Do My Homework / Homework Help Answers / Statistics Homework Help / Probability and Statistics Class

Probability and Statistics Class

Need help with this question or any other Statistics assignment help task?

See attached homework sets 1,2,3,4,5,6, which are also quizzes I need to complete (60 minutes limit). If this goes well then I will need help for an exam coming up in four weeks (120 - 180 minutes limit).
Additional Instructions:
Sheet1 Company Group Ratio 1 h 1.50 2 h 0.10 3 h 1.76 4 h 1.14 5 h 1.84 6 h 2.21 7 h 2.08 8 h 1.43 9 h 0.68 10 h 3.15 11 h 1.24 12 h 2.03 13 h 2.23 14 h 2.50 15 h 2.02 16 h 1.44 17 h 1.39 18 h 1.64 19 h 0.89 20 h 0.23 21 h 1.20 22 h 2.16 23 h 1.80 24 h 1.87 25 h 1.91 26 h 1.67 27 h 1.87 28 h 1.21 29 h 2.05 30 h 1.06 31 h 0.93 32 h 2.17 33 h 2.61 34 h 3.05 35 h 1.52 36 h 1.93 37 h 1.95 38 h 2.61 39 h 1.11 40 h 0.95 41 h 0.96 42 h 2.25 43 h 2.73 44 h 1.56 45 h 2.73 46 h 0.90 47 h 2.12 48 h 1.42 49 h 1.62 50 h 1.76 51 h 2.22 52 h 2.80 53 h 1.85 54 h 0.96 55 h 1.71 56 h 1.02 57 h 2.50 58 h 1.55 59 h 1.69 60 h 1.64 61 h 1.03 62 h 1.80 63 h 0.67 64 h 2.44 65 h 2.30 66 h 2.21 67 h 1.96 68 h 1.81 69 f 0.82 70 f 0.89 71 f 1.31 72 f 0.05 73 f 0.83 74 f 0.90 75 f 1.68 76 f 0.99 77 f 0.62 78 f 0.91 79 f 0.52 80 f 1.45 81 f 1.16 82 f 1.32 83 f 1.17 84 f 0.42 85 f 0.48 86 f 0.93 87 f 0.88 88 f 1.10 89 f 0.23 90 f 1.11 91 f 0.19 92 f 0.13 93 f 2.03 94 f 0.51 95 f 1.12 96 f 0.92 97 f 0.26 98 f 1.15 99 f 0.13 100 f 0.88 101 f 0.09
30-year mortgage rate 30-year mortgage rate (Aug 2020) 2.89 3.36 3.28 2.79 3.08 3.18 2.96 2.94 3.02
Covid-19 12-Aug-20 Deaths per 1M Tests per 1M Belgium 852 162,882 UK 688 277,778 Peru 651 79,115 Spain 611 159,806 Italy 583 121,909 Sweden 571 90,738 Chile 533 99,764 USA 509 202,846 Brazil 486 62,198 France 465 84,239 Mexico 418 8,615 Netherlands 359 69,751 Ecuador 337 15,206 Bolivia 322 16,952 Colombia 265 38,922 Canada 238 120,217 Switzerland 230 98,696 Iran 226 32,852 South Africa 185 55,486 Portugal 173 173,297 Honduras 153 11,934 Romania 146 74,269 Argentina 112 19,712 Germany 111 102,448 Russia 105 214,522 Turkey 70 64,601 Israel 69 217,118 Kazakhstan 67 116,123 Belarus 63 146,474 Hungary 63 37,935 Azerbaijan 49 79,266 Egypt 49 1,317 Poland 48 59,657 https://www.worldometers.info/coronavirus/country/belgium/https://www.worldometers.info/coronavirus/country/france/https://www.worldometers.info/coronavirus/country/mexico/https://www.worldometers.info/coronavirus/country/netherlands/https://www.worldometers.info/coronavirus/country/ecuador/https://www.worldometers.info/coronavirus/country/bolivia/https://www.worldometers.info/coronavirus/country/colombia/https://www.worldometers.info/coronavirus/country/canada/https://www.worldometers.info/coronavirus/country/switzerland/https://www.worldometers.info/coronavirus/country/iran/https://www.worldometers.info/coronavirus/country/south-africa/https://www.worldometers.info/coronavirus/country/uk/https://www.worldometers.info/coronavirus/country/portugal/https://www.worldometers.info/coronavirus/country/honduras/https://www.worldometers.info/coronavirus/country/romania/https://www.worldometers.info/coronavirus/country/argentina/https://www.worldometers.info/coronavirus/country/germany/https://www.worldometers.info/coronavirus/country/russia/https://www.worldometers.info/coronavirus/country/turkey/https://www.worldometers.info/coronavirus/country/israel/https://www.worldometers.info/coronavirus/country/kazakhstan/https://www.worldometers.info/coronavirus/country/belarus/https://www.worldometers.info/coronavirus/country/peru/https://www.worldometers.info/coronavirus/country/hungary/https://www.worldometers.info/coronavirus/country/azerbaijan/https://www.worldometers.info/coronavirus/country/egypt/https://www.worldometers.info/coronavirus/country/poland/https://www.worldometers.info/coronavirus/country/spain/https://www.worldometers.info/coronavirus/country/italy/https://www.worldometers.info/coronavirus/country/sweden/https://www.worldometers.info/coronavirus/country/chile/https://www.worldometers.info/coronavirus/country/us/https://www.worldometers.info/coronavirus/country/brazil/ Wages wage IQ 3845 93 4040 119 4125 108 3250 96 2810 74 7000 116 3000 91 5405 114 5770 111 5000 95 4650 132 4605 102 4500 125 6590 119 8960 118 4790 105 6800 109 4250 72 4150 105 2355 101 6375 123 8075 113 4365 95 10685 145 5265 114 8010 124 5940 93 4000 115 7085 125 3175 128 5000 103 7120 98 13340 108 3330 129 8895 132 3910 92 7860 108 6370 106 3570 105 5405 123 3460 108 6590 122 6195 109 5135 100 8740 125 4905 122 3850 105 5770 94 5775 102 4040 109 5500 105 5770 134 8745 108 5000 104 2310 112 3845 120 4375 124 6875 103 7260 115 4000 96 8740 123 5755 98 4200 96 4890 89 4815 109 3095 93 2210 82 3000 120 6830 122 8215 117 7275 109 11550 114 8410 126 6175 82 4275 119 5360 104 5200 115 5000 97 3375 105 5500 100 4980 114 3660 73 6000 96 8470 113 3430 106 3770 104 4285 80 4160 104 2895 122 3360 96 12500 95 5380 105 3750 94 5930 91 4165 96 3250 69 6250 110 5610 111 4325 110 4040 97 6495 125 4515 91 4500 86 3125 110 7930 92 4810 85 7695 120 5550 106 6410 112 3850 91 5000 90 4475 86 6025 86 3750 113 3270 111 3005 111 3000 106 2165 98 5940 105 3175 105 6125 118 5755 90 4325 95 5155 112 5245 120 5000 123 5525 103 9620 121 6730 90 4045 125 7475 109 6730 128 6000 97 2500 96 6625 97 4500 78 4000 112 4000 88 5170 97 4900 101 4420 106 2400 59 4615 105 2565 119 5525 93 5965 82 13855 134 3895 84 4750 98 6970 118 7475 113 3250 100 3350 93 5630 119 2250 67 5140 111 12020 106 9495 127 3785 113 6250 115 5810 102 5125 85 5500 117 3570 125 6590 118 7055 121 10810 115 6365 120 5700 109 2890 109 4710 85 5290 85 3750 106 5000 101 4755 116 3175 91 6250 91 3375 110 2000 75 2885 94 2950 109 4615 122 6250 102 5500 122 5650 105 3260 113 3090 104 4810 124 2645 110 4085 95 4810 120 4200 105 4330 121 12020 96 5630 110 5800 92 3615 98 8890 109 9515 118 5050 97 4855 95 2625 99 2625 106 3350 117 2500 90 5290 98 2750 95 2500 115 3635 106 4325 114 5405 92 6520 119 2875 85 3115 98 2575 114 6365 84 4950 108 3000 107 5800 84 2500 86 3975 88 2500 97 3700 84 6250 117 5070 110 6250 104 4565 110 6730 99 