#51935 Correlation between AQ, Myers-Briggs, and politics

[vorpal blade]

Waldorf said “Now it will make my day if a reader finds a correlation between Myers-Briggs, political leanings, and Asperger's syndrome.” The question interested me, so I attempted to find any correlations for Board Writers using previous data and the new data on Autism-Spectrum Quotient, or AQ.

The data base is really too small to make any significant correlations, but it is interesting to note the trends. There is virtually no correlation in left-right political leanings and AQ.

However, I did see some correlation between Authoritarian-Libertarian scores and AQ. Generally speaking, the higher the AQ score the more the writer leans towards Authoritarianism. Sauron, Tao, and Foreman had the highest AQ scores (27, 25, and 24, respectively), and they also had the highest Authoritarian scores (0.2, 0.21, and 0.26). As you go down in AQ from Kirke (22), to habiba (20), to Whistler (17), and finally to Black Sheep (14) you get progressively more Libertarian scores. Cognoscente and Waldorf are off the line (exceptions), with lower AQ scores than you might expect from their moderate Libertarian views.

Most of the Introverts scored high on the AQ, with the Extroverts scoring low, with the exception of Tao. Only the “N” types took the AQ test, so we can’t compare their scores with the “S” types. Similarly there appear to be no “P” types among the writers, so we can’t compare this personality type either.

There is a rather strong correlation between low AQ scores and the “F” or Feelings type of personality measure. All three of the “F” types (Black Sheep, Waldorf, and Cognoscente) scored low in the AQ test; in fact they have the three lowest AQ scores. So, it appears that Thinking types score above average in AQ, while Feeling types score below average.

I should perhaps point out that Sky Bones and Ineffable had low AQ scores, but I don't know how they scored on the political compass or MBTI tests. I could only attempt to find a correlation when I had the data.

[krebscout]

Hah. Oh boy.

High five for you.

[Damasta]

Got any statistical analyses for those?

[vorpal blade]

Got any statistical analyses for those?

Consider the relationship between AQ and the Authoritarian/Libertarian factor, AL. You can plot AQ on the x-axis and AL on the y-axis. Using a least-squares regression the slope is 0.228 and the y-intercept is -5.937. Hence the equation is AL = -5.937 + 0.228*AQ. The standard error associated with this linear regression is 1.326.

The Pearson's r correlation coefficient is 0.699.

I'm a little rusty with Mathcad, but that is what I get.

Is this the kind of statistical analysis you were looking for?

[vorpal blade]

If you eliminate the point that is most off the line (Waldorf) you get a correlation of 0.896 and a standard error of 0.887. The equation is then AL = -8.576 + 0.344*AQ.

[Whistler]

so... there's a .23 correlation between AQ scores and political compass AL scores?

[vorpal blade]

so... there's a .23 correlation between AQ scores and political compass AL scores?

No, I believe there is a 0.699 correlation between AQ scores and political compass AL scores for the nine Board writers.

Statistics is not my strong suit. I believe you are more familiar with statistical analyses of this kind of data. In math I was more interested in topology and number theory. What correlation do you find? Here are the (AQ,AL) data pairs I used:

(14, -4.16) Black Sheep
(27, 0.2) Sauron
(17, -3.95) Whistler
(20, -2.26) habiba
(12, -1) Waldorf
(22, -0.51) KirKe
(25, 0.21) Tao
(15, -2.05) Cognoscente
(24, 0.26) Foreman

[Whistler]

when you made equations for the lines, you found the slope, which is the same as a correlation... for example, if you plot two things that have a perfect correlation, the line of best fit will have a slope of 1. I am far too lazy to calculate the correlation myself, but I trust your equations.

[vorpal blade]

Correct me if I'm wrong, but it seems to me that the slope is not a good measure of correlation. For example, if I plot the height of a group of people in inches, against their height in centimeters, there will be a perfect correlation. Everyone's height in centimeters will be exactly 2.54 times their height in inches. But the slope of the line will be 2.54, not 1.

[Darth Fedora]

That's why you find correlation with standardized z-scores. Otherwise you could just change the units and make the correlation whatever you wanted.

[Whistler]

oh, yeah, that's important too.

[vorpal blade]

When I first read Darth Fedora's post I didn't know anything about standardized z-scores. I've since done a little reading on it.

I converted the AQ and AL data to z-scores using the formula that the z-score data is the difference between the raw data and the mean, divided by the standard deviation. Technically you should use the mean and standard deviation of the entire population. In this standardized data set you can then find a least squares fit line that runs through the data. The slope of that line is 0.699 in this case, which is exactly what I got using the Pearson correlation formula of Mathcad on the raw data.

So, if you standardize the data for z-scores, then the slope is the same as the correlation. If you don't standardize the data the slope of the line depends on the units used. However, the Pearson correlation is the same regardless of the units. You can calculate the Pearson correlation, without going through z-scores, by taking the covariance of the pair of data points and dividing by the product of the standard deviation for each data set.

Alternately you can find the correlation using just the raw data by multiplying the slope of the raw data (least squares line fit) by the standard deviation of the independent variable set (AQ in this case) and dividing by the standard deviation of the dependent variable set (AL).

I hope that is clear.