Re: Confidence interval is mathematically equivalent to hypothesis test

[John Darrington]

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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