ball, we found that Sanjays sample mean monthly salary is $10,722, which is an estimate of Sanjays expected monthly salary at Gentle Lentil. If Sanjay decides to accept the job as a consultant, he would make $6,667 per month, which is less than the expected salary at Gentle Lentil.(b) From the same Crystal Ball simulation, we found that the sample standard deviation is $8,796, which is an estimation of the true standard deviation of Sanjays monthly salary at Gentle Lentil. The true standard deviation is difficult to compute in this problem because we do not if the variables N, P, and L are independent of each other. Even if the variables are independent of each other, it is difficult to compute the standard deviation of S using the formula in terms of N, P, and L given above.(c) These are quantitative questions that are necessary to consider before making a decision in this case with and without the partnership option: What is the distribution of Sanjays salary? What is a 95% confidence interval for the expected monthly salary? What is P(S $6,667)?To answer those questions, we use Crystal Ball again. The distribution of Sanjays salary (without the partnership) can be estimated using the following chart:With the partnership, the distribution looks like this:A 95% confidence interval for the monthly salary (without the partnership) is $10,722 - (1.96)($8,796/1000), $10,722 + (1.96)($8,796/1000) = $10,177, $11,267. A 95% confidence interval for the monthly salary (with the partnership) is $7,551 - (1.96)($2,770/1000), $7,551 + (1.96)($2,770/1000) = $7,379, $7,723.Probability of making less than $5,000 per month is approximately 0.288, and the probability of making more than $6,667 per month is 0.649 for both alternatives (with and without the partnership).练习4.9 P194（a）本题所需的知识：课本第四章“4.6 总体比例的估计与置信区间”（P173）。相关公式如下：假设是观测样本比例。如果和满足：那么总体比例的一个的置信区间为：其中，数值满足下列等式：由题中提供资料可知：，由于和满足：，所以得置信区间为：（b）本题所需的知识：课本第四章“4.7 实验设计”“估计比例的实验设计”（P178-179）。相关公式如下：假设我们想要构造一个具有容忍水平的总体比例的的置信区间，也就是说，置信区间将由构成。那么，所要求的样本规模的一个估计值是：其中，数值满足下列等式：而Z服从标准正态分布。由题中资料可知：容忍水平，所以，可求得样本大小为：练习6.6（a）杰克的回归模型所产生的回归等式是：年销售额=-13707.1386+37.3444蛇麻草数量+1319.2696麦芽数量+0.0493年广告支出-63.1678啤酒的苦味分级+53.2305初始投资（b）95%的置信区间：，其中，为自变量的回归系统数，是回归系数的标准误差关于c的求解：注意：根据课本的方法，自由度dof=n-k-1=50-5-1=4430，应通过查标准正态分布的累计分布函数表（附表A.1）来确定c值，查表得c=1.96，计算结果见下表：截距及自变量bmsbmLower 95%Upper 95%Intercept-13707.1386 1368.4086 -16389.2195 -11025.0576 Hops (ounces per keg)37.3444 42.4970 -45.9498 120.6386 Malt (pounds per keg)1319.2696 161.0762 1003.5602 1634.9789 Annual Advertising ($)0.0493 0.0048 0.0398 0.0588 Bitterness Scale-63.1678 83.1086 -226.0606 99.7251 Initial Investment ($ million)53.2305 133.1423 -207.7284 314.1894 下表为EXCEL计算结果Coefficients标准误差Lower 95%Upper 95%Intercept-13707.138561368.408637-16464.98492-10949.2922Hops (ounces per keg)37.3443986842.49702237-48.30272103122.9915184Malt (pounds per keg)1319.269586161.0761968994.64184591643.897325Annual Advertising ($)0.0493142220.0048321170.039575730.059052714Bitterness Scale-63.1677599983.10859298-230.6621211104.3266012Initial Investment ($ million)53.23046901133.1422872-215.1001758321.5611138对比可见，两者结果相差很大。其原因在于，在t分布中，当n时，才完全服从标准正态分布。n30时，只是近似于标准正态分布，查附表A.1得到的c值与当dof=44时的t分布c值有较大出入。查t值表（附后），dof=44时，c=2.0162（内插求得），计算结果如下：截距及自变量bmsbmLower 95%Upper 95%Intercept-13707.1386 1368.4086 -16466.1241 -10948.1531 Hops (ounces per keg)37.3444 42.4970 -48.3381 123.0269 Malt (pounds per keg)1319.2696 161.0762 994.5078 1644.0314 Annual Advertising ($)0.0493 0.0048 0.0396 0.0591 Bitterness Scale-63.1678 83.1086 -230.7313 104.3958 Initial Investment ($ million)53.2305 133.1423 -215.2110 321.6719 （上表结果与EXCEL计算结果较接近，应为t分布c值查表与实际计算误差导致。）分析上表，发现Hops、Bitterness Scale、Initial Investment三项自变量的回归系数的95%置信区间均包含0，说明该三项回归系数不显著。（c）从表6.35中可知，R2=0.879，接近1，并且0.7，说明该模型总体是有效的。但是，通过上题分析，可以把Hops、Bitterness Scale、Initial Investment三项自变量从回规模型中去除，以重新构造一个更有效的模型。（d）去除Hops、Bitterness Scale、Initial Investment三项自变量后的回归结果如下：SUMMARY OUTPUT回归统计Multiple R0.9363 R Square0.8766 Adjusted R Square0.8713 标准误差546.1601 观测值50方差分析dfSSMSFSignificance F回归分析299567777.52 49783888.76 166.90 0.00 残差4714019672.48 298290.90 总计49113587450.00 Coefficients标准误差t StatP-valueLower 95%Upper 95%Intercept-14161.9844 1005.6258 -14.0828 0.0000 -16185.0425 -12138.9263 Malt (pounds per keg)1401.1300 114.2694 12.2616 0.0000 1171.2496 1631.0105 Annual Advertising ($)0.0495 0.0047 10.5236 0.0000 0.0400 0.0589 由上表，新的回归模型是：年销售额=-14161.9844+1401.1300麦芽数量+0.0495年广告支出上式中各回归系数均满足显著性要求；R2=0.8766，与最初模型R2值非常接近，说明模型较好。（e）把表6.37中每种啤酒中的麦芽数量和年广告支出两项数值代入上式中，即可求得该四种啤酒年销售额的预测结果。四种新啤酒麦芽 (磅/桶)年广告支出 (美元)预测年销售额(千美元)Great Ale81500004468.398953Fine Brau61550001913.517018HBC Porter81800005952.667589HBC Stout71500003067.268932注：上表中的数值与答案有出入，是由于计算时对回归系数保留的小数位不同造成的计算误差。（f）题中对新啤酒“亚瑟王最后的神剑”提供了麦芽和年广告支出两项的数值，所以可以用(d)中的回归模型预测其销售额。新啤酒的预测年销售额=-14161.9844+1401.13007+0.0495150000=3067.27千美元练习6.6附件：t值表nP(2)0.050.020.010.0050.0020.001P(1)50.0250.010.0050.00250.0010.000511.0000 3.0780 6.3140 12.7060 31.8210 63.6570 127.3210 318.3090 636.6190 20.8160 1.8860 2.9200 4.3030 6.9650 9.9250 14.0890 22.3270 31.5990 30.7650 1.6380 2.3530 3.1820 4.5410 5.8410 7.4530 10.2150 12.9240 40.7410 1.5330 2.1320 2.7760 3.7470 4.6040 5.5980 7.1730 8.6100 50.7270 1.4760 