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I have a question regarding the output from lm in this table below:

Variable              Est (SE)        t Value     Pr > |t|
Intercept             5.97(0.12)      49.83        <.0001 
β drug dose1         -0.93(0.14)      -6.71        <.0001
β drug dose2         -1.50(0.15)      -9.83        <.0001

If I want to adjust the Estimate and SE relative to the intercept, I know I can take 5.97 and add -0.93 to get 5.04 for the value of β drug dose1 Also, I can take 5.97 and add -1.50 to get 4.47 for the value of β drug dose2 So, then the table would look like this:

Variable              Est (SE)        t Value     Pr > |t|
Intercept             5.97(0.12)      49.83        <.0001 
β drug dose1          5.04(0.14)      -6.71        <.0001
β drug dose2          4.47(0.15)      -9.83        <.0001

Do I need to perform the same operation on the SE or are they correct the way they are?

Andy
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Vicki
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1 Answers1

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Regression estimators of the intercept and the slope coefficients are random variables.
We know that for a pair of random variables $X$, $Y$ the following holds:

$$\mathbb E(X+Y)=\mathbb E(X) + \mathbb E(Y)$$

and the same is true for the sample counterparts $\bar{x}$, $\bar{y}$, $\overline{x+y}$. Thus to obtain a point estimate of the sum of two regression coefficients (e.g. intercept and slope) you simply sum the two point estimates (just as you say you are doing).

Also,

$$\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)+2 \cdot \operatorname{Cov}(X,Y)$$

$$\operatorname{sd}(X+Y)=\sqrt{\operatorname{Var}(X+Y)}=\sqrt{\operatorname{Var}(X)+\operatorname{Var}(Y)+2 \cdot \operatorname{Cov}(X,Y)}$$

and $\operatorname{se}$ is analogous to $\operatorname{sd}$, just scaled by $\sqrt n$: $\operatorname{sd}=\frac{\operatorname{sd}}{\sqrt n}$. Hence,

$$\operatorname{se}(X+Y)=\frac{\sqrt{\operatorname{Var}(X)+\operatorname{Var}(Y)+2 \cdot \operatorname{Cov}(X,Y)}}{\sqrt n}$$

and for a particular sample you have

$$\widehat{\operatorname{se}}(X+Y)=\frac{\sqrt{\widehat{\operatorname{Var}}(X)+\widehat{\operatorname{Var}}(Y)+2 \cdot \widehat{\operatorname{Cov}}(X,Y)}}{\sqrt n}$$

Thus to obtain the standard error of a sum of two regression estimates (e.g. intercept and slope) you need to calculate it using the formula above.

(I am not quite sure if in the last formula scaling is by $\sqrt n$ or by $\sqrt{n-p}$ for some $p$ indicating the number of parameters being estimated.)

Also check this post.

Community
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Richard Hardy
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  • Thank you Richard. If I am understanding you, I must use the formula (dividing by the sqrt of the number of observations, n). – Vicki Apr 13 '15 at 07:42
  • Thank you Richard. If I am understanding you, I must use the formula either dividing by the sqrt of the number of observations, n, or else, subtract the number of parameters, p, in my case 2, from n. – Vicki Apr 13 '15 at 07:49
  • Yes, I am not exactly sure about that. It shouldn't be too difficult to find out. – Richard Hardy Apr 13 '15 at 15:16