Question 573830
Suppose a survey of 1,000 citizens finds that the proportion of respondents who report not having health insurance is .200. Would this sample result be unusual if the true proportion in the population is .158? 
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z(0.2) = (0.2-0.158)/sqrt[0.158*0.842/1000] = 3.6414
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Yes, 0.2 is unuaual.  It is over 3 standard deviations above the mean.
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cheers,
Stan H.
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