GRIP-APRIL21 @The Sparks Foundation

TASK 1 : PREDICTION USING SUPERVISIED ML

DATA SCIENCE AND BUSINESS ANALYTICS.
Task- OBJECTIVE: PREDICT THE PERCENTAGE OF AN STUDENT BASED ON THE NUMBER OF STUDY HOURS GIVEN THAT , TO SHOW WHAT WILL BE THE PREDICTED SCORE IF A STUDENT STUDIES FOR 9.25 HRS/DAY?

Step-1 Reading the Given Data Set.

# Reading The CSV Data

student_data <- read.csv("https://raw.githubusercontent.com/AdiPersonalWorks/Random/master/student_scores%20-%20student_scores.csv", header = TRUE)
head(student_data)

summary(student_data)

     Hours           Scores     
 Min.   :1.100   Min.   :17.00  
 1st Qu.:2.700   1st Qu.:30.00  
 Median :4.800   Median :47.00  
 Mean   :5.012   Mean   :51.48  
 3rd Qu.:7.400   3rd Qu.:75.00  
 Max.   :9.200   Max.   :95.00  

Step-2 Plotting the Given Data set.

# Plotting the given data

plot(x = student_data$Hours, y = student_data$Scores, xlab = "Hours", ylab = "Scores", main = "Score v/s Hours", col = "blue")

Step-3 Running Linear Regression on the data as there are only Two variables.

# Linear regression
student_data_regression <- lm(formula = Scores~Hours, data = student_data)
plot(x = student_data$Hours, y = student_data$Scores, xlab = "Hours", ylab = "Scores", main = "Score v/s Hours", col = "blue")
abline(student_data_regression, col= "Red")

summary(student_data_regression)

Call:
lm(formula = Scores ~ Hours, data = student_data)

Residuals:
    Min      1Q  Median      3Q     Max 
-10.578  -5.340   1.839   4.593   7.265 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   2.4837     2.5317   0.981    0.337    
Hours         9.7758     0.4529  21.583   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 5.603 on 23 degrees of freedom
Multiple R-squared:  0.9529,    Adjusted R-squared:  0.9509 
F-statistic: 465.8 on 1 and 23 DF,  p-value: < 2.2e-16

Step-4 Splitting Data into Test and Training Data

# Splitting Data Into Test and Training data
library(caTools) # package "caTools" is used for splitting the data

split = sample.split(Y = student_data$Scores, SplitRatio = 0.75)
training_set = subset(student_data, split == TRUE)
test_set = subset(student_data, split==FALSE)

training_set # checking the training data

test_set# checking the test data

Step-5 Training the dataset

# training 
result <- lm(formula = Scores~Hours, data = training_set)
summary(result)

Call:
lm(formula = Scores ~ Hours, data = training_set)

Residuals:
    Min      1Q  Median      3Q     Max 
-8.0607 -4.2990  0.3358  2.7462  9.4485 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   4.0205     2.7067   1.485    0.157    
Hours         9.2988     0.4954  18.769 2.54e-12 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 5.547 on 16 degrees of freedom
Multiple R-squared:  0.9566,    Adjusted R-squared:  0.9538 
F-statistic: 352.3 on 1 and 16 DF,  p-value: 2.543e-12

result$coefficients

(Intercept)       Hours 
   4.020513    9.298847 

Step-6 Using the Test Data to predict the outcome

pred <- predict(result, test_set)
head(pred) #printing the predicted result

       1        8       13       16       19       22 
27.26763 55.16417 45.86533 86.78025 60.74348 48.65498 

head(test_set) # printing the head of test set to compare with predicted values

Step-7 Calculating the Mean Absolute Error

#taking the head values of original test set
tset <- head(training_set)
tset

# taking the head values of predicted set
pset <- head(test_set)
pset

#Calculating Mean Absolute Error
library(ie2misc)

Registered S3 method overwritten by 'data.table':
  method           from
  print.data.table     

mae(pset$Hours, tset$Hours)

[1] 3.883333

Step-8 What will be predicted score if a student studies for 9.25 hrs/ day?

predicted_result <- predict(result, data.frame(Hours = 9.25))
predicted_result

       1 
90.03485 

So, the score of student will be 92.40105.