I am trying to convert -log (P-value) to normal P-values. I saw online that you can convert -log10(Pvalues) by taking the exponential as 10^-logPvalue, but Im not sure if it will apply on the -log values?
Maybe I can convert -log(P-values to -log10(P-value) by dividing with 2.303 and then take the exponential?
Yes it will apply. You really don't need to convert anything. Just use the scientific calculator on your smartphone. If you want examples, Google should supply plenty. Best, D. Booth
If s = -loge(p), then p = e-s
If s= -log10(p), then p = 10-s
If s= -logb(p), then p = b-s
Do you get the principle?
I agree with David that you should better not convert these -log(p) values. It usually makes sense if the statistical significances are provided this way. The negative logarithms express the significance as additive amounts of information the data provides relative to the null hypothesis (within the specified model). This is a more intuitive way to present the statistical discordance of the data with your model than the "simple p-value" and it allows to shows p values that are different by many orders of magnitude side by side in an easy readable way.The representation as -log(p) is tightly linked to information theory and they are often termed "surprisal values" (hence my choise to use "s" above):
https://lesslikely.com/statistics/s-values/
Article Valid P -Values Behave Exactly as They Should: Some Misleadi...
https://www.ime.usp.br/~patriota/p-value.pdf
Article Semantic and cognitive tools to aid statistical science: Rep...
PS: you would rarely think of calculating proton concentrations from pH values, or intensity ratios from OD values, or intensities from (field- or power-)levels.
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