An year ago, a group of students HD, RF, LD, and KC designed a project that would allow them to predict the best time taken to run the 5K. The description of the variables is given in the APPENDIX. a) (2) Generate the correlation between T and all the variables using EXCEL. Copy and paste the results here. b) (2) Generate the regression of T on all the independent variables in the data using EXCEL. Copy and paste the results here. This is MODEL I. c)(3) Interpret the estimated correlation coefficient between miles run per week and the best 5K time. d) (7) Test the claim that there is no correlation between the best time and the miles run against the claim that miles run and the best time for the 5K move inversely (as one goes up, the other falls). Use a 5% level of significance. Ho: Ha: Decision Rule (Draw the distribution, show the critical values, and shade the rejection region) Test statistic: (Show the calculated value formula and what you are using to generate the value) Decision : Do not reject Ho Reject Ho (Circle) Conclusion in the context of this problem: e) (4) Evaluate the explanatory power of MODEL 1. f) (2) Interpret the estimated coefficient of HSA (1 if the runner was a high school athlete when they achieved their best time and 0 if not) in MODEL I. g) (2) We have a runner (identified as female, G=0), who runs 35 miles per week, and ran her best time in college as a 20 year old after 9 years of competitive running. She completes a warm up routine (WU= 1) that is different from dynamic stretching (TW=0) warm up routine before a run. She runs to cool down and her strength training (STR) is focused on core/abs. 1 = 28.2129 - 0.0794M - 4.9236G - 2.1713CA-3.1165HSA -2.3771WU +2.3729TW + 3.1964CT - 2.9010STR-0.4130Y + 0.1404A. M= 35 G=0 CA = 1 HSA = 0 WU = 1 TW=0 CT = 1 STR = 1 Y = 9 A= 20 Her reported actual time is only 18.75 minutes. The residual shows that the uncontrolled factors made her faster than what we would have expected. T or F (7) Test the overall significance of MODEL 1. Use a 5% level of significance. Ho: Ha: Decision Rule (Use the p-value rule) Test statistic: (Report the calculated value and the p-value in the output) Decision : Do not reject Ho Reject Ho (Circle) Conclusion in the context of this problem: h) (7) Test the claim that holding all else constant, the type of strength training makes no difference to the best time. Use a 5% level of significance. Ho: Ha: Decision Rule (Use the p-value rule) Test statistic: (Report the calculated value and the p-value in the output) Decision : Do not reject Ho Reject Ho (Circle) Conclusion in the context of this problem: i) (4) Assume that the conclusion to (i) was that STR makes no difference. As a result, we can assume it is a redundant variable. Remove STR from your model and run a regression of T on all the other variables. Copy and ABC HD, RF, LD, and KC. 7 1 paste the output here. Did removing STR improve the explanatory power of the model? Please discuss this thoroughly using the concept of R2, the degrees of freedom and the adjusted R2. T M HSA WU TW CT Y STR CA T M G HSA WU TW 1 0.573380771 -0.429284183 -0.164130425 -0.20756139 0.092609625 -0.167764301 -0.369933808 0.207015267 0.026686857 -0.446429358 1 0.178611097 1 0.031925847 -0.02868 1 0.145215665 0.0041072 0.15107392 -0.048718937 0.089626 0.31501832 0.543828404 -0.02124 0.14106013 0.476292026 0.085607 0.1446786 -0.120867427 0.0999563 -0.32158 0.21966276 -0.410961 0.1823232 0.545863077 0.0657152 0.4364358 Y 1 0.293754852 1 0.252766289 0.0795724 1 -0.073990381 0.136033 0.1436118 1 -0.212830039 -0.231347 0.352596 0.3070905 1 0.012659242 0.2872972 0.2127624 0.075169 0.075169 0.179993 1 0.225961538 0.151074 0.3107782 0.5048241 -0.118682 0.2025479 A STR CA 1 SUMMARY OUTPUT Regression Statistics Multiple R 0.809130956 R Square 0.654692904 Adjusted R 0.601110768 Standard Er 3.154832977 Observatior 68 ANOVA df Significance F 1.43457E-10 9 Regression Residual Total SS MS F 1094.492694 121.6103 12.2184921 577.2723245 9.9529711 1671.765018 58 67 Intercept G HSA WU TW Coefficients 29.02667853 -5.578960378 -4.055170221 -1.980844488 2.42002585 1.949540774 -0.572326832 0.13670822 -3.793309552 -3.313471134 Standard Error t Stat P-value 2.254083617 12.877374 1.1829E-18 0.889337056 6.273168 4.8114E-08 1.138684783 -3.561276 0.00074515 1.068438293 -1.853962 0.06883182 0.952444235 2.5408583 0.01375594 0.917267706 2.1253782 0.03782548 0.160026695 -3.576446 0.00071066 0.054813771 2.4940488 0.01550178 1.526593034 -2.48482 0.01586871 1.556045746 2.129418 0.03747573 CT Y STR CA