statistics folly

I think the problem is more aimed at trying to get people to realize that a small sampling size (with large populations) can lead to outcomes with likelihoods may be further away from the "expected" 45% than one might intuit.

Although things like, "How many people in the room do you neeed to have a

50/50 chance of having the same birthday as someone else?" might be better for that.

Maybe they're hoping you'll start noticing the pattern of the returned values and recognize something that looks like a binomial distribution?

It does seem as though the original problem statement is aimed more at, e.g., liberal arts majors where they don't expect to teach them much actual math, but at least figure they should be able to come up with some sort of simulation that suggests one outcome or another. (...of course, one then might be worried when those folks then start building, e.g., climate "simulators"... :-) )

---Joel

Hey, not trying to ruffle your feathers. In a school with 1600 students, an underdog candidate with 45% support from the student body has about a .127 chance of winning an election that only 100 students vote in, a chance of .175 to at least tie. Of course his chances would be much worse in an election with high participation -- he'd be sure to lose. Interestingly, even if the school had a million students and 100 participated in the election, he'd still have only about .136 chance of winning (.183 of tying). So yeah, 1600 and a million aren't all that different. Okay, I admit I cheated -- I used an online calculator.

Actually, the US electors who got out to vote for Obama represent a very large "school", and the likelysampling error on the result was about 0.12% of his winning margin - 9,522,083 - out of 131,257,328 votes cast. The square root of 131,257,328 is about 11,457.

He won quite decisively - the biggest margin of any non-incumbent candidate so far.

-- Bill Sloman, Nijmegen

Oh boy, you're asking for it Bill. Jim's going to come out with his gun.

You're assuming that for *every* case of 100 randomly selected voters it is *impossible* for more than 45 of them to be underdog supporters. It is possible, albeit the probability is low, that 100% of them are underdog supporters.

By hand???? This is trivial using a spreadsheet.

My simulations show the underdog wins 44.96% of the time over 10,000 runs. The max votes he got on any one run was 66. Art

How many kids were in the school that you ran the simulation for? Underdog's maximum theoretical chances of winning (as the school population increases without bound) is about 13.5%. And it's less than that for any real school. You made a mistake somewhere.

^^^^^^^^ voters

-- Popular Vote -- --Electoral-- Obama = 69,438,983 = 52.87% 365 67.8% McCain= 59,028,439 = 48.27% 173 32.2% ---------- ----- --- +128,467,422 - 10,410,544 = 7.24% 192 35.6%

Year = 1980

-- Popular Vote -- --Electoral-- Reagan = 43,903,230 = 50.75% 489 90.9% Carter = 35,480,115 = 41.09% 49 9.1% ---------- +79,383,345 - 8,243,115 = 9.66% 440 81.8%

Reagan Obama Popular margin = 9.66% > 7.24% Electoral margin = 81.8% > 35.6%

IOW, you're full of shit, as usual.

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9,522,083 > 8,243,115 so Obama has the biggest absolute margin of any non-incumbent cabdidate so far. Expressing the margin as a proportion of the total vote makes Regan look as if he did better, but I just said "biggest", so you are full of shit - as if anybody cared. Since Regan had to dye his hair to do that well - the start of an eight year career of lying to the electorate - you'd have been well advised to shut up.

-- Bill Sloman, Nijmegen

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Not really a worry. I'm in Sydney at the moment, but since Jim has kill-filed my posts, he won't know that, and would be waddling around the Netherlands (where carrying a gun is illegal if you aren't a cop or en route to your shooting club) posing no great threeat to me, if something of a menace to all the other residents of the country.

-- Bill Sloman, Nijmegen

get

What an ass, Slowman. Figures don't lie but liars figure. That certainly describes you.

"Michael Robinson" wrote in message news:l6adnZjIINVRA93RnZ2dnUVZ snipped-for-privacy@giganews.com...

