tttjjj/elt_bench huggingface dataset:
Primer 5 is apparently software for analyzing and designing PCR primers. That part of the question may be better suited for BioStar SE. An extremely well thought out and developed post @whuber thanks for taking the time to write all of that up!
Not a dumb question since people upvoted it.
@Nick With respect to your comment about not using data that  has already been observed when making the 4:00 pm forecast. What happens is that if there is unusal activity (autonatically flagged by the software) for earlier time periods that day then this allert calls/suggests the need for  a management override. If on the other hand prior hours have been consistent with expectations then all is well and the forecast remains in place, as it should.
The number of missing values varies from subject to subject, from 0 to 2% of values. The missing values we suppose are simply due to the machine not taking the values. Being one value per second, one measurement doesn't influence much and for sure sure there are overestimation error together with NA values. That's why we are averaging by minute with SMA.
Are you looking for general aesthetic guidance?  Or statistical guidance on what is best to report? Or particular ways of doing it in a particular software package?  Please be more specific.
+1 Good answer, though I'd add the minor point that in both cases, *equality* corresponds to rejection as well -- and while it's not an issue with continuously-distributed test statistics, it matters when they're discrete.
@tmo `R` on this machine takes 18 seconds to obtain 1000 realizations of the null permutation distribution of the max correlation coefficient for a 300 by 100 matrix `x`: `correl <- function(x, k=1) {
  n <- dim(x)[2] * (dim(x)[2]-1) / 2;
  v <- cor(x); 
  sort(v[lower.tri(v)])[(n-k+1):n]
}; sim <- replicate(1000, correl(apply(x,2,sample)))`
Do you need intercept in the model ? Is intermediate value between 0 and 1 possible ? Is 0 < 1 or does not matter - they are just categories ?
Thank you very much! Your answer is so detailed and really really helpful! Thank you!
Well, at least he is pointing out a true problem...
Further, if data are dependent (akin to "paired") the number of available combinations is further reduced.
Do you mean intervals for each $\\mu_i$?  How are you calculating them?
@Robbie.  Apparently they are not, but they may be close enough for this not to matter much.  Try the qqplot.
I don't know what ERP is either ;-). If your likert scale is the sum of many likert items & people don't tend to be at the boundaries of the scale, the t-test is probably fine.
This pseudo-R^2 does not agree at all with the Nagelkerke R^2 of @chl's answer.
Here
is a very simple introduction to the Kalman filter,
to estimate the position of a robot (think of the position as the parameter you are trying to estimate):
https://sites.google.com/site/udacitymirrorcs373/cs-373/unit-2
(you may want to skip part of the begining, which is irrelevant,
and check the previous and next lectures, which present
non parametric alternatives to the Kalman filter: histogram filter and particle filter).
It looks to me that you want to know if the proportion of `BAD_RESULT`s differs by `GROUP`. I think you would be best to explore [logistic regression](http://en.wikipedia.org/wiki/Logistic_regression), instead of using Fisher's exact test. Out of curiosity, what book are you reading?
Check your calculations.
Because I need to know if GLOBALLY condition 1 is better, I don't want to know only the significance at stimulus level. Indeed I will have more than 3 stimuli in the real table I use. What do you mean with "making the design hierarchical"? Can you provide an example in R please, so I can understand what do you mean?
Which paper are you looking at?
Decode meaning understand. What do these files say? It looks like gibberish to me.
I want just to compute this expectation.
However, alternative estimator will be also a very interesting problem!
Also, I'm guessing you may have intended to use `\\cdot` instead of `\\dot`. If so, please go ahead and make those edits. Cheers. :)
How much depends on your purposes, the time you have to spend, etc. I would recall to R only if Python is too far from your ultimate goal. Perhaps if you list the algorithms you intend to use, someone will help you out.
@gung: This is $1.5 IQR$ beyond the quartile, so about $2 IQR$ from the median for a symmetric distribution.  It also depends on what you mean by "fairly often" and the distribution:  almost never for a sample from a uniform distribution; about 0.7% of a sample from a normal distribution; about 5% for a sample from an exponential distribution; about 16% for a sample from a Cauchy distribution.
