"Duane Allen" <[EMAIL PROTECTED]> wrote:
>Kelly <[EMAIL PROTECTED]> wrote in message
>[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
>> Hi, I'm trying to calculate sample size using the sample size formula
>> for simple random sampling, which requires an estimate of the
>> variance. But I don't know the variance, instead I wan't to use the
>> maximum varince the random variable can take. I know the range of the
>> varible say a to b. How can I use the range to calculate the max.
>> variance the random variable can possibly take?
>> Thanks in advance for your help.
>
>
>The bounding formulas are dependent on sample size.
>The bounding upper formulas are
>s^2 = [n/(4(n-1))]*R^2 for n even
OK, when applying the sample formula using (n-1).
The first part of your formula - in square brackets - can be rewritten as
{ n/(n-1) } * 1/4
>and
>s^2 =[(n+1)/(4n)]*R^2 for n odd
Here the first part can be rewritten as
{ (n+1)/n } * 1/4
Forget about the 1/4, that has to do with the denominator of the
deviation-scores and is perfectly allright.
I'm concerned about the first part of the two rewritten formulas, the part in
accolades.
Both the nominator and the denominator are different. Why is that?
The denominator in the second one should still be (n-1), as always in the
sample variance calculation. The nominator however should be (n-1) too,
instead of n+1.
With an even sample size, variance is maximal when half of the scores - n/2 -
equal the minimum, and the other half - n/2 - equal the maximum. The mean then
of course equals (a+b)/2. [a is minimum, b is maximum].
All deviation scores of course are equal - half is on the negative side, half
is on the positive side - and equal half the range (b-a)/2.
Squaring these deviation scores over n scores yields
n * {(b-a)/2}^2 ==> n * { (R^2)/4 }
where R is the range (b-a).
Subsequent division by (n-1) yields
n/(n-1) * 1/4 * R^2.
So far, so good.
With an odd sample size, the variance is maximal when _one_ score is exactly
in the middle, and half of the others - (n-1)/2 - equal the minimum, and the
others - also (n-1)/2 - equal the maximum.
The deviation scores of those scores that are on the extremes are of course
exactly the same as in the even-sized-sample example. The one in the middle is
by definition on the mean and consequently doesn't have a deviation score.
To calculate the variance, we have to sum the squared deviation scores.
Since the one in the middle doesn't have a deviation score, and doesn't add to
the variance, we now sum over (n-1) scores, yielding
(n-1) * {(b-a)/2}^2 ==> (n-1) * { (R^2)/4 }
Subsequent division by (n-1)
(n-1)/(n-1) * 1/4 * R^2 ==> 1/4 * R^2
Where in your derivation does the part (n+1)/n come from?
Chris
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