A researcher is interested in predicting the price of new homes based on a model
ID: 3431013 • Letter: A
Question
A researcher is interested in predicting the price of new homes based on a model with three variables; square footage, number of bedrooms and number of bathrooms. The ANOVA table below is the partial output for the Model Utility F Test from a multiple linear regression analysis.
Source
Degrees of freedom
Sum of Squares
Mean Squares
F
p-val
Regression
<0.0001
Residual
29900797
Total
119
56104751
Fill out the rest of the ANOVA table based on the provided information.
Use the table to calculate the coefficient of determination, R2. Interpret.
Use the ANOVA table to perform a Model Utility F test. Significance level = 0.01.
State the Null and Alternative Hypotheses.
State the F statistic along with the numerator and denominator degrees of freedom and p-value.
What can you conclude from the test?
Source
Degrees of freedom
Sum of Squares
Mean Squares
F
p-val
Regression
<0.0001
Residual
29900797
Total
119
56104751
Explanation / Answer
Fill out the rest of the ANOVA table based on the provided information.
-------------------------------------------------------------------------------------------------------------------------
Use the table to calculate the coefficient of determination, R2. Interpret.
R^2= 1-SSE/SST
=1-29900797/56104751
=0.4671
46.71% of variance can be explained by the regression line.
-------------------------------------------------------------------------------------------------------------------------
Use the ANOVA table to perform a Model Utility F test. Significance level = 0.01.
State the Null and Alternative Hypotheses.
Null hypothesis: all means are equal.
Alternative hypothesis: at least one mean is different.
-------------------------------------------------------------------------------------------------------------------------
State the F statistic along with the numerator and denominator degrees of freedom and p-value.
F statistic is 51.27
The numerator degrees of freedom =2
The denominator degrees of freedom =117
The p-value is closed to 0 (from F table)
-------------------------------------------------------------------------------------------------------------------------
What can you conclude from the test?
Since the p-value is less than 0.01, we reject the null hypothesis.
So we can conclude that at least one mean is different.
Degree of freedom Sum of Squares Mean Squares F p-value Regression 2 26203954 13101977 51.26724 <0.0001 Residual 117 29900797 255562.3675 Total 119 56104751Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.