tttjjj/elt_bench huggingface dataset:
How large is the training set? What algorithms do you want to apply? What software are you using? Are the three features categorical or continuous; if categorical, how many values can they each take on? Finally, are your 0/1 cases relatively balanced or is one class relatively rare? All these details may help people give you more useful advice.
It is customary to 'accept' an answer by clicking on the checkbox next to the most useful answer (in this case, there is only one). This helps other find answers to their questions more quickly, as well as letting people know that your question has been resolved. Note that you can go back and do this for your other questions as well, by clicking on your name at the top of the screen.
A couple comments. (1) The results of your current model are not sensible. Notice in the model output that all of the data are contained in a single "group", namely the Phase==F group. What is happening is that all the rows with missing data on the Intervention variable are still being dropped. (2) It is not clear that this is a multilevel problem in the first place. Although you have two "groups" of observations (baseline and non-baseline), considering groups to be a random factor is statistically and conceptually dubious, to say the least.
thank you for explanation
@JPT - I added my response above as Edit 2.
The answer depends on what "other analysis" you have in mind. Even when 10,000 values very closely approximate a Normal distribution, a distribution test (like the K-S or S-W) will reject the hypothesis of Normality, even though the data are *beautifully* behaved for other analyses. Perhaps you could edit this question to indicate the nature of the follow-on analyses you have in mind. In the meantime, searching our site will turn up dozens (if not hundreds) of discussions of this issue.
Do you have an answer to this yet?
Thank you for such an intuitive reply.
Do you really need to update centroids with every new patient arrival?
Welcome to our site! We would be honored if you would register your account and come visit once in a while. One small housekeeping note: You don't have to sign your posts, your identicon, with a link to your userpage, is automatically added to every answer you give.
I think you will need to show us the exact commands you used so that we can verify you applied them correctly. That will also help us determine whether this question needs a statistical answer or if it is only about how to use `R`, in which case it can be moved to a better forum on [SO].
It might also be good to mention that the gap between the frequentist and Bayesian approaches is not nearly as great on a practical level: any frequentist method that produces useful and self-consistent results can generally be given a Bayesian interpretation, and _vice versa_. In particular, recasting a frequentist calculation in Bayesian terms typically yields a rule for calculating the posterior _given some specific prior_. One can then ask "Well, is that prior actually a reasonable one to assume?"
My intuition says that it's ok, but here's a reference I dug up quickly that might help http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aos/1018031117
divide the domain of your non-bijective function into parts where the function is bijective and then apply change of variables. This might work.
@user023472 Suppose we have parameters $(\\theta_1, \\theta_2)$ and data $Y$. Gibbs sampling iterates between $p(\\theta_1 \\mid \\theta_2, y)$ and $p(\\theta_2 \\mid \\theta_1, y)$; if one cannot sample exactly from either of these two distributions, Gibbs sampling will not be feasible. But we can always apply the slice sampler, so we can replace the draws from the intractable distributions with draws from slice samplers which target to intractable distributions.
It will be er(i) and how do I plot it.
Many thanks. My model suggests that $X$ causes $Y$ and $Z$, and I don't know whether $Y$ causes $Z$. I wasn't sure whether I could regress $Z$ on $Y$ and $X$, since $X$ causes $Y$. The way I read your answer is that, this is ok. 

The estimates from this regression suggests $Z$ is related to both $Y$ and $X$. Since, my model predicts that $X$ causes $Z$, I take the significant coefficient of $X$ as consistent with this prediction. Finally, if I get you right, the significant coefficient of $Y$ suggests a relationship, but the direction of the causality (if there is) is not known.
Thanks very much for the recommendations. A link would've been useful, although I admit [finding the pdf](http://www.lacbiosafety.org/wp-content/uploads/2011/09/experimental-design-and-data-analysis-for-biologists1.pdf) was easy enough.
@C. Ross Unfortunately I do not think so. But, I suspect that the recursion should not be that hard as long as are dealing with reasonably small n and small s. You could just build-up a lookup table and use that repeatedly as needed.
