1. ## Sample size determination

the problem is to find the sample size needed. I'm given +or- 0.025 at the 95% confidence level and 58% of the population.
so the answer is given as follows:
a/2=0.025 z(a/2)=-1.96 [z(a/2)]^2=3.84 pie = 0.58 1-pie=0.42
I don't understand how to use this information to get the answer. All I've been able to find is that E=(z*stdev/sqrt n) which is n=(z(a/2)*stdev/E)^2.
What am I doing wrong and explain please.
Thanks
pugowner

2. Originally Posted by pugowner
the problem is to find the sample size needed. I'm given +or- 0.025 at the 95% confidence level and 58% of the population.
so the answer is given as follows:
a/2=0.025 z(a/2)=-1.96 [z(a/2)]^2=3.84 pie = 0.58 1-pie=0.42
I don't understand how to use this information to get the answer. All I've been able to find is that E=(z*stdev/sqrt n) which is n=(z(a/2)*stdev/E)^2.
What am I doing wrong and explain please.
Thanks
pugowner
Not a good description. I am not clear about E,margin of error ( 0.025 or 0.05)

Try to solve in this way.

Here z(a/2) = 1.96 and E =0.025
stdev =sqrt( pie *(1-pie) ) ( point binomail distribution)
Formula is same.
n=(z(a/2)*stdev/E)^2
= (1.96/0.025)^2*.58*.42
= 1498(approx )
if you give E =0.05 you will get n as 374.+

You can try the online sample size calculator.
http://www.berrie.dds.nl/calcss.htm
http://www.dimensionresearch.com/res...mple_size.html

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