This is a simple sampling problem. We can reconfigure the problem as in general there is a p chance of success and 1 - p chance of failure. We are looking at the statistics of a sample of n.

NOTE: IT IS POSSIBLE THAT THE WE COULD GET n FAILURE!!!!!! That is, 100% of the sample is all failures.

In your analysis you have misinterpreted the problem. In this case the number of kids in the school is infinite and the sample size is ALWAYS 100. You seem to be thinking the population size is 100 and the sample size varies from 1 to 100.

Think of having a large bin of components(after all, this is S.E.D.). We randomly choose n components from the bin. Note that the statistical parameters are given from the population(how we got them was by estimation and/or assuming certain things) and we want to use them to determine our sample statistics.

This is very basic statistics. sample mean vs population mean?

In the above case, what if p is the probability that a component works then

1 - p is the probability it is a failure.

NOW! What the probability that if we chose 50 or more components that we would have at least 50 sucesses(in the original situation, that the underdog will lose). It's possible we may get all failures(just highly unlikely).

Similarly you can think of flipping a coin(slightly unbalanced in this case). Whats the likelyhood we would get, out of a 100 flips, at least 50 H's in the flip(then H win's).

There are many ways to accomplish this.

The answer given by the book is correct.

You perform n independent trials and note the result. e.g., flip a coin 100 times and note the result. HHHHHTHTHHTHHTHTHTHHHHTHHTHHTTTTHTHTHTTTHTHHT. Count the number of heads. Note that I simply made up the sequence(it's not even 100 in it but lets assume). Note that I may have no used something that correlates to the 55/45 coin BUT if it truly were a sample from the population then we can assume(with a reasonable degree of confidence) that ON AVERAGE H should show up 55% of the time(we can seet he above is 26/45 ~=

57%. So I was actually pretty close(just coincidence).

I could easily have done HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH. and this would be 100% of the time! BUT, IF THIS IS A TRUE SAMPLE OF THE POPULATION IT IS A POSSIBILITY!!! (assuming the population is much larger than the sample we are taking)

Why is the answer in the book correct? Because it uses a RNG to flip a coin that is unbalanced. It does this by having the RND pick a number between 0 and 100 and then check and see if that number is N/2. (majority wins)

The average number of successes is given by the hypergeometric distribution(it is more general than the binomial).

So

Wins election = (n*m/N > N/2)

for N = 100 and n = 100, m = 55 we have

avg = 100*55/100 = 55 (note in this case "avg" is really exact)

so

Wins = (55 > 50) = true

In our problem n = 100 and is fixed(it is simply the sample size).

but m is the number of successes. They tell us on average we have 55 successes(or votes) BUT this is simply an approximation to m/N (this is where they got that number in the first place(if they actually didn't make it up but sampled a population).

100*m/N = 100*55 = 55. SO THE GUY WILL ALWAYS WIN BECAUSE 55 > 50. (on average of course but exactly in some cases).

Now m may not m exactly 0.55*N for each individual sample(for the same reasons I gave in the binomial case) BUT on average(and again, sometimes exactly) it will be 0.55*N and p = m/N = 0.55.

Statistics almost can never predict what will happen at any instant(as few things have probability 1) but it generally tells us what will happen on "average"(which in some sense never really happens). All I have done is shown you than on average the number of successes, given that the population has 55% successes, will hold approximately true for randomly chosen samples of that population (the law of large numbers).

It seems like we did a lot of work but we really didn't do anything. We were told the population statistics and I just showed that if N is relatively large then they hold for approximately for samples. This is almost a natural conclusion because this is the concept of a "population" in the first place. The things in a population have a correlation and it wouldn't make much sense if that didn't hold for large samples of it.

Main things with statistics:

1. Recognize what is being asked of.
2. Recognize what can be extrapolated from(this requires finding an appropriate model)
3. Extrapolate to the information wanted and give the information. Additionally give your confidence that the information is correct.