An year ago, a group of students HD, RF, LD, and KC designed a project that would allow them to predict the best time taken to run the 5K. The description of the variables is given in the APPENDIX. a) (2) Generate the correlation between T and all the variables using EXCEL. Copy and paste the results here. b) (2) Generate the regression of T on all the independent variables in the data using EXCEL. Copy and paste the results here. This is MODEL I. c)(3) Interpret the estimated correlation coefficient between miles run per week and the best 5K time. d) (7) Test the claim that there is no correlation between the best time and the miles run against the claim that miles run and the best time for the 5K move inversely (as one goes up, the other falls). Use a 5% level of significance. Ho: Ha: Decision Rule (Draw the distribution, show the critical values, and shade the rejection region) Test statistic: (Show the calculated value formula and what you are using to generate the value) Decision : Do not reject Ho Reject Ho (Circle) Conclusion in the context of this problem: e) (4) Evaluate the explanatory power of MODEL 1. f) (2) Interpret the estimated coefficient of HSA (1 if the runner was a high school athlete when they achieved their best time and 0 if not) in MODEL I. g) (2) We have a runner (identified as female, G=0), who runs 35 miles per week, and ran her best time in college as a 20 year old after 9 years of competitive running. She completes a warm up routine (WU= 1) that is different from dynamic stretching (TW=0) warm up routine before a run. She runs to cool down and her strength training (STR) is focused on core/abs. 1 = 28.2129 - 0.0794M - 4.9236G - 2.1713CA-3.1165HSA -2.3771WU +2.3729TW + 3.1964CT - 2.9010STR-0.4130Y + 0.1404A. M= 35 G=0 CA = 1 HSA = 0 WU = 1 TW=0 CT = 1 STR = 1 Y = 9 A= 20 Her reported actual time is only 18.75 minutes. The residual shows that the uncontrolled factors made her faster than what we would have expected. T or F (7) Test the overall significance of MODEL 1. Use a 5% level of significance. Ho: Ha: Decision Rule (Use the p-value rule) Test statistic: (Report the calculated value and the p-value in the output) Decision : Do not reject Ho Reject Ho (Circle) Conclusion in the context of this problem: h) (7) Test the claim that holding all else constant, the type of strength training makes no difference to the best time. Use a 5% level of significance. Ho: Ha: Decision Rule (Use the p-value rule) Test statistic: (Report the calculated value and the p-value in the output) Decision : Do not reject Ho Reject Ho (Circle) Conclusion in the context of this problem: i) (4) Assume that the conclusion to (i) was that STR makes no difference. As a result, we can assume it is a redundant variable. Remove STR from your model and run a regression of T on all the other variables. Copy and ABC HD, RF, LD, and KC. 7 1 paste the output here. Did removing STR improve the explanatory power of the model? Please discuss this thoroughly using the concept of R2, the degrees of freedom and the adjusted R2. T M HSA WU TW CT Y STR CA T M G HSA WU TW 1 0.573380771 -0.429284183 -0.164130425 -0.20756139 0.092609625 -0.167764301 -0.369933808 0.207015267 0.026686857 -0.446429358 1 0.178611097 1 0.031925847 -0.02868 1 0.145215665 0.0041072 0.15107392 -0.048718937 0.089626 0.31501832 0.543828404 -0.02124 0.14106013 0.476292026 0.085607 0.1446786 -0.120867427 0.0999563 -0.32158 0.21966276 -0.410961 0.1823232 0.545863077 0.0657152 0.4364358 Y 1 0.293754852 1 0.252766289 0.0795724 1 -0.073990381 0.136033 0.1436118 1 -0.212830039 -0.231347 0.352596 0.3070905 1 0.012659242 0.2872972 0.2127624 0.075169 0.075169 0.179993 1 0.225961538 0.151074 0.3107782 0.5048241 -0.118682 0.2025479 A STR CA 1 SUMMARY OUTPUT Regression Statistics Multiple R 0.809130956 R Square 0.654692904 Adjusted R 0.601110768 Standard Er 3.154832977 Observatior 68 ANOVA df Significance F 1.43457E-10 9 Regression Residual Total SS MS F 1094.492694 121.6103 12.2184921 577.2723245 9.9529711 1671.765018 58 67 Intercept G HSA WU TW Coefficients 29.02667853 -5.578960378 -4.055170221 -1.980844488 2.42002585 1.949540774 -0.572326832 0.13670822 -3.793309552 -3.313471134 Standard Error t Stat P-value 2.254083617 12.877374 1.1829E-18 0.889337056 6.273168 4.8114E-08 1.138684783 -3.561276 0.00074515 1.068438293 -1.853962 0.06883182 0.952444235 2.5408583 0.01375594 0.917267706 2.1253782 0.03782548 0.160026695 -3.576446 0.00071066 0.054813771 2.4940488 0.01550178 1.526593034 -2.48482 0.01586871 1.556045746 2.129418 0.03747573 CT Y STR CA