# Read in data
firstbase = read.csv("firstbasestats.csv")
str(firstbase)## 'data.frame': 23 obs. of 15 variables:
## $ Player : chr "Freddie Freeman" "Jose Abreu" "Nate Lowe" "Paul Goldschmidt" ...
## $ Pos : chr "1B" "1B" "1B" "1B" ...
## $ Team : chr "LAD" "CHW" "TEX" "STL" ...
## $ GP : int 159 157 157 151 160 140 160 145 146 143 ...
## $ AB : int 612 601 593 561 638 551 583 555 545 519 ...
## $ H : int 199 183 179 178 175 152 141 139 132 124 ...
## $ X2B : int 47 40 26 41 35 27 25 28 40 23 ...
## $ HR : int 21 15 27 35 32 20 36 22 8 18 ...
## $ RBI : int 100 75 76 115 97 84 94 85 53 63 ...
## $ AVG : num 0.325 0.305 0.302 0.317 0.274 0.276 0.242 0.251 0.242 0.239 ...
## $ OBP : num 0.407 0.379 0.358 0.404 0.339 0.34 0.327 0.305 0.288 0.319 ...
## $ SLG : num 0.511 0.446 0.492 0.578 0.48 0.437 0.477 0.423 0.36 0.391 ...
## $ OPS : num 0.918 0.824 0.851 0.981 0.818 0.777 0.804 0.729 0.647 0.71 ...
## $ WAR : num 5.77 4.19 3.21 7.86 3.85 3.07 5.05 1.32 -0.33 1.87 ...
## $ Payroll.Salary2023: num 27000000 19500000 4050000 26000000 14500000 ...We can see our first three columns are strings, then everything to RBI is an int and the rest are floars.
summary(firstbase)## Player Pos Team GP
## Length:23 Length:23 Length:23 Min. : 5.0
## Class :character Class :character Class :character 1st Qu.:105.5
## Mode :character Mode :character Mode :character Median :131.0
## Mean :120.2
## 3rd Qu.:152.0
## Max. :160.0
## AB H X2B HR
## Min. : 14.0 Min. : 3.0 Min. : 1.00 Min. : 0.00
## 1st Qu.:309.0 1st Qu.: 74.5 1st Qu.:13.50 1st Qu.: 8.00
## Median :465.0 Median :115.0 Median :23.00 Median :18.00
## Mean :426.9 Mean :110.0 Mean :22.39 Mean :17.09
## 3rd Qu.:558.0 3rd Qu.:146.5 3rd Qu.:28.00 3rd Qu.:24.50
## Max. :638.0 Max. :199.0 Max. :47.00 Max. :36.00
## RBI AVG OBP SLG
## Min. : 1.00 Min. :0.2020 Min. :0.2140 Min. :0.2860
## 1st Qu.: 27.00 1st Qu.:0.2180 1st Qu.:0.3030 1st Qu.:0.3505
## Median : 63.00 Median :0.2420 Median :0.3210 Median :0.4230
## Mean : 59.43 Mean :0.2499 Mean :0.3242 Mean :0.4106
## 3rd Qu.: 84.50 3rd Qu.:0.2750 3rd Qu.:0.3395 3rd Qu.:0.4690
## Max. :115.00 Max. :0.3250 Max. :0.4070 Max. :0.5780
## OPS WAR Payroll.Salary2023
## Min. :0.5000 Min. :-1.470 Min. : 720000
## 1st Qu.:0.6445 1st Qu.: 0.190 1st Qu.: 739200
## Median :0.7290 Median : 1.310 Median : 4050000
## Mean :0.7346 Mean : 1.788 Mean : 6972743
## 3rd Qu.:0.8175 3rd Qu.: 3.140 3rd Qu.: 8150000
## Max. :0.9810 Max. : 7.860 Max. :27000000We can see that the average salary is 697,2743
# Linear Regression (one variable)
model1 = lm(Payroll.Salary2023 ~ RBI, data=firstbase)
summary(model1)##
## Call:
## lm(formula = Payroll.Salary2023 ~ RBI, data = firstbase)
##
## Residuals:
## Min 1Q Median 3Q Max
## -10250331 -5220790 -843455 2386848 13654950
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2363744 2866320 -0.825 0.41883
## RBI 157088 42465 3.699 0.00133 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6516000 on 21 degrees of freedom
## Multiple R-squared: 0.3945, Adjusted R-squared: 0.3657
## F-statistic: 13.68 on 1 and 21 DF, p-value: 0.001331We can see that the p-value is low for the RBI predicting the salary.
# Sum of Squared Errors
model1$residuals## 1 2 3 4 5 6
## 13654950.2 10082148.6 -5524939.3 10298631.2 1626214.0 -6731642.8
## 7 8 9 10 11 12
## -5902522.2 -10250330.7 -4711916.8 -532796.1 -6667082.5 -6696203.1
## 13 14 15 16 17 18
## 7582148.6 -4916640.9 -1898125.3 -336532.3 -995042.5 -1311618.3
## 19 20 21 22 23
## -843454.5 8050721.3 1250336.9 1847040.4 2926656.0SSE = sum(model1$residuals^2)
SSE## [1] 8.914926e+14Lets see if adding the predictor AVG will improve this linear regression models ability to predict salary.