2225 80 1325 101 6250 117 8035 108 7260 69 6955 109 4105 85 3970 114 2500 90 2600 107 8650 117 9620 112 5775 106 10810 116 4615 123 5575 118 2245 91 7500 103 4130 114 4685 102 4890 104 5155 112 6360 107 5680 117 4000 110 6695 96 5315 90 4675 97 4040 99 1875 99 2685 97 5410 95 4650 126 5775 109 2740 86 3110 119 4205 102 3845 105 2935 117 9620 112 5290 89 2085 131 6010 114 5770 98 5350 124 6010 125 3555 100 6010 84 4250 113 5000 102 2450 82 5000 112 4325 111 6875 96 7930 127 8010 97 15390 120 4490 101 4530 120 4760 99 2855 78 2225 110 1445 94 7220 120 4810 108 5375 103 4545 97 6250 106 3100 99 5080 72 4000 88 5250 119 5395 87 3270 121 3905 112 5190 103 9620 99 6010 101 3330 117 4525 109 4450 73 4085 116 4060 98 2885 104 3780 78 5055 104 5775 85 5125 104 6750 107 5005 109 3980 119 6150 96 3770 93 3570 90 5000 103 10335 106 4560 96 3000 114 4755 86 3555 110 5755 87 5000 89 4205 106 2000 100 5875 111 6010 100 7210 114 2690 98 3905 89 3750 74 4205 110 3500 130 6730 108 4000 98 6250 125 5525 104 2375 98 3810 104 4810 120 3605 134 4000 92 3290 129 6350 118 6565 107 4120 92 7210 118 7000 108 5190 119 3340 109 5500 91 5000 118 2615 119 5555 103 4810 120 3645 122 3450 119 5050 104 3000 96 2980 83 4250 115 3350 88 3965 96 7210 123 3350 108 4380 116 4205 109 4875 97 6115 107 4550 95 2665 109 3750 104 6030 110 3725 94 4500 103 5850 123 2700 113 2750 110 3075 100 4545 95 3845 100 4920 108 4165 101 4395 90 5135 115 5000 114 2325 84 5500 92 3205 125 5175 107 6060 129 4750 98 4690 108 6250 114 2930 111 3465 101 2810 99 1875 75 3365 91 3270 113 3460 120 5555 98 6840 107 6410 121 6250 115 6730 134 7120 96 4270 120 4440 105 5805 121 2915 113 6300 106 4735 104 9250 80 7875 117 3790 126 7210 112 2445 104 5630 116 5000 70 2500 83 6000 84 2825 82 9600 109 3420 111 3870 94 1165 96 4875 101 6830 90 10685 120 3500 120 6000 108 5805 99 3645 80 3750 93 5130 85 5555 118 3125 115 6000 119 7705 94 5770 109 1550 104 3050 89 8745 101 5000 96 1750 109 3825 94 3950 115 4090 92 2385 71 4690 130 10495 119 1750 74 4700 104 6010 124 2250 94 5290 120 5000 96 1590 67 2780 69 4790 98 4975 92 8000 109 3030 100 2555 101 7055 111 6730 106 7610 102 5375 115 6000 99 6885 118 4370 132 3125 108 6250 98 5410 111 3465 79 3635 97 3075 99 4565 101 4420 120 3490 97 4000 106 4060 131 5000 93 2745 79 6500 100 4615 107 7695 112 3605 105 4075 114 3000 116 5500 104 5290 85 4810 93 2165 100 4700 71 7085 88 5000 93 2500 98 4000 110 3830 109 3750 94 2750 77 3975 89 3615 102 5195 113 1000 103 2000 90 2875 96 4250 97 2540 87 5810 89 5995 70 6350 104 1750 116 4815 85 7315 104 4010 99 3210 93 3755 114 2440 103 7000 92 4700 98 4330 114 3375 77 1875 89 1625 93 1730 110 7210 132 2800 112 2750 99 4750 96 3420 106 6610 128 8170 95 2085 80 2360 71 3330 89 8395 127 5195 114 6250 102 3800 88 5770 88 8010 110 7210 110 4325 130 3525 100 5000 101 4260 89 3850 80 5000 72 6645 115 8410 115 2885 108 3425 93 2885 111 5200 101 6250 92 3000 115 2400 76 5000 93 4940 91 2390 90 2200 84 2425 120 3525 89 7750 104 4800 107 5000 100 3125 129 6730 112 4805 76 2395 97 6005 98 6730 60 5610 132 2715 90 2250 63 2535 110 2735 88 2930 98 2310 61 3525 76 7830 127 8170 94 6410 97 2780 127 6000 109 4275 89 5265 90 5000 105 4500 100 8010 111 2540 125 7500 124 8410 107 3750 110 4215 102 2045 93 5020 81 3250 99 4040 93 10685 102 3805 95 3845 98 1725 84 2475 88 4935 109 12500 82 5490 105 6060 95 2885 92 1950 74 7500 123 2915 86 2300 92 4725 87 7210 117 6665 112 7500 119 6665 103 3500 74 4865 81 10810 124 3985 89 2000 88 5075 67 8720 106 3150 103 2225 91 3300 101 3895 85 1885 83 2800 70 5610 102 2265 94 6930 97 7695 115 5770 117 4810 105 3610 113 2400 83 4040 101 7210 113 5455 88 1750 93 2500 83 5130 113 2885 106 6665 106 4575 114 5525 119 4550 82 5000 89 5800 74 5005 77 3565 75 4645 97 2000 68 6205 83 5325 97 2015 98 4700 102 4060 81 3500 85 2875 86 2250 79 3105 84 2205 74 3125 92 3630 93 2500 69 5000 113 1965 92 3000 127 4810 109 4810 97 4325 66 5770 109 6930 109 3660 104 4325 85 3500 81 4875 87 6500 78 4500 54 4145 98 5000 98 4135 105 2500 87 5775 117 4750 85 3500 82 8550 114 2665 65 10020 105 4450 92 15390 107 7695 111 2540 73 6770 119 5715 97 4810 110 6250 96 4950 67 4525 102 4630 64 7795 103 6560 118 4615 118 4395 118 4000 75 5245 112 2750 95 5950 119 2915 94 6000 103 3985 90 6855 128 1800 110 6350 84 9160 97 4545 102 8730 131 2600 126 4040 107 10685 108 3460 92 4655 108 4060 115 4330 111 7210 115 2500 101 6920 114 2640 116 4405 90 5130 125 6060 98 8100 111 9215 109 8010 124 5000 121 3685 103 8495 96 8495 100 5125 104 5535 112 3125 73 3600 101 4275 99 6250 95 5650 109 4500 110 9620 98 4810 112 9370 64 7865 90 4700 102 4500 119 4410 96 8550 118 6300 79 3755 96 5485 80 5005 137 5770 116 5000 69 6250 78 4285 99 6055 92 4685 113 4520 98 4360 115 6000 96 6305 117 4750 96 575 108 3365 92 5420 95 5290 94 4000 115 1845 117 9620 105 4275 109 4320 87 5460 110 5125 110 4250 109 3365 85 4000 105 5245 122 4420 108 4000 122 6000 79 3320 105 5555 90 4250 111 5000 110 4375 92 6010 131 3330 104 4615 110 3940 99 4750 86 7210 105 4330 85 3125 72 3400 86 3465 68 4375 95 11540 116 5980 96 3500 85 3605 100 7500 101 5250 97 4005 100 3200 96 5990 117 1950 93 4445 116 5380 106 5635 98 3750 103 4275 118 2625 100 5520 96 2765 67 4000 96 6920 104 2125 106 2550 62 1850 73 2980 62 2010 76 2090 76 5665 106 3405 94 4750 81 3000 87 3305 96 4625 75 2750 95 3030 86 2125 86 1700 104 3460 95 2875 116 2855 66 4085 101 4935 96 3080 104 5130 105 4040 106 4040 101 3940 96 4250 105 4500 106 7090 102 1300 75 3310 99 2810 90 2810 104 1785 115 5045 118 7210 108 3255 76 3750 50 3770 55 3500 73 2515 84 4685 104 3120 92 3750 83 4500 98 2700 70 3210 87 2000 96 4500 80 2565 104 4470 104 6410 96 2425 89 1625 113 3845 105 3090 84 5200 108 3755 104 1900 106 1500 83 3765 70 5325 104 5350 91 7865 112 3250 97 3500 99 2470 74 4450 100 2600 86 4455 90 2850 101 7220 109 2405 98 2500 107 7365 110 4015 120 4810 97 5000 103 3000 114 2250 95 3145 93 2460 67 7810 97 1785 97 4800 83 2830 104 2405 82 7210 113 3225 93 3940 100 3220 101 2385 100 3320 82 2600 79 6010 102 2690 77 4365 109 5000 107 Amazon Date AMZN SP500 10-Year TB 1-Aug-20 3,167.46 3,351.28 0.56% 1-Jul-20 3,164.68 3,271.12 