I have calculated p-values of two independent groups (each group contains 17 samples and each sample is 100 points). I used scipy's . Further, I combined the p-values using  as suggested by one of my colleagues. However, the p-values are just too small to make any sense to me. Here is a sample output:  import scipy.stats as stats p_value_list = [3.3629559156133476e-56, 1.966307600030748e-254, 6.3484089727582271e-103,                  3.1221165197874102e-92, 5.8797795864262639e-128, 4.0807369798923832e-130,                  1.205918004734663e-187, 8.9953478731039129e-19, 2.1492251770664035e-127,                  0.11623743456915718, 6.182421882338173e-13, 3.7920992534448722e-112, 3.5888837913096105e-97,                  7.5565994692959197e-134, 0.0026669650138866794, 4.0844220972198632e-16, 4.583685848455044e-95] results= stats.combine_pvalues(p_value_list) print (results)   Which gives the following output (p is zero!)  (7172.864625387464, 0.0)   Now, my questions are:   What is the meaning of small p-values (something like 1.2e-187)? How to interpret a p-value that is zero? Since I am writing a paper based on this calculation, how can one report this in a peer-reviewed journal?
For some tests in R, there is a lower limit on the  calculations of $2.22 \cdot 10^{-16}$. I'm not sure why it's this number, if there is a good reason for it or if it's just arbitrary. A lot of other stats packages just go to 0.0001, so this is a much higher level of precision. But I haven't seen too many papers reporting $p &lt; 2.22\cdot 10^{-16}$ or $p = 2.22\cdot 10^{-16}$.   Is it a common/best practice to report this computed value or is it more typical to report something else (like p &lt; 0.000000000000001)?
Is the Nintendo Switch region locked? Many Nintendo devices often come with , which means if you purchase a device in one country, you won't be able to play games from another (generally).  Is the Nintendo Switch also region locked like many of its predecessors such as the Wii U and Wii? I accidentally opened Orca Screen Reader from the Dash when I was trying to open Screenshot. Now everything I type or click on is spoken out loud.  How do I make it stop?  For reference, here is a screenshot of Orca Screen Reader running: I've already defeated the Big Bug.   I see the hint that the small bugs disguised as cursors are vulnerable to the three words "Charity", "Uganda", and "Bram".  I think I have all keys available, I've tried undoing with the bugs on the three words, I've tried motions which include the three words in their range, I have each of the three words stored in a named register, but I haven't figured out how to defeat the three small bug minions underground in Level 14 of Vim Adventures.  I suspect I should have defeated them before defeating Big Bug, maybe even before finding 'undo'. Is there instead something I should do in the 'lorem' buffer where I found the three cursors (Blue, White, and Pink) at the beginning of the level?  How does one defeat the three small bug minions in Level 14? How do I convert a string to a byte[] in .NET (C#) without manually specifying a specific encoding?  I'm going to encrypt the string. I can encrypt it without converting, but I'd still like to know why encoding comes to play here.  Also, why should encoding even be taken into consideration? Can't I simply get what bytes the string has been stored in? 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Express $x^8-x$ as a product of irreducibles in $\Bbb Z_2[x]$ Express $x^8-x$ as a product of irreducibles in $\Bbb Z_2[x]$.  Work so far:  $$x^8 - x = x(x^7 - 1) = x(x - 1)(x^6+x^5+x^4+x^3+x^2+x+1)$$  From here, I think the Zeros of an Irreducible over a splitting field Theorem would be the method. I am unsure how to proceed Do natural weapons count for two weapon fighting? Do natural weapons such as those granted by being a Tabaxi or an Aarakocra count for the purposes of two weapon fighting? Would they have the light property to allow it and if not what about if you take the dual wielder feat that removes the restriction on them being light?  From a logical perspective I would think claws or talons would be light enough to count and with them being "natural weapons" this means they are definitely weapons so it would make sense for them to be considered light and one handed. Let $A,B \in \mathbb{M}_n (\mathbb{C})$. If $A^2B + BA^2 = 2ABA$ then exist $k \in \mathbb{N}$ where $(AB-BA)^k = 0$.  I tried solve with minimal polynomial, but I did not have much effect. proving roots of equation $x^2+x+4f=0$ are real. Let $a,b,c$ be three positive real numbers such that $a+b+c=1$. Let $f=\min(a^3+a^2bc, b^3+ab^2c,c^3+abc^2)$. Prove that the roots of the equation $x^2+x+4f=0$ are real. In $ \triangle ABC$ show that $ 1 \lt \cos A + \cos B + \cos C \le \frac 32$ What is the difference between String and string in C#? Check if a services is installed using C Where can I find the points per game throughout many NBA seasons? Where can I find the points per game (of both teams) throughout many NBA seasons?  To illustrate my point, what I mean is something like:   2000/01 average points per NBA game: 200; 2001/02 average points per NBA game: 210; 2002/03 average points per NBA game: 205;   so on... Display the full URL of the home page? 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Evaluate $\lim_{n \to \infty} \frac{1}{n}[(n +1)(n+2) \cdots (n+n)]^{\frac{1}{n}}$ Evaluate $$\lim_{n \to \infty} \frac{1}{n}[(n +1)(n+2) \cdots (n+n)]^{\frac{1}{n}}$$  We can write as $\displaystyle\lim_{n \to \infty} \frac{1}{n}[(n +1)(n+2) \cdots (n+n)]^{\frac{1}{n}}$ $$= \lim_{n \to \infty}\left[\frac{(n +1)(n+2) \cdots (n+n)}{n^n}\right]^{\frac{1}{n}} = \lim_{n \to \infty}\left[(1+\frac{1}{n})(1+\frac{2}{n})\cdots (1+\frac{n}{n})\right]^{\frac{1}{n}}$$  I have used this result if $\{a_n\}$ is a sequence of positive term which is convergent to $l$ and $b_n = (a_1. a_2 . \cdots a_n)^{1/n}$ ,thhen $b_n$ converges to $l$  here $a_n = 2$, but i think it is not right,how to use this result in this question. Give me hint how to proceed . Thank you How to create custom config section in app.config? Riddle (simple arithmetic problem/illusion) I'm not sure how well known this "riddle" is but here it goes.  3 people go to a restaurant, each buy food worth 10.00.  When they're done, they give 30.00 to the waitress.  She gives the money to the manager, manager says they paid too much, gives 5.00 back to the waitress.  She comes back to the table, gives a dollar back to each person and then puts the remaining 2.00 in her pocket for tip.  So now, each person has paid 9.00.  9.00 x 3 = 27.00.  Plus the 2.00 in the waitresses pocket is 29.00.  What happened to the 30th dollar?  So what's the issue with this way of calculating (since if you backward it works fine), that doesn't give us the correct result?