I'll denote corresponding new values of quantities with a prime ('). E.g. R
2' refers to the new R
2 value, whilst R
2 refers to the old R
2 value.
Proof why R2 is unchanged
(quick reason: R
2 is unitless so just depends on the actual data.)
proof of this:
,
where
is the sum of squared deviations of the actual data values and predicted data values (i.e. sum of squared residuals). w
i is a data value of weight (in lb, because I didn't put a prime) for a corresponding height measurement, and f
i is what the old model predicted would be the weight value for that height value. Also,
is the sum of squared distances of your weight data from the weight mean
. We show that
and
. In other words, we show that the NEW
and new
are both the same ratio compared to their old ones, so that when we plug them into the R
2 formula to get the new R
2, these common factors of b
2 will cancel and we will be left with our original R
2 value as our new one.
Now,
But this f
i is just our old model, that is, f
i = Mh + B, where M and B are given in your question.
So,
.
Now, for our new one,
. But
and
(i.e. our new model, found earlier in the Q).
Therefore,
, since we can factor out the b, and it becomes a b
2 since it was inside a square
, because h'/a = h (i.e. new h values are a times the old ones, by our definition of a)
, as required.
Now,
.
And
(because the new average is b times the old one, since multiplying a set of scores by a constant also does this to their average)
, as required.
.
So R
2 doesn't change.