Try to be more concise, like this. Otherwise you have a small chance that someone would try to decipher the question, and thus a little chance for an answer.
How much data do you have to work with?
A [top Google hit](http://people.few.eur.nl/mmartens/mdh_021017.pdf) that is open to all explains the *mixture of distributions* hypothesis (MDH).
@Fernando Your model seems very very complicated and probably the cause of your problem. If you wish to post the data from one of your time series and the auto results , I will try and help you.
Yup, I'm using that, thanks :)
It's a matter of [selection bias](http://en.wikipedia.org/wiki/Selection_bias).
Hint: By [the law of the unconscious statistician](http://en.wikipedia.org/wiki/Law_of_the_unconscious_statistician), $$E[e^{Y^2/2}] = \\int_{-\\infty}^{\\infty} e^{y^2/2}\\frac{1}{\\sqrt{2\\pi}}e^{-y^2/2}\\,\\mathrm dy = \\int_{-\\infty}^{\\infty} \\frac{1}{\\sqrt{2\\pi}}\\,\\mathrm dy = \\cdots$$
Thanks! I think I'm going to need some pointers before trying to implement this.
1) scipy use this W to compute the p_value. Actualy the process is the same as decribed in Wikipedia, except for the computation of W. 2) How can I check this?
And this one: http://stats.stackexchange.com/questions/3331/is-it-possible-to-do-time-series-clustering-based-on-curve-shape
(+1) I am sure there is. This one is the first thing that came to my mind (and the approach I would actually use). Checking that the hands are uniformly drawn from the set of possible hands sounds more perfect, but probably with much less power.
This question appears to be off-topic because it is about how to do something in SPSS.
This question appears to be off-topic because it is about how to use r without a reproducible example.
(I have version 7.) I have no problem loading the Statistics package. But what's the function in there called? Because I get the impression that this `Quantile` line will do the calculation manually instead of using a formula.
Why the down-vote? (Fair enough if it's for the strained humour.)
+1 Thank you for offering an actual answer :-).
But if authors want to put out material on the internet for free that is their choice. We all can benefit from that.
I think the colnames line above needs to be rewritten as `colnames(df)[1:2] <- c("variable","value")`?
The calculator was *NOT* telling you that the sd was 0.92, it actually tells you that something else is about 0.92 (the standard error of the difference).
@JandR I rechecked it today; see the corrections
You will need to add some representative values here; people aren't going to want to click a link and download an unknown file. I tried to edit your question, but it still isn't very clear. Can you clarify it further?
Could you also give us the output of `str(dataset)`
As I see it, there are four problems, 1) the choice of smoother (a LO(W)ESS smoother seems more suited), 2) the choice of bandwidth, 3) the autocorrelation, and 4) the use of prediction intervals instead of the more appropriate tolerance intervals. I would instead fit a *robust version* of LOESS, and just select reasonable limits by hand. Or *if* the resulting autocorrelation (excluding outliers) has a simple form, e.g. AR(1), take this into account when calculating the standard errors.
I am already using the ChiSquare (χ2) Test for the distinct (nominal) features.
But it cannot be applied to numeric data, can it?
Can you be more specific regarding how the new regression might differ from the old one? Eg, are you wondering if the slope of the relationship b/t X & Y differs by species, if the intercept does, or if both do? Something else?
"vanilla" seems an odd word to use about R; not clear on the translation in more usual R terminology. Any distinction between base R and extra contributed packages from CRAN really doesn't bite experienced users, or even novice ones, as they are equally free and equally accessible. My impression is that someone who had access to SPSS can say fairly easily that something isn't possible in SPSS at present without programming; to say the same about R requires familiarity with all the time series packages.
A couple points: 1) OLS regression does NOT require that the DV be normally distributed, it requires that the error, as estimated by the residuals, be normally distributed. 2) Look into Box-Cox transformations (lots of info on the web).