# Linear Regression (two variables)
model2 = lm(Payroll.Salary2023 ~ AVG + RBI, data=firstbase)
summary(model2)##
## Call:
## lm(formula = Payroll.Salary2023 ~ AVG + RBI, data = firstbase)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9097952 -4621582 -33233 3016541 10260245
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -18083756 9479036 -1.908 0.0709 .
## AVG 74374031 42934155 1.732 0.0986 .
## RBI 108850 49212 2.212 0.0388 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6226000 on 20 degrees of freedom
## Multiple R-squared: 0.4735, Adjusted R-squared: 0.4209
## F-statistic: 8.994 on 2 and 20 DF, p-value: 0.001636We can see tha that the adjusted r squared improved overall so this model is slighlty superior even though the p value is now higher on RBI and the AVG predictors pvalue is larger than 5%
# Sum of Squared Errors
SSE = sum(model2$residuals^2)
SSE## [1] 7.751841e+14We can also see that this model had a lower SSE which signifies that the distances between the data points and the fitted values are smaller. So this model is superior to the first that just had RBI.
Lets see if adding more predictors will improve this models ability to predict salary.
# Linear Regression (all variables)
model3 = lm(Payroll.Salary2023 ~ HR + RBI + AVG + OBP+ OPS, data=firstbase)
summary(model3)##
## Call:
## lm(formula = Payroll.Salary2023 ~ HR + RBI + AVG + OBP + OPS,
## data = firstbase)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9611440 -3338119 64016 4472451 9490309
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -31107858 11738494 -2.650 0.0168 *
## HR -341069 552069 -0.618 0.5449
## RBI 115786 113932 1.016 0.3237
## AVG -63824769 104544645 -0.611 0.5496
## OBP 27054948 131210166 0.206 0.8391
## OPS 60181012 95415131 0.631 0.5366
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6023000 on 17 degrees of freedom
## Multiple R-squared: 0.5811, Adjusted R-squared: 0.4579
## F-statistic: 4.717 on 5 and 17 DF, p-value: 0.006951The models p values are high for all the predictors, but the adjusted r squared is higer than the last at 0.46 which is promising.
# Sum of Squared Errors
SSE = sum(model3$residuals^2)
SSE## [1] 6.167793e+14This model does seem to be an imporvement with its lower SSE at 6.168 to the previous 7.75.
# Remove HR
model4 = lm(Payroll.Salary2023 ~ RBI + AVG + OBP+OPS, data=firstbase)
summary(model4)##
## Call:
## lm(formula = Payroll.Salary2023 ~ RBI + AVG + OBP + OPS, data = firstbase)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9399551 -3573842 98921 3979339 9263512
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -29466887 11235931 -2.623 0.0173 *
## RBI 71495 87015 0.822 0.4220
## AVG -11035457 59192453 -0.186 0.8542
## OBP 86360720 87899074 0.982 0.3389
## OPS 9464546 47788458 0.198 0.8452
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5919000 on 18 degrees of freedom
## Multiple R-squared: 0.5717, Adjusted R-squared: 0.4765
## F-statistic: 6.007 on 4 and 18 DF, p-value: 0.00298Again a small improvement to the adjusted r squared.
Lets inspect the corelation between the RBI feature and the Salary2023 feature.
firstbase<-firstbase[,-(1:3)]
# Correlations
cor(firstbase$RBI, firstbase$Payroll.Salary2023)## [1] 0.6281239Lets inspect the corelation between the AVG feature and the OBP feature.
cor(firstbase$AVG, firstbase$OBP)## [1] 0.8028894Lets inspect the corelation between all of the features.
