I just responded to your statements about the relations between CIs and
hypothesis test that a CI is *not* always associated with a hypothesis.
The equations I mentioned were only examples for a confidence interval
and its equivalent hypothesis test.
BTW: It's not safe to always use z instead of t. If your sample size is
small and you don't know the population variance, it's better to use t
instead of z.
Am 12.10.2018 um 16:56 schrieb Mark Hancock:
This is a good point, yes. I'm not the original requester, but I think
they were really asking for a simple way to get a CI when reporting
summary/descriptive statistics (without having a second mean to
compare to). In SPSS you can do this:
https://en.wikibooks.org/wiki/Using_SPSS_and_PASW/Confidence_Intervals
Maybe this is just my misunderstanding of AGGREGATE and PSPP syntax,
but my point was just that there's nothing inherent about the question
that should require a t-test - i.e., you can use z by default (and
t-tests are really just extensions of z-scores anyway). z=1.96 works
for 95% CIs, and Alan's suggestion does what I think the original
requester was asking.
Pointing to t-tests isn't a bad idea either, though, and maybe
providing syntax for how to reduce it to a z-score would help the
original requester (though I don't think they have another mean or
value to compare it to).
On Fri, Oct 12, 2018 at 9:33 AM Dr. Oliver Walter
<o.wal...@psychometrie-online.de
<mailto:o.wal...@psychometrie-online.de>> wrote:
A confidence interval is mathematically equivalent to its
corresponding hypothesis test. The hypothesis test is significant
if the corresponding confidence interval does not contain the
parameter value of the null hypothesis. The confidence interval
does not contain the parameter value of the null hypothesis if the
hypothesis test is significant. Hence, wether you calculate the
confidence interval or conduct the hypothesis test, doesn't really
matter.
mean(X) +/- t * sd/sqrt(n): confidence interval for the expected
value of X, mu, X normally distributed with unknown population
variance
t = (mean - mü0)/ (sd/sqrt(n)) : test statistic for testing if mu
equals the value in the null hypothesis, mu0, X normally
distributed with unknown population variance
If mü0 is not contained in the confidence interval, the hypothesis
test is significant.
Dr. Oliver Walter
Am 12.10.2018 um 15:01 schrieb Mark Hancock:
I unfortunately don't know enough about PSPP syntax to suggest
how to do this, but a CI is *not* always associated with a
hypothesis and can be calculated from just a mean and SD (and a
cumulative distribution function, which is typically the normal
one). Typically the formula is something like:
mean ± z(SD/sqrt(n)), where z is from the CDF.
On Fri, Oct 12, 2018 at 6:29 AM John Darrington
<j...@darrington.wattle.id.au
<mailto:j...@darrington.wattle.id.au>> wrote:
The confidence interval is a concept associated with a
hypothesis.
If it's the confidence interval on the test for a mean value,
typically you
would get that by using a T-Test.
On Fri, Oct 12, 2018 at 10:40:22AM +0200, Werner LEMBERG wrote:
Folks,
I would like to get a 95% confidence interval so that I
could use it
in AGGREGATE, e.g.,
AGGREGATE OUTFILE * MODE ADDVARIABLES
/BREAK=...
/Mean = mean(V)
/CI = ci(V, 0.95)
What must I do to get the result of my hypothetical `ci'
function?
I'm a PSPP novice, so maybe there is a better solution
than AGGREGATE
??? what I ultimately want is to emit the confidence
interval of a
variable to a CSV file using SAVE TRANSLATE.
Werner
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