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I have a group of bearings that appear to be clustered toward the SW. I wanted to compare this to a uniform distribution, so I selected a Rayleigh test.

Here is my code in R:

library(CircStats)

bear <- c(-113.055485, -113.055485, -113.055485, -113.055485, 
          -117.775314, -118.902297, -113.055485, -113.055485, 
          -117.775314, -121.597818, 5.130404, 5.130404, 
          -113.055485, -113.055485, -113.055485, -113.055485, 
          -178.019797, -118.902297, -118.947140, -118.947140, 
          -117.824638, -127.296215, -159.028166, -126.898379,
          -159.028166, -159.028166, -117.150693, -125.275715, 
          -117.824638, -118.258142, -159.028166, -124.370972, 
          -118.096576, -118.096576, -118.382553, -118.096576, 
          -118.096576, -118.096576, 6.297989, -121.789656, 
          -121.896598, -126.883761, -117.150693, -117.150693, 
          -122.929838, -122.929838, -176.720148, -120.156298, 
          -127.981467, -127.981467, -119.707813, -119.707813, 
          -121.064324, -119.707813, -106.097798, -105.572360, 
          17.880409, -11.125599, 91.056381, -121.599492)

r.test(bear, degree = TRUE)

$r.bar
[1] 0.8054295

$p.value
[1] 1.247252e-17

My interpretation of this result is that the results are significantly oriented in a specific direction (rather than randomly around the circle). Can anyone add on that? What does the r.bar statistic indicate?

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$\bar{R}$, or r.bar, is a measure of spread around the circle. It should be noted that:

  • If $\bar{R} = 0, $ then the data is completely spread around the circle.

  • If $\bar{R} = 1, $ the data is completely concentrated on one point.

$\bar{R}$ and sample size $n$ together determine the p-value of the Rayleigh test.

In your case, the $\bar{R} = .81$ makes it quite likely that the data is not generated from the uniform distribution around the circle.

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  • $\begingroup$ Thanks @Kees Mulder. That was a very helpful explanation and matches with my interpretation of the figures. $\endgroup$ – tnt Oct 29 '19 at 4:11

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