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sim0nz12345 · Oct 1, 2009, 01:50 AM

Hi I am having particular difficulty with this question.

A financial analyst wants to determine the mean annual return on mutual funds. A random sample of 60 returns shows a mean of 12%, Assume that the population is normally distributed and the population standard deviation is 4%.

The question is:
What sample size should we use if we want to estimate the mean annual return with (I) a 95% confidence level and (ii) the width of the interval to be 4%?

Does the width of the interval of 4% mean that the standard error "e" is 4? And how would you do this question?

Thanks.

galactus · Oct 2, 2009, 03:29 AM

Minimum sample size when dealing with proportions can be gotten from the formula:

$n=pq\left(\frac{z}{E}\right)^{2}$

$p=.12, \;\ q=.88, \;\ E=.04, \;\ z=.95$ .

Just plug and chug.

sim0nz12345 · Oct 2, 2009, 07:03 AM

Thanks. Much appreciated

I get n = 59.565
Therefore n =60
However, I still don't trully understand what the question means by the width of the interval to be 4%.

galactus · Oct 2, 2009, 07:58 AM

That's the margin of error is within 4% one way or the other.

sim0nz12345 · Oct 2, 2009, 09:24 AM

Yeah that what I thought as well and was taught, but got confused on this wording. Thank you very much for your help and confirmation.