tttjjj/elt_bench huggingface dataset:
We can't read your mind. If you don't tell us what you've got, all we can say is "yes, that formula is correct for sensitivity" and you can find one for specificity.

odds ratios, though, are not percentages and do not range from 0 to 100. They range from 0 to infinity
In your case, you know that the completion times cannot be negative, yes. So a linear model is maybe not appropriate? Try some other model, survival analysis maybe, or some GLM with a distribution on the positive real line (Gamma distribution, weibull distribution ...)
(ctd) ... You don't need even to be able to compute that distribution under the null, if you can simulate it. This leaves you with the opportunity to invent statistics with the properties you want in very specific situations. When their assumptions hold, LR tests tend to have desirable properties that make them worth knowing about.
@BR I guess the issue is how do you prove that the parameter $c'$ is zero rather than just close to zero?
I was thinking that this wouldn't be appropriate because the two would actually be highly correlated.  As the center is far away so will the near tend to be.
I can't answer it. Just can link to how SPSS (for instance) computes Savage ranks: http://pic.dhe.ibm.com/infocenter/spssstat/v20r0m0/index.jsp?topic=%2Fcom.ibm.spss.statistics.help%2Falg_rank_proportion.htm
That did it. Thank you very much.
Just set 'axes=FALSE' when you call '(box)plot' and then set your custom axes with 'axis' (look around the 'at' and 'labels' params).
This would benefit from a fuller explanation and also complete and precise references.
This isn't actually a research level question (it's not even a question).  Perhaps you should buy a book on the topic (I can recommend AIMA by Russell and Norvig), or at least read a Wikipedia article.  There is, by the, way not one hidden neuron in the hidden layer, that wouldn't do much.
It may not be quite as mathematical as you want, but I highly recommend Harrell's *[Regression Modeling Strategies](http://www.amazon.com/Regression-Modeling-Strategies-Applications-Statistics/dp/0387952322)*.
Consider a 2x2 matrix x with x[1,1] = x[2,2] = 1 and x[1,2] = x[2,1] = 100.  It is symmetric but not positive definite. The diagonal needs to be "big enough" so that you get a positive definite covariance matrix.  You could find an upper bound on your parameter c by trying out various values for c in R and checking the eigenvalues of the resulting matrix (eigen(va)$values); they should all be positive.  For good numerical properties, you don't want the ratio between the largest and smallest to be too big, either; less than 10^6 would be good.
Try centering the continuous variable, that often helps.
I meant to say Pearson Chi square, I've edited the post.
You cannot accurately transform one such model into the other: they express different relationships among the variables altogether.  You need to re-estimate the model based on the log data.
Can you make your question more concise?
Agreed -- it is folly to assign a probability to so-called Fourth Quadrant events such as this, or to even think we can predict them. All we can do is to make the system robust to their negative effects through redundancies etc.
OK, then it's important to disclose the criteria you have used to identify this sequence.  For instance, did it stand out among other sequences as having more variation in its frequency within the string?  It will also help to say something about how the string is generated, because serial correlation among the characters can account for such variation.
Without undue punning, much depends on what is meant by "exist" and indeed of "population". It is too late, but I would have asked your teacher what meaning is to be attached to inference with experimental data if no population exists. Even in simple cases such as tossing coins or throwing dice, the population is an abstraction rather than something that can be identified concretely.
Thank you for the point - I'll remember that in the future.
+1 to you and thanks for the compliment
Hey thanks! What happens if the variables do not have a bivariate normal distribution?
The mother of all nonlinear machine learning tools is probably neural networks.
If your data is so that k-means works reasonably well, a local minimum will be about as good as the global minimum. When you get very different results by running k-means multiple times, then your data does not fit the algorithm. (For example, you might be trying to use k-means on binary data. Don't.)
In addition to questions of interpretability, what matters are the linear relationships among the columns of the design matrix, whereas what you are focusing on here are the *rows* of that matrix.  And don't forget that usually $(1,1,\\ldots,1)'$ is a column of the design matrix!
