MATH 399:Course Project Part I & II Applied Managerial Statistics

Course Project Part I & II MATH 399: Applied Managerial Statistics Most young athletic students wrestle with the question of which career path to pursue and one of the big influencers of their choices is notably the financial gain involved in being a professional athletic over academia. This topic got me to ask the question; what amount of money are we really talking about? For example, when considering pursuing a career in basketball over some academic program. The table below reveals the salaries payed to the top 50 ranked NBA players in the recent basketball season. (ESPN, 2019). R K SALARY R K SALARY R K SALARY R K SALARY R K SALARY 1 $37,457,1 54 11 $30,000,0 00 21 $25,434,2 63 31 $24,107,2 58 41 $20,445,7 79 2 $35,654,1 50 12 $29,230,7 69 22 $25,434,2 63 32 $24,000,0 00 42 $20,421,5 46 3 $35,654,1 50 13 $28,928,7 09 23 $25,434,2 63 33 $23,241,5 73 43 $20,099,1 89 4 $35,654,1 50 14 $27,977,6 89 24 $25,434,2 62 34 $23,114,0 67 44 $20,000,0 00 5 $32,088,9 32 15 $27,739,9 75 25 $24,605,1 81 35 $22,897,2 00 45 $19,500,0 0 6 $31,214,2 95 16 $26,011,9 13 26 $24,157,3 03 36 $22,347,0 15 46 $19,356,9 32 7 $31,200,0 00 17 $25,976,1 11 27 $24,157,3 03 37 $21,666,6 67 47 $19,200,1 27 8 $30,560,7 00 18 $25,759,7 66 28 $24,119,0 25 38 $21,590,9 09 48 $19,169,8 00 9 $30,521,1 15 19 $25,467,2 50 29 $24,119,0 25 39 $21,587,5 79 49 $19,000,0 00 10 $30,421,8 20 $25,467,2 30 $24,107,2 40 $21,000,0 50 $19,000,0 54 50 58 00 00 The sample mean of the above data is $25,044,674.4 and the sample standard deviation is $4,816,519.842 a) When confidence level is 80%, Maximum error (E)= CONFIDENCE.NOM (0.2, 4,816,519,842.50) = 872,940.09 - Therefore, the confidence interval (Mean +/- E = 25,044,674.4 +/- 872,940.09) : ($24,171,734.31, $25,917,614.49) b) When confidence level is 95 %, E = CONFIDENCE.NOM (0.05, 4,816,519.842,50) = 1,335,046.65 - Therefore, the confidence intervals: ($23,709,627.75, $26,379,721.05) c) When confidence level is 99%, (E) = CONFIDENCE.NOM (0.01, 4,816,519.842,50) = .72 - Therefore, confidence intervals: ($23,290,125.68, $26,799,223.12) d) When the confidence level is 93% - E = CONFIDENCE.NOM (α, δ, n) - α = 1 – 0.93 = 0.07 - δ = $4,816,519,842.5 - n = 50 - Therefore E = CONFIDENCE.NOM (0.07, 4,816,519.842, 50) - E = $.84 - Margins of Error (E) = +/- $.84 - Confidence interval: (µ - E, µ + E) - µ-E = $25,044,674.4 - $.84 = 23,810,475.56 - µ+E = $25,044,674.4 + $.84 = $26,278,873.24 Therefore, confidence interval: ($23,.56, $26,278,873.24) As the confidence level rises, the margin of error increases because the larger the expected proportion of intervals that will contain the parameter, the larger the margin of error. At 80% confidence level, the margin of error is plus or minus 872,940.09. This increases to .72 for 99%, almost double that at 80%. And as margin of error expands, so does the intervals that will contain a given parameter. Therefore, as the confidence level increase, so does the confidence interval expand. In this case the confidence interval at 80% confidence level is (($24,171,734.31, $25,917,614.49) as oppose to ($23,290,125.68, $26,799,223.12) at 99% confidence level. We are 80% confident the interval ($24,171,734.31, $25,917,614.49) captures the salary that an NBA player in the top 50 made in season. We are 93% confident the interval ($23,.56, $26,278,873.24) will contain the salary earned by an NBA player in the top 50 rank in season. We are 95 % sure that the interval ($23,709,627.75, $26,379,721.05) contains the salary paid to an NBA top 50 player in the season. Lastly, we can assert with 99 % confidence that the interval ($23,290,125.68, $26,799,223.12) captures the salary of an NBA top 50 player in season. This project gave me an opportunity to construct my own confidence interval. In the process, I had to master the