<PRE><B>Question 1113821</B>
In a survey of 5100 T.V. viewers, 40% said they watch network news programs. Find the margin of error for this survey if we want 95% confidence in our estimate of the percentage of T.V. viewers who watch network news programs.
ME = z*sqrt(p*q/n)
ME = 1.96*sqrt(0.4*0.6/5100) = 1.34% 

A) 2.00% 
B) 1.34% 
C) 1.54% 
D) 1.76% 
E) 1.12
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Cheers,
Stan H.
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