Wk 5 - Apply: Regression Modeling [due Day 7] Wk 5 - Apply: Regression Modeling [due Day 7] Assignment Content Purpose This assignment provides an opportunity to develop, evaluate, and apply...

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Answered Same DaySep 28, 2021

Answer To: Wk 5 - Apply: Regression Modeling [due Day 7] Wk 5 - Apply: Regression Modeling [due Day 7] ...

Shubham answered on Sep 28 2021
143 Votes
Linear Regression (Bivariate and Multivariate)
This assignment aims to develop, evaluate, and apply bivariate and multivariate linear
regression models on the provided data.
The given database contains the information about the tax assessment value assigned to medical
office buildings in a city. The following
is a list of the variables in the database:
· FloorArea: square feet of floor space
· Offices: number of offices in the building
· Entrances: number of customer entrances
· Age: age of the building (years)
· AssessedValue: tax assessment value (thousands of dollars)
Using the data, we have to visualize and construct the models that predict the tax assessment
value assigned to medical office buildings with specific characteristics.
Tabulations and Findings
In the given data, we have four independent variables namely FloorArea, Offices,
Entrances and Age. AssessedValue is the dependent variable.
    Before applying any regression model, first we will visualize the relationship of
independent variables with the dependent variable. By doing so, we can get an idea, whether a
given characteristic is linearly related to the dependent variable or not.
Scatter plot with FloorArea as the independent variable and AssessedValue as the dependent variable:
From the above graph we can clearly observe a strong positive linear relationship
between FloorArea and AssessedValue. It means, as the FloorArea increases, AssessedValue will
also increase. The bivariate linear equation is:
AssessedValue = 0.3067*FloorArea + 162.66
The above equation shows that if FloorArea is increased by one unit (Sq. ft.) then the
AssessedValue will increase by 306.7$. Further, the R2 (0.9377) shows that FloorArea alone, is
able to explain around 94% of variability in the AssessedValue, which is very significant.
Regression analysis of FloorArea and AssessedValue:
    SUMMARY OUTPUT
    
    
    
    
    
    
    
    
    
    
    
    
    
    Regression Statistics
    
    
    
    
    
    Multiple R
    0.968358209
    
    
    
    
    
    R Square
    0.937717621
    
    
    
    
    
    Adjusted R Square
    0.935641541
    
    
    
    
    
    Standard Error
    115.5993039
    
    
    
    
    
    Observations
    32
    
    
    
    
    
    
    
    
    
    
    
    
    
ANOVA
    
    
    
    
    
    
     
    df
    SS
    MS
    F
    Significance F
    
    Regression
    1
    6035851.903
    6035851.903
    451.6771673
    1.22548E-19
    
    Residual
    30
    400895.9721
    13363.19907
    
    
    
    Total
    31
    6436747.875
     
     
     
    
    
    
    
    
    
    
    
     
    Coefficients
    Standard Error
    t Stat
    P-value
    Lower 95%
    Upper 95%
    Intercept
    162.6627673
    54.47856531
    2.985812243
    0.005585612
    51.40269388
    273.9228407
    FloorArea (Sq.Ft.)
    0.306732084
    0.014432619
    21.25269788
    1.22548E-19
    0.277256744
    0.336207424
Hypothesis: H0 (Null Hypothesis): FloorArea is not significant.
H1 (Alternate Hypothesis): FloorArea is significant.
P-value for the FloorArea is 1.22548E-19, which is clearly less than 0.05 (level of
significance) therefore, we can reject the null hypothesis with 95% confidence. Hence, we can
conclude that FloorArea is a significant predictor of AssessedValue. It...
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