Group Assignment 5
Vanessa Wasveiler, Keydy Sanchez
#install.packages("tidyverse")
#install.packages("forecast")
#install.packages("ggcorrplot")
library(tidyverse)## ── Attaching packages ──────────────────────────────────── tidyverse 1.3.0 ──## ✓ ggplot2 3.3.2 ✓ purrr 0.3.3
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## ✓ tidyr 1.0.2 ✓ stringr 1.4.0
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## x dplyr::lag() masks stats::lag()library(forecast)## Registered S3 method overwritten by 'quantmod':
## method from
## as.zoo.data.frame zoolibrary(ggcorrplot)
library(plotly)##
## Attaching package: 'plotly'## The following object is masked from 'package:ggplot2':
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## last_plot## The following object is masked from 'package:stats':
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## filter## The following object is masked from 'package:graphics':
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## layoutlibrary(e1071)
library(caret)## Loading required package: lattice##
## Attaching package: 'caret'## The following object is masked from 'package:purrr':
##
## liftvehicles <- read_csv("vehicles.csv")## Parsed with column specification:
## cols(
## MPG = col_double(),
## Cylinders = col_double(),
## Engine_Size = col_double(),
## Horse_Power = col_double(),
## Vweight = col_double(),
## Acceleration = col_double(),
## Origin = col_double()
## )#Data from data mining for the masses
View(vehicles)#Question #1 - Partition your data #60% training, 40% validation split.
#1 of 237 rows, and 1 of 158 rows.
set.seed(1)
#row.names(vehicles)
#nrow(vehicles)
training_set_rows_vehicles <- sample(rownames(vehicles), nrow(vehicles)*0.6)
validation_set_rows <- sample(setdiff(rownames(vehicles), training_set_rows_vehicles), nrow(vehicles)*0.4)
#View(training_set_rows)
#View(validation_set_rows)
training_data <- vehicles[training_set_rows_vehicles, ]
validation_data <- vehicles[validation_set_rows, ]
View(training_data)
View(validation_data)#MODELING #Question #2 - Create the model and identify where it can be improved. Write a brief summary of 2-3 sentences on what you see with the model summary output.
training_lm_model <- lm(MPG ~ ., data = vehicles, subset = training_set_rows_vehicles)
#iew(training_lm_model)
training_residuals <- data.frame(training_data$MPG, training_lm_model$fitted.values, training_lm_model$residuals)
View(training_residuals)
summary(training_lm_model)##
## Call:
## lm(formula = MPG ~ ., data = vehicles, subset = training_set_rows_vehicles)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.6515 -3.1288 -0.2349 2.3266 13.6203
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 44.046228 3.639154 12.103 < 2e-16 ***
## Cylinders -0.989061 0.540854 -1.829 0.06874 .
## Engine_Size 0.026204 0.013913 1.883 0.06090 .
## Horse_Power -0.072093 0.024087 -2.993 0.00306 **
## Vweight -0.005034 0.001287 -3.912 0.00012 ***
## Acceleration -0.061426 0.160098 -0.384 0.70157
## Origin 2.001437 0.465542 4.299 2.53e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.278 on 230 degrees of freedom
## Multiple R-squared: 0.7187, Adjusted R-squared: 0.7114
## F-statistic: 97.95 on 6 and 230 DF, p-value: < 2.2e-16print(training_residuals)## training_data.MPG training_lm_model.fitted.values
## 324 43.4 29.779739
## 167 20.0 23.330437
## 129 32.0 32.735700
## 299 34.2 27.909927
## 270 23.8 24.466147
## 187 15.5 16.650151
## 307 41.5 29.485157
## 85 13.0 15.977608
## 277 29.5 31.991941
## 362 22.4 20.073369
## 330 29.8 31.727043
## 263 17.5 14.993375
## 329 33.8 31.901490
## 79 28.0 29.440667
## 213 13.0 15.890179
## 37 14.0 12.820810
## 105 18.0 24.021158
## 217 25.5 25.836256
## 366 28.0 24.363632
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## 290 18.5 16.120834
## 387 26.0 25.642214
## 89 13.0 14.592433
## 289 19.2 17.049818
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## 326 30.0 25.387682
## 386 38.0 24.628074
## 42 13.0 9.386870
## 111 21.0 23.156082
## 20 25.0 26.937677
## 44 22.0 27.279475
## 343 35.1 34.041812
## 70 15.0 14.925750
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## 172 18.0 21.687982
## 25 11.0 8.438742
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## 331 32.7 23.651937
## 13 22.0 23.238450
## 296 23.0 17.592225
## 379 36.0 31.129706
## 176 23.0 24.962319
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## 351 33.0 29.611908
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## 287 16.9 13.267182
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## 327 44.6 33.487390
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## training_lm_model.residuals
## 324 13.62026091
## 167 -3.33043657
## 129 -0.73569991
## 299 6.29007307
## 270 -0.66614733
## 187 -1.15015112
## 307 12.01484283
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## 362 2.32663087
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## 79 -1.44066656
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## 165 -2.31670154
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## 387 0.35778629
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## 289 2.15018229
## 340 3.22617748
## 326 4.61231765
## 386 13.37192581
## 42 3.61312973
## 111 -2.15608181
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## 44 -5.27947520
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## 378 8.14777497# The model below creates multiple linear regression model under the name of Training residual which is the predicted value minus the actual value in order for us to perform different models with our result, we can also see how it generates a standard error for our predictions. #Question 3 - Predict using the validation data and your model for comparison purposes and compare the results from that to the actual values to see what the residuals are.
