Stan,
Thanks for the detailed explanation. I have one follwoup ?. You say,
"If the original population is normally distributed, the sample means
will also be normally distributed. Even if the original population
is skewed, the sample means will still be approximately normally
distributed given some assumptions, such as that the sample size
(81) is small compared to the population size (unknown). I don't
know enough to state all the conditions precisely."
Asssume the facts already given. Suppose the population was some demographic
aspect of each person living in India, - n = 1 billion. The _mean_ of
sample means stays at 3. If the one billion population was badly skewed, is
it possible that a sample size of 81 would NOT result in a normal
distribution and would require a larger sample size.
"Stan Brown" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> I'm just a journeyman in this area, but I'm going to presume to
> answer in hopes that if I make any errors the real gurus will
> correct me and the shame will facilitate my learning. :-)
>
> @Home <[EMAIL PROTECTED]> wrote in sci.stat.edu:
> >I am trying to solve a ? which basically gives the following facts:
> >population of unknown number
> >popu std dev of 27
> >pop mean of 78
> >sample of size n=81
> >2000 random samples
> >
> >what is the sample mean?
> >what is the std error (std dev of sample means)
> >what shape would the histogram be?
> >
> >The sample mean is obviously 78 and I calculate the std error of the
sample
> >means to be 3.
>
> I think you mean the _mean_ of sample means? The mean of one sample
> could obviously be anything, though we expect it to be within 78-2*3
> to 78+2*3 about 95% of the time.
>
> I calculate the standard error or the mean (sigma-sub-xbar) the same
> way you do, as sigma/sqrt(n) or 27/9 = 3.
>
> >However I can't put the whole picture together. I suspect the distrib
would
> >be normal given the 81 samples, but is 3 a low number for a std error.
>
> If the original population is normally distributed, the sample means
> will also be normally distributed. Even if the original population
> is skewed, the sample means will still be approximately normally
> distributed given some assumptions, such as that the sample size
> (81) is small compared to the population size (unknown). I don't
> know enough to state all the conditions precisely.
>
> >Is it possible to translate it into a z score without any addtional data.
>
> If the population mean and standard deviation are known, that's all
> you need for a z score. The formula is
> z = [ xbar - mu ] / [ SEM ]
> For your scenario,
> z = (xbar-78)/3
>
> A sample mean of 60 has a z score of -6, so it is quite unlikely
> that you'd draw a sample with a mean of 60. (My TI-83 says that the
> area in the tail past z=-6 is just under 10^-9.)
>
> >In other words is the std deve of 27 and mean of 81 in any way predictive
of
> >what a histogram of a distribution would look like?
>
> I assume you meant to say "mean of 78 and sample size of 81"?
> Assuming that, the histogram of sample means should be normal or
> nearly so, with mean (mu-sub-xbar) 78 (same as population mean) and
> standard deviation (standard error of the mean, sigma-sub-xbar) 3.
>
> >Finally what difference does it make how many random samples you take
(ie.
> >100 or 1000). What statistic or parameter does this speak to?
>
> None that I know, in a formal sense. If you take 100 random samples
> of size 81, or 100,000 random samples of size 81, your histogram of
> sample means will have the same shape, though the curve will be a
> bit smoother with 100,000 samples.
>
> --
> Stan Brown, Oak Road Systems, Cortland County, New York, USA
> http://oakroadsystems.com
> My reply address is correct as is. The courtesy of providing a correct
> reply address is more important to me than time spent deleting spam.
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