Confidence limits, error bars and method comparison in molecular modeling. Part 2: comparing methods.

\sqrt {var} }} = \frac{0.373}{0.206} = 1.81 < t_{95\% } \left( { = 2.01\;for\;N = 50} \right) \hfill \\ \end{aligned}$$\end{document} v a r F 0.9 - F 0.8 = 1 N - 3 + 1 N - 3 = 2 47 = 0.0425 F 0.9 - F 0.8 v a r = 0.373 0.206 = 1.81 < t 95 % = 2.01 f o r N = 50 Therefore the difference in r values is not quite statistically significant at the 95 % level, as we surmised from the combining of the asymmetric error bars for r .

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