The origin of the term is pretty old, and I believe goes back to the origins of inferential statistics in analysis of experiments; in particular, I think it referred to the way that the X-matrix related to the actual experimental design (the specific settings of the $x$-values). If I can find a specific reference I'll post an answer.
I figured out that random effect should be used. Thanks again.
Hello gung, thanks for the comment.
The question is primarily about the ideas involved and not the coding as far as I can see. I edited the question to make it clearer.
Definitely take a look at the scale of the data - it may be that one is simply huge (eg, weight measured in grams, vs height measured in kilometers).
A decibel is on a log scale, which immediately suggests looking at the log of the other variable.  Any other consideration, to be applicable and relevant, would have to consider the nature of your data and the objective of your comparison.
Either will work with only one parameter to optimize.
Hi @whuber, it is distributed as a $\\chi^2$ because it is the sum of squared independent normal variables, see for example [here](https://en.wikipedia.org/wiki/Chi-squared_distribution#Applications). Probably you are referring to Pearson's $\\chi^2$ test?
@ttnphns: I would consider two cases similar if they differ in only one attribute (except "work status") and if the age is $\\pm 1$ or $\\pm 2$ classes away. Do you consider this insufficient?
Thanks. I've been feeling that there are many cases in statistics where the boundary of right and wrong is vague, and they are making me so confused. As you say, I understand that the language is not too important here and there's always possibility of making incorrect decisions with some probability, so my question does not have much practical importance, yet it was my attempt to make one of the simplest vague cases clear.
I don't know why this is tagged with Bayesian - most research in nonparametric methods is frequentist. And isn't the success of the SVM an argument that we can do nonparametric stuff with provable empirical risk bounds? Neural nets are also, in spirit, nonparametric models (universal approximation theorem) with some clever regularization schemes, and these are state-of-the-art for many problems.
If the $X_i$ are really six orders of magnitude apart, it sounds like they are different things altogether or at least have been expressed in different units of measurement. Wouldn't that make the slopes automatically uncomparable? (2) Why are you not just doing a multiple regression of $Y$ versus $X_1$ and $X_2$? Is it (perhaps) because you have no data for which $X_1$ and $X_2$ are simultaneously available? (3) Exactly what F-test are you referring to?
I agree wholeheartedly about the use of simulation for explaining this.  But just a small note on the example at the end: I find that people (not just students) do find it difficult to distinguish for any particular distributional assumption, e.g. the poisson, between being *marginally* poisson distributed and being *conditionally* poisson distributed.  Since only the latter matters for a regression model, a bunch of dependent variable values that aren't poisson need not necessarily be any cause for concern.
I functionalized what you gave me this way: 
`myperm <- function(limit) {
myindex <- seq(0,limit)
endDim<-factorial(limit)
permutations<-sample(myindex)
while(is.null(dim(unique(permutations))) || dim(unique(permutations))[1]!=endDim) {
    permutations <- rbind(permutations,sample(myindex))
}
return(unique(permutations))
}'  
It works, but while I can do limit=6, limit=7 makes my computer overheat.  =P  I think there must still be a way to subsample this...
Just because you don't see the need for it doesn't mean that there's no reason to ask for it; there's several reasons why it might help @Peter formulate an answer. What you should do is put the information into your question, and if layout is important, indent the entire thing by 4 spaces (by clicking the $\\{\\}$ button with the entire section of output selected).
Thank you for the answer and welcome to Cross Validated (CV for short)! I am afraid, that both cases has that one-to-one match. My question is about something more subtle: about the assumption, that in Pearson's correlation test all cases must be independent. The more my data are time series, the less this assumption is satisfied.
@varinsacha This is not a trivial question and cannot be answered in the comments. Please ask a new question for this and make sure to include a [minimal working example](http://stackoverflow.com/q/5963269/289572). This is probably necessary.
