I am running a logistic regression in SAS where the outcome variable is a binary on sale success (IVs are: location of the customer, Vendor, handling time, experience level of sales person etc.). Either an offer was applied (sale) or it wasn't (no sale). However, the cases with the event of a sale amount to 500 and the one with no sale amount to 10k. Is that a problem? And if it is, how does site affect my estimates/odd ratios and what are possible solutions?
Okay, so I think I have a decent enough sample, taking into account the 20:1 rule of thumb: a fairly large sample (N=374) for a total of 7 candidate predictor variables.   My problem is the following: whatever set of predictor variables I use, the classifications never get better than a specificity of 100% and a sensitivity of 0%. However unsatisfactory, this could actually be the best possible result, given the set of candidate predictor variables (from which I can't deviate).   But, I couldn't help but think I could do better, so I noticed that the categories of the dependent variable were quite unevenly balanced, almost 4:1. Could a more balanced subsample improve classifications?
I want to save the list of packages that have installed. I figured it might be the best to list all and diff it with the default set.  Edit: Clarification: I don't want to get the list of installed packages that are installed at the moment. Rather I want to get the list of packages that I added since the default install. Facebook must have changed something again because I cannot find out how to view the list of subscribers.  In Googling for this, I came upon this question:    However, it doesn't take me to my list os subscribers.  Where do I view this list now?  Please include a screenshot and highlight the link I need to click on.  This way, when Facebook changes their layout again, your screenshot will at least clue future inquirers in on this fact so they know they're not going crazy.  Edit  Here's what I see when I log into Facebook: Is there a way to find the number of accepted answers you have, other than manually counting each one? Problems with pie charts There seems to be in increasing discussion about pie charts.   The main arguments against it seem to be:   Area is perceived with less power than length. Pie charts have very low data-point-to-pixel ratio   However, I think they can be somehow useful when portraying proportions. I agree to use a table in most cases but when you are writing a business report and you've just included hundreds of tables why not having a pie chart?  I'm curious about what the community thinks about this topic. Further references are welcome.  I include a couple of links:         In order to conclude this question I decided to build an example of pie-chart vs waffle-chart. Uninstalling Homebrew messed up git I uninstalled Homebrew using .  After running the above commands, I can't find .bashrc and .bash_profle. All files are gone from Documents , Desktop and Downloads :'(  When I do git status on /Users/ava it shows:  # On branch master # Untracked files: #   (use "git add &lt;file&gt;..." to include in what will be committed)  ... all the files which are in /Users/ava   history   113  cd `brew --prefix`   114  rm -rf Cellar   115  brew prune   116  rm `git ls-files`   117  rm -r Library/Homebrew Library/Aliases Library/Formula Library/Contributions   118  rm -rf .git   119  rm -rf ~/Library/Caches/Homebrew   120  sudo chown -R $USER:admin /usr/local   121  ruby -e "$(curl -fsSL https://raw.github.com/mxcl/homebrew/go)"   122  brew doctor   What just happened? Determine if $\sum_{n=1}^\infty \frac{n+4^n}{n+6^n}$ converges This code in JS gives me a popup saying "i think null is a number", which I find slightly disturbing. What am I missing?    if (isNaN(null)) {   alert("null is not a number"); } else {   alert("i think null is a number"); }     I'm using Firefox 3. Is that a browser bug?  Other tests:    console.log(null == NaN);   // false console.log(isNaN("text")); // true console.log(NaN == "text"); // false     So, the problem seems not to be an exact comparison with NaN?  Edit: Now the question has been answered, I have cleaned up my post to have a better version for the archive. However, this renders some comments and even some answers a little incomprehensible. Don't blame their authors. Among the things I changed was:   Removed a note saying that I had screwed up the headline in the first place by reverting its meaning Earlier answers showed that I didn't state clearly enough why I thought the behaviour was weird, so I added the examples that check a string and do a manual comparison. Why did Slytherin House not get disbanded? Find $\lfloor {\alpha}^6 \rfloor$ If $\alpha$ is a real root of the equation   $$x^5-x^3+x-2=0$$   find the value of $\lfloor {\alpha}^6 \rfloor$.      This one totally stumped me. We are asked to calculate $\lfloor {\alpha}^6 \rfloor$ without actually calculating the root or using wolfram alpha or any other calculator. I found that the above equation has only one real root by sketching its graph using calculus. I was also able to use the intermediate value theorem to conclude that $1&lt;\alpha&lt;2$ , but this is of little use while calculating $\lfloor {\alpha}^6 \rfloor$. Please help! How to prove that minimum of two exponential random variables is another exponential random variable? How can I prove that the minimum of two exponential random variables is another exponential random variable, i.e. Z = min(X,Y) The limit of a complex integral If $f(z)$ is continuous in the neighborhood of the point $z=a$, then $$\lim_{r\to 0^+} \int_{\vert z-a\vert =r} \frac{f(z) dz}{z-a}=2\pi i\,f(a)\tag{1}$$  In my desperate intents of proving the above exercise, I tried to prove a 'particular' case of it, which is when $z=0$.   