LinearRegressionAssignment
Pawan Srivastav
30 June 2018
Simple Linear Regression Assignment
1 - Calories_consumed-> predict weight gained using calories consumed
Reding the data from data file and saving into a variable
Calories_consumed <- read.csv("C:/Users/Pawan Srivastav/Desktop/Data Science/Data Sets/Data Sets/Simple Linear Regression/calories_consumed.csv") Getting Summary of Import Data
summary(Calories_consumed)## Weight.gained..grams. Calories.Consumed
## Min. : 62.0 Min. :1400
## 1st Qu.: 114.5 1st Qu.:1728
## Median : 200.0 Median :2250
## Mean : 357.7 Mean :2341
## 3rd Qu.: 537.5 3rd Qu.:2775
## Max. :1100.0 Max. :3900# Variance and Standard deviation of Calories.Consumed column
var(Calories_consumed$Calories.Consumed)## [1] 565668.7sd(Calories_consumed$Calories.Consumed)## [1] 752.1095# Variance and Standard deviation of Weight.gained..grams. column
var(Calories_consumed$Weight.gained..grams.)## [1] 111350.7sd(Calories_consumed$Weight.gained..grams.)## [1] 333.6925Creating Linear Model for weight gain
WeightGainModel <- lm(Weight.gained..grams. ~ Calories.Consumed, data = Calories_consumed)
summary(WeightGainModel)##
## Call:
## lm(formula = Weight.gained..grams. ~ Calories.Consumed, data = Calories_consumed)
##
## Residuals:
## Min 1Q Median 3Q Max
## -158.67 -107.56 36.70 81.68 165.53
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -625.75236 100.82293 -6.206 4.54e-05 ***
## Calories.Consumed 0.42016 0.04115 10.211 2.86e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 111.6 on 12 degrees of freedom
## Multiple R-squared: 0.8968, Adjusted R-squared: 0.8882
## F-statistic: 104.3 on 1 and 12 DF, p-value: 2.856e-07plot(Calories_consumed)
Hence the P-value is less than 0.05. So X varibale is significance and also Multiple R-Square value is 0.8968. That’s mean this model will predict the output 89.68% time correct
2 - Delivery_time -> Predict delivery time using sorting time
Reding the data from data file and saving into a variable
delivery_time <- read.csv("C:/Users/Pawan Srivastav/Desktop/Data Science/Data Sets/Data Sets/Simple Linear Regression/delivery_time.csv") summary(delivery_time)## Delivery.Time Sorting.Time
## Min. : 8.00 Min. : 2.00
## 1st Qu.:13.50 1st Qu.: 4.00
## Median :17.83 Median : 6.00
## Mean :16.79 Mean : 6.19
## 3rd Qu.:19.75 3rd Qu.: 8.00
## Max. :29.00 Max. :10.00# Variance and Standard deviation of Delivery.Time column
var(delivery_time$Delivery.Time)## [1] 25.75462sd(delivery_time$Delivery.Time)## [1] 5.074901# Variance and Standard deviation of Sorting.Time column
var(delivery_time$Sorting.Time)## [1] 6.461905sd(delivery_time$Sorting.Time)## [1] 2.542028Creating Linear Model for delivery time
deliverTimeModel <- lm(Delivery.Time ~ Sorting.Time, data = delivery_time)
summary(deliverTimeModel)##
## Call:
## lm(formula = Delivery.Time ~ Sorting.Time, data = delivery_time)
##
## Residuals:
## Min 1Q Median 3Q Max
## -5.1729 -2.0298 -0.0298 0.8741 6.6722
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.5827 1.7217 3.823 0.00115 **
## Sorting.Time 1.6490 0.2582 6.387 3.98e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.935 on 19 degrees of freedom
## Multiple R-squared: 0.6823, Adjusted R-squared: 0.6655
## F-statistic: 40.8 on 1 and 19 DF, p-value: 3.983e-06plot(deliverTimeModel)
Hence the P-value is less than 0.05. So X varibale is significance and also Multiple R-Square value is 0.6823. That’s mean this model will predict the output 68.23% time correct
For Increasing R squared value
Using mvinfluence in Linear Model to find the point which are creating problems
library(mvinfluence)## Loading required package: car## Loading required package: carData## Loading required package: heplotsinfluenceIndexPlot(deliverTimeModel)
deliverTimeModel <- lm(Delivery.Time ~ Sorting.Time, data = delivery_time[c(-5,-9,-21),])
summary(deliverTimeModel)##
