log(x) - log(y) ~ %$delta$
So log(x) - log(y) + 1.96$\sigma$ = 1.1 implies that $\sigma$ = 1.78.
With standard deviation of log heights at 5(%), R2 = 1 - 1.78 /5 = .64
log(x) - log(y) ~ %$delta$
So log(x) - log(y) + 1.96$\sigma$ = 1.1 implies that $\sigma$ = 1.78.
With standard deviation of log heights at 5(%), R2 = 1 - 1.78 /5 = .64