Dear Nick, you mean calculate the explained deviance? I've tought of that but as you say my variables are not orthogonal, so I am not sure how to compare the explained deviance with the one calculated by PCA? I wonder if there is any sort of data transformation that will make both measurments compatible?
You might look into discrete choice modeling.
@whuber: Thanks for the edit.
@andand I have tried least squares method, but the problem with this method is that it is not robust to outliers.
Germination and viability are independent of eachother and are not subsets. Viability pertains to the proper development of the embryo inside the seed, whereas germination involves the emergence of the seedling from the seed. The two are tested in completely different ways, and indicate different aspects of seed health. It is most likely impossible that a seed lot would be 0% viable but still germinate (at any given percentage). I apologize if some of the biology is over your head!
W/ so many free parameters (due to the multiple hidden layers) you will need *a lot* of data or the "accuracy" will only be due to overfitting. You will need to cross validate to get an unbiased estimate of the accuracy you really have.
If two variables are normally distributed, their ratio probably won't be, so normality is unlikely to help here. You can still apply ANOVA with a third, categorical, variable; just don't call it "correlation". That terminology is likely to mislead. Most importantly, watch the distribution of the ratio. It's quite possible you are better off with a transformation. Note that if $Z = X/Y$ and $X > 0, Y > 0$ then $\\log Z = \\log X - \\log Y$ can be used and and more generally ratios are often better logged. Alternatively, think of a generalized linear model.
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this is a very good idea but for my reasons I would not like to go through articles distance. I want the probability/weight not for the actual tag suggestion which it can be done as you suggested, but as a solid tag score that I will use in my system for many other purposes.
Ok thanks! Thats what I thought too..:)
Thanks a lot for your answer. The thing is that I don't want to check if a stimulus is preferred rather than an other. Instead I want to know if a Condition is preferred compared to the other two. How can I do this? Should I just swap the Stimulus and Condition, so I get the trasposed version of the matrix?
I'll have to read it more thoroughly later on, but yes, this package is definitely addressing the kinds of problems I'm facing here.  Thanks!  And also, thanks for the kind words about the plots ;p
I am not referring to your link.
Data mining is an extremely broad field, you might get more informative responses if you specify particular areas that you'd like to learn more about.
Just wondering, are you working within a specific framework, like JAGS or STAN?
Although you list some goals, I am still not really clear what the point of your study is. Is this just initial-phase exploratory work? Are you trying to test some hypothesis taken from a theory? What would the "consistency" of their ratings mean? Are they rating the same violins more than once? For research like this, you may want to look into [psychometrics](http://en.wikipedia.org/wiki/Psychometrics), a discipline related to statistics; I added the psychometrics tag to your Q.
What is a model you are fitting?
It is including the different variance in the model. The output (which is difficult to read in a comment) tells you that the residual variance of sex 1 is 0.669 times the residual variance of sex 0. The value of 0.669 is fitted.
While respondents have listed some good resources, it is a difficult question to answer in this format, and (many) books have been devoted solely to this topic. There is no cook book, nor should there be given the potential variety of situations that linear regression could encompass.
Fine description of LOOCV, but I can't agree with your last few statements (unless I'm misinterpreting). Normally what one would do is average all the errors to get an estimate of generalization error. CV is not for estimating parameters. It's to asses predictive performance, compare models, select hyper-parameters, and whatnot. Using CV as you describe in your last sentence can get you into real trouble (especially with LOOCV). Imagine you have a duplicate datapoint. If you have a flexible enough model, the lowest generalization error is likely to be the fold with this point left out.
Pretty much. My 'dictionary' is the vector of numbers 1:27511. I take two samples without replacement - one of size 23706 and one of size 14557. I then compute the size of the overlap. The 27511 is the 23706+14557 (A+B) minus 10752 (A&B) because otherwise you count the intersection twice. Draw a Venn diagram.
Fixed the typo ... now it makes more sense.