0.54% 1-Jun-20 2,758.82 3,100.29 0.65% 1-May-20 2,442.37 3,044.31 0.65% 1-Apr-20 2,474.00 2,912.43 0.62% 1-Mar-20 1,949.72 2,584.59 0.70% 1-Feb-20 1,883.75 2,954.22 1.13% 1-Jan-20 2,008.72 3,225.52 1.52% 1-Dec-19 1,847.84 3,230.78 1.92% 1-Nov-19 1,800.80 3,140.98 1.78% 1-Oct-19 1,776.66 3,037.56 1.69% 1-Sep-19 1,735.91 2,976.74 1.68% 1-Aug-19 1,776.29 2,926.46 1.51% 1-Jul-19 1,866.78 2,980.38 2.02% 1-Jun-19 1,893.63 2,941.76 2% 1-May-19 1,775.07 2,752.06 2.14% 1-Apr-19 1,926.52 2,945.83 2.51% 1-Mar-19 1,780.75 2,834.40 2.41% 1-Feb-19 1,639.83 2,784.49 2.71% 1-Jan-19 1,718.73 2,704.10 2.64% 1-Dec-18 1,501.97 2,506.85 2.69% 1-Nov-18 1,690.17 2,760.17 3.01% 1-Oct-18 1,598.01 2,711.74 3.16% 1-Sep-18 2,003.00 2,913.98 3.06% 1-Aug-18 2012.709961 2901.52002 2.85% 1-Jul-18 1777.439941 2816.290039 2.96% 1-Jun-18 1699.800049 2718.370117 2.85% 1-May-18 1629.619995 2705.27002 2.82% 1-Apr-18 1566.130005 2648.050049 2.94% 1-Mar-18 1447.339966 2640.870117 2.74% 1-Feb-18 1512.449951 2713.830078 2.87% 1-Jan-18 1450.890015 2823.810059 2.72% 1-Dec-17 1169.469971 2673.610107 2.40% 1-Nov-17 1176.75 2584.840088 2.42% 1-Oct-17 1105.280029 2575.26001 2.38% 1-Sep-17 961.349976 2519.360107 2.33% 1-Aug-17 980.599976 2471.649902 2.12% 1-Jul-17 987.780029 2470.300049 2.29% 1-Jun-17 968 2423.409912 2.30% 1-May-17 994.619995 2411.800049 2.20% 1-Apr-17 924.98999 2384.199951 2.28% 1-Mar-17 886.539978 2362.719971 2.40% 1-Feb-17 845.039978 2363.639893 2.36% 1-Jan-17 823.47998 2278.870117 2.45% 1-Dec-16 768.659973 2238.830078 2.48% 1-Nov-16 750.570007 2198.810059 2.37% 1-Oct-16 789.820007 2126.149902 1.83% 1-Sep-16 837.309998 2168.27002 1.61% 1-Aug-16 769.16 2170.95 1.56% 1-Jul-16 758.81 2173.6 1.45% 1-Jun-16 715.62 2098.86 1.48% 1-May-16 722.79 2096.95 1.85% 1-Apr-16 659.59 2065.3 1.84% 1-Mar-16 593.64 2059.74 1.77% 1-Feb-16 552.52 1932.23 1.74% 1-Jan-16 580 1940.24 1.92% 1-Dec-15 675.89 2043.94 2.27% 1-Nov-15 664.8 2080.41 2.21% 1-Oct-15 625.9 2079.36 2.15% 1-Sep-15 511.89 1920.03 2.04% 1-Aug-15 512.89 1972.18 2.21% 1-Jul-15 536.15 2103.84 2.19% 1-Jun-15 434.09 2063.11 2.35% 1-May-15 429.23 2107.39 2.13% 1-Apr-15 421.78 2085.51 2.04% 1-Mar-15 372.1 2067.89 1.93% 1-Feb-15 380.16 2104.5 2.00% 1-Jan-15 354.53 1994.99 1.64% 1-Dec-14 310.35 2058.9 2.17% 1-Nov-14 338.64 2067.56 2.17% 1-Oct-14 305.46 2018.05 2.34% 1-Sep-14 322.44 1972.29 2.49% 1-Aug-14 339.04 2,003.37 2.42% 1-Jul-14 312.99 1,930.67 2.55% 2-Jun-14 324.78 1,960.23 2.60% 1-May-14 312.55 1,923.57 2.56% 1-Apr-14 304.13 1,883.95 2.70% 3-Mar-14 336.37 1,872.34 2.72% 3-Feb-14 362.1 1,859.45 2.71% 2-Jan-14 358.69 1,782.59 2.86% 2-Dec-13 398.79 1,848.36 2.90% 1-Nov-13 393.62 1,805.81 2.72% 1-Oct-13 364.03 1,756.54 2.61% 3-Sep-13 312.64 1,681.55 2.81% 1-Aug-13 280.98 1,632.97 2.74% 1-Jul-13 301.22 1,685.73 2.59% 3-Jun-13 277.69 1,606.28 2.30% 1-May-13 269.2 1,630.74 1.94% 1-Apr-13 253.81 1,597.57 1.75% 1-Mar-13 266.49 1,569.19 1.96% 1-Feb-13 264.27 1,514.68 1.98% 2-Jan-13 265.5 1,498.11 1.91% 3-Dec-12 250.87 1,426.19 1.72% 1-Nov-12 252.05 1,416.18 1.65% 1-Oct-12 232.89 1,412.16 1.75% 4-Sep-12 254.32 1,440.67 1.72% 1-Aug-12 248.27 1,406.58 1.68% 2-Jul-12 233.3 1,379.32 1.53% 1-Jun-12 228.35 1,362.16 1.63% 1-May-12 212.91 1,310.33 1.80% 2-Apr-12 231.9 1,397.91 2.05% 1-Mar-12 202.51 1,408.47 2.18% 1-Feb-12 179.69 1,365.68 1.97% 3-Jan-12 194.44 1,312.41 1.97% 1-Dec-11 173.1 1,257.60 1.97% 1-Nov-11 192.29 1,246.96 2.01% 3-Oct-11 213.51 1,253.30 2.15% 1-Sep-11 216.23 1,131.42 1.97% 1-Aug-11 215.23 1,218.89 2.28% 1-Jul-11 222.52 1,292.28 2.99% 1-Jun-11 204.49 1,320.64 3.00% 2-May-11 196.69 1,345.20 3.17% 1-Apr-11 195.81 1,363.61 3.45% 1-Mar-11 180.13 1,325.83 3.41% 1-Feb-11 173.29 1,327.22 3.58% 3-Jan-11 169.64 1,286.12 3.39% 1-Dec-10 180 1,257.64 3.31% 1-Nov-10 175.4 1,180.55 2.77% 1-Oct-10 165.23 1,183.26 2.54% 1-Sep-10 157.06 1,141.20 2.65%
45-750 PROBABILITY AND STATISTICS PROBLEM SET 1: BASIC PROBABILITY, DISCRETE RANDOM VARIABLES This first assignment tests your understanding of the basic probability models, the various rules for computing probabilities of events represented as sets, joint probability tables, conditional probabil- ity and independence, the total probability rule, variance, correlation, and the binomial and Poisson distributions. Problem 1 (Probability rules). In the USA, the result of presidential elections often depends on a small number of swing states in which the winner takes all the electoral votes in that state. Can you compute the chances that presidential candidate X will win the elections in the following hypothetical situation: Candidate X needs 37 electoral votes from 4 swing states to win the presidential election. Swing State Electoral votes F (Florida) 29 M (Michigan) 16 P (Pennsylvania) 20 W (Wisconsin) 10 (1.1) Easy. The various combinations of swing states that could be won by candidate X represent the basic outcomes in the sample space. For example FP represents a basic outcome where X wins Florida and Pennsylvania, and looses Michigan and Wisconsin. As an other example, ∅ represents a basic outcome where candidate X looses all four swing states. What is the size of the sample space? (1.2) Moderate. Represent the event ”X wins the election” as the union of all the basic outcomes contributing to this event. (1.3) Moderate. The probability that candidate X wins a swing state is 0.5, and the four events ”X wins F”, ”X wins M”, ”X wins P”, ”X wins W”, are pairwise independent. What is the probability that candidate X wins the elections? (1.4) Challenging. Polls provide the following independent probabilities for each of the four swing states. State Probability that X wins the state F 0.6 M 0.8 P 0.7 W 0.6 What is the probability that candidate X will win the presidential elections? 