I see. Then my answer is quite a bit off indeed, I did not read your question carefully enough. Let me think if I can meaningfully edit my answer to address your question better.
the wikipedia article on [deep belief networks](http://en.wikipedia.org/wiki/Deep_belief_network) is fairly clear although it would be useful/insightful to have a bigger picture of the etymology/history of the terms. basically a deep belief network is fairly analogous to a deep neural network from the probabilistic pov, and deep boltzmann machines are one algorithm used to implement a deep belief network. apparently all ANNs have probabilistic interpretations/models but they are not as easily/directly obtained as some bayesian/probabilistic-oriented "belief" models.
For comparing many groups against one control group, you can perform an ANOVA followed by a post-hoc mean comparison using [Dunnett's adjustment](http://www.jstor.org/stable/2281208). Most common statistical software packages allow that option.
Got it. Thanks! You could also bootstrap the hypothesis for the next experiment with the first, descriptive experiment, obviously.
Cross-posted at http://stackoverflow.com/questions/14037939/python-scipy-chisquare-test-returns-different-p-value-from-excel-and-libreoffice.
@Penguin_Knight Thanks, I know how to search for packages (at http://rseek.org/ and other places). The point is that *people should not have to search the Web in order to be able to read questions here!*  If they have to, the question needs clarification.
+1 There are also several R packages, including `HMM` and `RHMM` that might be helpful.
Hmmm I actually like the KS test answer better than mine !
The multicore won't speed up the single model as it is doing computations that are *not* parallel. My point was, if you have 4 cores, you can reduce your compute time by about 4 over the entire set of 50-100 models by doing them 4 at a time at the weekend, not one at a time as R only uses a single core. I fitted several thousand complex gamm() timeseries models with pretty large data sets and left my workstation going for a week on 3 cores to do the fitting. I'd worked out the code needed first on a small data set, then set it to work on the main problem.
+1 but what would be the rationale for considering a rotation when the GRM actually assumes an unidimensional scale?
Hi Pat. I'm talking about traffic jams or speed cams to make it more clear.
1. The "k" in k-means clustering stands for the number of clusters. In my case I would have one cluster for each location, whereas I don't know how many different locations there are. 
2. Yes, they also include a timestamp. But I don't understand why that should be important, because I only care about the position.
@PeterFlom Yes, primarily STEPWISE but I also use FORWARD and BACKWARD selection to check which variables are chosen. I also look for multicollinearity using VIF (PROC REG) and generally ensure that the final variables have a value less than 2 (3 at max).
I think both are too mathematical. Unfortunately, there is really no middle ground between say Cochran (or even Lohr), on one hand, and the "softer" methodology on [survey methodology](http://www.amazon.com/Survey-Methodology-Wiley/dp/0470465468/). You can look at what other people are doing in the grey literature of method reports -- if you want a serious smack on your head with tons of details, you can read the one from [California Health Interview Survey](http://healthpolicy.ucla.edu/Documents/Newsroom%20PDF/CHIS2009_method5.pdf).
Hi rdittmer, welcome to the site! As this is a self-study question, could you please add the `self-study` tag? Thank you.
2) To incorporate information about the type of list (not just the name), would I add those variables, for example listType which may be the demographic profile of the list by doing the following, assuming the existing individual level data construct and listType as a column in the data frame:   mod<-lmer(outcome~ listType+ (1|listName),data=Lists,family=binomial)?
I am not quite sure I understand what kind of correlations you want to compute. It makes no sense to compute correlations by pairing the first observation of group 1 with the first observation in groups 2, 3 and 4, as the order of observations in cross-section data is usually arbitrary... unless you have a time series data, and these observations relate to the identical time periods.
@whuber Sorry, I meant lower-case, i.e. all possible joint realizations of the $Z$ random variables.
I respond to your questions. The dependent value is slightly skewed towards the left. The scatter respect to the discrete explanatory variable is not necessarily linear, and it does display some isolated large variations. The other explanatory is categorical, with two categories. 

Thanks Peter, as it is a count, I will look at Poisson or negative binomial.
A uniform prior assumes that all possibles issues have the same chance to occur: it is not noninformative. Moreover, as noted by @timbp, you include a constraint on the range. It is better to use Jeffreys' prior: see http://stats.stackexchange.com/questions/33382/how-do-i-calculate-a-confidence-interval-for-the-mean-of-a-log-normal-data-set/33387#33387
I would consult http://www.amstat.org/about/ethicalguidelines.cfm which seems to bear directly on what you propose.
