If sample standard deviation is biased, why do we use it in typical mean tests?
Why does this excerpt say that unbiased estimation of standard deviation usually isn't relevant?
What do 'real', 'user' and 'sys' mean in the output of time(1)? $ time foo real        0m0.003s user        0m0.000s sys         0m0.004s $   What do 'real', 'user' and 'sys' mean in the output of time?  Which one is meaningful when benchmarking my app? Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ I was messing around with the definition of the derivative, trying to work out the formulas for the common functions using limits. I hit a roadblock, however, while trying to find the derivative of $e^x$. The process went something like this:  $$\begin{align} (e^x)' &amp;= \lim_{h \to 0} \frac{e^{x+h}-e^x}{h} \\ &amp;= \lim_{h \to 0} \frac{e^xe^h-e^x}{h} \\ &amp;= \lim_{h \to 0} e^x\frac{e^{h}-1}{h} \\ &amp;= e^x \lim_{h \to 0}\frac{e^h-1}{h} \end{align} $$  I can show that $\lim_{h\to 0} \frac{e^h-1}{h} = 1$ using L'Hôpital's, but it kind of defeats the purpose of working out the derivative, so I want to prove it in some other way. I've been trying, but I can't work anything out. Could someone give a hint? Node access based on view mode I downloaded a file from the internet (I can link to it here, but in the interests of not annoying other users, I won't post the link), and I can't delete the file with the message:   You need permission to perform this action  You require permission from the computer's administrator to make changes to this file   Trouble is, I am the administrator and UAC is completely off.  If I go into the Properties/Security tab, I get the message   You do not have permission to view or edit this object's permission settings   Clicking "Advanced"/Owner, it says:   Current owner: Unable to display current owner.   If i try to change the owner, I get the same message:   Unable to set new owner on file  Access is denied. Why GCP Kubernetes Node host location for europe-north1 seems to be in US? Why do we have an elementary charge $e$ in physics but no elementary mass? Is an elementary mass ruled out by experiment or is an elementary mass forbidden by some theoretical reason? I am currently writing an academic text and need to directly quote/cite text from a paper. What is best practice when I need to add a remark/note to the quote?  For instance, I would do it like this: The authors claim that "we do the best work ever in the area of CS (note: Computer Science)".  So in the above case the actually quoted text is "we do the best work ever in the area of CS", however, I want to add a note which explains the abbreviation "CS".  How should this be done?  Related questions: Find orphaned users In SQL Server 2005, is there a way to find users that either don't exist at the server level (an account that was deleted at server level but wasn't disassociated from databases before it was deleted) or accounts that aren't linked (an account may have been deleted at the server level but not db level, then readded but the db level was never cleaned up).  I've got a very messy server and it would be awesome if there was a query to run to find these. I want to have a page-filling and -breaking longtable, so I decided to try xltabular. Unfortunately xltabular seems to screw up the booktabs rules in the normal tabular environments, see:  \documentclass{article} \usepackage{booktabs, xltabular} \usepackage[active,tightpage]{preview} \PreviewEnvironment{tabular} \begin{document} \begin{tabular}{ccc}     \toprule     col1 &amp; col2 &amp; col3 \\     \midrule     row 2: &amp; more and moooooooooooore &amp; test \\     \bottomrule \end{tabular} \end{document}     Not loading xltabular resolves the issue.  Did I miss some package option or some incompatibilities? Is there a better package than xltabular which doesn't break booktabs?  (I am using TeXLive 2019 and compile with lualatex)    EDIT (by Skillmon)  I reduced the bug to the following MWE (which doesn't make much sense on its own), it looks like booktabs tries to be smart detecting something, but fails badly.  \documentclass{article} \usepackage{longtable} \let\xltabular\relax \usepackage{booktabs}  \begin{document} \begin{tabular}{ccc}     \toprule     col1 &amp; col2 &amp; col3 \\     \midrule     row 2: &amp; more and moooooooooooore &amp; test \\     \bottomrule \end{tabular} \end{document}   It seems to suffice that longtable is loaded and \xltabular is defined (even if it's just \relax). Until now, I have been compiling my LaTeX documents with pdfLaTeX. My standard document preamble includes the "usual suspects" for non-English speakers (, , ) and a bunch of other packages. Assuming I were to convert to XeLaTeX, what modifications of my preamble are advisable? I'm aware that the expansion=true option of the  package is not (yet) available for XeLaTeX, and that  is sort of a default package for OpenType fonts. What other packages (and package options) should be removed and, vice versa, incorporated when switching from pdfLaTeX to XeLaTeX? Every finite subset of $\mathbb R$ contains its supremum (and its infimum)   Proof Let $A=\{a_1,...,a_n\}$ be a finite subset of $\mathbb{R}$. Since it is non-empty and it is bounded ($\max A$ is an upper bound), it has supremum, that is $\exists \sup A$ and by definition $\forall a \in A \;\, a \leq \sup A$. Let's suppose that $\sup A  \not\in A$ then, since $\max A \in A$ we have that $\max A &lt; \sup A$. But considering that $\mathbb Q$ is dense in $\mathbb R$ we can conclude that $\exists r \in \mathbb Q$ s.t. $\max A &lt; r &lt; \sup A$, but this is absurd since $r$ is an upper bound of $A$ and it is lower than the supremum. Necessarily, $\sup A\in A$.  Is there anything wrong? Is there any way to prove this without using density of $\mathbb Q$ or another property? Thanks in advance. I have a WordPress install on a dedicated server with decent memory (upwards of 8GB).  The WordPress install includes many plugins - 50 active plugins. Most important to note is the  plugin to make WordPress multilingual. The theme is .  The stack is LAMP - CentOS 6.3, Apache 2.4.4, Percona MySQL, PHP-FPM.  I have been quite stupid, and didn't set up a test or staging server.  Everything was running fine till I updated to WordPress 3.8, and also updated several plugins at the same time.  I didn't pay attention at the time, but all my widgets got messed up, and so did any sort of navigation etc.  I enabled error_log for a bit, i.e.  ini_set('log_errors',TRUE); ini_set('error_reporting', E_ALL); ini_set('error_log', dirname(__FILE__) . '/error_log.txt');   This resulted in an immense file - basically the error_log grows by about a GB every 4 minutes or so.  The primary error I found in the error log was database server disconnect, better known as      MySQL server has gone away   The first priority was to patch the problem. So I installed  plugin. This mostly, though not completely solved the issue. Many of the widgets came back.  However, the special (non-English) characters are all messed up (converted to question marks). I tried to fix this by opening wp-mysql-ping.php and commenting out   if ( !DB_CHARSET &amp;&amp; version_compare(mysql_get_server_info($this-&gt;dbh), '4.1.0', '&gt;='))                                     {                                         $this-&gt;query("SET NAMES '" . DB_CHARSET . "'");                                     }   Didn't help.  Besides, the plugin is not a permanent or complete solution. So leaving the site in this messed up state, I started working on other possibilities:  As suggested at , I edited my.cnf to add  wait_timeout = 60   Didn't help.  Also added caching to my.cnf  query_cache_type = 1 query_cache_limit = 8M query_cache_size = 512M   Didn't help.  (and yes, I restarted mysql every time I edited my.cnf)  Also tried the method : In wp-db.php before  $this-&gt;select( $this-&gt;dbname, $this-&gt;dbh );   I added  $this-&gt;query("set session wait_timeout=600" );   Didn't help.  I'm at my wit's end. Please help! singlespacing in table with no indent color code in SharePoint calendar I want to Setup GIMP 2.10 on Ubuntu 16.04 (Xenial Xerus) without flatpak.  can it be possible now or on future?? Is it possible to resize an image in an answer? For example,  has  that should have the image in the answer, but it's too big.  