I encourage you to include them even if they only go in an online appendix.  Our transparency should no longer be limited by the space constraints of a journal!  In that spirit, I'd encourage you to consider going further and include a matrix of scatterplots, or grouped box-plots, etc.
And i can make a consensus prediction, like..
If the precision of Exp 1 = 95% and exp 2 = 95% the final precision is 97,5%.
If final precision is 97,5% and a third experiment have a precision of 95%, The final precision becomes 98,2%, and so on ...
Binary search is a searching algorithm, not a sorting algorithm. Your question is if there are "sorting algorithms" with sub-linear time (specifically $O(logn)$). The answer is no.
Thanks Peter. So an estimate with lower margin of error is more 'precise'. Is there a way to also know how biased it is from just X% +/- x1% error margin? I guess that's not possible unless you know individual sample's preferences, right?
Let me add to Greg Snow's answer that, doing pairwise comparisons, you ought probably to recognize psychological exponentiality of ranking. Usually, drop from 1st place to 2nd is steeper than from 2nd to 3rd, and so on. So, in ranking D,B,C,A you might let distance D-B be 1, B-C be, say, 0.5, C-A be 0.25.
Complicated territory, but if you want or need a test, go with the existing machinery; don't invent a new procedure ad hoc. (For example, successive quantiles are necessarily dependent.) Naturally, almost nothing is guaranteed to work well with small samples.
Not sure what's 'only' about being a PhD student, that's usually hard work.
Asking if some data “are significant” is not a well-defined problem. If you are doing a test at all (it's not the only way to analyze quantitative data), you would typically test a specific hypothesis, perhaps compare several subgroups or test if there is a particular relationship between several variables. You need to tell us what you are trying to find out in this study.
One small addition - the null hypothesis should actually mean something outside the context of the current data analysis for standard HT to apply.  It shouldn't be "invented" so that you have something to reject in favour of your theory/finding.
I just wanted to add a side comment - from what I've seen, often in non-parametric literature the curse of dimensionality is 'ignored' by assuming that the number of derivatives the target function has grows with the dimension. This obviously is a pretty huge assumption to make. Thanks for the corrections to my fortmatting user603.
Did you really mean to say "many residuals" and not "large size of residuals" or something like that? After all, by definition, there will *always* be exactly as many residuals as there are data being fitted. I am also curious what you mean by a "scatter plot for the whole model," given that at least five variables are now in evidence: three predictors, a predictand, and the residuals. Which two of those do you propose to show in the scatterplot?
The univariate median has a lot of desirable properties. It has many generalization to higher dimensions as well (which is what you are really after here). None of these generalization combine all the advantages of the univariate median, however. Depending on what property of the univariate median you are interested in, we could suggest a multivariate estimator with said property.
Can you clarify what you mean by problematic post-treatment confounders? I thought injury and the like were outcome variables. Also remember that confounders are those variables which affect selection into the treatment.
(see answer below for clarification)
I think any markov chain with $\\alpha <1 $ will struggle as there are multiple modes that need to be visited and this gibbs sampler just isn't able to find them by making "local"  moves.  Better to control the number of topics with T directly
This theory is incomplete and depends on parameter ordering, choice of compact region, and the likelihood function.  So it doesn't obey the likelihood principle for example.  Also, it is difficult to apply to non-independent data.  Further, Bernardo's theory is only complete for 1-d parameter problems.  It is probably the best method available currently though.  A good competitor is the transformation group approach of Jaynes.
@levon9 - I was able to generate all possible subsets using 50 variables in SAS.  I do not believe there is any hard limitation other than memory and CPU speed.\\n\\n-Ralph Winters
"use estimated se"... for what purpose? Without some context about goals, your question cannot be answered.
What is the measurements were much closer together?
thanks wayne. I will reinterpret my data using a Poisson model. I think that would be a better option. @Irish answer seems good but I'm afraid it surpasses my knowledge by a lot. At least I know a bit about Poisson.
What do you mean by "[...] predict the **collectability of debts of a given debtor**" ? Is this variable binary?
Using weights with OLS? Heteroskedasticity could well be an issue.