@BrashEquilibrium I have monthly data over 8 years
Thank you! That makes a lot of sense. I don't use stata but can easily change the code to R. Will have a look at the referenced literature after work.
All my features are floating point numbers.
The answer depends on the definition of "white noise": by some definitions it is (trivially) ergodic whereas other broader definitions allow for it to be non-ergodic.  *What is your definition*?
what is an outlier? The answer heavily depends on the definition. For example 2008 crisis will be reflected by large drop in quite a few macroeconomic indicators and according to a conventional definitions of outlier (the value which "is far" from "the center") this drop will be an outlier, but definitely not the kind you want to throw out.
Not only the point estimate is biased, but the standard error that's reported by default is biased, as well. If you are not surprised that the OLS-reported standard error is biased when the data violate the homoskedasticity assumption of OLS, you should not be surprised to hear that the standard errors are wrong when the strong instrument assumption of IV is violated.
Also I not sure if this question is more at home here or at Math.SE
I realize this was migrated from SO, where one would not normally look for information on these statistical terms.  You have a great resource here!  For example, see what you can learn from a search on some of your terms, like [AIC](http://stats.stackexchange.com/search?tab=votes&q=aic).  A little time spent doing this should either fully answer your question or at least guide you to asking a more specific one.
Actually I dont have to drop observations when using Random Effects model. But even when I am using the same sample size I get different results? And yes, the sample size is the same
I added more detail to my post.  I don't think there is a separation issue in the data.  The results are about what I'd expect from comparing the descriptive statistics of the the 2 groups.
@Deathkill14 Could you explain: 'then cluster variables that are grouped together by the PCA loadings' please? The loadings have the same dimensions as the variables right? Why would you cluster these variables by PCA loadings?
In broad terms, chi-square tests are possible, but the sample size of 15 is likely to bite in one or two ways (1) your expected frequencies may be small (2) you need strong effects to establish significant differences. Also, the chi-square tests take no account of the Likert scale. Those are reasons why many people would prefer Mann-Whitney-Wilcoxon or ordinal logit here, although as said your sample size is small. In broad terms, if sample size is small for one test it is small for another and using a different test is not a solution. Using the best test for the data is always a good idea.
In R that's var.test(q1, q2)
I will have a look at this books. I also know R by the way. Thanks!
Thank you, I find his book packed with that kind of nonsense.
Nothing need be wrong. There is no reason this can't happen. A 3 way ANOVA asks a different question than a 2 way ANOVA, and when you ask a different question, you shouldn't be surprised at a different answer.

In particular, adding a factor controls for that factor.
@JamesStanley Thank you! Yes I'm having a covariate in the ANCOVA. But I am still wondering about the second problem: since the F-value I get from the ANCOVA isn't corrected for inhomogeneity of variance it's actually useless to report it, isn't it? Is there an option to chose so I get the corrected value (like in ANOVA the Welch-value)?
You are correct that using time series data would need to be taken into account, but that is not your main problem. You looked at the data and want to establish whether the highest mean is significantly different. That's a hypothesis suggested by the data. You'd be better off with a model with day of week as a control and testing parameter differences systematically.
@Conjgate They appear to be ordinal, If they were purely attributes/classes/labels , ARIMA processes would seem to be inapplicable.
Linear interpolation of the 2-tailed p-value for $z=\\sqrt{0.1185}$ gives $0.73$ which is fully accurate to two d.p.  (In R this can be done as `(sqrt(0.1185)-.39)/(.25-.39) * (.80-.70) + 0.70`; the numbers 0.39, 0.25, 0.80, and 0.70 come from your teacher's table.)  Alternatively, applying Simpson's Rule (`x<-0.1185;1 - sqrt(x)*(2*exp(-x/2) + 4)/(3*sqrt(2*pi))`) gives 0.7306, which is within 0.0001 of the truth: no table is needed at all.
Should be completely identical with two treatment groups and two time periods.
