# Linear regression - confidence interval for expected difference in $Y$ with respect to unknown values of $X$

Suppose I am given all of the necessary parameters about some linear model, but not the data itself. Namely, I am given $$\hat{\beta}_1,\hat{\beta_0},\bar X, S_e, r^2$$, etc. Also, I know that $$X_1,\dots,X_n$$ are all within the range of $$[40,70]$$. I'm being asked to construct a confidence interval for the expected difference in $$Y$$ over two units of $$X$$. What I'm not sure about is the parameter to be estimated.

I am guessing that a C.I. for $$\mathbb{E}[Y|X=2]$$ is not a good idea, because 2 is not within the range and the intercept will twist the results. I thought about estimating C.I. for $$\mathbb{E}[Y|X=40+2]$$ or $$\mathbb{E}[Y|X=\bar{X}+2]$$, the last one seems more reasonable but I can't think of any justification for it, let alone know whether this is the right approach at all.

Would appreciate any help.

• Imagine $X$ were measured in, say, yards. Then a change in two units of $X$ would be a change of one fathom. Thus, if you were simply to change all units to fathoms, then your question would be the standard one of finding a CI for $\hat\beta_1.$ Now, in changing the units to fathoms, $\hat\beta_1$ is doubled (because it is units proportional to 1/yards) and so is its standard error (ditto). The math stays the same even if the units don't happen to be called "yards" and "fathoms."
– whuber
Feb 5, 2023 at 16:58

• That turns out to be $\mathbb{E}[2\cdot \hat{\beta}_1]$, so the appropriate C.I. and SE would be for the slope, not for some predicted value, right? the SE, I believe, would be just the same as for estimating $\hat{\beta}_1$, it's just the center of the interval that has doubled. Am I correct? Feb 17, 2019 at 12:31