concept of confidence level and how it translates to confident intervals. I became very familiar with the mathematical formula and operations to obtain any required value concerning this topic. And having to repeatedly calculate confidence interval has improve my understanding of this concept and elevated my ability to look at statistics problem as it relates to practical examples in our day to day life. Summary Table for Deaths, September 1996 Vol. 45, No. 9 + March 31, 1997 Mean 3524. Median 2479 Standard Deviation 3691.23816 Minimum 194 Maximum 18229 Summary Table for Live Births, September 1996 Vol. 45, No. 9 + March 31, 1997 Mean 6652. Median 4310.5 Standard Deviation 8690. Minimum 519 Maximum 54211 Summary Table for Marriages, September 1996 Vol. 45, No. 9 + March 31, 1997 Mean 4172. Median 2882 Standard Deviation 4482. Minimum 0 Maximum 19560 Summary Table for Divorces, September 1996 Vol. 45, No. 9 + March 31, 1997 Mean 1464. Median 1093.5 Standard Deviation 1544. Minimum 0 Maximum 7757 3 a. Determine if there is sufficient evidence to conclude the average amount of births is over 8000 in the United States and territories at the 0.05 level of significance. This claim is alternate. And it is a right tail test. Null hypothesis Ho: µ = µo Alternate hypothesis Ha: µ > µo The region is based on α = 0.05 Reject Ho if the P value is lower than α = 0.05. Fail to reject Ho if P value is higher. α = 0.05, µo = 8000, x ≈ 6652.81, n= 52, δ ≈ 8690.01 Test Statistic: Z = (X - µo) / (δ/√n) Z ≈ -1.12 P- value for z = 1.12: (0.1312) The region is based on α = 0.05 and (p = 0.1312 > α = 0.05) Therefore, we fail to reject Ho and say that there isn’t sufficient evidence to conclude that the average amount of births is over 8000 in the United States and territories. b. Determine if there is sufficient evidence to conclude the average amount of deaths is below is less than or equal to 6000 in the United States and territories at the 0.10 level of significance. This claim is Null. And it is a right tail test. Null hypothesis Ho: µ ≤ µo Alternate hypothesis Ha: µ > µo The region is based on α = 0.1 Reject Ho if the P value is lower than α = 0.1. Fail to reject Ho if P value is higher. α = 0.01, µo = 6000, x ≈ 3524.44, n= 52, δ ≈ 3691.24 Test Statistic: Z = (X - µo) / (δ/√n) Z ≈ -4.836 P- value for Z = -4.836: (≈ 0.00001) The region is based on α = 0.1 and (P = 0.00001 < α = 0.1) Therefore, we reject Ho and say that there is sufficient evidence to conclude that the average amount of deaths is below is less than or equal to 6000 in the United States and territories at the 0.10 level of significance. c. Determine if there is sufficient evidence to conclude the average amount of marriages is greater or equal to 7000 in the United States and territories at the 0.05 level of significance This claim is Null. And it is a left tail test. Null hypothesis Ho: µ ≥ µo Alternate hypothesis Ha: µ< µo The region is based on α = 0.05 Reject Ho if the P value is lower than α = 0.1. Fail to reject Ho if P value is higher. α = 0.05, µo = 7000, X ≈ 4172.35, n= 52, δ ≈ 4482.17 Test Statistic: Z = (X - µo) / (δ/√n) Z ≈ -4.549 P- value for Z = -4.549: (< 0.00001) The region is based on α = 0.05 and (P = 0.00001 < α = 0.05) Therefore, we reject Ho and say that there is sufficient evidence to conclude the average amount of marriages is greater or equal to 7000 in the United States and territories at the 0.05 level of significance. d. Determine if there is sufficient evidence to conclude the average amount of divorces is less than or equal to 4000 in the United States and territories at the 0.1 level of significance. This claim is Null. And it is a right tail test. Null hypothesis Ho: µ ≤ µo Alternate hypothesis Ha: µ > µo The region is based on α = 0.1 Reject Ho if the P value is lower than α = 0.1. Fail to reject Ho if P value is higher. α = 0.1, µo = 4000, x ≈ 1464.94, n= 52, δ ≈ 1544.71 Test Statistic: Z = (X - µo) / (δ/√n) Z ≈ -11.834 P- value for Z = -11.834: (< 0.00001) The region is based on α = 0.1 and (P = 0.00001 < α = 0.1) Therefore, we reject Ho and say that there is sufficient evidence to conclude the average amount of divorces is less than or equal to 4000 in the United States and territories at the 0.1 level of significance. 