I'm working on a model for something and at the moment I prefer working solely in Excel. I've been double checking the results of the linear model in JMP, Minitab, and Statistica, and (more or less) been getting the same answers.

One thing that's coming out odd though is my standardized residuals, I'm getting much different answers than Excel's regression routine, and I know it has to do with how I am calculating them:

The standard deviation of our population varies relative to the output, so we work in terms of the relative standard deviation. We have an assumed %RSD of 5% (based on a lot of previous work, we also have reason to assume normality). From this I standardize the residuals by saying $\frac{(x-u)}{u\cdot RSD}$ where x = the observed value and u = the predicted value, so x-u = the residual.

Note that $u\cdot RSD = s$. Simple z-score. Problem is that the values Excel is giving me for the standardized residuals are much different than mine. This isn't exactly surprising since I am using a varying standard deviation. But their values don't seem to be tied to the reality of the data. One observation could be off by as much as 50% (around 6 standard deviations away) and the standardized residuals I'm given are only like 2 or 3.

Anyways, I'm having a really hard time finding out exactly how the residuals are standardized in a linear regression. Any help would be appreciated

  • $\begingroup$ Which version of Excel and what implementation of regression (LINEST? The analysis toolpack? Something else)? Are these standardized or studentized residuals? Could you explain what it means to "work in terms of the relative standard deviation"? How exactly is that modifying the usual least squares regression model? $\endgroup$
    – whuber
    Commented Aug 10, 2015 at 18:50
  • $\begingroup$ using the analysis toolpack. Standardized residuals (z not t). My understanding was that many systems assumed a static standard deviation that was independent of the scale. So say you have a stdev = 500, it would be 500 if your u was 2000 or 20,000. Is that how it works in linear regression? Honestly I don't know. I don't have the best background in stats: a decent knowledge of it but just enough to get me in trouble. ed: is this related to homoscedasticity and heteroscedasticity? $\endgroup$ Commented Aug 10, 2015 at 18:55
  • $\begingroup$ It's ok not to have a stats background, but you still ought to read the replies at stats.stackexchange.com/questions/3392 concerning using Excel. Your question uncovers a particularly egregious error in the Analysis ToolPak (which has wholly inadequate documentation). $\endgroup$
    – whuber
    Commented Aug 10, 2015 at 19:16
  • 1
    $\begingroup$ I didn't know that Analysis ToolPak actually had documentation. $\endgroup$
    – Aksakal
    Commented Aug 10, 2015 at 19:29
  • $\begingroup$ @Aksakal To paraphrase an early '70's meme, having no documentation means never having to admit you're wrong ;-). $\endgroup$
    – whuber
    Commented Aug 10, 2015 at 19:52

2 Answers 2


The statistical tools in Excel have always been black boxes. There's nothing for it but to do some forensic reverse-engineering. By performing a simple regression in Excel 2013, involving the data $x=(1,2,3,4,5,6,7,8,9)$ and $y=(2,1,4,3,6,5,9,8,7)$, and requesting "standardized residuals" in the dialog, I obtained output that states

  • The "Standard Error" is $1.3723\ldots$.

  • There are $9$ observations.

  • The residuals $r_i$ are $(0.5333\ldots, -1.35, \ldots, 0.35, -1.533\ldots)$.

  • The corresponding "Standard Residuals" are $(0.4154\ldots, -1.0516\ldots, \ldots, 0.2726\ldots, -1.1944\ldots)$.

Since "standardized" values are typically numbers divided by some estimate of their standard error, I compared these "Standard Residuals" to the residuals and to the "Standard Error." Knowing that various formulas for variances are sums of squares of residuals $r_i$ divided variously by $n$ (the number of data) or $n-p$ (the number of data reduced by the number of variables, in this case two: one for the intercept and a second for the slope), I squared everything in sight. It became immediately obvious that Excel is computing the "Standard Residual" as

$$\frac{r_i}{\sqrt{\frac{1}{n-1}\sum_{i=1}^n r_i^2}}.$$

This formula reproduced Excel's output exactly--not even a trace of floating point roundoff error.

The denominator is what would be computed by Excel's STDEV function. For residuals from a mean, it is an unbiased estimate of their variance. For residuals in a regression, however, it has no standard meaning or value. It's garbage! But now you know how to compute it... .

  • $\begingroup$ I double checked this and you're absolutely right. But is it completely without value? Lets assume a system is homoscedastic. So I position myself so that I am looking down the line of fit, if that makes sense, so that I've reduced a dimension from the system and now I have a mean, the line of fit, and the 1-dimensional scatter. In that sense how is it different from a normal standard deviation? This assumes homoscedacity, if the standard deviation is dependent on the scale of the result then you couldn't do this. I'm not a statistician by any means so I could be way off. $\endgroup$ Commented Aug 11, 2015 at 13:52
  • 1
    $\begingroup$ It is not "completely without value," but it is misleading. Residuals do not behave like "normal" data because they are mutually (negatively) correlated. When $n-p$ is large, this correlation is of little import, so the Excel calculation yields almost the same results as a correct calculation would. But worse, the residuals are not homoscedastic even when the error model is! The standard error of a residual depends on the values of its independent variables. The further those values are from their means, the greater is the SE. Excel has no excuse for not doing valid calculations. $\endgroup$
    – whuber
    Commented Aug 11, 2015 at 14:01

In R :

    modeGlob <- lm(rnorm(100)~ abs(rnorm(100))) #Your model.

    hii <- hatvalues(modeGlob) # hat matrix.

    rst <- modeGlob$residuals / (summary(modeGlob)$sigma * sqrt(1-hii)) # manually calculate standardized residuals.

    identical(rstandard(modeGlob) , rst) # check, this must be TRUE.

    plot(rstandard(modeGlob) , rst) # check it graphically.
  • 4
    $\begingroup$ Welcome to the site. Be aware that this is not an R Q&A site, but a statistics one. Not everyone will use, or even be able to read R. In light of that, can you add some text to explain this? $\endgroup$ Commented Feb 22, 2018 at 13:49

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