Question

2.-Interpret the following regression model Call: lm(formula = Sale.Price ~ Lot.Size + Square.Feet + Num.Baths + API.2011 + dis_coast + I(dis_fwy * dis_down * dis_coast) + Pool, data = Train...

2.-Interpret the following regression model
Call:
lm(formula = Sale.Price ~ Lot.Size + Square.Feet + Num.Baths + 
    API.2011 + dis_coast + I(dis_fwy * dis_down * dis_coast) + 
    Pool, data = Training)

Residuals:
    Min      1Q  Median      3Q     Max 
-920838  -84637  -19943   68311  745239 

Coefficients:
                                    Estimate Std. Error t value Pr(>|t|)    
(Intercept)                       -7.375e+05  7.138e+04 -10.332  < 2e-16 ***
Lot.Size                          -5.217e-01  1.139e-01  -4.581 5.34e-06 ***
Square.Feet                        1.124e+02  1.086e+01  10.349  < 2e-16 ***
Num.Baths                          3.063e+04  9.635e+03   3.179  0.00153 ** 
API.2011                           1.246e+03  8.650e+01  14.405  < 2e-16 ***
dis_coast                         -5.164e+00  1.017e+00  -5.078 4.70e-07 ***
I(dis_fwy * dis_down * dis_coast)  1.098e-08  4.236e-09   2.591  0.00973 ** 
Pool                               1.048e+05  2.010e+04   5.211 2.37e-07 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 141400 on 834 degrees of freedom
Multiple R-squared:  0.5527,    Adjusted R-squared:  0.549 
F-statistic: 147.2 on 7 and 834 DF,  p-value: < 2.2e-16
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2.-Interpret the following regression model Call: lm(formula = Sale.Price ~ Lot.Size + Square.Feet + Num.Baths + API.2011 + dis_coast + I(dis_fwy * dis_down * dis_coast) + Pool, data = Train...
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