ANOVA & Least Squares

[Name of the Author]

Table of Contents

Case Assignment1

Case 11

Case 24

Case 37

Case 49

ANOVA Single Factor9

References11

Session Long Project (SLP)12

References13

Case Assignment

Case 1

SUMMARY OUTPUT

Regression Statistics

Multiple R

0.966977263

R Square

0.935045027

Adjusted R Square

0.926925656

Standard Error

1.900607384

Observations

10

 

Coefficients

Standard Error

t Stat

P-value

Intercept

-1.10153256

0.60561287

-1.818872455

0.106435653

X

1.996168582

0.186012514

10.73136714

4.99937E-06

Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0%

-2.49807834

0.295013213

-2.498078347

0.295013213

1.567222956

2.425114209

1.567222956

2.425114209

From the above output the equation for regression can be obtained. The linear regression model is given by

Y = a + ß * X

Where,

a = intercept

ß= slope

x = Explanatory Variable

y = Response variable

Here the intercept of the model a = -1.101, and the slope of the model ß=1.99, so the linear regression model to explain and predict the response variable (Y) through the relationship between X and Y variables is given by

Y= -1.101 + (1.99*X)

This equation explains that if the value of X increases by 1 then the value of Y will decrease by -1.101. If the value of X is 0 then the Y variable will be -1.101. These predictions of response variable (Y) lie in the range of ß (1.567 - 2.425) and a (-2.498 - 0.295), with a 95% confidence interval. This equation is also the line of best fit for the variables.

The coefficient of determination provides information about the capacity of the explanatory variable (X) to explain the response variable (Y). In this linear regression the coefficient of determination =0.93. This means that approximately 93% of variability in the response variable (Y) is caused by the explanatory variable (X). The coefficient is not statistically significant so it is likely that this correlation between the variable has occurred by chance.

The value of Y, the dependant variable, when X, the independent variable, is equals to -2 is found to be

Y = -1.101 + ( 1.99*-2 )

Y = -5.093

This predicted value of Y (when X=-2, Y=-5.093) is -2.9 more than the actual value. So the residual or error in the prediction of Y at x = -2 has turned out to be -2.9. This is because the coefficients of the regression line have a range, with 95% confidence interval.

The value of Y, the dependant variable, when X, the independent variable, is equals to -2 is found to be

Y = -1.101 + (1.99*4)

Y = 6.883

This is the predicted value of Y when X=4.

One of the limitations of the linear regression model is that it does not predict the actual value of a response variable at a given value of explanatory variable. Also, linear regression only predicts numerical values, i.e. it cannot be used for a rating scale or qualitative data. For linear regression model, the variables should have a linear relationship. It can be determined by correlation analysis and scatter plots.

Case 2

SUMMARY OUTPUT

Regression Statistics

Multiple R

0.850929358

R Square

0.724080772

Adjusted R Square

0.689590868

Standard Error

6.017840039

Observations

10

 

Coefficients

Standard Error

t Stat

P-value

Intercept

66.39873418

4.882348385

13.5997534

8.21754E-07

Hours (X)

4.905063291

1.070525701

4.58192016

0.00179748

Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0%

55.14001862

77.65744973

55.1400182

77.6574497

2.4364266

7.373699982

2.4364266

7.37369998

From the above output the equation for regression can be obtained. The linear regression model is given by

Y = a + ß * X

Where,

a = intercept

ß= slope

Hours (X) = Explanatory Variable

Exam Scores (Y) = Response variable

Here the intercept of the model a = 66.3987, and the slope of the model ß=4.905, so the linear regression model to explain and predict the response ...