<PRE><B>Question 1029050</B>
This has to be +/-0.04
use t(0.975 with df=unknown)
t*0.24/sqrt(n)=0.04
square everything
t^2*0.0576/n=0.0016
t^2*0.0576/0.0016=n; I have to approximate t and then use that to get n.
t^2*36=n
will use 1.96 to approximate t.  That squared is 3.84
t=3.84*36=138.24 or 139.  Use calculator for that
t(0.975, df=139) is +/-0.0403
for n=140, the margin of error is +/-0.04
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95 are expected.  This is another definition of a confidence interval.  One does not know which 95 will contain it, only that 95 will be expected to and 95% with an infinite number.
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