For your problem, you have given us a 100% confidence that the underdog will lose if N = 100(n = 100, p = 55%. (this is so obvious it's stupid... because if N = 100 and p = 55% they are simply saying the underdog only got 45 votes)).

The hypergeometric distribution says on average(which may or may not be exact) the underdog gets 45% of the votes. The binomial distribution says the same thing. So we should say, on average the undedog should lose. We can get an estimation of the variance and such to refine our information

So all we can realy say is that on average the population statistics is what we use(that is all we know about the situation). If we new more information then we could do more with it. If we looked at the psychology of the voters and history we might be able to make better guesses. BUT the problem is innately stated using a binomial distribution(p = 55%) and which is why the hypergeometric analysis doesn't give better results(it would if we knew what m was and simply didn't approximate it as p*N).

In any case I wrote a lot of crap and hopefully you got something out of it. Your not necessarily wrong in your analysis but your approach is wrong as you didn't recognize the simplicity of the problem and tried to answer the wrong question. The good thing is that you went above and beyond what was expected which shows you have the capability to solve the problem in more detail than what the "average" person probably can. The bad thing is that you didn't answer their question(well, your answer was not really correct as you talked about complex simulations and such as this is not really what was needed) ;)

The problem here is that you don't want to get in the habit of always making things overly complex as you end up missing the obvious.

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Do keep trying. It will makes you look even dumber than you do at the moment - not that that is going to affect your status as a hopeless right-wing half-wit - but it will give a few more lurkers an opportunity to appreciate your place in the scheme of things as part of the group's non-contributing wall-paper (and not the most attractive of the patterns on offer).

-- Bill Sloman, Nijmegen

The unacceptable answer is the fact that the "sample size" is insufficient for any MEANINGFUL results and thus ANY "calculations" are moot. But...these are idiots that lie with numbers and actually get away with it. The TRUE technical answer is "insufficient sample size for any meaningful answer. Case closed, period.".

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Sorry, Slowman, I've made my point. You're dumber than a stump, but trying to pass yourself off as the master of all information. Sorry, chump. You got called on your asinine statement and lost, as usual.

About your answer above - how did you arrive at the conclusion that the sample size is insufficient? I'm missing something.

Ed

The whole population is too small (a thousand if i remember correctly), so any sample size is too small - even if ALL were sampled, which is absurd. Also, the sample must be "chosen" in a totally random manner, which is next to impossible for a population of a mere 1000; the smaller the total population, the worse it gets. There was one idiot that lied using "statistics", sampling, etc with a total population of TEN - and using a lot of bs blather with the numbers in a long report, got away with the lies.

Oh, BTW, if, in a LARGE population the probability of such-and such is X, that means that any ONE randomly chosen sample fits that sec, and the next sample fits, etc. Meaning if the probability of a coin toss landing on edge is 0.0001 DOES NOT mean that the coin will not land on edge 2 times (or 2,000 times) in a row; each try is independent of all others (in a fair game).

It's a good thing for BS that his computer runs on methane.

--
Anyone wanting to run for any political office in the US should have to
have a DD214, and a honorable discharge.

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Really? Care to remind us of the "asinine" statement?

From my point of view, you were trying to pass off Regan's proportionally larger margin over Carter as as the "biggest" margin of a non-incumbent, when "big" is an expression of an absolute - not proportional - comparison.

You'd be the ass, the more so for persisting with the kind of proposition that only a fat-headed right-winger could see as persuasive.

-- Bill Sloman, Nijmegen

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I know you're not bright enough to read, but it's right up there.

That is the vote *margin* dumbass! Yes, idiot, you did say "margin".

"...the biggest margin..."

One would think English wasn't your first language, but we all know that you're just STUPID.

...unspeakably stupid, Slowman.

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If I'm not bright enough to read, I'm clearly not bright enough to respond to your latest fatuous claim.

Are you planning on taking a course in joined-up logic sometime soon?

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