Re: chi-square test goodness of fit

From: D. Touie (dtouie_at_tscnet.com)
Date: 01/05/05 

Date: Wed, 05 Jan 2005 17:51:53 GMT

On Tue, 04 Jan 2005 08:44:48 -0500, Bruce Weaver
<bweaver@lakeheadu.ca> wrote:

>D. Touie wrote:

>> On Mon, 03 Jan 2005 11:35:49 -0500, Bruce Weaver
>> <bweaver@lakeheadu.ca> wrote:

>>>What do you disagree with? I was not suggesting that smaller values of
>>>chi-square and larger values of p *than reported by the OP* represent
>>>good fits. It was simply a general statement: the smaller the value of
>>>chi-square (and the larger the value of p), the better the fit.
 
>> You seem to believe Pearson's Chi-square is a one-sided measurement
>> construct. It is two-sided. A probability of .01 is the same as a
>> probability of .99. These two probabilities usually have a different
>> causality, but the exact same probability meaning.

>Can you provide any examples of this? I'm afraid you lost me.

The problem posed by the original poster in this thread: Chi-square
sum of 24 with 28 degrees of freedom = about .68 probability evaluated
the "professional" way. This result is precisely equivalent to 1 - .68
= about .32 probability for goodness-of-fit testing purposes. In a
goodness-of-fit test, we aim for the middle ground surrounding .50
probability; not the probability extremes, either high or low.
 
>> I think part of the confusion here is the traditional misnomer
>> "expected" value. It could more properly be described as a
>> "comparative" value.
>>
>> For instance, if I flip a fair coin 100 times and record the outcomes
>> faithfully, I do not "expect" 50 heads and 50 tails to result. That
>> would be an unexpectedly rare result. What I would expect is about 45
>> heads and 55 tails to result, or equally 45 tails and 55 heads.
 
>You've lost me again. If the coin is fair (which is the hypothesis
>being tested), the probability of exactly 50 heads and 50 tails would be
>about 0.08, which is not *that* rare. And it is the single outcome with
>the highest probability. The probability of exactly 45 heads (and 55
>tails), for example, is about 0.05. We all know that the probability of
>any single outcome will not be extremely high (especially as N gets
>large). But the expected value of a binomial distribution is still N*p.

Apparently you buy into that fair coin teaching nonsense from
beginning college statistics. I do not. I would not think of testing
the hypothesis you hypothesize above.

I see this kind of testing as a fundamental "ordinary sample, or
unusual sample" opportunity. Also it provides me with useful grounding
in my larger goal of measuring randomness.

I made a table of all the outcomes from a 100 coin-flips trial. Here
is a partial summary of that table:

49/51 split = .156 probability
48/52 " = .147
47/53 " = .133
46/54 " = .116
45/55 " = .097

50/50 " = .080

44/56 " = .078

Sixth most common predicted result is rare enough for government work.

I did the 45/55 split estimate off the top of my head from a very old
memory. It turns out I was one off in that estimate. A 46/54 split is
the typical mid-value split I would anticipate from an ordinary 100
coin-flips trial.

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