My mean is 3.454992884900722e+008, and my confidence intervals from three distributions (when compared to the given mean) are:

CI for dist. 1: 1.0e+008 * [4.084733001497999 4.097677503988565]

CI for dist. 2: 1.0e+008 * [5.424396063219890 5.586301025525149]

CI for dist. 3: 1.0e+008 * [2.429145282593182 2.838897116739112]

EDIT: More information as follow:

For Dist. 1: p = 8.094614835195452e-130 and h = 1

For Dist. 3: p = 2.824626709966993e-072 and h = 1

For Dist. 3: p = 3.054667629953656e-012 and h = 1

Can you explain which distribution among the three should be picked when compared to the mean on the basis of CI, and why?

I hope my question is clear. Feel free to edit/suggest anything regarding question. Thanks!

  • $\begingroup$ What's p? what's h? Start at the beginning and just say what you're measuring, why the distributions are different, and what you want to accomplish. $\endgroup$ – John Sep 28 '10 at 23:28
  • $\begingroup$ Perhaps if I guess some things it would help?? Are the p's the probability of the mean showing up in each distribution? You want to say what one is farthest fromthe mean?... closest to it? $\endgroup$ – John Sep 28 '10 at 23:46
  • $\begingroup$ @John: This is my question (stats.stackexchange.com/questions/2639/…) where I have defined my problem about what I want to accomplish. Thanks. $\endgroup$ – user1102 Sep 28 '10 at 23:55
  • $\begingroup$ None of this answers why you're comparing these distributions to a mean and why simple analysis of the confidence interval could possibly help. You don't compare a distribution to a mean just because it has some numeric property. That's why you can't get an answer for your question. $\endgroup$ – John Sep 30 '10 at 10:43

I am afraid that there is something you are missing. The documentation says, p is p-value (which is definitely a probability under the null, but the probability of observing extreme observations if the null holds) and h=1 means you are rejecting the null in favor of the alternative hypothesis.

Here is a good quote from Bert Gunter borrowed from R mailing list.

If you are willing to trust p-values which are as low as yours, I have a bridge to sell.

Also, going by the value of your mean, its better that you look at the data and see what distribution your data follows and check if the assumptions of 1 sample t-test hold. I have a feeling that probably you can't use ttest here. I say this because your p-values are unbelievably small. (I understand your means are large ... etc ..., but still ttest can't be applied without verifying assumptions etc.)



  • 2
    $\begingroup$ @Harpreet When constructing confidence intervals, say a 95% CI for a difference of means, you are explicitely assuming a sampling distribution (t or z, i.e. that of your test statistic under $H_0$). $\endgroup$ – chl Sep 29 '10 at 16:28

Reporting a CI around a mean is not reporting the distribution of values, only an estimate of how well you captured that mean value. It will always get smaller as n goes up. It's NOT what you want because you want to see how well a point fits into a distribution.

With your fairly large N's you might be able to do a nice plot of the distributions, all together on one plot overlayed in different colours and then place your point estimate. You may be able be able to see from what group it seems most likely to come from. Then estimate the z-score to the point value from each distribution. Whichever is the lowest is the one I'd be most inclined to pick but this would depend a great deal on how these looked in terms of distribution. For example, I'm guessing dist.3 is going to be the flattest and so it's going to have a very low z-score there.

Plot it out and see what it looks like. (after dividing everything by 1e8 :) )


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