## Call:
## lm(formula = Delivery.Time ~ Sorting.Time, data = delivery_time[c(-5,
## -9, -21), ])
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.3407 -1.5027 0.2275 0.9328 3.6815
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.0240 1.1751 5.126 0.000102 ***
## Sorting.Time 1.6741 0.1872 8.941 1.27e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.839 on 16 degrees of freedom
## Multiple R-squared: 0.8332, Adjusted R-squared: 0.8228
## F-statistic: 79.94 on 1 and 16 DF, p-value: 1.273e-07plot(deliverTimeModel)
After removing 3 points Multiple R-Square value is increased to 0.8332. That’s mean this model will predict the output 83.32% time correct
3 - Emp_data -> Build a prediction model for Churn_out_rate
Reding the data from data file and saving into a variable
Emp_data <- read.csv("C:/Users/Pawan Srivastav/Desktop/Data Science/Data Sets/Data Sets/Simple Linear Regression/emp_data.csv") Getting Summary of Import Data
summary(Emp_data)## Salary_hike Churn_out_rate
## Min. :1580 Min. :60.00
## 1st Qu.:1618 1st Qu.:65.75
## Median :1675 Median :71.00
## Mean :1689 Mean :72.90
## 3rd Qu.:1724 3rd Qu.:78.75
## Max. :1870 Max. :92.00# Variance and Standard deviation of Salary_hike column
var(Emp_data$Salary_hike)## [1] 8481.822sd(Emp_data$Salary_hike)## [1] 92.09681# Variance and Standard deviation of Churn_out_rate column
var(Emp_data$Churn_out_rate)## [1] 105.2111sd(Emp_data$Churn_out_rate)## [1] 10.25725Creating Linear Model for Churn_out_rate
Churn_out_rate_Model <- lm(Churn_out_rate ~ Salary_hike, data = Emp_data)
summary(Churn_out_rate_Model)##
## Call:
## lm(formula = Churn_out_rate ~ Salary_hike, data = Emp_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.804 -3.059 -1.819 2.430 8.072
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 244.36491 27.35194 8.934 1.96e-05 ***
## Salary_hike -0.10154 0.01618 -6.277 0.000239 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.469 on 8 degrees of freedom
## Multiple R-squared: 0.8312, Adjusted R-squared: 0.8101
## F-statistic: 39.4 on 1 and 8 DF, p-value: 0.0002386plot(Churn_out_rate_Model)
Hence the P-value is less than 0.05. So X varibale is significance and also Multiple R-Square value is 0.8312 That’s mean this model will predict the output 83.12% time correct
4 - Salary_hike -> Build a prediction model for Salary_hike
Reding the data from data file and saving into a variable
Salary_hike <- read.csv("C:/Users/Pawan Srivastav/Desktop/Data Science/Data Sets/Data Sets/Simple Linear Regression/Salary_Data.csv") Getting Summary of Import Data
summary(Salary_hike)## YearsExperience Salary
## Min. : 1.100 Min. : 37731
## 1st Qu.: 3.200 1st Qu.: 56721
## Median : 4.700 Median : 65237
## Mean : 5.313 Mean : 76003
## 3rd Qu.: 7.700 3rd Qu.:100545
## Max. :10.500 Max. :122391# Variance and Standard deviation of Salary_hike column
var(Salary_hike$YearsExperience)## [1] 8.053609sd(Salary_hike$YearsExperience)## [1] 2.837888# Variance and Standard deviation of Churn_out_rate column
var(Salary_hike$Salary)## [1] 751550960sd(Salary_hike$Salary)## [1] 27414.43Creating Linear Model for Salary_hike
Salary_hike_Model <- lm(Salary ~ YearsExperience, data = Salary_hike)
summary(Salary_hike_Model)##
## Call:
## lm(formula = Salary ~ YearsExperience, data = Salary_hike)
##
## Residuals:
## Min 1Q Median 3Q Max
## -7958.0 -4088.5 -459.9 3372.6 11448.0
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 25792.2 2273.1 11.35 5.51e-12 ***
## YearsExperience 9450.0 378.8 24.95 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5788 on 28 degrees of freedom
## Multiple R-squared: 0.957, Adjusted R-squared: 0.9554
## F-statistic: 622.5 on 1 and 28 DF, p-value: < 2.2e-16plot(Salary_hike_Model)
Hence the P-value is less than 0.05. So X varibale is significance and also Multiple R-Square value is 0.957 That’s mean this model will predict the output 95.7% time correct