1 2 PROBLEM SET 1: BASIC PROBABILITY, DISCRETE RANDOM VARIABLES Problem 2 (Probability table, conditional probability and independence). Beth flies from Miami to Pittsburgh with a 25-minute layover in Atlanta. If Beth misses her connection, she will have to spend the night in Atlanta. The probability that Beth’s flight from Miami to Atlanta arrives on time is 0.7; the probability that it is delayed is 0.3. If Beth arrives on time in Atlanta, the probability that she makes the connection to Pittsburgh is 1, If she is delayed arriving in Atlanta, the probability that she makes her connection to Pittsburgh is 0.4. (2.1) Easy. Consider the following events: O = Flight from Miami arrives ON TIME in Atlanta; D = Flight from Miami is DELAYED arriving in Atlanta; M = Beth MAKES the connection in Atlanta, N = Beth does NOT make the connection and therefore spends the night in Atlanta. Fill in the probability table. M N O 0.7 D 0.3 1 (2.2) Easy. What is the probability that Beth’s flight is delayed arriving in Atlanta AND she makes her connection to Pittsburgh. (2.3) Easy. What is the probability that Beth makes her connection in Atlanta? If Beth arrives on time in Altanta, the probability that her luggage makes the connection with her is 0.8. If Beth’s flight is delayed arriving in Atlanta, the probability that both Beth and her luggage make the connection is 0.1. (2.4) Moderate. What is the probability that both Beth and her luggage make it to Pittsburgh on her scheduled flight? (2.5) Moderate. Are the events O = ”Beth’s flight from Miami arrives ON TIME in Atlanta” and L = ”Beth makes the connection and her LUGGAGE is there at her arrival in Pittsburgh” statistically independent? (2.6) Challenging. Given that Beth made it to Pittsburgh on her scheduled flight, what is the probability that her luggage is there at her arrival in Pittsburgh? (2.7) Challenging. Beth made it to Pittsburgh on her scheduled flight and her luggage was not there at her arrival. What is the probability that Beth’s flight from Miami was delayed arriving in Atlanta? [Hint: Let K = ”Beth made it to Pittsburgh on her scheduled flight and her luggage was not there at her arrival”. You need to compute P (D|K). Note that M = L ∪K and that L and K are mutually exclusive.] Problem 3 (Correlation). Many companies such as Walmart, Costco, and Amazon take advantage of ”risk pooling” to reduce their inventories. To illustrate how risk pooling works, consider a company with warehouses in Pittsburgh and Philadelphia to serve its customers in these two cities. Many different products are held in inventory. The idea of risk pooling will become clear by focusing on one such product. Weekly demand for one of the company’s products is 40 units on average in Pittsburgh with a standard deviation of 20 units, and 80 units on average in Philadelphia with a standard deviation of 30 units. The correlation between these two demands is ρ = 0.1. To protect against stockouts, the company maintains an inventory at each warehouse equal to the expected weekly demand locally plus two standard deviations of this weekly demand. The company is considering replacing its warehouses in Pittsburgh and Philadelphia by a more centrally located warehouse in Harrisburg. The inventory level at the Harrisburg warehouse will be based on the total demand in Pittsburgh and Philadelphia. Let us call D this total weekly (random) demand. Inventory at the Harrisburg warehouse will be equal to the expected value of D plus two standard deviations of D. (3.1) Easy. What are the current inventory levels at the Pittsburgh and Philadelphia warehouses? (3.2) Challenging. What would be the inventory needed at the Harrisburg warehouse? What would be the decrease in inventory achieved by switching to a single warehouse in Harrisburg as compared to (3.1)? 45-750 PROBABILITY AND STATISTICS 3 Problem 4 (Independence?). A fair coin is tossed 10 times. Let X denote the random variable representing the number of heads and Y denote the number of tails. (4.1) Easy. Compute the variance of X. (4.2) Easy. Note that Y = 10−X. Use this formula to compute the variance of Y . (4.3) Moderate. Compute the variance of X + Y . [Hint: Recall that Y = 10−X.] (4.4) Challenging. Use (4.3) to compute the correlation between X and Y . Are X and Y indepen- dent random variables? (4.5) Challenging. Compute the variance of 2X − Y . Problem 5 (Binomial Distribution). Airlines regularly overbook flights to compensate for no-show passengers. In doing so, airlines are balancing the risk of having to compensate bumped passengers against lost revenue associated with empty seats. In a USA Today analysis of airline statistics, it was found that the average no-show rate was 12%. An airline books 110 passengers on a 100-seat plane. Assume a 12% probability that a passenger will not show up for boarding. (5.1) Easy. What is the probability that no passenger will be bumped? (5.2) Easy. What is the probability that the plane will be full? (5.3) Challenging. The airline makes a profit of $100 per passenger on this flight, and it estimates the cost of a bumped passenger to be $1000 (this includes cash awards to bumped passengers, free air vouchers, loss of goodwill, etc.). For this reason, the airline would like a probability of at least 0.9 that no passenger will be bumped. How many tickets should the airline sell on this 100-seat plane? Problem 6 (Poisson Distribution). The neighborhood coffee shop Moonbucks sells three types of coffees: regular, Americano and caffe latte and charges $2, $2.50 and $4 respectively. Orders for regular coffees occur according to a Poisson distribution with a mean of 20 per hour during the 9-10am time period. Orders for Americanos occur according to a Poisson distribution with a mean of 8 per hour during the same period. Orders for caffe lattes occur according to a Poisson distribution with a mean of 10 per hour during this period. (6.1) Easy. What is the expected number of regular coffees that Moonbucks sells between 9AM to 9:15AM. (6.2) Easy. What is the probability that Moonbucks sells at least one regular coffee between 9AM and 9:05AM? (6.3) Easy. It is 9:05AM. Moonbucks had no client for regular coffees from 9AM to 9:05AM. What is the probability that it has no client for regular coffee between 9:05AM and 9:10AM? (6.4) Moderate. What is Moonbucks’ expected revenue between 9AM and 10AM? (6.5) Moderate. What is the probability that Moonbucks sells 35 coffees or more (all three types combined) between 9AM and 10AM?
