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[miyuru]

Hopefully somebody on the board can help me out...

Anyway at work I'm doing research in Cardiology, and I don't know why but my boss assigned my first study to be - "does the lunar cycle have an effect on heart attacks".

So basically I maintain a database of all patients that enter the CCU (cardiac care unit). I filtered it so only the heart attack (MI) patients remained. Then beside their admission date (i.e. date of MI), I put a rating in another field which meant how close they were to the closest full moon (that's all I'm really looking at).

The ratings went from -14 (14 days before the closest full moon) to 0 days (the day of the full moon) to +15 days (15 days after the closest full moon). The +15 only happens about half as much as any rating from -14 to +14 though, so I adjusted all the results in the end by dividing the frequency by how many times each rating occured from the time period I analysed.

Anyway in the end, I have a table of simple data, basically the frequency of MI's for each rating (and there was no correlation in case you're interested ).

So from here I have to show that it isn't statistically significant, etc...

What kind of test do I run the data through?

:-\

I'm desperate

 

[slayn]

wow its been a long time since stats... but I don't recall there ever being a way to prove that two things had no correlation... I only remember tests like:

my hypothesis is that patients within 5 days of a full moon are 30% more likely to to have a heart attack than those further away than 5 days, and then you can disprove that hypothesis with a certain... I forget the term... some percentage...thingy... or something.

theres like t tests... and z tests... and um... yeah.

I'm no help am I.

 

[Dilbert]

Stats is NOT my strong point...but I'll try to at least give you an idea on how to start.

First of all (although it might be interesting to investigate), there is no difference between +n and -n days away from a full moon in your simple analysis. You can bin your results by |n|, with n ranging from 0 to 14.

If the rate of MI -- call it R(MI) -- is dependent on the n, the number of days away from the full moon, then you would expect that it would be a function of n. The most general form would be:

R(MI) = kn^p + C

k is a scaling constant, p is the power of n (linear dependence? quadratic dependence?), and C is an overall constant.

I believe that you can perform a least-squares analysis of some kind to prove that there is no dependence between R(MI) and n, but it's been a while since I've done one.

Or, this may be a useless suggestion. Everything I learned about stats was in the context of solving real-world problems, and this sounds more like a textbook stats problem. Sorry I can't be more help.

 

[miyuru]

First of all (although it might be interesting to investigate), there is no difference between +n and -n days away from a full moon in your simple analysis. You can bin your results by |n|, with n ranging from 0 to 14.

Yeah, I already did that. Unfortunately the program at work sucks so I wrote out the new combined results out by hand

I'll look up least-squares, but I was wondering if there was another way to do it. Excuse my idiocy in stats...but:

After combining the +/- results, you get a distribution from 0 to 15. We'll omit 15 days from a full moon since really it doesn't occur very often. So, 0 to 14.

Now think of those z-curves (I think?). You know, those bell curves.

For the null hypothesis, we can say: there is no relationship between the full moon and MI.

For our experimental hypothesis, we can say: there is a relationship between the full moon and MI.

Thus for the experimental hypothesis, we will assume there to be a normal distribution (bell curve) since most MI's will occur on day 0, the full moon.

For the null hypothesis, we assume a uniform distribution, as the MI dates will occur between 0 to 14 evenly.

I don't really know where to go from here...

In the meantime I'll read up on the least-squares method. Just from the way you described it though, but I have a bad feeling about it Please keep posting, I'll cut you a cheque!

 

[miyuru]

Crap, I guess nobody's online that can help me...bump :-\

I think chi-square might work.