## Astrology

From this skeptical analysis of some astrology data, listing the numbers of famous rich people in each sign, we see the use of the chi-squared goodness of fit test. The data are:Sign | Number of People |

Aries | 95 |

Taurus | 104 |

Gemini | 110 |

Cancer | 80 |

Leo | 84 |

Virgo | 88 |

Libra | 87 |

Scorpio | 79 |

Sagittarius | 84 |

Capricorn | 92 |

Aquarius | 91 |

Pisces | 73 |

Total | 1067 |

To apply the chi-squared test, we simply compare the above numbers to the expected numbers if completely random, which is 1067 people/12=88.9 people according to:

where O are the observed data and E are the expected counts. Once we have the chi-square value and the degrees of freedom (11 in this case), we can look up in tables to get the p-value:

Normally, this might be the end of the story, given that there is not even close to a significant value (usual cut-off around p=0.05).

### Subset of the Data

So, if we only take the extreme values, say:

Sign | Number of People |

Gemini | 110 |

Pisces | 73 |

Total | 183 |

then we calculate a different chi-squared, with 1 degree of freedom, and get

Now this is pretty silly: of course, if you take the extreme values of 12 numbers, and pretend that they came from a 2-category situation, then it'll appear more significant. What about lumping 6 points together, say Capricorn to Gemini (the first part of the year) and the second part. In this case we aren't cherry picking, and the sums should be less significant than the individual data. We then have:

Sign | Number of People |

Capricorn-Gemini | 565 |

Cancer-Sagittarius | 502 |

Total | 1067 |

And we expect 533.5 people in each category. Notice that we went from (the most extreme) 20 person difference from expected in about 100 to a 30 person difference in 500...closer to the expected. What do we get from our chi-squared test?

The test says that this is significantly different from random, more than the individual data! At least the goodness of fit measure, chi-squared value, went down to denote a closer fit to expected but the reduction in the number of data points changes the test quite a lot.

### A different measure

E.T. Jaynes suggests in his book to use a different measure of goodness of fit, the &psi measure closely related to the log-likelihood

Using this measure on the above examples, we get

- All data: &psi = 28.9

- Extreme data: &psi = 39.1

- Lumped data: &psi = 8.1

**compare values of the goodness-of-fit measure to different models on the same data**. It makes no sense, if you have only one model, to reject it by a statistical test...reject it in favor of what? If you have only one model, say Newton's Laws, and you have data that are extremely unlikely given that model, say the odd orbit of Mercury, you don't simply reject Newton's Laws until you have something else to put on the table. The either-or thinking of orthodox statistical tests is very similar to the either-or thinking of the pseudoscientist: either it is random, or it is due to some spiritual, metaphysical, astrological effect. You reject random, and thus you are forced to accept the only alternative put forward. I am not implying that all statisticians are supportive of pseudo-science, and they are often the first to say that you can only reject hypotheses not confirm them. However, since the method of using statistical tests does not stress the searching for alternatives, or better, the

*necessity*for alternatives, it is conducive to these kinds of either-or logical fallacies. An example of a model comparison, from a Bayesian perspective, on a problem suffering from either-or fallacies can be found in the non-psychic octopus post I did earlier.

An elementary question about the motivation for the chi-square formula:

ReplyDeleteIf I were to create something like the chi-squared statistic from scratch, I'd base it on the square of the "z-score" of the observed data in a bin. If the there are N observations and each has a probability of p of landing in the bin then the z-score when O observations are in the bin would be (O - Np)/ sqrt ( Np(1-p)) and the square of that would be (O - E)^2/ (E(1-p)) where E = Np is the expected number of observations in the bin. So there would be a factor of (1-p) in the denominator that is missing from the chi-squared statistic. is my algebra wrong? - or Is there an intuitive explanation of why the factor of (1-p) is left out?