4. a. Determine if there is sufficient evidence to conclude the average amount of births is under 8000 in the United States and territories at the 0.01 level of significance. This claim is alternate. And it is a left tail test. Null hypothesis Ho: µ = µo Alternate hypothesis Ha: µ < µo The region is based on α = 0.01 Reject Ho if the P value is lower than α = 0.01. Fail to reject Ho if P value is higher. α = 0.0, µo = 8000, x ≈ 6652.81, n= 52, δ ≈ 8690.01 Test Statistic: Z = (X - µo) / (δ/√n) Z ≈ -1.12 P- value for Z = 1.12: (0.1312) The region is based on α = 0.01 and (p = 0.1312 > α = 0.01) Therefore, we fail to reject Ho and say that there isn’t sufficient evidence to conclude that the average amount of births is under 8000 in the United States and territories. b. Determine if there is sufficient evidence to conclude the average amount of deaths is below is greater than or equal to 6000 in the United States and territories at the 0.05 level of significance. This claim is Null. And it is a right tail test. Null hypothesis Ho: µ ≥ µo Alternate hypothesis Ha: µ > µo The region is based on α = 0.05 Reject Ho if the P value is lower than α = 0.05. Fail to reject Ho if P value is higher. α = 0.05, µo = 6000, x ≈ 3524.44, n= 52, δ ≈ 3691.24 Test Statistic: Z = (X - µo) / (δ/√n) Z ≈ -4.836 P- value for Z = -4.836: (< 0.00001) The region is based on α = 0.05 and (P = 0.00001 < α = 0.05) Therefore, we reject Ho and say that there is sufficient evidence to conclude that the average amount of deaths is below is greater than or equal to 6000 in the United States and territories at the 0.05 level of significance. c. Determine if there is sufficient evidence to conclude the average amount of marriages is less or equal to 7000 in the United States and territories at the 0.1 level of significance This claim is Null. And it is a left tail test. Null hypothesis Ho: µ ≤ µo Alternate hypothesis Ha: µ > µo The region is based on α = 0.1 Reject Ho if the P value is lower than α = 0.1. Fail to reject Ho if P value is higher. α = 0.1, µo = 7000, X ≈ 4172.35, n= 52, δ ≈ 4482.17 Test Statistic: Z = (X - µo) / (δ/√n) Z ≈ -4.549 P- value for Z = -4.549: (< 0.00001) The region is based on α = 0.05 and (P = 0.00001 < α = 0.05) Therefore, we reject Ho and say that there is sufficient evidence to conclude the average amount of marriages is less or equal to 7000 in the United States and territories at the 0.1 level of significance. d. Determine if there is sufficient evidence to conclude the average amount of divorces is greater than or equal to 4000 in the United States and territories at the 0.01 level of significance. This claim is Null. And it is a right tail test. Null hypothesis Ho: µ ≥ µo Alternate hypothesis Ha: µ < µo The region is based on α = 0.01 Reject Ho if the P value is lower than α = 0.01. Fail to reject Ho if P value is higher. α = 0.01, µo = 4000, x ≈ 1464.94, n= 52, δ ≈ 1544.71 Test Statistic: Z = (X - µo) / (δ/√n) Z ≈ -11.834 P- value for Z = -11.834: (< 0.00001) The region is based on α = 0.01 and (P = 0.00001 < α = 0.1) Therefore, we reject Ho and say that there is sufficient evidence to conclude the average amount of divorces is greater than or equal to 4000 in the United States and territories at the 0.01 level of significance. References ESPN. (2019). NBA Player Salaries - . Retrieved June 17, 2019, from Center for Disease Control & Prevention. (2019). Monthly Vital Statistics Reports. Vol. 45, No. 9 + March 31, 1997 Retrieved from