Q:

Confidence Intervals in case of transformed Y

Suppose in Multiple Linear regression 
I have transformed Y and taken the square of Y, referring to it as Ysq
Now the SE for this lm is say 2 
So if for a given set of new Xs my estimate of Ysq is 25 then the CI will 
25-(1.96)*2 to 25+1.96*2 that is 21.08 to 28.92
Estimate of Ysq is 25 and CI 21.08 to 28.92
but all this is in the units of Ysq
and if we have convert it Ysq to its orgnal units then the estimate will be 5 (in our example we have only +ve values of Y from 1 to 10 (Y is continuous) 
and the range will be from sqrt(21.08) to sqrt (28.92)
so we have estimate of 5 and CI of 4.59 to 5.37 
but mathematically now 5 is not in the centre of 4.59 and 5.37 
The derived estimate of 5 and CI of 4.59 to 5.37 - Is this correct?

A:

You say that $X > 0$ and you believe that $$P(21.08 \le Y = X^2 \le 28.92) = 0.95.$$ Then you must believe that $$P(4.591296 \le X \le 5.377732) = 0.95.$$
You have performed a non-linear, monotonic transformation (as @David commented). so the inequality in the event remains valid. However, the midpoint of transformed
interval need not be the same as the transform of the midpoint.
Just consider: $1 \le 5 \le 9$ has evenly spaced numbers, but $1 \le 2.236 \le 3$ does not.
Also, even though it is often the case that the point estimate of
a parameter lies at the midpoint of a confidence interval for that
parameter, there are many instances in which that is not true.

Note: For a statistician, such transformations are familiar territory. This same procedure of taking the square root is used to get a confidence interval for the population standard deviation $\sigma.$ If you have $n = 20$ observations from a normal population with sample variance $S^2 = 25,$ then a 95% confidence interval for the population variance $\sigma^2$ is $(14.459, 52.533)$ and
a 95% confidence interval for the population standard deviation $\sigma$ is $(3.802, 7.303).$
This is based on the fact that $(n-1)S^2/\sigma^2 \sim \mathsf{Chisq}(n-1).$ In case you happen to be interested in the details of the computation of this CI for $\sigma,$ the R code is shown below:
19*25/qchisq(c(.975,.025),19)
[1] 14.45864 53.33174
sqrt(19*25/qchisq(c(.975,.025),19))
[1] 3.802452 7.302858