cor(firstbase)## GP AB H X2B HR RBI
## GP 1.0000000 0.9779421 0.9056508 0.8446267 0.7432552 0.8813917
## AB 0.9779421 1.0000000 0.9516701 0.8924632 0.7721339 0.9125839
## H 0.9056508 0.9516701 1.0000000 0.9308318 0.7155225 0.9068893
## X2B 0.8446267 0.8924632 0.9308318 1.0000000 0.5889699 0.8485911
## HR 0.7432552 0.7721339 0.7155225 0.5889699 1.0000000 0.8929048
## RBI 0.8813917 0.9125839 0.9068893 0.8485911 0.8929048 1.0000000
## AVG 0.4430808 0.5126292 0.7393167 0.6613085 0.3444242 0.5658479
## OBP 0.4841583 0.5026125 0.6560021 0.5466537 0.4603408 0.5704463
## SLG 0.6875270 0.7471949 0.8211406 0.7211259 0.8681501 0.8824090
## OPS 0.6504483 0.6980141 0.8069779 0.6966830 0.7638721 0.8156612
## WAR 0.5645243 0.6211558 0.7688712 0.6757470 0.6897677 0.7885666
## Payroll.Salary2023 0.4614889 0.5018820 0.6249911 0.6450730 0.5317619 0.6281239
## AVG OBP SLG OPS WAR
## GP 0.4430808 0.4841583 0.6875270 0.6504483 0.5645243
## AB 0.5126292 0.5026125 0.7471949 0.6980141 0.6211558
## H 0.7393167 0.6560021 0.8211406 0.8069779 0.7688712
## X2B 0.6613085 0.5466537 0.7211259 0.6966830 0.6757470
## HR 0.3444242 0.4603408 0.8681501 0.7638721 0.6897677
## RBI 0.5658479 0.5704463 0.8824090 0.8156612 0.7885666
## AVG 1.0000000 0.8028894 0.7254274 0.7989005 0.7855945
## OBP 0.8028894 1.0000000 0.7617499 0.8987390 0.7766375
## SLG 0.7254274 0.7617499 1.0000000 0.9686752 0.8611140
## OPS 0.7989005 0.8987390 0.9686752 1.0000000 0.8799893
## WAR 0.7855945 0.7766375 0.8611140 0.8799893 1.0000000
## Payroll.Salary2023 0.5871543 0.7025979 0.6974086 0.7394981 0.8086359
## Payroll.Salary2023
## GP 0.4614889
## AB 0.5018820
## H 0.6249911
## X2B 0.6450730
## HR 0.5317619
## RBI 0.6281239
## AVG 0.5871543
## OBP 0.7025979
## SLG 0.6974086
## OPS 0.7394981
## WAR 0.8086359
## Payroll.Salary2023 1.0000000Now lets make a linear regression model using the predictors RBI, OBP, and OPS
#Removing AVG
model5 = lm(Payroll.Salary2023 ~ RBI + OBP+OPS, data=firstbase)
summary(model5)##
## Call:
## lm(formula = Payroll.Salary2023 ~ RBI + OBP + OPS, data = firstbase)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9465449 -3411234 259746 4102864 8876798
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -29737007 10855411 -2.739 0.013 *
## RBI 72393 84646 0.855 0.403
## OBP 82751360 83534224 0.991 0.334
## OPS 7598051 45525575 0.167 0.869
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5767000 on 19 degrees of freedom
## Multiple R-squared: 0.5709, Adjusted R-squared: 0.5031
## F-statistic: 8.426 on 3 and 19 DF, p-value: 0.000913Now lets make a linear regression model using the predictors RBI, and OBP
model6 = lm(Payroll.Salary2023 ~ RBI + OBP, data=firstbase)
summary(model6)##
## Call:
## lm(formula = Payroll.Salary2023 ~ RBI + OBP, data = firstbase)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9045497 -3487008 139497 4084739 9190185
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -28984802 9632560 -3.009 0.00693 **
## RBI 84278 44634 1.888 0.07360 .
## OBP 95468873 33385182 2.860 0.00969 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5625000 on 20 degrees of freedom
## Multiple R-squared: 0.5703, Adjusted R-squared: 0.5273
## F-statistic: 13.27 on 2 and 20 DF, p-value: 0.0002149Lets Import the test dataset of the first base stats.
# Read in test set
firstbaseTest = read.csv("firstbasestats_test.csv")
str(firstbaseTest)## 'data.frame': 2 obs. of 15 variables:
## $ Player : chr "Matt Olson" "Josh Bell"
## $ Pos : chr "1B" "1B"
## $ Team : chr "ATL" "SD"
## $ GP : int 162 156
## $ AB : int 616 552
## $ H : int 148 147
## $ X2B : int 44 29
## $ HR : int 34 17
## $ RBI : int 103 71
## $ AVG : num 0.24 0.266
## $ OBP : num 0.325 0.362
## $ SLG : num 0.477 0.422
## $ OPS : num 0.802 0.784
## $ WAR : num 3.29 3.5
## $ Payroll.Salary2023: num 21000000 16500000We can see it only has two records One for Matt Olson and the other for Josh Bell
Lets make a prediction for salary using our last model which was uses the predictors RBI, and OBP
# Make test set predictions
predictTest = predict(model6, newdata=firstbaseTest)
predictTest## 1 2
## 10723186 11558647We can see that the first predictin is off by around 50% and teh second is off by ~50,000
# Compute R-squared
SSE = sum((firstbaseTest$Payroll.Salary2023 - predictTest)^2)
SST = sum((firstbaseTest$Payroll.Salary2023 - mean(firstbase$Payroll.Salary2023))^2)
1 - SSE/SST## [1] 0.5477734We can see that te r squared is the highest thus far so this is the best model thus far. We will also see that the SSE for this model was the lowest so far so I would say model 6 has been our best performer yet with only two predictor variables RBI and OBP.
SSE## [1] 1.300299e+14