Basically, you _must_ specify a covariance structure in GLM. If by "assuming no covariance", you mean "all off-diagonal entries in the covariance matrix are zero", then all you did was assume one very specific covariance structure. (You could be even more specific, e.g., by assuming that all variances are equal.) Bottom line: I am unsure whether the question makes sense in its current form. (I upvoted anyway, because I think I'll learn a lot when the gurus start talking.) Or am I misunderstanding you?
Please make the question self-contained. I do not have access to the book.
First versus second order accuracy is not about the width of the confidence interval; it is about the coverage probability. The usual 95% confidence interval is such that $P(\\mu \\in \\hat{C}) = 0.95 + O(1/\\sqrt{n})$. A $BC_\\alpha$ 95% interval is such that $P(\\mu \\in \\hat{C}_{BC_\\alpha}) = 0.95 + O(1/n)$. Both intervals will have widths proportional to $n^{-1/2}$.
@screechOwl Glad to hear it! I love R `data.frame`s.
I repeat: Cronbach's alpha is only appropriate for unidimensional scales. Alpha values are meaningless otherwise.
That was it! 95% accuracy with 50 trees and 600k training examples.
Is this question perhaps the same as the one at http://stats.stackexchange.com/questions/2275/how-can-i-determine-accuracy-of-past-probability-calculations?
Yes, probably. But be aware that *neither* solution is correct. In fact, I wouldn't know which one happens to be more accurate. It is not uncommon to discard/round up eigenvalues below a given threshold $t$ (for example $t=10^{-12}$.
Thanks for info, good to know.
Aren't the rows you showed about industrial sectors, and not occupations? In any case, a common distinction between data format is the [wide and long format](http://stats.stackexchange.com/q/7439/930) but I think it all depends on the software you use and what statistic(s) you want to produce.
Mods: If you would like to migrate this to stats.SE we would merge it with the version posted there.  Alternatively, we would be happy to migrate our version here, but the question looks more statistic-y than math-y.
I think that you have to differentiate between the question: What is the probability that I will hit some point? If we agree that you always throw a dart and it always hits somewhere along the line, that probability is 1. Also, I am not just saying you won't hit 0. I am saying that the probability that you hit ANY point you pick BEFORE throwing the dart is 0. In fact you can pick any finite set of points and the probability will still be 0.
Maybe you're thinking of contingency tables:http://en.wikipedia.org/wiki/Contingency_table ?
@Leo One (speculative) interpretation is that the O.P. seeks a subset of $S$ of {cat1, cat2, cat3} such that $\\max(S)$ and cat4 differ in as few cases as possible.  With $p$ independent categories a brute-force solution takes $O(2^p)$ effort, which is terrible.  Can one do better?  I hope @McP will correct any errors in this interpretation or confirm it is what he intended.
If you want more details on the last part of @gui11aume's answer, you can check out the answer to here (http://stats.stackexchange.com/questions/44262/t-distribution-confidence-intervals-for-non-gaussian-data-but-large-n/44280#44280).
+1 for exactly how I would do it down to the colors in the plot. One thing: you should consider jittering your points (or at least suggesting the option to OP). Look at point (3,5): it is both a start and an end point, which makes the transition it takes look strange, like a two step move when it's really two single step moves (presumably).
(+1) Great that you add some pictures to illustrate the whole thing!
1/365?.........
Ah. My bad. I just sort of read 'I want to calculate...' in your question. Well, for the rest, you can find $a$ as $P(S1 \\& S2 \\& D)$, and similarly for the others. Using Bayes again on these forms (and doing some lifting on the negations) should enable you to calculate them all.
There is no one-dimensional sufficient statistics for this model.
@AndyW Re "not sure how you would do your $h$-block for pair-wise distances", you can think of us as removing them from the training set before computing the pair-wise distances. Really what we do is just ignore those cells of the already computed matrix when choosing the $k$ nearest neighbours.

I'll take a look at Lahiri; it appears we have a copy in the library here. Thanks again for your comments/thoughts.
Yes, you are right, Greg. I should have give a more precise description what I want. I was thinking more of a graphical selection tool which GGobi offers.
