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While you don't give a lot of background information, most likely this is a one-proportion z-test, and most likely you're assuming that in a random sample, 50 out of 100 would change their minds.

I put those numbers into the statistical tests menu on a TI-83 graphing calculator and came up with p = .02275

2007-01-11 02:52:48 · answer #1 · answered by dmb 5 · 0 0

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