Re: A question about confidential interval

Thank you very much for your comments.

For these 10 values, the mean = 0.12; the s.d.=1.35-4e;
So, the standard error is 1.35-4e/sqrt(10); and the confidential
interval is (0.12 - 0.000135, 0.12 + 0.000135).

Is this right? If yes, then the confidential interval is very small. Do
you think it is still reasonable to show this interval?Thank you very
much.

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