prediction_lm_model_vehicles <- predict(training_lm_model, newdata = validation_data)
validation_residuals_vehicles <- data.frame(validation_data$'MPG' - prediction_lm_model_vehicles, residuals = validation_data$'MPG' - prediction_lm_model_vehicles)
ggplot(data = training_lm_model) +
aes(x = training_lm_model$residuals) +
geom_histogram(bins = 20)
ggplot(data = validation_residuals_vehicles) +
aes(x = residuals) +
geom_histogram(bins = 20)
#Question #4 - Change the model formula to not include Acceleration to see what the impact is, if any. Describe what has changed in the summary results between your previous model and this new one (1-2 sentences).
training_lm_model2 <- lm(MPG ~ Cylinders + Engine_Size + Horse_Power + Vweight + Origin, data = vehicles, subset = training_set_rows_vehicles)
summary(training_lm_model2)##
## Call:
## lm(formula = MPG ~ Cylinders + Engine_Size + Horse_Power + Vweight +
## Origin, data = vehicles, subset = training_set_rows_vehicles)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.8514 -3.0766 -0.1866 2.3508 13.3867
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 42.999302 2.403441 17.891 < 2e-16 ***
## Cylinders -0.987226 0.539833 -1.829 0.068725 .
## Engine_Size 0.026878 0.013776 1.951 0.052256 .
## Horse_Power -0.066110 0.018323 -3.608 0.000378 ***
## Vweight -0.005260 0.001142 -4.605 6.81e-06 ***
## Origin 1.999769 0.464661 4.304 2.48e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.27 on 231 degrees of freedom
## Multiple R-squared: 0.7185, Adjusted R-squared: 0.7124
## F-statistic: 117.9 on 5 and 231 DF, p-value: < 2.2e-16#We can see how the following increase, Residual standard, degrees of freedom, Multiple R-squared, Adjusted R-squared, F-statisticand the p-value as we change the model formula.#Question #5 - Using one of the models that you created, and the corresponding multiple linear regression formula, predict the MPG of a vehicle with the following characteristics: #4 cylinders #120 engine size #130 horse power #2200 vweight #3 origin
MPG1 <- (-0.989061*4) + (0.026204*120) + (-0.072093*130) + (-0.005034*2200) + (2.001437*3) + 44.045228
print(MPG1)## [1] 28.79089Part 2 - Classification Model
Question #1 - Using the pima native americans dataset, make a naive bayes model but with the NA values present instead of median imputation. Make a confusion matrix for the validation set. Answer the following - which provides better positive predictive value of your two models (the median or the NA)? No explanation needed - just answer that question. You should include the code that you made that led to the answer from above.
pima <- read_csv("pima_diabetes.csv")## Parsed with column specification:
## cols(
## Pregnancies = col_double(),
## Glucose = col_double(),
## BloodPressure = col_double(),
## SkinThickness = col_double(),
## Insulin = col_double(),
## BMI = col_double(),
## DiabetesPedigreeFunction = col_double(),
## Age = col_double(),
## Outcome = col_double()
## )view(pima)pima_cleaned_na <- pima %>%
mutate(Glucose = ifelse(Glucose == 0, NA, Glucose),
BloodPressure = ifelse(BloodPressure == 0, NA, BloodPressure),
SkinThickness = ifelse(SkinThickness == 0, NA, SkinThickness),
Insulin = ifelse(Insulin == 0, NA, Insulin),
BMI = ifelse(BMI == 0, NA, BMI))
pima_cleaned_na$Outcome <- factor(pima_cleaned_na$Outcome, levels = c(1, 0), labels = c("Positive", "Negative"))
set.seed(1)
training_partition_na <- createDataPartition(y = pima_cleaned_na$Outcome, p = 0.6, list = FALSE)
training_pima_new_na <- pima_cleaned_na[training_partition_na, ] ## Warning: The `i` argument of ``[`()` can't be a matrix as of tibble 3.0.0.
## Convert to a vector.
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_warnings()` to see where this warning was generated.validation_pima_new_na <- pima_cleaned_na[-training_partition_na, ]
prop.table(table(pima_cleaned_na$Outcome)) * 100 ##
## Positive Negative
## 34.89583 65.10417prop.table(table(training_pima_new_na$Outcome)) *100##
## Positive Negative
## 34.92408 65.07592prop.table(table(validation_pima_new_na$Outcome)) *100##
## Positive Negative
## 34.85342 65.14658predictors_na <- training_pima_new_na[, 1:8]
response_na <- training_pima_new_na$OutcomeThe Median provides a better positive predictive value.