"Jo" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED]
> The time that students spend at part time jobs per week is approximately
> Normally distributed with standard deviation 8 hours. A random sample of 130
> students is taken and the sample mean is 11.4 hours.
>
> How do I find the 90% lower confidence limit for the true mean (mu)?
>
> How do I find the 90% upper confidence limit for the true mean (mu)?
Suppose
M = sample mean
s = standard deviation
n = sample size
There is a theorem that says that in this case the thing
( M - mu ) / (s/sqrt(n))
is pretty Standard Normally distributed.
So, when we abbreviate it like for instance:
u = ( M - mu ) / (s/sqrt(n)) ,
then you have to find u1 and u2 such that
Prob[ u < u1 ] = 0.10
and
Prob[ u > u2 ] = 0.10
which is equivalent with
Prob[ u < u2 ] = 0.90
You find
u1 = -1.28
u2 = 1.28
So you have
u1 = ( M - mu1 ) / (s/sqrt(n))
u2 = ( M - mu2 ) / (s/sqrt(n))
from which you can easily algebrate and calculate mu1 and mu2 ;-)
Dirk Vdm
.
.
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