Regression
- Thread starter darshang3
- Start date
jvillatuya
New Member
hi,
i think this can be solved by recalling what correlation is, it is actually the square root of the "R squared"..and R squared is calculated as "Explained sum of squares divided by total sum of squares". In the problem:
Total sum of squares would have to be SD(Y)^2 or 0.0676
Unexplained sum of squares (error term or noise term) of the model is the SD(E)^2 or 0.01
Given what we have, we can calculate the percentage of the variance in the regression that cannot be explained our independent variable as follows:
SD(E)^2 divided by SD(Y)^2 equals 0.147928 (or roughly 14.79%)
Hence, the explained percentage of the regression variance would have to be 100% less 14.79% is 85.21%.
So the R squared is 85.21% or 0.8521..and the correlation estimate is the square root of 0.8521 which is 0.9231.
i think this can be solved by recalling what correlation is, it is actually the square root of the "R squared"..and R squared is calculated as "Explained sum of squares divided by total sum of squares". In the problem:
Total sum of squares would have to be SD(Y)^2 or 0.0676
Unexplained sum of squares (error term or noise term) of the model is the SD(E)^2 or 0.01
Given what we have, we can calculate the percentage of the variance in the regression that cannot be explained our independent variable as follows:
SD(E)^2 divided by SD(Y)^2 equals 0.147928 (or roughly 14.79%)
Hence, the explained percentage of the regression variance would have to be 100% less 14.79% is 85.21%.
So the R squared is 85.21% or 0.8521..and the correlation estimate is the square root of 0.8521 which is 0.9231.
Bom,
your effort is valiant but: 1. unexplained sum of squares (RSS) = SER^2*d.f., so you'd need still need the d.f.?
and 2. SD(Y)^2 doesn't even approximate TSS.
...it seems to me that, maybe if we were given (n) we could maybe use the d.f. to infer RSS/TSS...
...but that also seems ridiculously hard
is there any chance that SD(e)=0.1 is really supposed to be "SD(X) = 0.1"
in that case, we could use beta (b) = correlation*cross volatility = correlation * SD(X)/SD(Y)
David
your effort is valiant but: 1. unexplained sum of squares (RSS) = SER^2*d.f., so you'd need still need the d.f.?
and 2. SD(Y)^2 doesn't even approximate TSS.
...it seems to me that, maybe if we were given (n) we could maybe use the d.f. to infer RSS/TSS...
...but that also seems ridiculously hard
is there any chance that SD(e)=0.1 is really supposed to be "SD(X) = 0.1"
in that case, we could use beta (b) = correlation*cross volatility = correlation * SD(X)/SD(Y)
David
Similar threads
