1) A) Interpret the coefficients of the independent variables in the following l

Question

1)
A) Interpret the coefficients of the independent variables in the following linear regression      
log(price) = 2.1 + 0.1*distance + 1.5*value
  
B) Interpret the coefficients of the independent variables in the following linear regression
price = 2.1 + 0.1*distance + 1.55*value – 0.8*rowhouse

C) Interpret the coefficients of the independent variables in the following linear regression log(price) = 2.1 + 0.1*log(distance) + 1.55*value

price=house price($10,000)

distance=miles from downtown

value=home value ($10,000)

Rowhouse=Is the home a rowhouse? (yes=1, no=0)

Explanation / Answer

I am unable to discriminate between the two variables "home value" and "house price" . For explaining purpose,i am assuming home value is the value of making an home and house price is the price demanded by customers for buying the house. Also,it is given that unit of home value is $10,000 and the unit of house price is also same.

Result:- If z=c+ax+by, 'a' represents change in the variable z for unit change in variable x,keeping y constant.The change is positive(or negative) if a has positive(or negative) sign. Similarly, 'b' represents change in the variable z for unit change in variable y,keeping x constant.The change is positive(or negative) if a has positive(or negative) sign. c implies the value of z when both x and y vanishes.

A) log(price) = 2.1 + 0.1*distance + 1.5*value . Comparing we get z=log(price),c=2.1,x=distance,y=value,a=0.1,b=1.5.

So, c=2.1 implies,when the distance is zero miles(the home is in the downtown only) and home value for that place is $0(it means that if there is no home built),then price of a new house at that place will be constant at exp(c)=exp(2.1) =$(8.166*10000)=$81660.

a=0.1 implies that if we assume home value of a house is known to us,then if the distance of the home from downtown is 1 mile more than its current distance ,its price will be exp(0.1)=1.1052 times the price at current location.(log(new price)-log(old price)=0.1*change in distance,so,new price/old price=exp(0.1)).Here,by current distance,we dont mean that the distance is known to use,we mean that if we think of a constant distance,then if the home was 1 km more than that distance,its price would be 1.1052 times of its price for being situated at that particular constant distance..

So, 0.1 is increase in logarithmic value of price per 1 mile increase in distance

b=1.5 implies,at a known constant distnce from downtown,if home value increases by 1 unit,the price for that house will be exp(1.5)=4.4187 times the price corresponding to the old home value.( log(new price)-log(old price)=1.5=>new price/old price=exp(1.5)=> new price=exp(1.5)*old price.)

Here,1.5 is increase in logarithmic value of house price per $10,000 increase in home value

B) price = 2.1 + 0.1*distance + 1.55*value – 0.8*rowhouse

We interpret them almost similar way.

Interpretation of 0.1:- Suppose the type of the house(rowhouse or not) and its home value is known to us.Then if the distance of the home from downtown is 2.718282(vale of e,we assume log is of base e) mile more than its current distance ,its price will be 0.1unit(0.1*$10000=$1000) more than the price at current location.(new price-old price=0.1*change in distance,so,new price=old price+0.1)

Interpretation of 1.55:- Suppose the type of the house(rowhouse or not) and its distance from downtown is known to us.If home value of the house increases by 1 unit($10,000),the price for that house will be 1.55unit(1.55*$10,000)=$15500) more than the price corresponding to the old home value.

Interpretation of 0.8 :- Suppose,we dont know about the type of the house. We only know about its home value and its distance from downtown. Then, we can interpret that if the house is a rowhouse,then its price will be exp(-0.8)=0.44933 times the price,had it not been a rowhouse.

C)I think it is the same question of A.

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