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Many studies only report the relationship between two variables (e.g. linear or logistic equation), $n$, and $r^2$. I want to use these reported statistics to reproduce this relationship with its variation. Most statistical software will generate a parameter distribution from a mean and standard error. Assuming a normal distribution, can the standard error of the parameter estimates be calculated with just these three statistics? Essentially, can I get a standard error from $r^2$?

Or will I need to do some kind of bootstrapping procedure to generate a distribution that has the same $r^2$ and then calculate the standard error? if so are there better ones for linear vs. nonlinear equations?

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  • $\begingroup$ Sorry for the typo, it should be correlation coefficient, not correction coefficient. $\endgroup$ – janice Oct 23 '13 at 19:05
  • $\begingroup$ Welcome to our site, Janice! We encourage you to continue improving your question, which includes editing it for typographical errors. More information is available at our help center. $\endgroup$ – whuber Oct 23 '13 at 20:14
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If you look at the Wikipedia page for the Pearson product-moment correlation, you will find sections that describe how confidence intervals can be calculated. Typically, people will use Fisher's $z$-transformation (arctan) to turn the $r$ into a variable that is approximately normally distributed:
$$ z_r = \frac 1 2 \ln \frac{1 + r}{1 - r} $$ Having applied this transformation, the standard error will be approximately $^1/_{\sqrt{(N-3)}}$. With this you can form whatever length confidence interval you like. Once you've found the confidence limits you want, you can back-transform them to the original $r$ scale (i.e., $[-1, 1]$) like so:
$$ \text{CI limit}_r = \frac{\exp(2z) - 1}{\exp(2z) + 1} $$ In other words, you can form a confidence interval for $r$ without the original data, so long as you have the original $N$.

Notes: This approach is an approximation, there are exact formulae listed on the Wikipedia page, but they are harder to use. Although it doesn't say on the Wikipedia page, there are several conditions you want to meet in order for this approximation to be reasonable. The $N$ should be at least $30$ (IIRC), and the marginal distributions (i.e., the univariate distributions of the two variables being correlated) should be normal. For example, I'm not sure that this will be accurate if the correlation were composed of two vectors of $1$s and $0$s. However, higher $N$ should allow you to compensate for minor non-normality.

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  • $\begingroup$ Where is the mean or parameter estimate used? The example in Wikipedia only uses r and n. If I am trying to get a SE or CI of an intercept or slope of the line (using a linear function), wouldn't I need to use the parameter estimate at some point. Maybe my question wasn't clear, I need the SE of the estimate for slope and intercept using r and n, not the SE of r. $\endgroup$ – janice Oct 24 '13 at 20:54
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    $\begingroup$ $r$ is the slope of the regression line when both variables (ie, $x$ & $y$) have been standardized first. If all you have is $r$ & $N$, that is the best you can do. $\endgroup$ – gung - Reinstate Monica Oct 24 '13 at 21:30
  • $\begingroup$ Would it be possible to simulate the variance of the relationiship until I get the same r, then use that set of data to calculate SE? $\endgroup$ – janice Oct 25 '13 at 14:13
  • $\begingroup$ Basically, no. Without more information than just $r$ & $N$ you won't be able to do better. $\endgroup$ – gung - Reinstate Monica Oct 25 '13 at 15:37
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To add to gung's answer, one can also use the a lazy approach of directly calculating the standard error for the correlation. This will produce inaccurate results in some cases and may produce impossible out of range confidence intervals. But for most cases, it's fine. The equation is:

enter image description here

Example calculation of confidence interval

Assume that n=200, r=.3. I use the CIr function from psychometric to get the CIs based on Fisher's Z transformation. Then I calculate the standard error of the correlation based on the direct approach and find the same CI (95%):

> psychometric::CIr(.3, 200)
[1] 0.17 0.42
> sqrt((1-.3^2)/(200-2))
[1] 0.068
> .3 - 1.96 * 0.068
[1] 0.17
> .3 + 1.96 * 0.068
[1] 0.43

.17-.42 vs. .17-.43. Thus, we see that the approaches yield only a minor difference.

The quick method gets more inaccurate the closer the |correlation| gets to 1 and with small n's. To illustrate, assume now that n=20, r=.9. Then:

> psychometric::CIr(.9, 20)
[1] 0.76 0.96
> sqrt((1-.9^2)/(20-2))
[1] 0.1
> .9 + 1.96 * 0.1
[1] 1.1
> .9 - 1.96 * 0.1
[1] 0.7

So, here the results are markedly different: .76-.96 vs. .7-.1.1! The latter is impossible, so we could reduce to .7-1.0. The two plots below show the difference in the lower and upper limits, respectively:

enter image description here enter image description here

So, blue indicates that the quick method produced too low values, red that it produced too high values and green when it gave correct answers. My takeaway is that when n is below 100, the quick method gives pretty imprecise results, but for larger n's, it doesn't matter so much.

The equation is given in e.g.:

Cohen, J., & Cohen, J. (Eds.). (2003). Applied multiple regression/correlation analysis for the behavioral sciences (3rd ed). Mahwah, N.J: L. Erlbaum Associates.

See also this question on SO.

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  • $\begingroup$ Although many sources give that formula, it is not correct. First, the standard error of a statistic cannot be defined in terms of a statistic. It must be defined in terms of parameter. I believe this site mathworks.com/matlabcentral/newsreader/view_thread/107045.html? is correct. $\endgroup$ – David Lane Feb 20 '17 at 3:25
  • $\begingroup$ As you can see from my simulation, it is an approximation and gives correct results in many cases. $\endgroup$ – Deleet Feb 20 '17 at 16:45

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