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Prediction using Supervised Machine Learning using Simple Linear Regression

In this task we have to find the students scores based on their study hours. This is a simple Regression problem type because it has only two variables. 

Sample Dataset- you can download a csv file of dataset

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import pandas as pd

data = pd.read_csv("E:\python practice data\StudentHoursScores.csv")

# In[3]:

print(data.head())

print(data.tail())

print(data.shape)

print(data.dtypes)

print(data.columns)

print(data.corr())

# In[4]:

data.describe()

# In[5]:

type(data)

# In[13]:

x = data.iloc[:,0:-1]

y = data.iloc[:,-1]

print(x.head())

print(y.head())

print(x.shape,y.shape)

# In[14]:

from sklearn.model_selection import train_test_split

x_train,x_test,y_train,y_test = train_test_split(x,y,test_size=.2,random_state=1)

print(x_train.shape)

print(x_test.shape)

print(y_train.shape)

print(y_test.shape)

# In[15]:

from sklearn.linear_model import LinearRegression

std_model  = LinearRegression()

std_model.fit(x_train,y_train)

# In[17]:

y_pred = std_model.predict(x_test)

print(y_pred)

print(y_test)

print(x_test)

# In[22]:

print(std_model.coef_)

print(std_model.intercept_)

from sklearn.metrics import r2_score, mean_squared_error

print("accuracy is: ", r2_score(y_test,y_pred))

print(mean_squared_error(y_test,y_pred))

# In[24]:

vif = 1/1-(r2_score(y_test,y_pred))

vif

coef:  [9.9073553]
intercept: 0.2569567372371395
accuracy is:  0.9311770139374417
mean squared error: 3.7164412473781487

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