Is there a way to handle resizing? I want to build an uncountable well-ordered set within the theory $Z^\textbf-$. So, I take $A=\omega$ (exists by infinity axiom) and define $$W:=\{(X,R)\in\mathcal{P}(A)\times\mathcal{P}(A\times A):\langle X,R\rangle\ \text{is a well-ordered set}\}$$  With this I consider the set $T:=W/\cong$ (where $\cong$ is isomorphism relation). Note that $W$ and $T$ are sets by power set axiom. So I define for each equivalence class $[x],[y]\in T$ the order  $$ [x]\leq_T[y]\Leftrightarrow \text{type}(x, R_x)\leq \text{type}(y,R_y) $$  Here, $R_x$ means the order of $x$ that make it belongs to $[x]$.   So, $T$ is well ordered by $\leq_T$ and it is uncountable.  Question: I don't pretty sure if I can build $\leq_ T$ without replacement axiom. Another doubt is, Can I take the $R_x$ without AC? What the #$@&%*! is that called? Consider repeated measures ANOVA (RM-ANOVA) with one within-subject factor A and several measurements per subject for each level of A.  It is closely related to two-way ANOVA with two factors: A and subject. They use identical decomposition of the sum of squares into four parts: A, subject, A⋅subject, and residual. However, two-way ANOVA tests the effect of A by comparing SS of A with the residual SS, while RM-ANOVA tests the effect of A by comparing SS of A with the A$\cdot$subject interaction SS.   Why the difference?    Does this difference automatically follow from the repeated-measures structure of the data, or is it some convention?  Does this difference between two-way ANOVA and RM-ANOVA correspond to testing two different nulls? If so, what exactly are they and why would we use different nulls in these two cases? Two-way ANOVA's test can be understood as an F-test between two nested models: the full model, and the model without A. Can RM-ANOVA be understood in a similar way?   (If there is only one measurement per subject for each level of A, then the distinction sort of disappears because A$\cdot$subject and residual variation cannot be disentangled: )    Demonstration  I will use toy data d2 generated in . The same webpage shows correct syntax for RM-ANOVA.  # Discarding between-subject factors and leaving only one within-subject factor d = d2[d2$Xb1=='CG' &amp; d2$Xb2 == 'f', c(1,4,6)]   (See reproducible version .) Data look like that:       id Xw1     Y 1    s1   A  28.6 2    s1   A  96.6 3    s1   A  64.8 4    s1   B 107.5 5    s1   B  77.3 6    s1   B 120.9 7    s1   C 141.2 8    s1   C 124.1 9    s1   C  88.0 10   s2   A  86.7 ...   Here is two-way ANOVA: summary(aov(Y ~ Xw1*id, d))               Df Sum Sq Mean Sq F value   Pr(&gt;F)     Xw1           2  95274   47637  16.789 3.73e-07 *** id           19  31359    1650   0.582    0.913     Xw1:id       38  71151    1872   0.660    0.929     Residuals   120 340490    2837                    Here is RM-ANOVA: summary(aov(Y ~ Xw1 + Error(id/Xw1), d))  Error: id           Df Sum Sq Mean Sq F value Pr(&gt;F) Residuals 19  31359    1650                 Error: id:Xw1           Df Sum Sq Mean Sq F value   Pr(&gt;F)     Xw1        2  95274   47637   25.44 9.73e-08 *** Residuals 38  71151    1872                       Error: Within            Df Sum Sq Mean Sq F value Pr(&gt;F) Residuals 120 340490    2837               Note the identical SS decomposition, but two-way ANOVA tests Xw1 against the residual, while RM-ANOVA tests Xw1 against the Xw1:id interaction.  Why?  This question is related to . If we try using Error(id) instead of Error(id/Xw1) in the example above, then Xw1 will get tested against Xw1:id interaction lumped together with the residual variation.  (The same issue arises in factorial RM-ANOVA with multiple within-subject factors, where each factor or interaction is tested against its own "error term" aka "error stratum". These error strata are always given by the corresponding interaction with the block/plot/subject variable id.) Conditional clause (then) In this first instance the main clause only stands alone if you  omit (then).  But in the other example it clearly has a main clause. So would this be classed as an embedded main clause as it only works as a independent clause if it's adjusted and or if it's joined with a dependent clause.  If I have enough money, then I will buy a new car.  If it rains, the picnic will be cancelled.  The picnic will be cancelled if it rains.