Assignment #2 (adv data)-2

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University of Toronto *

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343

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Statistics

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Feb 20, 2024

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Uploaded by MagistrateRookMaster3708

Question 1 : The mean for player 1 = 89.4 and the mean for player 2 = 84.025 > meanP1=mean(FreeThrow$Player1) > meanP1 [1] 89.4 > meanP2=mean(FreeThrow$Player2) > meanP2 [1] 84.025 Question 2: >DevP1= (FreeThrow$Player1-meanP1) >DevP1 [1] 2.7 1.2 -4.4 3.0 1.7 -10.4 4.8 2.4 -4.4 1.6 3.0 0.4 [13] 1.6 -1.8 -5.3 2.3 -6.9 0.7 2.2 5.6 >DevP2= (FreeThrow$Player2-meanP2) >DevP2 [1] -19.725 6.675 -0.325 0.375 4.275 -27.325 -0.625 12.775 3.375 [10] 6.375 2.175 -2.125 -2.925 -4.425 0.275 14.275 -2.725 -1.925 [19] 12.175 -0.625 Question 3: I don’t think that my calculations from question 2 best represent how well the mean fits the data. To further examine the data I calculated the sum of squares, the variance and the standard deviation for player 1 and 2. This is because just calculating how much the shooting percentage from each practice deviates only shows us the difference between the mean and a single data point. Moreover, to best determine how well the mean fits the data and the variability of the data we can use the sum of squares, the variance and the standard deviation which ultimately give us the same thing. Now instead of only looking at the difference between the mean and a data point we can now understand how well the mean fits the data. > sumsq1= sum((DevP1)^2) > sumsq1 [1] 331.5 > sumsq2= sum((DevP2)^2)

>sumsq2 [1] 1815.4175 > VP1=var(FreeThrow$Player1) > VP1 [1] 17.44737 > VP2=var(FreeThrow$Player2) > VP2 [1] 95.54829 > stdevP1=sqrt(var(FreeThrow$Player1)) > stdevP1 [1] 4.177005 > stdevP2=sqrt(var(FreeThrow$Player2)) > stdevP2 [1] 9.774881 Question 4: > cohen.d(FreeThrow$Player1,FreeThrow$Player2) Cohen's d d estimate: 0.7150931 (medium) 95 percent confidence interval: lower upper 0.05478061 1.37540560 Question 5: I would select player 1 because he had a higher mean shooting percentage of 89.4 in comparison to player 2 who had a mean shooting percentage of 84.025. Additionally, player 1 had a lower standard deviation. Player 1’s lower standard deviation of 4.177005 shows that player 1’s data points (his shooting percentage from each practice) are clustered closely to the mean. With this information we can conclude that player 1 not only had a higher mean shooting percentage but they were also a lot more consistent/reliable than player 2 who had a standard deviation of 9.774881. Lastly, there is a medium-sized difference in free throw shooting percentages between player 1 and 2, with a Cohen's d estimate of 0.7150931 and a 95 percent confidence interval. With this information I can conclude that player 1 is the better option for the team and should be the selected player.

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