Finding that the plots produced by the commands above are slightly different from each other, I tried this:


I got a slightly less than perfect line. I'd have tried regressing the result of qnorm(seq(80)/81) on the variable that was plotted on the $x$-axis and plotting residuals against the predictor, expecting to see some graceful curve, but for the fact that I don't know what to use as the predictor. Possibly such a residual plot would reveal more than just the graceful S-shaped thing I'd anticipate.

So my question is this: if the thing on the $x$-axis in the plot produced by qqnorm is not what I get from qnorm(seq(80)/81) (and what I did shows that indeed it is not), then what is it?

  • $\begingroup$ One other thing I could imagine it might be is the expected values of the standard normal order statistics. But I don't know an efficient way to compute those or any standard commands to get them, unless that's what I actually get from qqnorm. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 2 '15 at 21:05
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    $\begingroup$ This seems to be a question about R code, not the nature of qq-plots themselves. If this is actually a statistical question, please edit it to make that aspect more obvious. $\endgroup$ – gung - Reinstate Monica Aug 2 '15 at 21:11
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    $\begingroup$ See the help page for ppoints, as referenced in that for qqnorm. $\endgroup$ – Scortchi - Reinstate Monica Aug 2 '15 at 21:52
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    $\begingroup$ As Scortchi suggests, the precise details of how the quantiles are calculated are in ppoints. They're not quite expected quantiles, but the use of a=3/8 for n<10 is a good approximation suggested by Blom(1958); I am not sure why the function changes to a=1/2 above that but it only makes visually noticeable difference at the two extreme points and for n past about 30 or so not even then. $\endgroup$ – Glen_b -Reinstate Monica Aug 3 '15 at 0:35
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    $\begingroup$ See this question and also this plot of expected normal order statistics against Blom's approximation for n=10 with a line through $O$ with slope 1 in red. $\endgroup$ – Glen_b -Reinstate Monica Aug 3 '15 at 0:59

Based on comments above and pages to which they link, it appears that the command


gives us this:

\begin{align} & \left( \frac{1}{2\cdot 80},\ \frac{3}{2\cdot 80},\ \frac{5}{2\cdot 80},\ \frac{7}{2\cdot 80},\ \ldots,\ \frac{159}{2\cdot 80} \right) \\[10pt] = {} &\left( \ldots,\ \frac{2i-1}{2\cdot 80},\ \ldots : i=1,\ldots,80 \right) \tag 1 \end{align} and


gives us precisely what is plotted on the $x$-axis when one uses the command


where $x$ is a vector with $80$ components.

My first guess had been that instead of $(1)$ we would have \begin{align} & \left( \frac 1 {81},\ \frac 2 {81},\ \frac 3 {81},\ \ldots,\ \frac{80}{81} \right) \\[10pt] = {} & \left( \ldots,\ \frac i {80+1},\ \ldots\ : i=1,\ldots 80 \right). \end{align}

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