Ada Ma
>
> I've been staring at the results for days and read the
> related part of the
> Stata manual 10 times or more and I'm still confused. I
> want to find out
> whether the results tell me that I can reject these hypostheses:
>
> H0: mean(he) = mean(hexo) [HA: mean(he)~=mean(hexo)]
> H0: sd(he) = sd(hexo) [HA: sd(he)~=sd(hexo)]
>
> Does the following results tell me that I can can reject
> both HA and thus
> accept H0? Is it that the larger is the P figure, the more
> likely I'll have
> to accept HA?
>
>
> . ttest LFShe=LFShePOT if regwk==1, unpaired
>
> Two-sample t test with equal variances
>
> ------------------------------------------------------------
> ------------------
> Variable | Obs Mean Std. Err. Std. Dev.
> [95% Conf.
> Interval]
> ---------+--------------------------------------------------
> ------------------
> LFShe | 560 9.520129 .2146156 5.078732
> 9.098578
> 9.941681
> LFShePOT | 560 8.862436 .2016839 4.772713
> 8.466285
> 9.258587
> ---------+--------------------------------------------------
> ------------------
> combined | 1120 9.191283 .1475172 4.93687
> 8.901841
> 9.480724
> ---------+--------------------------------------------------
> ------------------
> diff | .6576928 .2945102
> .0798378
> 1.235548
> ------------------------------------------------------------
> ------------------
> Degrees of freedom: 1118
>
> Ho: mean(LFShe) - mean(LFShePOT) = diff = 0
>
> Ha: diff < 0 Ha: diff ~= 0
> Ha: diff > 0
> t = 2.2332 t = 2.2332
> t = 2.2332
> P < t = 0.9871 P > |t| = 0.0257 P
> > t = 0.0129
> . sdtest LFShe=LFShePOT if regwk==1
>
> Variance ratio test
>
> ------------------------------------------------------------
> ------------------
> Variable | Obs Mean Std. Err. Std. Dev.
> [95% Conf.
> Interval]
> ---------+--------------------------------------------------
> ------------------
> LFShe | 560 9.520129 .2146156 5.078732
> 9.098578
> 9.941681
> LFShePOT | 560 8.862436 .2016839 4.772713
> 8.466285
> 9.258587
> ---------+--------------------------------------------------
> ------------------
> combined | 1120 9.191283 .1475172 4.93687
> 8.901841
> 9.480724
> ------------------------------------------------------------
> ------------------
>
> Ho: sd(LFShe) = sd(LFShePOT)
>
> F(559,559) observed = F_obs = 1.132
> F(559,559) lower tail = F_L = 1/F_obs = 0.883
> F(559,559) upper tail = F_U = F_obs = 1.132
>
> Ha: sd(1) < sd(2) Ha: sd(1) ~= sd(2)
> Ha: sd(1) > sd(2)
> P < F_obs = 0.9290 P < F_L + P > F_U = 0.1420 P >
> F_obs = 0.0710
I am not clear whether you intend some kind of joint test or
if one test is considered as prerequisite to another.
Setting that aside, according to many statisticians,
your question(s) cannot be answered because you
don't tell us what your alternative hypothesis was
(e.g.) before you carried out the t test, i.e.
two-tailed or one-tailed, etc. And according to the
same conservative view your test is dubious if not
meaningless without that being sorted out in
advance.
However, informally, I imagine most users
would feel encouraged, if not obliged, to reject the
null hypothesis of no difference between means and accept
instead the alternative hypothesis of a positive difference, on this
evidence, and at a level of 0.05. Clearly, if you use a different
threshold, the decision may vary (notably, one of 0.01).
This year marks the 50th anniversary of George
Edward Pelham Box's paper in Biometrika which pointed
out that the t test for means is in
practice much more robust than a test comparing variances.
He used some more colourful language: if I recall
correctly, he compared the common practice of F test
before t test to putting out
to sea in a dinghy to see if it was safe for an
ocean liner to leave port.
Having said that,
1. Looking at confidence intervals is often preferable.
2. I'd still look at a graph to compare the
whole of the distributions, e.g. -qqplot-.
Not what you asked, but possibly relevant
to the scientific problem which presumably
underlies this.
Nick
[email protected]
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