2.0150 2.5710 3.3650 4.0320 4.7730 5.8930 6.8690 60.7180 1.4400 1.9430 2.4470 3.1430 3.7070 4.3170 5.2080 5.9590 70.7110 1.4150 1.8950 2.3650 2.9980 3.4990 4.0290 4.7850 5.4080 80.7060 1.3970 1.8600 2.3060 2.8960 3.3550 3.8330 4.5010 5.0410 90.7030 1.3830 1.8330 2.2620 2.8210 3.2500 3.6900 4.2970 4.7810 100.7000 1.3720 1.8120 2.2280 2.7640 3.1690 3.5810 4.1440 4.5870 110.6970 1.3630 1.7960 2.2010 2.7180 3.1060 3.4970 4.0250 4.4370 120.6950 1.3560 1.7820 2.1790 2.6810 3.0550 3.4280 3.9300 4.3180 130.6940 1.3500 1.7710 2.1600 2.6500 3.0120 3.3720 3.8520 4.2210 140.6920 1.3450 1.7610 2.1450 2.6240 2.9770 3.3260 3.7870 4.1400 150.6910 1.3410 1.7530 2.1310 2.6020 2.9470 3.2860 3.7330 4.0730 160.6900 1.3370 1.7460 2.1200 2.5830 2.9210 3.2520 3.6860 4.0150 170.6890 1.3330 1.7400 2.1100 2.5670 2.8980 3.2220 3.6460 3.9650 180.6880 1.3300 1.7340 2.1010 2.5520 2.8780 3.1970 3.6100 3.9220 190.6880 1.3280 1.7290 2.0930 2.5390 2.8610 3.1740 3.5790 3.8830 200.6870 1.3250 1.7250 2.0860 2.5280 2.8450 3.1530 3.5520 3.8500 210.6860 1.3230 1.7210 2.0800 2.5180 2.8310 3.1350 3.5270 3.8190 220.6860 1.3210 1.7170 2.0740 2.5080 2.8190 3.1190 3.5050 3.7920 230.6850 1.3190 1.7140 2.0690 2.5000 2.8070 3.1040 3.4850 3.7680 240.6850 1.3180 1.7110 2.0640 2.4920 2.7970 3.0910 3.4670 3.7450 250.6840 1.3160 1.7080 2.0600 2.4850 2.7870 3.0780 3.4500 3.7250 260.6840 1.3150 1.7060 2.0560 2.4790 2.7790 3.0670 3.4350 3.7070 270.6840 1.3140 1.7030 2.0520 2.4730 2.7710 3.0570 3.4210 3.6900 280.6830 1.3130 1.7010 2.0480 2.4670 2.7630 3.0470 3.4080 3.6740 290.6830 1.3110 1.6990 2.0450 2.4620 2.7560 3.0380 3.3960 3.6590 300.6830 1.3100 1.6970 2.0420 2.4570 2.7500 3.0300 3.3850 3.6460 310.6820 1.3090 1.6960 2.0400 2.4530 2.7440 3.0220 3.3750 3.6330 320.6820 1.3090 1.6940 2.0370 2.4490 2.7380 3.0150 3.3650 3.6220 330.6820 1.3080 1.6920 2.0350 2.4450 2.7330 3.0080 3.3560 3.6110 340.6820 1.3070 1.0910 2.0320 2.4410 2.7280 3.0020 3.3480 3.6010 350.6820 1.3060 1.6900 2.0300 2.4380 2.7240 2.9960 3.3400 3.5910 360.6810 1.3060 1.6880 2.0280 2.4340 2.7190 2.9900 3.3330 3.5820 370.6810 1.3050 1.6870 2.0260 2.4310 2.7150 2.9850 3.3260 3.5740 380.6810 1.3040 1.6860 2.0240 2.4290 2.7120 2.9800 3.3190 3.5660 390.6810 1.3040 1.6850 2.0230 2.4260 2.7080 2.9760 3.3130 3.5580 400.6810 1.3030 1.6840 2.0210 2.4230 2.7040 2.9710 3.3070 3.5510 500.6790 1.2990 1.6760 2.0090 2.4030 2.6780 2.9370 3.2610 3.4960 600.6790 1.2960 1.6710 2.0000 2.3900 2.6600 2.9150 3.2320 3.4600 700.6780 1.2940 1.6670 1.9940 2.3810 2.6480 2.8990 3.2110 3.4360 800.6780 1.2920 1.6640 1.9900 2.3740 2.6390 2.8870 3.1950 3.4160 900.6770 1.2910 1.6