The posterior distribution has the form given  in the above Wiki link. You don't need to use R to find the posterior, although you can download the [Normal-Gamma Package](http://cran.r-project.org/web/packages/NormalGamma/NormalGamma.pdf) if you need to have the normal-gamma PDF or CDF in R.
... then the answer may be more useful to you (& perhaps  to others, perhaps even to the answerer) than one that trots out the usual explanation.
Thank You! currently looking into this.
Every algorithm can overfit, as this problem also depends on input data and the full training methodology. You can do a lot to prevent it (watch for assumptions, do not cherry-pick parameters or settings, think of consistency of the data, etc.), but the most important thing is to test the whole methodology well.
While I haven't heard of *matrix regression* I have heard of *principal component regression* and of *2D principal component analysis*. I guess you could easily calculate scores from the 2D PCA on your images to reduce the dimensionality of your sample greatly. Then analyzing those PC score vectors instead of your original image would be straight-forward and easy to achieve even with standard univariate techniques as those score vectors would be orthogonal to each other.
The problem with PCA as I see it is that part of the rotation is due to the array-like structure of the data.  If PCA is applied to a rectangular array with random counts it will rotate it and if it is applied to an upper triangular array with random counts it will rotate it to the diagonal, that being the rotation to explain most of the variance.

Because of this and since I know that (0,0) and (7,8) are worst/best responses, I was thinking of the non-linear projection.  The data are almost constrained to the upper triangular, but there is more structure to them than random counts.
Why exactly do you think this is a problem?
In all parametric statistics there is a direct functional link between the test statistic (F in this case) and the p-value.  These have been put into table for convenience, but can also be computed directly.  You can either use alpha to find the cut-off for a critical region to compare the test statistic to (which I think is more intuitive) or use the computed test statistic to find the p-value to compare to alpha.  In either case we start with an alpha level and a test statistic formula that follows a given distribution when the null is true.
`gls()` in **nlme** does allow you to alter the off-diagonal elements of the variance covariance matrix and Fox's (2008) Applied Regression Analysis & GLMs has GLS very much at the centre of a chapter on time series regression. The details are in Pinhiero & Bates on mixed models in S
I'm using KPSS, ADF, PP and ERS unit root test, So I think i'm using all the test about unit roots, the problem is that the series have not unit root, ok, but have strange different volatility during all the period... maybe an homoscedastic test to filter the series that have different variance(residuals) ?
Between-matrix `B` should be the same size as Within-matrix `W`. Because `B=T-W` where `T` is Total scatter matrix (matrix for the whole sample).
@Bogdanovist, your comment makes it sound like "frequentist" tests use prior distributions.  This makes the entire comment confusing.
Ridiculous, really!
I think at best you are going to find an approximate value. `?pvalues` is probably relevant here.
I tried TinnR and didn't like it much, it seemed unpolished and clumsy.
It might be better to delete the data for that date, if you can justify doing so.
I think the time-series plot makes it pretty clear that the nonstationarity is not going to be remedied by anything on the order of 24 lags. I suspect the "structure" you see at around 24 lags is actually the high frequency oscillations also very apparent in the first plot. Indeed, as a coarse estimate, I counted the visible troughs between index 3500 and 4000 and I see 20 of them. If a simple lag-1 difference were to take care of it, you'd probably see a fairly pronounced 1/f like decay in the ACF coefficients. It doesn't immediately look like that to me, but there are very few lags plotted.
@mark999: Added some examples and a suggestion of how you can do the ones by hand.
The notion / notation "p > cdf(x)" is unfamiliar to me. Can you clarify your situation? Are you thinking in terms of a strategy to simulate Bernoulli data by drawing a normal random variate & assigning a $1$ if it's > some threshold & $0$ otherwise?
If that is a screenshot of the pdf, it will get saved as png or jpeg, and be in RGB like the others. But the actual pdf *is* gray. Perhaps take a look at your computer's color settings? On mac this is the "Color" tab under the "Displays" system pref pane. Also check the color settings and color profile that you're using to print. If either of those are off, there could be color casts like you're getting.
@rolando2 I don't think so.  In mixture designs the treatment is a mixture.  By contrast, here it is the outcome that is a mixture.  We want to test the effect of treatment on the mixture proportions.
