Topic: So....who's good at statistics?

[gpw11]

Because I'm currently enrolled in a stats class and could probably use some help with the nuances of probability equations.

[ren]

Stats is the worst. I'm usually pretty good at math but with stats I'm utterly hopeless.

So what's your question?

[iPPi]

I did statistics 6 years ago?  I don't remember it much; I played cards with friends in the lecture all throughout the semester.  Still got an A- though so I was happy about that. 

[idolminds]

4 out of 5 doctors agree that 53% of the time I am pretty OK at statistics.

[Ghandi]

I got an A in it back in the day, I don't know if I would be any good at it now though.

[gpw11]

I've done Stats before but didn't get credit for the course in my current program, meaning I have to take it again.  While you may think this might have made it easier, it's actually about 3 times harder since I remember just enough to be dangerous and get confused. It also doesn't help that I haven't been putting as much time into it as I should, it's a night class, and it's only once a week (four fucking hours).

Stats is the worst. I'm usually pretty good at math but with stats I'm utterly hopeless.

So what's your question?

I don't have it on me right now, but it concerns probability and if there's a shortcut for something (which I know there is...just not what it is).  The example that I couldn't figure it out on was a question regarding quality control, and finding the probability of random sampiling causing a batch of something to fail when testing without replacement. 

 So, if you had a batch of, say, 4000 Ipods you were testing, knew the failure rate was 12%, and wanted to know what the chances were of finding a failure if you were going to randomly sample 100 of them without replacement.  I know you find the probability of failure by taking the inverse of the probability of finding no failures over the whole series (P[at least one failure] = 1-P[Nofailure]).

Now, finding the probability of that seems like a long and drawn out process of multiplying all the probabilities of each test together (since there is no replacement).  So, if the chance of no failure on the first test is (3520/4000), you'd looking at finding P[nofailure] of the whole series by doing the following:

P[nofailure]= (3520/4000) * (3519/3999) * (3518/3998)....... for the entire 100 test series. I know that there's a shortcut for this, but can't remember what it is at all.
 

Conversely, I'm wondering if you could find the probability of finding a failure (480/4000) and multiply it by the chances of finding a failure in all the subsequent tests of the same batch.  I.e. (480/4000)*(479/4000)*(478/4000).......for the entire series and do it that way.  The common denominator makes it easier to solve as a factorial (I think).  Does that make sense as a way to solve for this sort of thing?  I don't really have a way of testing for this at the moment.

I can post the actual question when I have my books on the weekend.

[Pugnate]

I feel so dumb.

[Cobra951]

It's been way too long for me.  I do remember that factorials were involved, and I can see that being useful in your problem, somehow.

Edit:  What I mean is that, for example, this can be expressed with factorials:

P[nofailure]= (3520/4000) * (3519/3999) * (3518/3998) * . . . * (3421/3901)

I'll probably screw this up, but you'll get the gist of it:

P[nofailure]= (3520!/4000!) / (3420!/3900!)

I think . . .  That's for a 100-element series.  Now all you need is a calculator that can do ginormous factorials. 

Edit 2:  Numerators and denominators are throwing me here.  I can't quite wrap my head around it right now.  So do not quote me when it matters.   

Edit 3:  Damn you.  Now I can't quit thinking about this.  Separating out numerators and denominators.  Multiply out numerators, then multiply out denominators, then divide results, then use factorials for conciseness . . .

P[nofailure]= (3520!/3420!) / (4000!/3900!)

Are those 2 equivalent?  Bah!  I suck.  I'm more confident about this last way to associate terms, though.

[gpw11]

I think you're definitely onto it, and I THINK the key somehow lies in keeping the denominators the same...or at least part of it.  Ideally, your first solution should work but I don't think it's possible for a calculator to do that without running into a memory error....leading me to believe that there's another method.  I don't know, I plan on investing some time into researching it over the next day or so...let you know what I find.


Oh, and no, the professor did not help at all.  He doesn't answer questions about assignments until after they are marked...and no one else I talked to in the class seemed to remember either.

[Cobra951]

The reason I like my last attempt is that it deals with the multiplication of a series of fractions correctly: numerators * numerators, denoms * denoms, and finally division of numerator product by denominator product.  In other words, I re-associated your series this way:

(3520 * 3519 * 3518 * . . . * 3421) / (4000 * 3999 * 3998 * . . . * 3901)

Doing that allows me to abbreviate each series with factorials.  If the series went down to 1, it would be real simple:

(3520!) / (4000!)

But since they stop after 100 elements apiece, you have to divide out the remainders:

(3520!/3420!) / (4000!/3900!)

Are there no stat calculators or programs that can handle factorials?  This has to be handled with floating-point math, of course: mantissa and exponent.  And yes, there may be some other shorthand way of doing this, like integrals simplify summations so much in calc.  That I can't help with at all.

[gpw11]

Hmmm, I see what you're saying and agree that it should be the way of going about it and finding the answer....if not the way I was particularity looking for.  I don't see why it wouldn't work. It leads to the same problem as my first method, however, where there's really no way to figure it out (quickly) in a test environment.  Pretty much all stat programs and calculators will handle factorials, but trying to calculate a series of 100 for such a large number leads to a buffer overflow on every calculator I've tried (including Ti 8X graphing calculators).

Thinking about it just as I was writing this, however, I think I was using the calculator wrong. I was putting in the numerator/denominator/fraction and hitting the factorial button ("n!").  There was no option to define the series of 100.  There must be another function....npr (permutation) or NcR (combination) might be it...I really can't remember. I'm just about to head out but I'll look into it more tomorrow.