Hi @Demian, I'm not sure that even a boxplot will be helpful when one group has a sample size of 2. Otherwise, yes I think boxplots in particular are very helpful in visualising continuous data across groups.
The situation is not quite clear from your description. Can you add some more specific details please, both of the variables and how they enter you model or models?
Probably better than using Cairo to produce a bitmap image is to save graphs as vectors, which are resolution independent, so look better when resized, and always suit print resolution - I like svg, which can easily be edited with [inkscape](http://www.inkscape.org).
Cross-validation will tell you that you paid a price for model uncertainty.  IMHO it is better to use a semiparametric model to completely avoid the $Y$-transform issue, and use a regression spline for $X$ and pay for $X$-transformation uncertainty according to the degrees of freedom in the spline function.
(1) If you use an R package, at least tell us which one you used. Is this the `lsqnonneg` function in `pracma` or something else? (2) Even if your solution were determined by your equations, you still wouldn't expect identical sums of squared residuals, since the two won't by default use the same stopping criteria, nor likely even the same algorithms.
Note that there is *no* difference between the uniform case and the general (absolutely continuous) case. Though it deals with the *two-sample* version, some hints as to why this is true can be found in the comments [here](http://stats.stackexchange.com/questions/17495/kolmogorov-smirnov-2-sample-p-values). That is, the K-S statistic is invariant under the transform $F$.
You may also want to check out kernlab and caret for SVMs.  They're interesting (thought not necessarily better) alternatives.
Ok, thanks. So whenever it's not 1 and there's only one x the probability of that to happen is 0, hence the result from gtool. Did I get this right?
Thanks for your thoughts (+1). I was using the uniform prior and I think I will keep that... My real problem is more complicated that the one exposed here, I have correlations between the $\\alpha_i$'s. This "random walk prior" in intriguing, I’ll have a look.
Please spell GLM in full right in your question
It is zero because the "degrees of freedom" in the two models are the same. In the case that this happens just pick the one with the smallest AIC/BIC/RSS/ what have you. If you believe they shouldn't be the same, check your model specifications and see why they come out looking the same. You practically do a log-likelihood ratio test with `anova()` but I don't think you can access effectively over-fitting within a single model with it. Can you provide code reproducing your problem?
The replicate measurements give you the *Type A uncertainty* whereas the specs from the manufacturer give the *Type B uncertainty*. The two are combined to give the *combined standard uncertainty* for the measurement. The Wikipedia entry on "Measurement Uncertainty" gives an overview of this process and gives references to the relevant ISO Standards.
Also closely related: http://stats.stackexchange.com/questions/166, http://stats.stackexchange.com/questions/5158
+1.  The remark about final presentations brings to mind, as a notable counterexample, Hans Rosling's famous 2006 TED talk (http://www.ted.com/talks/hans_rosling_shows_the_best_stats_you_ve_ever_seen.html ).  Re: the point about examining "far more," I am reminded of how a lawyer in a deposition asked me about how I had examined the data that backed up my testimony and how her face fell when she learned the work had been done interactively and therefore nothing was printed or saved (which she could then subpoena, examine, and try to impugn). ;-)
The most important thing to learn is WHEN to draw ROC curves. 95% of the examples that I have seen are unnecessary and do not represent the best approach to the problem.
Sounds like you're after some sort of [diversity index](http://en.wikipedia.org/wiki/Diversity_index)
What is your noise model? Presumably your $X$ values are fixed and known, rather than random. If the noise is Gaussian, e.g., you know the distribution of the test statistic exactly under the null hypothesis and for each fixed value of the parameter under the alternative. Everything should be able to be written pretty explicitly under such assumptions.
I asked for specifics because it depends on algorithm, purpose and usage. Inference most certainly needs assumptions, simple prediction may not. Also there are many ways to estimate a linear model. What do you mean by "does everything work"?  Furthermore, why do you have the impression that "people are not worried"?
