larger sample size or more accurate measurement? This question is completely theoretical in the sense that I don't have any real data.
Suppose I am designing a study, in which I will go measure N pairs of (X,Y), then the goal is to build a predictive model (say linear regression) between X and Y. i.e. I want good predictive accuracy, but it will also be good to have a prediction interval.
Given the limited resources, I have two choices:


*

*Increase N, but X will be measured less accurately; or 

*Decrease N but measure X more accurately.


The question is, how can I find the balance between the 2 cases?
I understand I haven't provided real data, which is why I am not asking for specific answer. I just want to know what's the line of thought you would follow? What's a good way to approach this question? etc.... 
Clarification:
Don't know why I put it the other way around. X in the dot points above should be Y. I am predicting Y using X. My choice is between increasing N or measuring Y more precisely.
Added:
It will be good if your response considers both small sample size and adequate sample size situations.
 A: I'll assume for the moment you're predicting X from Y and it's only the variable being predicted that's measured with error (see the warning in my comment if this is not the case). I'll interchange the roles of X and Y to be the usual way around. That is, in what follows $Y$ is the variable being predicted, and is measured with error, while $X$ is the predictor or predictors.
The calculation is reasonably straightforward.
The variance of a confidence interval is $\sigma^2\mathbf{x^*}(X^TX)^{-1}\mathbf{x^*}^T$ *  where $\mathbf{x^*}$ is the row vector of predictors. If the rows of $X$ are random draws from some population of predictors, than as $n$ grows,  $n(X^TX)^{-1}$ approaches a constant.
*(while that for a prediction interval is $\sigma^2(1+\mathbf{x^*}(X^TX)^{-1}\mathbf{x^*}^T)$)
That is, there's a $\sigma/\sqrt{n}$ type effect in the standard error of the mean forecast at $\mathbf{x^*}$ (just as with an average).
You need 4 times the data to make the standard error halve, but you only need to  halve the average error (make measurements with twice the precision) to have the same effect.
If you can make the measurements have 90% of the error, that would be equivalent to having ($1/0.9^2 \approx 1.235$) 23.5% more observations.
Because $\sigma$ should be proportional to the average error, while $1/\sqrt{n}$ decreases as the square of the sample size, the tradeoff is pretty easy to work with.
If you're talking prediction intervals rather than confidence intervals, larger $n$ is unlikely to be worthwhile; it soon gives almost no return.