45-750 PROBABILITY AND STATISTICS PROBLEM SET 2: DECISION TREES, CONTINUOUS RANDOM VARIABLES This assignment tests your understanding of decision theory and continuous random variables. The learning objectives are: constructing decision trees, calculating the value of information when making decisions under uncertainty, working with common continuous distributions (uniform, exponential), combining continuous and discrete distributions. Problem 1 (Decision Trees and Value of Information). NovelTech is a high-tech company dedicated to technological innovations. The company is trying to decide whether or not to develop a new product. It is estimated that, with a probability of 40%, this product will be a major success with an estimated profit of $400 million; with a probability of 30% the product will break even; and with a probability of 30% it will be a total failure and will cost the company $600 million. (1.1) Easy. Draw a decision tree for the problem of deciding whether to develop the new product or not. Solve the decision tree assuming the goal is to maximize the expected payoff. Should the company develop the new product? (1.2) Easy. What is the expected value of perfect information (EVPI)? Explain what it means. The company can hire an expert that will design an experiment to test the new product. Based on previous experience with the expert, the following conditional probabilities are realistic estimates of the expert’s evaluation accuracy. Pr(experiment is positive | major success) = 0.8 Pr(experiment is positive | break even) = 0.6 Pr(experiment is positive | failure) = 0.3 (1.3) Easy. Use the law of total probability to compute the probability that the experiment is positive. (1.4) Moderate. What is the probability of a major success given that the experiment is positive? What is the probability of breaking even given that the experiment is positive? What is the probability of a failure given that the experiment is positive? What is the probability of a major success given that the experiment is negative? What is the probability of breaking even given that the experiment is negative? What is the probability of a failure given that the experiment is negative? (1.5) Moderate. Construct a decision tree for this problem. Solve the decision tree assuming that the goal is to maximize the expected payoff. What is the expected value of sample information (EVSI)? What is the optimal decision strategy? The expert charges $1 million for her services. Should you hire the expert? (1.6) Challenging. What is the standard deviation of the payoff of the optimal decision strategy? Solve with and without including the expert’s fee. What is the difference? Problem 2 (pdfs and cdfs). Let X be a continuous random variable with pdf f(x) and cdf F (x). (2.1) Easy Can f(x) ever be negative? (2.2) Easy Can F (x) ever be negative? (2.3) Easy Can F (x) ever be greater than 1? (2.4) Moderate Can f(x) ever be greater than 1? (2.5) Moderate Suppose that f(x) = 0 when x ≤ 0, f(x) = x when 0 < x < a, and f(x) = 0 when x ≥ a. Compute a. (2.6) Moderate Consider the pdf of question (2.5). Compute F (1). 1 2 PROBLEM SET 2: DECISION TREES, CONTINUOUS RANDOM VARIABLES (2.7) Challenging Suppose that f(x) = 0 when x ≤ 0, f(x) = x when 0 < x ≤ 1, f(x) = 2−x when 1 < x < 2, and f(x) = 0 when x ≥ 2. Compute F (x) in each of the ranges x ≤ 0, 0 < x ≤ 1, 1 < x < 2, and x ≥ 2. Problem 3 (Uniform Distribution). You arrive at a bus stop at 8 o’ clock, knowing that the bus will arrive at some time uniformly distributed between 8am and 8:20am. (3.1) Easy What is the expected value of your waiting time at the bus stop? (3.2) Easy What is the probability that you will have to wait at least 15 minutes? (3.3) Easy What is the median of your waiting time at the bus stop? (3.4) Moderate If at 8:05am, the bus has not yet arrived, what is the probability that you will have to wait at least an additional 10 minutes? (3.5) Challenging You arrive at a bus stop at 8 o’ clock, but suppose now that there are three different buses that can bring you to your destination. Each bus will arrive at the bus stop at some time uniformly distributed between 8 and 8:20am and these arrivals are independent. You take the first bus that shows up. What is the probability that you will have to wait no more than 10 minutes? Problem 4 (Exponential Distribution). A small private ambulance service in Western Pennsylvania has determined that the time between emergency calls is exponentially distributed with a mean of 120 minutes. When a unit goes on call, it is busy for 50 minutes. If a unit is busy when an emergency call is received, the call is immediately routed to another service. The company is considering buying a second ambulance. However, before doing so, the owner would like to have answers to the following questions. (4.1) Easy What is the probability that an emergency call is received when the unit is busy? (4.2) Easy What is the probability that at least two hours elapse between an emergency call and the following one? (4.3) Easy The unit just went on call. What is the probability that the next emergency call will arrive between 50 minutes and two hours? (4.4) Moderate What is the distribution of the number of emergency calls received in 24 hours? What is the expected number of emergency calls received in 24 hours? Problem 5 (Combining Continuous and Discrete Distributions). The life of a certain type of automobile tire, called MX tire, is uniformly distributed in the range 36, 000 to 48, 000 miles. (5.1) Easy Compute the standard deviation of the life of an MX tire. Consider a car equipped with 4 MX tires. The lives of the 4 tires are assumed to be independent. (5.2) Moderate What is the probability that the car can drive at least 40, 000 miles without changing any tire? (5.3) Moderate What is the probability that at least one of the tires has to be changed before the car reaches 42, 000 miles? (5.4) Challenging What is the probability that at least 2 tires have to be changed before the car reaches 42, 000 miles?
