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September 10th, 2011, 12:03 AM   #1
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Statistics/Probability Problem (dichotomous dataset)

Hi everyone,

I am taking a statistics class this semester and I haven't done math since 12th grade calculus. I have one problem I seem unable to solve, and I am sure it will be easily solved by the geniuses around here.

A ballot is being held. 9000 votes are cast. Of these 9000 votes 52% vote yes, the bill is therefore passed. The question is, if a sample of random 200 people's vote was taken, what is the chance that the correct result (i.e. >50%) would be found?

I tried the following.

52% is the mean

Standard Deviation of dichotomous data: (1-0) x root of(.52)(.4 = .....
Standard error of the sample: = SD/root of 200

SE is something like .035327...

I then take the z-score of a random number slightly above 50%. e.g. : (.500001 - .52) / .035327 = -566 z-score.

Therefore, the probability of the number being 50% or higher would be 1- the area below that z-score, which is miniscule. This probability, however, seems way too high -.-. What am I doing, or thinking wrong ?
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September 10th, 2011, 11:02 AM   #2
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Re: Statistics/Probability Problem (dichotomous dataset)

I would use binomial distribution: what's the chance of getting 100 or more "no" votes which have 48% probability each? That comes to

If I was going to use a normal approximation of the above, I would set stdev = sqrt(200 * .52 * .4.
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September 10th, 2011, 11:15 AM   #3
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Re: Statistics/Probability Problem (dichotomous dataset)

Thank you so much for the response. I have to admit I do not fully get it, because I only have a limited number of formulas available that I can use for this problem. When trying to solve this using google I came across the binomial distribution, but I am not allowed to use it. At least this way I know the correct answer .

Perhaps you can walk me through like an idiot.
I can use the following:

SD for a dichotomous dataset: (smaller - bigger) * sqrt(p(smaller)*q(bigger), so that would be 1* sqrt(.52*.4, = .4995998399
The mean in any case is 52%, .52.

If a sample out of those 9000, of 200 is taken the Standard Error for that sample should be: SD/sqrt(200) correct?, I get .035327
How would I proceed from here, cause my z-score attempt is wrong..., but I am pretty sure I have to use z-scores somehow.

I also have these formulas for samples: expected value of the sum: n * mean
standard error of expected value of the sum: sqrt(n) * SD

Sorry I am hard of understanding, it's been four years..
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September 10th, 2011, 03:14 PM   #4
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Re: Statistics/Probability Problem (dichotomous dataset)

Quote:
Originally Posted by Deeper
Thank you so much for the response. I have to admit I do not fully get it, because I only have a limited number of formulas available that I can use for this problem.
Yeah... I literally had to leave right then, so my answer was incomplete. (I delayed leaving just long enough to finish the sentence.)

Quote:
Originally Posted by Deeper
At least this way I know the correct answer .
This answer will be more accurate, actually, then the answer you'll turn in.

Quote:
Originally Posted by Deeper
If a sample out of those 9000, of 200 is taken the Standard Error for that sample should be: SD/sqrt(200) correct?
...no?

A simple thought experiment: if the sample was the whole 9000, the standard deviation wouldn't be .4996/sqrt(9000). It should be sqrt(n * .48 * .52) = 0.4995... * sqrt(n).
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September 10th, 2011, 04:24 PM   #5
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Re: Statistics/Probability Problem (dichotomous dataset)

Well, I looked it up more and I know how to do it now.

It would be the following:

Standard Error of the Expected value of the sum which in this case is sqrt(200) * .52*.48 = 7.065...

Then I just calculate the probability which would be: (50.000001 - 52)/7.065.... =.28

It's kinda correct, but I don't understand it , at all. Why do I have to use the Standard Error of the Expected value of the sum and not the standard error of the expected value of the average? Why do I get the probability from this: (50.000001 - 52)/7.065.... =.28 instead of a z-score??

I had a similar Homework problem which I solved correctly, which I thought was basically the same thing and it worked:

out of 16000 people, 26% are black. If 100 are chosen randomly, what is the probability of only 8 out of those being black?

mean = .26 N = 100
SD = sqrt(.26*.74) = .4386

Standard Error of the sample = .439/sqrt(100) = .0439

z-score for 8 out of 100 is (.08-.26)/.0439 = -.41 which is correct. I don't see how this is different from my first problem, which I keep doing wrong
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September 10th, 2011, 09:01 PM   #6
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Re: Statistics/Probability Problem (dichotomous dataset)

Quote:
Originally Posted by Deeper
I had a similar Homework problem which I solved correctly, which I thought was basically the same thing and it worked:

out of 16000 people, 26% are black. If 100 are chosen randomly, what is the probability of only 8 out of those being black?

mean = .26 N = 100
SD = sqrt(.26*.74) = .4386

Standard Error of the sample = .439/sqrt(100) = .0439

z-score for 8 out of 100 is (.08-.26)/.0439 = -.41 which is correct. I don't see how this is different from my first problem, which I keep doing wrong
I don't see how that could possibly be correct. The probability that exactly 8 people in the sample are black would be only 0.00036%; this increases to 0.00047% if you allow 8 or fewer.
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September 10th, 2011, 10:54 PM   #7
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Re: Statistics/Probability Problem (dichotomous dataset)

My professor went over it. The -.41 is the z-score which translates into an about 0 % probability.

I still don't get the first one really, cause I don't see how it should not be the same calculation .
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September 10th, 2011, 11:21 PM   #8
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Re: Statistics/Probability Problem (dichotomous dataset)

Quote:
Originally Posted by Deeper
My professor went over it. The -.41 is the z-score which translates into an about 0 % probability.
?

Left of -.41 is 34%; right of -.41 is 66%; between -.41 and 0 is 16%.

Quote:
Originally Posted by Deeper
I still don't get the first one really, cause I don't see how it should not be the same calculation .
The first one is the right calculation. If anything, question the second.
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September 10th, 2011, 11:31 PM   #9
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Re: Statistics/Probability Problem (dichotomous dataset)

I just messed up the decimal position. The z-score was -4.1, which is correct.

I am not questioning that what you did is correct. I just said that I cannot do it that way, I have to figure it out with the means available on my sheet notes so far, and there I don't get why it is different than the second one, which is correct.
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September 10th, 2011, 11:41 PM   #10
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Re: Statistics/Probability Problem (dichotomous dataset)

Quote:
Originally Posted by Deeper
I just messed up the decimal position. The z-score was -4.1, which is correct.
I don't know... that gives an error of 340% to 470% depending on which interpretation you take.

The calculation that you showed involved dividing by the square root of n, and I can't think of a time that that would be the right thing to do.
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