I added the answers above in my original response. See if they are good enough.
I agree that it's likely that bayesian or not doesn't matter. I just said it was Bayesian just in case it matters.
Thanks gpoo. I actually increased the max iter to 1000 and the tolerance as well.
"Holdout" and "cross-validation" are slightly contradictory.  Most people who use the term "holdout sample" are done 1-fold cross-validation, a highly inefficient approach.  Please clarify.  100 repeats of 10-fold cross-validation, or the bootstrap, would be good approaches.
Dear FMZ, thank you for your kind reply, but could you please confirm me when I use proportions as independent variables then "per unit change" in the independent variable means an increase by 0.01. I was guessing it, but as it is a continuous variable just like the other continuous variables (which would generally mean an increase by 1), so I was afraid if a unit change meant '+1' or '+0.01' in this case! Probably very immature question, but I would like you to explain this a little.
I am having difficulties to understand your question. Why does your title refer to "multivariate ANOVA" but your question not? You might want to give more details and then we might be able to answer your question.
(+1) GLM might be the best way to disentangle different meanings. It should also be noted that in the history of ANOVA calculation procedures were used that obscure the relation betweeen OLS and ANOVA. The nomenclature might therefore be justified by historical reasons.
@Artem I didn't have enough reputation to make such a small edit.  (Low rep users are only allowed substantial edits.)
(+1) Thanks for stopping by and providing that link, Sean!
This issue has already been addressed. See answer under [http://stats.stackexchange.com/questions/37526/problems-with-groups-in-traminer][1]\\n\\n\\n  [1]: http://stats.stackexchange.com/questions/37526/problems-with-groups-in-traminer
Thanks for your advice! I'm unsure how to do a quantile regression unfortunately.  However as I just told Hernan I have noted that that outlying group at the left side of the residual plot is coming from data near the start of the time frame I am studying (1961 to 2004). However I don't know how to interpret this finding.
If you want some idea of power at some *particular* value of Kendall correlation (or at some Sen slope), you can do it with a variety of more or less plausible parametric models. If they all give the same answer, you can probably work out why it doesn't depend on the model. If they give different answers, that's a rather important thing to know.
+1 Interesting read!
Please note that you are also in a gray zone of this site's scope; this seems like an R programming Q which might be asked on StackOverflow. Anyway, it would be easier to answer if you would make it reproducible, i.e. add information how you have calculated `sensitivity` and `oneMinusSpecificity` from the decision and raw biomarker level.
Some more intuition: Statistical learning theory leverages the fact that w.h.p your iid training and test sets will be representative of the distribution from which they were drawn. Hence low training error and high test error implies a poor classifier rather than some unlucky pick of the test points. Therefore, you should trust the data and not try to massage the test error.
Durable, resale is an option. Basically, imagine a monopolist's market with complete leverage to set prices and determine supply, but the demand for any one good is very finite.
Hi Santiagozky: there is no single "best" tree that the random forest algorithm produces. In fact this algorithm uses multiple trees, and hence the name "forest". For details explanation of the algorithm, i suggest the following [link](http://stat-www.berkeley.edu/users/breiman/RandomForests/)
**Hint**: Analyze $Y_t = e^{-B_t + \\frac12 t} X_t$.
+1.  "...choosing a statistic is equivalent to defining the alternative hypothesis that you are considering" strikes me as a deep insight.
Strictly speaking, that it is a $-5/3$ slope
Octave is basically Matlab, but slightly different. The regress command in Matlab is one option that works on my version of Octave. You probably need more data to make any decent forecast. Also, the day 7 thing would be more like interpolation than extrapolation (so interp1 would work if you want linear or spline interpolation).
Would you please give a link or two to explanation of data standardization?
You might find the answers in the following question helpful: http://stats.stackexchange.com/questions/2306/feature-selection-for-final-model-when-performing-cross-validation-in-machine?rq=1
Usually, some hypothesis test relating to that variable is rejected (e.g. when a test of whether a regression coefficient is zero is rejected). A complete answer would require a book, since there are too many contexts in which a person might use such a phrase (not always correctly)