45-750 PROBABILITY AND STATISTICS PROBLEM SET 3: NORMAL RANDOM VARIABLES, SAMPLING DISTRIBUTIONS This assignment tests your understanding of normal random variables, and of sampling distributions. The learning objectives are: working with the probability distribution of a normal random variable, com- bining independent normal random variables, the central limit theorem, and the sampling distribution of the sample mean. Problem 1 (Value at Risk). In finance, VaR does not stand for variance but for “value at risk.” It is another measure of risk. It was introduced by J.P. Morgan in the 1980s as a way to answer a question that was often asked by investors: “How much might I loose?” Let X be a random variable representing the profit on an investment at the end of some time horizon, such as a year. If X is negative the investment results in a loss. Choose a confidence level α (α = 95% is a commonly used value). The α-VaR is the value x for which P (X ≤ x) = 1 − α. This means that the profit on the investment will be greater than x with probability α, and it will be x or smaller with probability 1-α. In particular, if x is negative and α = 95%, a loss of |x| or greater will occur with probability 5%. (1.1) Easy Investment A has a profit at the end of a year that is normally distributed with mean $2 million and standard deviation $1.5 million. What is the 95%-VaR of investment A? (1.2) Moderate The investor owning A considers the potential loss calculated in (1.1) to be excessive. Risk might be reduced using a hedging strategy. Also available is investment B, with a profit at the end of a year that has mean -0.3 million dollars and standard deviation $ 1 million. The correlation between the profits on investments A and B is -0.8. What is the expected profit of investment A+B at the end of a year? What is the standard deviation of the profit on investment A+B at the end of a year? Assuming that the profit on investment A+B is normally distributed, compute the 95%-VaR of investment A+B. (1.3) Challenging The investor is considering hedged portfolios of the form A + t B for t > 0. Can you find the value of t that minimizes the variance of this portfolio? Assuming that the profit on investment A + t B is normally distributed, compute the 95%-VaR of investment A + t B for the value of t found above. Problem 2 (Difference of Independent Normal Random Variables). The nation of Somonga, located in the South Pacific, has asked you to analyze its trade balance (the trade balance is the difference between the total revenue from exports and the total cost of imports in a year). Somonga’s only export is coconut oil. It exports 18,000 metric tons of coconut oil per year. The price of coconut oil in the world market is normally distributed with mean $920 per metric ton and standard deviation $160. Somonga’s total cost of imports in a year is also normally distributed, with mean $16,500,000 and standard deviation $1,600,000. Total cost of imports is independent of the price of coconut oil in the world market. (2.1) Easy What is the mean of the trade balance? (2.2) Easy What is the standard deviation of the trade balance? (2.3) Moderate What is the probability that the trade balance is negative? (2.4) Challenging What is the probability that the price of coconut oil in the world market is greater than $1000 given that it is greater than $900 ? 1 2 PROBLEM SET 3: NORMAL RANDOM VARIABLES, SAMPLING DISTRIBUTIONS Problem 3 (Independence versus Dependence). Electronix produces several products, one of which is called ElecPlus. The production cost C of one unit of ElecPlus is a normally distributed random variable with mean 32 and standard deviation 6. The sales price P for a unit of ElecPlus is a normally distributed random variable with mean 40 and standard deviation 8. The random variables C and P are independent of each other. (3.1) Easy What is the distribution of the profit P − C made on one unit of ElecPlus? (3.2) Easy What is the probability of a positive profit on one unit of ElecPlus? (3.3) Challenging What is the distribution of the profit made on ten units of ElecPlus? Assume independence of the random variables associated with different units. What is the probability of a positive profit on ten units of ElecPlus? (3.4) Challenging What is the distribution of the profit made on ten units of ElecPlus? Do not assume independence of the random variables associated with different units. Instead assume that all ten units have the same production cost C, which is a normally distributed random variable with mean 32 and standard deviation 6, and all ten units have the same sales price P , which is a normally distributed random variable with mean 40 and standard deviation 8. As earlier the random variables C and P are independent of each other. What is the probability of a positive profit on ten units of ElecPlus? Problem 4 (Sample mean). The operations manager at a cookie factory is responsible for process adherence. Each cookie has an average weight of 15 g. Because of the irregularity in the number of chocolate chips in each cookie, the weight of an individual cookie varies, with a standard deviation of 1 g. The weights of the cookies are independent of each other. The cookies come in standard packs of 30 and jumbo packs of 60. In both packs, the label states that the cookies have an average weight of 15 g each. Federal guidelines require that the weight stated on the packaging be consistent with the actual weight. (4.1) Easy. Based on the information given above is it possible to compute the probability that a randomly selected cookie has a weight less than 14 g? (4.2) Moderate. What is the probability that the average cookie weight in a pack of 30 is less than 14.8 g? (4.3) Moderate. What is the probability that the average cookie weight in a pack of 60 is less than 14.8 g? Explain how increasing pack size affects this probability. (4.4) Moderate. What is the probability that the average cookie weight in a pack of 60 is between 14.8g and 15.2 g? (4.5) Challenging. Find an interval symmetrically distributed around the population mean that will include 90% of the sample means, for the packs of 30 cookies. Problem 5 (Central Limit Theorem). Are the following statements true or false? (5.1) Easy. Based on the Central Limit Theorem, the sample mean can be used as a good estimator of the population mean, assuming that the sample size, n, is sufficiently large. (5.2) Easy. When applying the Central Limit Theorem, one usually considers that a sample size n ≥ 30 is sufficiently large. (5.3) Easy. The Central Limit Theorem can be applied to both discrete and continuous random variables. (5.4) Easy. Assuming that the sample size, n, is sufficiently large, the Central Limit Theorem permits to draw conclusions about the population based strictly on sample data, and without having any knowledge about the distribution of the underlying population. (5.5) Moderate. Assuming that the population is normally distributed, the sampling distribution of the sample mean is normally distributed for samples of all sizes.
45-750 PROBABILITY AND STATISTICS PROBLEM SET 4 This assignment tests your understanding of Interval Estimation and Hypothesis Testing. The assign- ment aims to walk you through the following concepts: computing confidence intervals for population means with known and unknown variance, and for population proportions with sufficiently large sam- ples; phrasing hypotheses tests that rely on these confidence interval computations, and understanding the difference between using z-stats and t-stats. Problem 1 ( Confidence Interval Estimation with Known Variance). According to USA Today, customers are not settling for automobiles straight off the production lines. They typically get accessories that add a significant amount over the base sticker price. For BMW 3 Series with a sticker price of $40,250, the cost of accessories has an unknown mean µ and a known standard deviation σ = $3,450. A recent sample of 75 purchases yielded a sample mean of $8,200 above the sticker price as the cost of the accessories. (1.1) Easy Calculate the 90%, 95% and 99% confidence intervals for the average cost of accessories customers get on 3 Series. (1.2) Easy For the 95% confidence interval, determine the margin of error in estimating the average cost of accessories customers get on 3 Series. (1.3) Moderate What sample size would be required to reduce the margin of error by 50% (i.e. to half of its current size)? (1.4) Challenging Based on the above sample of 75 purchases, a statistician calculates a confidence interval of 8200 ± 400. What is the confidence level 1 − α associated with this interval. Problem 2 (Confidence Interval Estimation with Unknown Variance). Executive compensation has risen dramatically beyond the rising levels of an average worker’s wage over the years. Sarah is an MBA student who decides to use her statistical skills to estimate the mean CEO compensation for all large companies in the United States. She takes a random sample of eight CEO compensations in 2019. Firm Compensation ($ Million) eBay 34.8 Walt Disney 65.6 Wyndham 21.5 T-Mobile 66.5 Amex 24.2 AT&T 29.1 Williams-Sonoma 50.8 JPMorgan Chase 30.0 (2.1) Easy How will Sarah use the above information to provide a 95% confidence interval of the mean CEO compensation at large companies in the United States. (2.2) Easy What assumptions did Sarah make for deriving the interval estimate in (3.1)? (2.3) Easy How can Sarah reduce the margin of error? Problem 3 (Confidence Interval Estimation for Population Proportion). The coronavirus pandemic is altering the way students attend school. The admission’s office at Carnegie Mellon University would like to have a sense of how this will affect its graduate programs. A sample of 500 students was surveyed. Among them 145 enrolled in an online program. (3.1) Easy Construct a 95% confidence interval for the proportion of graduate students enrolled in an online program. 