"Dirk Van de moortel" <[EMAIL PROTECTED]> wrote
in message news:[EMAIL PROTECTED]
>
> "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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------------------
I see great confusion here. Confidence intervals have been incorrectly
stated.

There are three possibilities:

Known Information
1. Population Mean, Population Variance
2. Population Variance, Sample Mean
3. Sample Mean, Sample Variance

Tables to use:
For 1, normal table (z)
For 2, normal table (z)
For 3, t table (t)

Confidence Interval is centered about
For 1, population mean
For 2, sample mean
For 3, sample mean

Confidence Interval is to Include
For 1, future sample means
For 2, population mean
For 3, population mean

Standard Deviation to be used
For 1, Stated value or square root of population variance
For 2, Stated value or square root of population variance
For 3, Square root of unbiased variance

Interval is , + or -, where n is sample size.
For 1, z * population stdev / square root (n)
For 2, z * population stdev / square root (n)
For 3, t * sample stdev / square root (n)

Valid only for normal populations.

Other possibilities can be considered, but lead to incorrect interval
values.

I just completed a Monte Carlo study on this using a set of 3.2 million
random standard normal deviates, and can only verify the three intervals for
correct P values. For 400,000 sets of 8 values (n=8) I hit the table P
values within 0.02%.

David Heiser


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