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I'm wondering: Is there a statistical test for determining "no correlation"? I would not want to use Fisher's transformation.

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  • $\begingroup$ In general, you can't "prove" a point null for a continuous parameter (since you can get arbitrarily close alternatives), nor can you flip null and alternative (you can't calculate the distribution under the original unspecified alternative -- now the null, and the limiting case is actually the alternative!) . This is the subject of many questions on site. $\endgroup$
    – Glen_b
    Jul 26, 2016 at 23:57
  • $\begingroup$ e.g. See stats.stackexchange.com/questions/6225/… $\endgroup$
    – Glen_b
    Jul 27, 2016 at 0:06
  • $\begingroup$ A possibility is to abandon the hopeless task of trying to prove a point null for a continuous parameter and look instead at equivalence testing (of which there are a number of posts on site). In essence this would require you to define an interval of small correlations you regard as "effectively equivalent to zero" $\endgroup$
    – Glen_b
    Jul 27, 2016 at 0:09
  • $\begingroup$ If you believe the marked duplicate doesn't answer your question, please more carefully differentiate it from that one $\endgroup$
    – Glen_b
    Jul 27, 2016 at 0:10
  • $\begingroup$ thanks glen, i wish the model powering question recommendations improve $\endgroup$
    – user46925
    Jul 27, 2016 at 0:33

2 Answers 2

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"No correlation" means r=0.00000000000000000... that is almost never the case and never of practical interest. You could define, how small an r is "as small as to be zero for practical purposes". Computing correlation indices in statistics software will give you a confidence interval for the correlation coefficient.

If you say, that abs(r)<.10 you can compute the confidence interval and see, if it includes only values that meet that requirement.

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  • $\begingroup$ yea, compute CI and if 0.0 in CI - conclude no correlation: makes sense $\endgroup$
    – user46925
    Jul 26, 2016 at 20:01
  • $\begingroup$ @sponge If 0 is in the CI and you conclude that the correlation is 0 on that basis, how is that substantively different from failing to reject in a hypothesis test? $\endgroup$
    – Glen_b
    Jul 27, 2016 at 0:07
  • $\begingroup$ CIs measure how confident that 0 would appear given an infinite number of reptitions of the same exp. We're not testing a hypothesis here. $\endgroup$
    – user46925
    Jul 27, 2016 at 0:30
  • $\begingroup$ @sponge 1. if you use the CI to conclude the correlation is 0 (in your own words, "no correlation") that's using the CI to do a hypothesis test. $\:$ 2. If you're not actually interested in testing, why does your title say "test for no correlation"? ... perhaps you should clarify your question. $\endgroup$
    – Glen_b
    Jul 27, 2016 at 0:45
  • $\begingroup$ My proposal is not to check, whether zero is within the CI, but whether only small correlations are within the CI. $\endgroup$
    – Bernhard
    Jul 27, 2016 at 11:30
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R values range from -1 to 1. A value of -1 indicates a completely negative correlation, and 1 indicates a completely positive correlation. The distribution for the correlation coefficient is not normally distributed, and its variance is not constant. Fisher's transformation attempts to solve this problem, which allows us to compute the confidence interval. Determining correlation and no correlation would depend on your confidence interval. This interval defines a range that estimates that values of r that are likely to contain unknown. This is useful because it provides a range of potential values for r given an unknown. Otherwise you are simply left with a number from -1 to 1, and you have to make a personal decision if it correlated. Is .5 a correlation, what about .4? I would consider that something interesting when examining data, but its not enough to definitely say yes this is the golden relationship. Confidence interval gives us a mathematical way to say yes or no under given criteria which is what is lacking in the correlation equation.

The CI is calculated using the mean and standard deviation. For variance uses size and sample size and for this, CI cannot be calculated directly. Fisher's transformation provides an indirect calculation that can be be made for CI. This is why most of the Fischer problems from your stats books solve for or give CI.

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