Data programming issues are off-topic on this site and belong on SO, see our Meta suggestion on [How to ask question related to the use of R?](http://meta.stats.stackexchange.com/q/1335/930) What statistical method (e.g., multiple imputation) to choose in a particular context would be ok there. In either case, however, you'll need to focus on specific and answerable issues ("please, do this task for me" does generally not fall under this category). If you can focus on specific question, we will be happy to answer or migrate to Stack Overflow.
Actually, we *can* put a minus sign on the result from step 1... note that $-(\\theta-1) = (1-\\theta)$, and you can cancel out the exponent in the denominator's $(1-\\theta)$.  (It happens to the best of us sometimes.)
Consumer psychology can play an important part in this type of analysis. Back to spatial factors, is there anything about the roading (e.g. traffic patterns, congestion, roundabouts vs. uncontrolled intersections vs. traffic lights) that could help explain?
This assertion is false unless a particularly narrow interpretation of "symmetric" is adopted (and even then @Cardinal's admonition must be heeded): see http://stats.stackexchange.com/a/29010 for more about this. For *familiar* counterexamples in Cardinal's spirit you may contemplate what happens with any Student $t$ distribution with $\\nu$ degrees of freedom and you use any odd power greater than or equal to $\\nu$.
Can you address part (c) in my comment above? More specifically, this looks like a standard textbook question, and as such it probably should have the `self-study` tag. We treat those question [differently](http://stats.stackexchange.com/tags/self-study/info), providing hints in return for a clear explanation of what you've already tried and where you got stuck. If this is for some subject, or even just an exercise for personal study, you should add the tag.
Naively, it appears that Y_(t-1) could be moved into an expanded theta_t and with an expanded Phi (include F, essentially), could keep updating the Y_(t-1), as a seasonal ARIMA would work when translated into DLM... Except that it would not include e_t, while the actual Y_t does. :-(
Can you elaborate a little more on what your concerns are? For example, why are you looking at watermarking versus, e.g., calculating and storing robust checksums of the data?
I am using power in that sense.  I realized after I wrote this that one generally thinks of power calculations as frequentist, and many ML techniques have Bayesian analogues.  So maybe power calculations don't make any sense in this world and I should just classify a few times and see whether the results are good enough.
Could you please tell us what you would have expected to see after running your code?  What exactly were you hoping to achieve?
@ArunJose Your tree uses a single predictor variable? It looks like this. I managed to show that after scaling the best split for the same variable, scaled and unscaled is the same. The question is how scaling affects the interactions between variables. I still consider that not a proof. And again, impurity functions like gini and brothers are not used in regression trees.
In SPSS you can go to Help...Tutorial or Help...Index (and once you've chosen a topic, click Show Me to get the tutorial for that topic).  A couple of hours spent on this should answer most of your questions in a visual and/or verbal way.
@Jacob--The problem is that RF has lots of decision trees. I'd love to report a single formula (< a few lines long if possible) that predicts nearly as accurately as RF. Since I'm publishing my work to an audience of modest statistical sophistication, I think exporting pages upon pages of trees would severely limit the probability of my findings becoming implemented in clinical settings.
I understand you have n problems grouped by difficulties. But what do you mean when you say “each benchmark”? By “interval averages” do you mean averages within a difficulty-group?
It looks like one answer to your question might be Fisher's exact test.  But I'm having trouble understanding your question. You use unusual phrases like "universe values" and I'm not sure what that means.  What do your data look like?
If you don't mind relaxing the R/python preference, may be this work of mine can help?  http://goo.gl/l7SLlB  Some of the advantages of this method are that you don't need to worry about types of features, normalizations and more.
Thank you very much for this insight.
I went through the papers again and none of them standardize their coefficients or manipulate them in any other way except for transformations of the variables as was described in the previous comment.
sorry for the unclear question. The dependent variable DV is CO2 production (values from 10 to 30 before the experiment has started; I should not have called these ID in the first place). Later, the DV is called y. In the experiment, I want to have one treatment group and one control group which is represented by the factor x. These two groups should be as equal as possible before the experiment starts. I will edit the question.