On Sun, Oct 14, 2018 at 09:28:47AM +0200, Dr. Oliver Walter wrote:
Am 14.10.2018 um 08:46 schrieb John Darrington:
> AGGREGATE OUTFILE * MODE ADDVARIABLES
> /BREAK=g
> /Mean = mean(V)
> /sd = sd(v)
> /n = n(v)
> .
>
> compute ci_upper=mean + sd/sqrt(n).
> compute ci_lower=mean - sd/sqrt(n).
>
> list.
Sorry for interrupting, but this doesn't give a 95% (or 90%) CI, but only
mean +/- one standard error which is a 68%-CI if X is normally
distributed and sd equals the population variance or an approximate 68%
CI if the sample size goes to infinity (is large). You have to include a
t value into the equation for calculating a 95% (or 90%) CI. If your
sample sizes are small and differ from each other you should use
different t values for each CI and each group. If you sample size is
large you could use one z value (1.96) for all groups, but this is not
appropriate in this case (n1 = n2 = 15, sample sizes are too small for
this standard normal approximation).
You are right. Which is why I suggested using one of the CDF functions.
There is no T function, but there is a F function, which I think is the
same if you set DF2 to 1. But you probably know better than me about
those details. Perhaps IDF.F (0.05, N -1, 1) is what Werner wants (I
haven't tried it)?
J'
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