A quant jock from your firm used a linear demand specificati
Solution
summary output
R square = (multiple R square)^2
= 0.38*0.38= 0.1444
Adjusted R square = (1-R square)*(N-1) / (N-p-1)
Where N= Total number of observation (=150) and p= Total number of estimators (=2)
Thus Adjusted R square =(1-0.1444)8(150-1) / (150-2-1)
=0.8672
Analysis of variance table
Total degree of freedom= regression degree of freedom + residual degree of freedom
= 147+2
=149
Sum of square for rgression = total sum of square -residual sum of square
=73807.49-63408.62
=10398.87
Cofficient table
standard error for intercept = estimated value of intercept / t statistic value corrosponding to intercept
= 58.87 / 3.84
=15.33073
t statistic value for price estimate = coefficient value of price / standard error of price estimate
=-1.64/0.85
=-1.92941
cofficient value of income = t statistic value corrosponding to income *standard error of income
=4.64*0.24
=1.1136
a)
Regression equation for demand can be summarized as
Qxd= 58.87 - 1.64 * Px + 1.11 * M
Where
Px = variable represents price
M= variable represents income
b)
If the p value observed for the coefficient value is less than 0.05, variable corrosponding to the perticular coefficient will be statistically significant.
As p value observed for intercept coefficent and income coefficient is less than 0.05, thus it can be said that income and intercet value are statistically significannt.