Thanks for pointing out the syntax issue - yes, I meant $y_1$ takes that distribution. Will edit.
If the means are zero, this requires (virtually) *no* calculation at all. Are you interested in the general case or, perhaps, this specific one?
Thank you Sergio. That clarified it for me.
It's ok provided $E(e_tv_t)=0$
I am sorry but there is not much I can help you with... 1) E(V) denotes the expected value of V. exp(V) denotes the exponential of V. 2) The uniform density over (a,b) is constant over (a,b); as it integrates to 1, it has to equal 1/(b-a).
Evidently step.plr uses effect coding
@NickAdams: yes. But I'm a little bit a suspicious person in such questions, so I'd actually have a quick look whether they are the same...
How is 'expression' measured?
You haven't clearly responded to the direct question @Quantibex asked. You used $\\epsilon_i$ twice. Is it meant to be the same value both times? It *matters*, since the answer to your question changes depending on what the answer is. If you are not clearly explicit when asked, you cannot hope to get good answers. As you have written it, it's the same value both times and the answer to your question is "No, they're different" ... though the margins are the same. Look at the correlations between them.
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Don't I have a single binary and really four rank-order variables because I have the mood and the assessments in both the plant and no-plant condition?
It's a very interesting question. Seems like they knew that this can be done or has been implemented in the literature. This one is for a psycometric guy I guess... I'm curious to read the answers they would give you.
Many thanks for these quick clarifications!
Sorry, but that's against my statistical religion. The post hoc analysis you most need is to think about what the results mean scientifically. Others might well give different answers.
I guess sharing and variation at different levels can be used for non statisticians.
It would be the mean when `wool=="A"` AND `tension=="L"` if you also included the interaction between `wool` and `tension` in the model. Without that, I think you're just left with an estimate of the mean, assuming the two factors have independent effects.
Can you clarify what your $a_n \\sim b_n$ notation means? It cannot be the standard asymptotic notation indicating $a_n / b_n \\to 1$. The left hand side is huge and the right hand side is tiny! Maybe once this is sorted out, it will not surprise @Michael so much since convergence in distribution (and even a *local limit law* like this) is simply an analytical statement about certain sequences of monotonically nondecreasing functions and so, at its heart, is not tied to any "probabilistic notions".
Take observation $j$ of group $i$: $Y_{ij}$. Because $Y_{ij}$ belongs to group $i$, subtract the estimated mean of group $i$: $Y_{ij} - \\bar{Y}_{i\\cdot}$. Square: $(Y_{ij} - \\bar{Y}_{i\\cdot})^2$. Repeat for all $j$ and $i$ and sum: $\\sum_i \\sum_j (Y_{ij} - \\bar{Y}_{i\\cdot})^2$.
The logistic equation has a range that is continuous, yet it is valid for modeling binomial data.  Why would the transition from a continuous domain to a binomial one not be appropriate?
None of the answers below are written in terms of finite population inference. The word *finite* is absolutely crucial here; that's what Kish's book is about (and whoever was saying "The book is wrong" simply don't know enough about finite population surveys and samples). The quotient $N-1$ instead of $N$ just makes computations nicer and obviates the need to haul around factors like $1-1/N$. The full answer to this question would have to introduce the sampling inference where the sample indicators are random, and the values of observed characteristics $y$ are FIXED. Non-random. Set in stone.
:Carlos Unfortunately you are still assuming the form of the model i.e. an equally weighted average of the last N data points. The optimization of this is to determine BOTH the best "N" and the string of weights to be applied . To assume the weights are equal may appear to be "fair" but it is presumptive.
It would be great if you could explain why b is used generally. Maybe give some references, or just give a 2 sentence explanation.
@ttnphns gosh, I'm so stupid. First I used "N", but to avoid confusion (sic) I changed it to Q... but forgot to change it in all instances. I am not going to give a real world example, amongst other reasons because the data set is so large. However this is a problem similar to enrichment analysis of GO terms, except that I cannot define a "foreground" and "background" set for a hypergeometric test.