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September 10th, 2011, 12:03 AM  #1 
Newbie Joined: Sep 2011 Posts: 7 Thanks: 0  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: (10) x root of(.52)(.4 = ..... Standard error of the sample: = SD/root of 200 SE is something like .035327... I then take the zscore of a random number slightly above 50%. e.g. : (.500001  .52) / .035327 = 566 zscore. Therefore, the probability of the number being 50% or higher would be 1 the area below that zscore, which is miniscule. This probability, however, seems way too high .. What am I doing, or thinking wrong ? 
September 10th, 2011, 11:02 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  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. 
September 10th, 2011, 11:15 AM  #3 
Newbie Joined: Sep 2011 Posts: 7 Thanks: 0  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 zscore attempt is wrong..., but I am pretty sure I have to use zscores 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.. 
September 10th, 2011, 03:14 PM  #4  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Statistics/Probability Problem (dichotomous dataset) Quote:
Quote:
Quote:
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).  
September 10th, 2011, 04:24 PM  #5 
Newbie Joined: Sep 2011 Posts: 7 Thanks: 0  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 zscore?? 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 zscore 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 
September 10th, 2011, 09:01 PM  #6  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Statistics/Probability Problem (dichotomous dataset) Quote:
 
September 10th, 2011, 10:54 PM  #7 
Newbie Joined: Sep 2011 Posts: 7 Thanks: 0  Re: Statistics/Probability Problem (dichotomous dataset)
My professor went over it. The .41 is the zscore 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 . 
September 10th, 2011, 11:21 PM  #8  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Statistics/Probability Problem (dichotomous dataset) Quote:
Left of .41 is 34%; right of .41 is 66%; between .41 and 0 is 16%. Quote:
 
September 10th, 2011, 11:31 PM  #9 
Newbie Joined: Sep 2011 Posts: 7 Thanks: 0  Re: Statistics/Probability Problem (dichotomous dataset)
I just messed up the decimal position. The zscore 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. 
September 10th, 2011, 11:41 PM  #10  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Statistics/Probability Problem (dichotomous dataset) Quote:
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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