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Topic: [ap-stat] Mike's 2-Sample Proportion Problem
Replies: 1   Last Post: Mar 21, 2007 10:26 PM

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David Bee

Posts: 4,194
Registered: 12/6/04
[ap-stat] Mike's 2-Sample Proportion Problem
Posted: Mar 20, 2007 7:55 PM
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Mike, Corey, and Others:

Mike writes:

> I posed this question to my students today. A survey of 1200 potential Democrat
> primary voters asked who they would vote for. 444 said Clinton, 390 said Obama.
> (these are fabricated numbers). Is there a significant difference in the
> proportions? We treated this as a 2-sided 2-proportion test and got a p-value
> of .0206, therefore concluding that there is pretty strong evidence that Clinton
> is ahead. My question is whether this is valid in light of the condition that
> the SRS's must be independent. I'm not sure how the independence condition is
> met.
> Thanks in Advance,
> Mike McCarvill
> Blind Brook High School
> Rye Brook, NY
> ====


Let's summarize this as follows:

(1) Clinton: pcap1= 444/1200 = 0.370

(2) Obama: pcap2= 390/1200 = 0.325

(3) Others/Undecided: 366/1200 = 0.305

Not only are these not from independent samples but since we have three categories (and so the 2 df as Corey indicated) this is really a multinomial (trinomial) situation (but, unlike Corey, will ignore using a chi-square test here). For such, the exact [oh, here we go with this again!] standard error of the difference pcap1-pcap2 is

sqrt[Var(pcap1) + Var(pcap2) - 2Cov(pcap1, pcap2)] = sqrt[(p1)(1-p1)/n + (p2)(1-p2)/n + 2(p1)(p2)/n]

where that last term under the square root comes from the covariance for such a multinomial (trinomial) distribution. Thus, using the usual point estimates for p1 and p2, namely pcap1 and pcap2, we have for the estimated standard error

sqrt[(0.370)(0.630)/1200 + (0.325)(0.675)/1200 + 2(0.370)(0.325)/1200]

which gives 0.0235

compared with 0.0194 sans the covariance term. Thus, Mike should be using 0.0235 in the denominator of the test stat instead of 0.0194.

Hope this was helpful.


-- David Bee


PS: If we were to form 95 percent interval estimates for the difference p1 - p2 instead, then with the covariance it would be (-0.001, 0.091) and incorrectly without it (0.006, 0.084) [assuming my arithmetic is correct].
PPS: So, which one would Hillary want and which one would Barack want reported in the press?
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