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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.

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  • $\begingroup$ 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." $\endgroup$
    – whuber
    Feb 5, 2023 at 16:58

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I'd say you want to estimate E[f(x) - f(x+2)] ? Since it's a linear model you only need the slope to know how much will y change for 2 units of x. For the confidence interval, I am not sure, but there's where the SE may come into play

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  • $\begingroup$ 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? $\endgroup$
    – gbi1977
    Feb 17, 2019 at 12:31
  • $\begingroup$ From the ci on the slope you can then deduce the ci on y, right? $\endgroup$
    – gsanroma
    Feb 17, 2019 at 12:44
  • $\begingroup$ I'm not sure, you mean consturcting the C.I. for the slope in the standard way, and then double the two ends of the interval? its a different result from the one I mentioned in the previous comment, but here the estimator in the center of the C.I remains unbiased, the manipulation is performed only once the interval is constructed - so it makes more sense, but I want to be sure. $\endgroup$
    – gbi1977
    Feb 17, 2019 at 13:17

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