1 2 PROBLEM SET 4 (3.2) Moderate What will the confidence level be if we want our confidence interval to have a margin of error of 2%? Problem 4 (Hypothesis Testing for Population Mean Using Two Approaches). Who is the author? Statistics can help decide the authorship of literary works. Sonnets by a famous Elizabethan poet are known to contain an average of µ = 10.4 new words (words not used in the poet’s other work). The number of new words in this poet’s sonnets is normally distributed with standard deviation σ = 2.7. Now a manuscript with 5 new sonnets has come to light. Several scholars claim that these sonnets are the work of the famous Elizabethan poet. The new sonnets contain an average of X̄ = 12.1 words not used in the poet’s known works. Is this sufficiently strong evidence to reject the scholars’ claim that the manuscript is by the Elizabethan poet? We will carry out a one-tailed hypothesis test. (4.1) Easy Clearly write out the null hypothesis and the alternative hypothesis. (4.2) Easy Give the Z-test statistic and its p-value. (4.3) Moderate Test the null hypothesis using both the critical value approach (also called the Z- test) and the p-value method, and confirm that they lead to the same conclusion. Use α = 0.05. What do you conclude about the authorship of the new poems? Would your conclusion change if you were using a significance level α = 0.1? Problem 5 (Hypothesis Testing for Population Proportion). Before a new drug enters the market, it needs to get approval by the Food and Drug Administration (FDA) to guarantee that it is safe. Only if a vaccine’s benefits are found to outweigh its potential risks does the FDA grant a license for the vaccine, allowing it to be used by the public. The new vaccine Novatrax will get approved only if at most 5% of the patients will have side effects after taking it. The pharmaceutical lab that developed Novatrax claims that it meets this criterion. The FDA administered Novatrax to a test group of 1000 randomly selected patients, and found that 62 of them had side effects. Is this strong enough evidence for the FDA to deny approval of the vaccine? (5.1) Easy At a 5% level of significance, test whether at most 5% of the patients taking the new drug will have side effects: clearly write out the null and alternative hypothesis (explain why a one- sided test is appropriate here), construct the test statistic, compute its p-value, and compare against the required level of significance. (5.2) Easy Would your conclusion change if you repeated the analysis at the 1% significance level? Problem 6. Hypothesis Testing for Mean with unknown variance (30-Year Fixed Mortgage rate): Rates on 30-year fixed mortgages continue to be at historic lows. According to Freddie Mac, the average rate for 30-year fixed loans in the last week of July 2020 was 2.99%. An economist wants to test if there is any change in the mortgage rates since then. She searches for 30-year fixed loans on google.com and reports the rates offered by 9 banks, which is given in Excel workbook titled ‘mortgage’ in the sheet ‘HW4data2020.xslx’. Assume that rates are normally distributed. (6.1) Moderate. Does the average mortgage rate differ from 2.99%? Use a t-test and α = 0.10. Clearly state your null hypothesis, the test statistic, the critical value, your decision rule and conclusion. (6.2) Challenging. Report the 90% confidence interval for the average mortgage rate. Use this confidence interval to confirm your conclusion in part (6.1). (6.3) Challenging. Finally, carry out a hypothesis test against the null that the average mortgage rate is at most 2.99%. Clearly state your hypotheses, the test statistic, the critical value, your decision rule and conclusion. Use a t-test and α = 0.10. Does your conclusion change if you use α = 0.05?
45-750 PROBABILITY AND STATISTICS PROBLEM SET 5 This assignment tests your understanding of Simple Regression. The assignment aims to walk you through the following concepts: doing a hypothesis test for zero correlation, carrying out a simple one- variable regression and interpreting its results, conducting hypothesis tests for the value of the slope and intercept coefficients in a simple regression, and computing prediction intervals for the value of the dependent variable for a new data point. Problem 1 (Testing for Covid-19). The data set in sheet Covid-19 in the Excel document HW5data2020 contains information on the amount of testing for Covid-19 (column C) and the number of deaths per million (column B) in various countries since the beginning of the pandemic. We would like to know whether there is a relationship between these two variables. (1.1) Easy. Compute the sample correlation between ”deaths per 1M” and ”tests per 1M”. Test the hypothesis that there is no correlation between ”deaths per 1M” and ”tests per 1M” at the 5% significance level. (1.2) Moderate. Estimate the simple regression using ”tests per 1M” as the independent variable and ”deaths per 1M” as the ”dependent” variable. Interpret the slope coefficient. What is the R2 statistic for this regression? deathsi = β0 + β1 testsi + �i (1.3) Moderate. Compute a 95% confidence interval for the slope β1. Test the hypothesis that β1 = 0 at the 5% significance level. Problem 2 (Wages and IQ). For this problem use the data in sheet Wages in the Excel file HW5data2020 to estimate a simple regression explaining monthly salary (wage) in terms of IQ score (IQ). (2.1) Easy. Using the data provided, find the sample mean and sample standard deviation for IQ. (2.2) Moderate. IQ tests are standardized so that the average in the population is 100 with a standard deviation 15. Is the sample mean found (2.1) fairly ”average” or somewhat ”extreme”? To quantify this, compute the p-value of getting a sample mean as extreme or more extreme than the one computed in (2.1). Assume a 2-sided test. (2.3) Easy. Estimate the simple regression. wagei = β0 + β1IQi + �i (2.4) Easy. Calculate a 95% confidence interval for β1. (2.5) Easy. What is your best estimate for the monthly salary of an individual with an IQ of 120? (2.6) Challenging. Give a 95% prediction interval for the monthly salary of an individual with an IQ of 120? (2.7) Moderate. Does IQ explain most of the variation in monthly salary? Problem 3 (CAPM Model in Finance). In this problem, you will investigate the static capital asset pricing model (CAPM) of finance. This model holds that the excess return (i.e. the return in excess of the risk-free rate) on an asset is made up of two components, a systematic component and a diversifiable component. Let r̃m be a random variable representing the excess return on the market, and r̃ be a random variable representing the excess return on some asset. Then the CAPM says that: (1) r̃ = α+ βr̃m + � 1 2 PROBLEM SET 5 In writing equation (1), I have followed the finance convention of denoting the intercept as α and the slope as β. In this equation, � is the diversifiable component – a random variable that cannot be predicted with information prior to date t. The CAPM presumes that the � for a stock is not correlated with the � for any other stock and hence is not correlated with r̃m. The variation in the return on the market, r̃m, is driven by changes in the economic environment. The systematic component of the return to the asset is the component associated with the market return. Coefficient β in equation (1) is the asset’s ”beta” in finance parlance. It is a measure of the systematic risk of the asset. For a given total expected return, an asset with a high beta is considered riskier than an asset with a low beta. Why? Suppose you buy equal amounts of shares in a large number of assets. Since the idiosyncratic return (�’s) on these assets are uncorrelated, the variance of the idiosyncratic component of your portfolio will go to zero as the number of assets held gets large – thanks to the law of large numbers. In short, the risk on the � component of the asset return r̃ can be diversified away. This implies that the variance of the non-diversifiable component of r̃ is just β2 times the variance of r̃m (Recall that Var(α+ βr̃m) = β 2 Var(r̃m) always holds when α and β are constants and r̃m is a random variable). As a consequence, assets with a larger beta have a larger non-diversifiable risk and should therefore have a larger expected return. We will compute alpha and beta for Amazon, using the data in the Excel spreadsheet Amazon in HW5data2020. (3.1) Challenging. Using the data provided, compute the monthly rates of return on Amazon’s stock. Note that Amazon does not pay dividends, so the monthly rates of return can be computed directly from the stock prices in consecutive months (column B in the spreadsheet). For example, the monthly rate of return on 1-Aug-20 is =(B2-B3)/B3. Enter this formula in cell F2. Complete the computations by copying this formula down column F. You should now have 119 monthly rates of return (from 1-Oct-10 to 1-Aug-20). In the next column, compute the monthly excess rates of return on Amazon’s stock. Use the 10-year Treasury Bill given in column D as the yearly risk-free rate. Divide by 12 to get the monthly risk-free rate. Thus Amazon’s excess rate of return on 1-Aug-20 is =F2-D2/12. Enter this in cell G2. Complete the computations by copying this formula down column G. Repeat the same steps for the market’s rates of return. Use the SP500 as the market; the monthly values of this index are given in column C. In cell I2, write the formula =(C2-C3)/C3 and copy the formula down column I; in cell J2, write the formula =I2-D2/12 and copy it down column J. Run a linear regression where the independent variable is the market’s excess rate of return and the dependent variable is Amazon’s excess rate of return. What is the beta of Amazon? (3.2) Moderate. Provide a 95% confidence interval for the beta of Amazon. Yahoo.finance claims that Amazon’s beta is 1.33. Test this hypothesis at the 5% significance level. (3.3) Moderate. A prediction about the CAPM that will be developed in your finance class is that the intercept α = 0 for any asset. What value of α did you find in the regression performed in (3.1)? Test the hypothesis that α = 0 at a 5% significance level. Is the prediction α = 0 supported? (3.4) Moderate. Interpret the coefficient of determination R2 in terms of the systematic and diver- sifiable components of the risk.
