Is “In any case, with 99.9% probability, …” correct?

I'm wondering whether the meaning of the idiom "in any case" still has a hint of "in every single case".

I would like to say

We expect an R² of 0.79 (in any case within 0.75 ± 0.15, with 99.9 % probability).

But I would never say

We expect an R² of 0.79 (in every single case within 0.75 ± 0.15, with 30 % probability).

After all, 30% and 99.9% neither mean always nor in every single case.

Does the first sentence strike as strange?

• It atleast seems odd, if not contradictory, to me to say "in any case, ..., with 99.9% probability". Why not say "We expect an R² of 0.79 (or atlest with a 99.9% probability within 0.75 +/- 0.15). Or do you aim to express 99.9% probability in a different way? As far as I understand "in any case" means "regardless of what may happen" and thus equal how. – AverageGatsby Jan 23 '15 at 13:28
• "In any case" is an idiom with a couple of subtly different meanings. In one sense it means roughly "tl;dr". In the above first sentence, however, it means "considering the broader case of 0.75 +/- 0.15". The ")" is misplaced. – Hot Licks Jan 23 '15 at 13:31
• At least in the spoken language, "in any case" can mean something along the lines of "So anyway" as in "So anyway, the R² is within 0.75 ± 0.15 with 99.9% probability". – Pertinax Jan 23 '15 at 13:39
• @HotLicks correct me if I am wrong but I do not see "in any case" in the sense of "tl;dr". If you have a fairly long text/conversation and you come to a point where you say "in any case, I am still of the opinion that.." it seems just to say "doesn't matter"/"regardless of all this". "tl;dr" does not mean the same. Thunder Chimp has expressed the same less clumsily. – AverageGatsby Jan 23 '15 at 13:41
• @AverageGatsby What I want to say is "We expect an R² of 0.79 since, on average, the R² is of 0.79. We are also 99.9% sure that the R² is within 0.75 ± 0.15, since 99.9% of tested R² fall within that range." – Pertinax Jan 23 '15 at 13:46

This has to do with confidence intervals.

When you say something like 0.75 ± 0.15 with 99.9% probability what you mean to say is that on repeating the experiment many times, the range reported by the experiment (e.g. 0.75 ± 0.15) will contain the true population mean 99.9% of the time. Note that when the experiment is repeated the reported mean and its error may not be the same (they are a function of the data).

In other words, in repeated experiments, C% of confidence intervals will contain the true population mean.

The confidence interval (most commonly 95% CIs are used, by convention) is often misinterpreted as the probability that the population mean falls within the reported range. This is not strictly true, since the population mean is a constant and either does or does not fall within the range.

You would never use with 30% probability (although you can) because 30% is too low.

In polling data the verbiage is "19 times out of 20". For example, you might hear "75% ± 3 people agree, 19 times out of 20". This corresponds to a 19/20 = 0.95 (95%) confidence interval. In other words, if the same poll were to be repeated many many times, the population fraction of those who agree would be spanned by the interval 19/20 times.

The appropriate way to report such results is 0.75 ± 0.15 (95%CI).