So if $z=0$ and as $f$ is continuous on a neighborhood of zero, given $\epsilon&gt;0$, there exists $\delta&gt;0$ such that $|f(re^{i \theta})-f(0)|\le \epsilon$ for $r\in[0,\delta)$ and $\theta\in\mathbb R$. Hence,  $\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right| &lt;2\pi\epsilon \Rightarrow \limsup_{r\to0}\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right| \le2\pi\epsilon.$  If $\epsilon\to0$, then $\limsup_{r\to0}\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right|=0 \Rightarrow \lim_{r\to0}\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta=0.$  Is this proof correct?  How can I make the proof for $(1)$ following the 'particular proof'? Can someone help me? I wonder if it is possible to consider $\mathbb R$ as a Cartesian product $X\times X$ for some set $X$. From the point of view of the dimensionality, there are spaces with a Hausdorff dimension $1/2$ (sort of Cantor sets), but I guess there are other problems in this construction  unrelated to the dimension.  Edited: replying on the comment by Asaf. I want that for all $r\in\mathbb R$ there exists a unique representation $r=\langle x,y\rangle$ where $x\in X$ and $y\in X$. Also, what if we want to have $X$ to be a topological space and $h:X\times X\to\mathbb R$ to be a homeomorphism? How to 'sum' a standard deviation? I have a monthly average for a value and a standard deviation corresponding to that average. I am now computing the annual average as the sum of monthly averages, how can I represent the standard deviation for the summed average ?  For example considering output from a wind farm:  Month        MWh     StdDev January      927     333  February     1234    250 March        1032    301 April        876     204 May          865     165 June         750     263 July         780     280 August       690     98 September    730     76 October      821     240 November     803     178 December     850     250   We can say that in the average year the wind farm produces 10,358 MWh, but what is the standard deviation corresponding to this figure ? Using `M` instead of `k` when rep is >= 1M on https://chat.stackexchange.com/users I read on  ()    Shouldn't it be 1.03m or (if we really follow the conventions of the metric system) 1.03M instead? Ubuntu 14.04 Wireless not working: no network interface. Atheros [168c:003e] device I just installed Ubuntu 14.04.3 LTS after installing windows 10 in my new VN7-571G and the wireless is not working. It never had: even during the installation it was not recognized. I run a script to gather the data , which is . lspci -knn | grep Net -A2 output 03:00.0 Network controller [0280]: Qualcomm Atheros Device [168c:003e] (rev 20)         Subsystem: Lite-On Communications Inc Device [11ad:0804]  The OS behaves like it has no wifi card. I do not see wifi settings in network settings I'm not a linux power user, so I do not know what to do to solve the issue. What should I do to make it work? I have not idea how to approach this differential equation.   $$y'-\int_0^xy(t)dt=2$$.  Basically, I did,  $$F''(t)-F(x)+F(0)=2 \;\;\;\;\;\;\; F'=y$$  I am stuck.  Thank You. Downloading from Play Store gives error 491 When I try to download anything from the Play Store it shows error 491 and can't finish the download. I have a Samsung Galaxy S 7582. How do I fix this? If $f:U\to V$ is holomorphic and $f'(z)\neq 0$ for all $z\in U$, then$f$ is locally bijective. I am trying to solve the following problem:     Let $f:U\to V$ be a holomorphic function such that $f'(z)\neq 0$ for all $z\in U$. Show that for all $z_0\in U$, there exists a disc $D_\varepsilon(z_0)\subseteq U$ such that $f:D_\varepsilon(z_0)\to f(D_\varepsilon(z_0))$ is bijective.   Atempt:  I am trying to use Rouche's Theorem, but I am stuck on a particular step. Let $z_0\in U$. Since $f'(z_0)\neq 0$, there is a disc $D_r(z_0)\subseteq U$ such that $$f(z)=f(z_0)+(z-z_0)h(z),\quad\forall z\in D_r(z_0)$$ where $h$ is holomorphic on $D_r(z_0)$ and $h(z)\neq 0$ for all $z\in D_r(z_0)$.   Let $0&lt;\varepsilon&lt;r$ to be defined later and fix $w\in D_\varepsilon(z_0)$. To show that $f$ is injective in $D_\varepsilon(z_0)$ is to show that the function $f(z)-f(w)$ has exactly one zero in $D_\varepsilon(z_0)$. But $$F(z):=(z-z_0)h(z)$$ has exactly one zero in $D_\varepsilon(z_0)$, so we might want to apply Rouche's Theorem with $F$ and $$G(z):=f(z)-f(w)-(z-z_0)h(z)=f(z_0)-f(w)=-(w-z_0)h(w)$$ to conclude that $F$ and $F+G$ have the same number of zeros and hence $f$ is injective. But that means that we need to find $0&lt;\varepsilon&lt;r$ such that $|G(z)|&lt;|F(z)|$ on $\partial D_\varepsilon(z_0)$. That is, $$|w-z_0||h(w)|&lt;|z-z_0||h(z)|$$ for all $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$.  Can we find such $\varepsilon&gt;0$?  It seems intuitive since if we expand $h$ in a power series at $z_0$, $h(z)=\sum_{n=0}^\infty a_n(z-z_0)^n$, then $$|w-z_0||h(w)| = |w-z_0|\sum_{n=0}^\infty a_n|w-z_0|^n &lt; |z-z_0|\sum_{n=0}^\infty a_n|z-z_0|^n.$$ for $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$. But this is of course not a proof. Why do physicists think that the electron is an elementary particle? When may adjectives precede determiners? (E.g. too difficult a task) The expression too difficult a task sounds a little pompous, but it doesn't sound ungrammatical. According to my folk-beliefs about English grammar, determiners precede adjectives. However, the expression too difficult a task seems to controvert that belief.  Additionally, the expression too difficult of a task seems grammatical too.    Under what conditions may determiners precede adjectives? Are there conditions under which too difficult of a task would be a more grammatical expression than too difficult a task would be?   (This question does not ask what the question “How big of a problem” vs. “how big a problem” asks. That question does not pertain to determiners.)