45-750 PROBABILITY AND STATISTICS PROBLEM SET 6: ADVANCED HYPOTHESIS TESTS AND REVIEW OF EARLIER MATERIAL The first part of this assignment tests your understanding of hypothesis testing for the difference between two populations. The second part of the assignment tests your understanding of statistical inference concerning variance. The third part is a review of earlier material from the course. Problem 1 (Difference in means). Healthy Companies versus Failed Companies: In what ways are companies that fail different from those that continue to do business? To answer this question, one study compared various characteristics of 68 healthy and 33 failed firms. One of the variables was the ratio of current assets to current liabilities. Roughly speaking, this is the amount that the firm is worth divided by what it owes. The data is given in excel file ‘companies.xlsx’. Take Group 1 to be the firms that are healthy and Group 2 to be those that fail. The question of interest is whether or not the mean ratio of current assets to current liabilities is different for the two groups. (1.1) Easy. Write down the appropriate null and alternative hypotheses. (1.2) Challenging. Assuming equal population standard deviations for the two groups, give a 95% confidence interval for the difference in mean current assets to current liabilities ratios for healthy versus failed firms. (1.3) Moderate. Compute the test statistic to test the null hypothesis in (1.1). Test your hypothesis at the significance level α = 0.05. Assume equal population standard deviations for the two groups. Problem 2 (Testing for Difference in Proportion). “No Sweat” Garment Labels: A survey of U.S. residents aged 18 or older asked a series of questions about how likely they would be to purchase a garment under various conditions. For some conditions, it was stated that the garment had a “No Sweat” label; for others there was no mention of such a label. On the basis of the responses, each person was classified as a “label user” or a “label nonuser”. One purpose of the study was to describe the demographic characteristics of label users and nonusers. Here is a summary of the data. We let X denote the number of label users. Population n X p̂ = X/n Women 296 63 0.213 Men 251 27 0.108 (2.1) Easy. What is the fraction of ”label users” overall (aggregating over men and women). (2.2) Easy. Estimate the gender difference in label use. (2.3) Moderate. Calculate the 95% margin of error for the estimated gender difference in label usage as well as the 95% confidence interval for gender difference in label usage. (2.4) Challenging. We would like to test the hypothesis that men and women are equally likely to be label users, using the p-value. State the null hypothesis. Compute the Z-statistic under the null hypothesis. Compute the p-value. Test the null hypothesis at a significance level of 1%. Test the null hypothesis at a significance level of 0.1%. Problem 3 (Hypothesis testing for Variance). Rentals: We would like to determine whether the standard deviation on monthly rents for one-bedroom apartments is smaller than $200 in Pittsburgh. That is, we hope to reject the null hypothesis that the standard deviation is $200 or greater. The monthly rents on a random sample of 7 one-bedroom apartments are listed below (in $’s). Assume these monthly rents are normally distributed. 725 652 662 723 926 867 901 1 2 PROBLEM SET 6: ADVANCED HYPOTHESIS TESTS AND REVIEW OF EARLIER MATERIAL (3.1) Easy. Obtain an estimate of the standard deviation given the data. (3.2) Moderate. State the appropriate null and alternative hypotheses for the test and compute the value of the test statistic. (3.3) Moderate. Test the null hypothesis using critical value at α = 0.01. (3.4) Moderate. Repeat the test at α = 0.10. Problem 4 (Testing for equality of spread). Healthy vs. Failed Firms contd. Recall that in Problem 1 part (1.2) above you ran the hypothesis test for difference in means assuming that the population standard deviation for both the groups is identical. Let’s test that assumption. (4.1) Easy. What statistic would you use to test that the population standard deviations for the two types of firms is equal? Clearly write out the hypothesis and calculate the relevant statistic. (4.2) Moderate. What are the exact degrees of freedom associated with the F-statistic in this case? (4.3) Challenging. Test the hypothesis in part (4.1) at the 5% level. Does your conclusion justify the assumption made in Problem 1? Problem 5 (Decision Analysis). You are considering investing $500,000 in a real estate project. There is an 80% chance that you will make a profit of $300,000. But there is also a 20% chance that you will loose $100,000. (5.1) Easy. What is your expected profit if you invest? (5.2) Moderate. Suppose you have the option of hiring the research firm RELCO, that can accurately predict if the project will be successful, for a fee of $10,000. Should you hire RELCO? Suppose RELCO is not always able to predict accurately. Based on historical performance, RELCO predicts correctly 90% of the time when a project is successful; RELCO also predicts correctly 90% of the time that a project is unsuccessful when this is the case. (5.3) Moderate. Compute the probability that RELCO predicts a successful project. Compute the probability that the project will be successful, given that RELCO predicts it will be successful. Compute the probability that the project will be successful, given that RELCO predicts it will be unsuccessful. (5.4) Challenging. Would you hire RELCO for $10,000? Problem 6 (Normal Distribution). Mike is in charge of producing a sophisticated electronic device and the deadline is coming up. The specifications for the device are stringent. In particular, the device requires 60 capacitors which must operate within ±0.50 pF from a specified standard of 28 pFs. The problem is that the supplier can provide capacitors that operate according to a normal dis- tribution, with a mean of 28 pFs and a standard deviation of 1 pF (the capacitor characteristics are independent random variables). Therefore, to have 60 capacitors that Mike can use in the electronic device, he will have to order more than 60 from the supplier. These items are expensive, and Mike wants to order the smallest batch that will have at least a 95% chance of containing enough usable capacitors. Mike needs your help in figuring out the size of the order he should place. (6.1) Easy. What is the probability that a capacitor provided by the supplier will meet the specifi- cation needed for the electronic device? (6.2) Challenging. What is the size of the order that Mike should place?
There are no answers to this question.
Login to buy an answer or post yours. You can also vote on other others

Get Help With a similar task to - Probability and Statistics Class

Related Questions

Similar orders to Probability and Statistics Class
Popular Services
Tutlance Experts offer help in a wide range of topics. Here are some of our top services: