tttjjj/elt_bench huggingface dataset:
In addition to any answers below, I find Faraway's Practical Regression and ANOVA using R to be a concise and helpful reference for regression basics (available here: http://cran.r-project.org/other-docs.html). You can either go through the examples in R, or translate them to SPSS. I get the impression that your overarching question is, "How do I perform and interpret regression analyses". A good question, but one that most here studied in semester-long university courses, and some have spent careers studying.
Thank you for your suggestion. I already knew about that course, but it is too specific in my opinion. I am looking for something that extends section 17 of this very well known online course: https://www.coursera.org/course/ml (you can click on preview lectures to check it out).
@Brandon I am glad to hear that this is not a post-hoc test.  You don't need R to implement it: the second paragraph of my response contains all the calculations needed!  In effect, this well-researched pattern would have a 1/512 chance of occurring by chance in the case that all nine possibilities (increase/decrease) were independent and equally likely.  Because 1/512 is so small (less than the standard threshold of 1/20), you have a valid claim that it is evidence that the pattern is not due to chance.
Doesn't an unbalanced design lower the power, versus a balanced design, for the same total number of samples?
You can test those ICCs by simply testing the respective variance components in the numerator (i.e., $\\tau^2_1$ and $\\tau^2_2$). In other words, fit two models where you leave out one of the two components and then compare those reduced models against the full model using the ``anova`` function.
This question appears to be off-topic because it is about a search for information on coding projects. Did you Google or try searching among the resources for your favourite software? (NB it is always polite and helpful to spell out what that is.) Please see the Help Center for advice on software-related questions.
thank you for the very helpful answer. The one other area where I get confused is that if "Estimates using type I SS tell you how much of the variability in Y can be explained by A, how much of the residual variability can be explained by B, how much of the remaining residual variability can be explained by the interaction, and so on, in order," then why does the stats associate with A change when additional variables are added? Is it because the tests are based on SS(A)/SS(error) and the error is a function of all the terms in the model? Am I thinking of this correctly?
You might be interested in `help("pvalues")`. However, a p-value is nothing without a hypothesis. You should clearly state the hypothesis you want to test.
Traditionally, $\\alpha = 0.05$ rather than $\\alpha = 0.005$.
+1 for the appropriate citation
@mbq - I don't see how cross validation overcomes the "un-representativeness" problem.  If your training data is un-representative of the data you will receive in the future, you can cross-validate all you want, but it will be unrepresentative of the "generalisation error" that you are actually going to be facing (as "the true" new data is not represented by the non-modeled part of the training data).  Getting a representative data set is vital if you are to make good predictions.
well... i have reviewed the Ruscio & Kaczetow (2008) article. sadly it's just another (yet more flexible) implementation of the NORTA (NORmal To Anything) family of algorithms which is known to not work well with multivariate 3rd & 4th moments.                        i guess i'm back to square one on this one.
"optimal" in what sense?
@MarcShivers It is a bad tool 1) because the data will not be independent. 2) The best predictor of whether it is raining at 12:15 in NYC is whether it was raining at 12:00.
Ask yourself (and tell us ?): What is the desired property of the ranking ? Without a metric, any ranking is fine.
I agree with @RussLenth. You should look at this [answer](http://stats.stackexchange.com/questions/111900/multiple-measurements-of-a-single-datapoint-what-to-do-with-variability/111999#111999), I think it will help you understand.
It does matter, but it depends on your data. I wouldn't know what are the models in your field
I can understand why this urban legend. When estimating the reliability of a questionnaire, it turns out that just by making the questionnaire longer (e.g. through putting in some more parallel questions), one can boost up the reliability. If someone has some less than perfect questions, and would like to make them into a tool with reliability of 0.8 or above, then I can see why that researcher will advocate for a range of 80-110. In reality, though, I don't think asking any teenager to sit down and go through 100 questions an appealing method.
Check out the `window()` function.
Shouldn't $P(A_1 \\cup A_2) * P(A_3 \\cup A_4)$ be $P((A_1 \\cup A_2) \\cap (A_3 \\cup A_4))$, since your events are non-mutually exclusive? Then you have $P((A_1 \\cap A_3) \\cup (A_2 \\cap A_3) \\cup (A_1 \\cap A_4) \\cup (A_2 \\cap A_4))$, so I don't think you can trivially extend the algorithm...
It's still not clear what you mean. Do you mean the number of points which are different? Do you mean the percent difference in the averages? Do you mean the mean percent difference for each of the pairs of points? There's no obvious way to interpret what you're asking for. Percentage difference is not something people normally think about for sets of points. Percent difference can fully specify how two individual points are different, but you need a lot more information to fully specify how different two sets of points are.
Thank you Andy, I will look more into dynamic time warping, I was thinking about applying it to my false ring problem, where some of the years make two rings instead of one, but maybe I can apply it to my current problem.
Hi Arbit, welcome to CV! Thanks for adding the `self-study` tag to your question. This page offers info on homework questions: http://stats.stackexchange.com/tags/self-study/info One of the things you should add to your question is what you've tried so far and where you get stuck (as of now, you've merely copy-pasted the question without showing an attempt at solving it)
Hello IMA. I would appreciate if you have any advice about my question here http://stats.stackexchange.com/questions/59326/breakpoint-for-bivariate-data
@curious_cat: Yes they are. I commented on this in an answer (but he apparaently deleted if since it was wrong).
Here you go, I am also Heinlein's fan :-)
This question appears to be off-topic because it is about probability/statistics and has been flagged for migration to Cross Validated.
Answering my own question, at least to an extent. I think I was seeing a problem where there wasn't one, perhaps because Minitab gave me an "insufficient rank" error when fitting a nested model to data containing several 'Additive = None, Level = 0' rows. R gave me an answer with `Y ~ Level %in% Additive`. Perhaps the way to think of it is to lose the categorical 'Additive' column and have a different continuous 'Level' column for each Additive; no additive is just zero in all the Level columns. I think Minitab may be able to fit such a model by multiple linear regression; I'll try on Monday.
Can you elaborate further? Are you looking for pitfalls of p-values in the context of bioinformatics?
@mpiktas Thanks for the hint. I clarify my question. I mean stationary time-series, i.e. $\\rho < 1$, but not a simple normally distributed random variable. Sorry for the confusion.
I have a indicator variable ky which takes values 1 and 2, but if I try to do ["x:ky1"] it says that object does not exist. What am I missing?
Related question on the site in reference to this answer, [What does linear stand for in linear regression?](http://stats.stackexchange.com/a/8706/1036)
Could you be a little more specific in terms of "code a pretty basic version"? Did you simulate from a Hidden Markov process, or did you code the Viterbi, forward, or Baum–Welch algorithms? (The last three would be used to compute the most-likely corresponding sequence of states, the probability of the sequence of observations, or the starting probabilities, transition function, and observation function of a hidden Markov model, respectively.)
It's pretty much the same as what I've written above. $X-Y$ is the same as $X+(-Y)$. The distribution of $-Y$ is the same as the distribution of $Y$ except that it is mirrored over the vertical line at 0.
Welcome to the site.  Ummm.... you should at least consider the possibility that there genuinely is no significant evidence that your explanatory variables are related to the response.  However, also have a read of some of the answers on this site about model selection.  I'm afraid that beyond that it will be very difficult for anyone to answer your question in a helpful way.
`Gower's similarity coefficient is generally not compatible with Ward's method` You might have also read that it is not comatible with k-means as well. Because k-means and Ward are relatives. Won't you prefer k-medoids in place of k-means?
Migrating the Math question here is strongly preferable because it would collect all the answers in one thread rather than scattering them across two sites. In the meantime, I respect the thought behind this question and agree that the text and the existing answers are inadequate (although correct): probability calculations are notoriously tricky and it's easy to get subtly wrong answers by invoking the "obvious" leaps of intuition exhibited here. A *good* answer will rely only on basic axioms and proven results of elementary probability theory, justifying every step with rigor.
Hi, thanks for replying. I'm using multinomial logistic regression
@DimitriyV.Masterov The Data Q&A site would be a very good fit but it is not open yet. Our only options right now seems to be (1) answer the question here (2) close/delete it entirely
Why would anyone want to distinguish the priors?   Why not instead start with a prior something like $\\lambda \\sim \\text{Ga}(\\alpha+\\frac{1}{4}, \\beta)$ and then update it with the observed data to give a posterior distribution?
Following up on Wolfgang's comment, if a question you have asked has received a good answer, then [as described here](http://stats.stackexchange.com/faq#howtoask), you should click the check next to the answer that you find most helpful.
Split your data set in a training set and a test set. Start from very few training records and then continue to add records. For every iteration compute the training set error with the records you have used to train your classifier and then compute the test set error always with all test records. I know this is a standard method used in common practise. It would be interesting to see your results! Cheers, Simone.
Look at the works of [Alain Celisse](http://math.univ-lille1.fr/~celisse/). His work as far as I read (too little alas) is about merits of cross-validation.
(+1) Please note that `nls()` is in the base `stats` package. There exist, however, the [nls2](http://cran.r-project.org/web/packages/nls2/index.html) and [nlstools](http://cran.r-project.org/web/packages/nlstools/index.html) packages on CRAN.
There's nothing wrong with the model.  This posterior is meaningful in the sense that if you receive another observation, you can multiply it in and possibly get back to a proper posterior.  So it's not like a NaN, upon which all further operations are NaN.
Thank you, very clear answer.
It's hard to say, just from this, whether you should be concerned. Can you post the crosstabulation of L1 and the DV?
+1 Elegant, clear presentation.
The means are not that important in this case it's the shape of the distribution that is important.  The kurtosis, skew and variance seem to be the most important moments in this analysis.  I need something like a repeated measures for a test of distribution shapes (I doubt such an animal exists).
+1 for glmnet. I believe you can also force inclusion of certain variables, and it will do variable selection/shrinkage on the rest.  It's also very fast.
OK, well, what I would do is generate a bunch of random numbers for all the components of your model, do the calculations, and then run the model and see what happens. Then say "how do I tell if something real is happening or if it's just an effect of how the variables are calculated?"
Essentially my query is that even though I surveyed the entire population, I am, through non response, left with a sample. Can I apply the conclusions of non response literature which seem to indicate even much lower rates of response provide accurate population estimates even though the situation isn't exactly analogous as I mentioned. And further, can CLT be held to operate on this type of variable and data point it generates?
Why $z=exp(sT)$?
+1 The usual chi-squared statistic works well.  I examined the distributions of bootstrapped p-values for zero-mean *shifted* Poisson distributions ($X + \\lfloor\\lambda\\rfloor \\sim \\text{Poisson}(\\lambda)$) and found good power even with moderately small sample sizes.  E.g., with two datasets of $100$ values each, $\\lambda=1$ is discriminated from $\\lambda=1.4$ with 50% power at $\\alpha=.05$. These chi-squared statistics do not appear to have chi-squared distributions, whence the need to bootstrap the p-values.
Explaining a bit more. Suppose two variables, $x$ and $y$ are linearly related as, $y=mx$. So $y$ is always $m$ times of $x$. Now we transform $y$ as $y/x$ so that new variable $z=y/x=m$. This $z$ is then independent of $x$ and will have $0$ correlations. Please let me know if it is still not clear.
OK, Inspector, why don't you add this information to your question?  Specifically, it now sounds like you have data telling us *how* good the predictors are.  If you put that into evidence--or at least tell us its nature--we would have a better chance to provide objective, justified answers rather than just guessing at things that might work.  (I think AVB is on the right track--so to speak--with the Kalman filter suggestion.)
#1 made by @charles still applies. It seems that you are a student of economics or some related subject, and the choice of dependent variable must still make economic sense.
I haven't tried that yet but it can be done. There are two ways to do this - evaluationScheme creates an object of class evaluationScheme. The code for how it does that is https://r-forge.r-project.org/scm/viewvc.php/pkg/recommenderlab/R/evaluationScheme.R?view=markup&revision=62&root=recommenderlab You can overwrite the default method, create your own or simply overwrite the slots of the object (not recommended as it is unsafe but easy to do). The other way is to divide into test and train matrices ahead of time and do your own evaluation. Another tool is VowpalW (See Matrix Factorization)
(+1) I hope you don't mind the typo correction and small tweak. :)
The link is not available. Please include your response in its entirety here, or leave a link as a comment to the question.
In this article no, these two quantities are different.
@Dail, if you know that in your price data some changes in explanatory variables happened (new products, brands, shops, other prices, income etc.), the process will still be mean-reverting, however to a different mean, since the latter could change due to the mentioned facts. If not and picture of $R_t$ looks stationary, then (A)DF will be ok. Well Engle-Granger goes for cointegration, it's a bit different story.
I think what you're saying applies here. Especially the time to heal. The fact of the matter is, that the mean is being massively afffected by those patients who had long healing times. One can see this in the histogram. Perhaps I should just look at it as continous variable and log-transform it. Seems to do the job. @Glen_b: Because of the censoring effect (when someone's healed they drop out) I've been thinking of using a cox-proportional hazards model for the regression analysis.
Nine out of 38 years exhibit "unusual" activities?  In a model with (apparently) 25 parameters?!  Something's wrong with that interpretation.  Apart from all these pulses and the level shift, have you found the *nonlinear* component yet in the trend of cohort b?
"Using the covariance matrix standardises the data" - I think you mean correlation matrix here...cheers
Depends on the kind of Bayesian estimate. MAP? Posterior mean? The result of minimizing Bayes risk for some loss function? Each of the above? Something else?
It is focused on Machine-Learning though, which can be seen as a related field of data-mining, just like AI. Although commonly it is use synonymous, as "prediction" is one of the key challenges in data mining. But there is more than "learning" to data mining.
Is this homework?  If so, please add the "homework" tag... we'll still help guide you to an answer (rather than just giving it to you.)
http://en.wikipedia.org/wiki/Kernel_density_estimation introduces a common technique in this area with many links.
Clearly $E[\\max\\{X,Y\\}]$ is not $\\max\\{E[X],E[Y]\\}$ in general; e.g. consider $X,Y$ being the values on a pair of (distinguishable) dice. $\\max(E(X),E(Y))=3.5$, yet $E(\\max(X,Y))>3.5$.
I do not think you can correlate data from two different surveys. Canonical correlation analysis is not the right tool for that, nor do I think there is any other.
Can you elaborate a little bit about the purpose/context of this particular analysis, or why do you seek an alternative measure of association?
@Wayne, I don't ignore anything, the OP was about the future and not about what is now. In 5 years, we might see many more libraries for stats in Julia than there are now for R. And this, just because Julia has a good chance to be a much better language.
He has X, pretty important info.
@whuber : As I describe in the question, delta, sigma, and N are known, so the variable of integration is x. I didn't know how to write this in notation, but it should be integrated from 0 to 1. As for it being correct, try it yourself, using the program R:
T <- function(x) {
pt(((7*sqrt(108))/20)-qt(1-(0.5*x),499),499)
}
integrate(T,0,1)$value
Where, delta is 4, sigma is 20, and N is 108. Compare the results with this code to the results you get from the code I gave previously. It does work!
The argument for delaying travel is the same one appearing in the old joke about how the TSA found a statistician with a bomb on board an airplane. When asked to explain herself, the statistician said "Well, the odds of one person having a bomb are small yet not small enough for comfort, but the odds of *two* people having a bomb are infinitesimal. Therefore when *I* bring a bomb, there is almost no chance there will be two bombs and we will be perfectly safe."
The more control the better. That is, if I can control the number of vertices and edges, it would be nice. The graph needs to be connected.
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"Because of the constraint imposed, and the nature of the problem, the set of 4 probability estimates would be drawn from a Dirichlet distribution." Where does the above statement fit in your math above? Also, another probably novice question: what does variance or confidence interval around probability mean. I thought if I have mean of something I would like to know the 95% confidence interval for mean to be bounded. Thanks for your time.
is this tl;dr version correct-ish: "Shalizi's paper is just a specialized restatement of Maxwell's demon"?
There's nothing problematic with a negative lower bound for a non-negative variable from a mathematical point of view. The important question is whether this is evidence that the prediction interval procedure in use might be a poor one in general or inappropriate for this phenomenon in particular. Have you performed the usual regression diagnostics, including goodness of fit and distributional evaluation of the residuals?
I am not sure why that situation would be different from others with multiple comparisons
@Ringold, what worked? The prior on sigma or the update? Or both? Have you tested them separately?
Yes, you're right.  It is possible to generate a similar looking plot with random data drawn from a normal distribution.
Or do you mean that a single datum can belong to more than one category?
Thank you for your contribution.  It seems to me it fails to address the question, though, because (1) the O.P. is interested in the CDF, not the PDF.  (2) He asks for a quantitative answer.
Thanks, they're just Excel-but with absolutely no defaults left over.
I cannot install copula, so cannot tell what the issue is but indeed the notion is to use the conditional for $X_t$ given $X_{t-1}$ derived from the copula: compute $u_{t-1}=F(x_{t-1})$, generate $u_t|u_{t-1}$ from the copula, computer $x_t=F^{-1}(u_t)$.
-1: Thank you for the insight! However, this does not answer my question. Because it's relevant commentary/discussion on a statement I made in the background to my question, your answer probably belongs in a comment, instead of in an answer.
The "treatment" in this case is the wheather. How would you manipulate the wheather for your experiment?
I was not completeley correct in the formuation. i meant the foolowing. The regression would be always the same (in sense of $\\mathbf{R^2}$) if you perform only linear transformations of the data. But if you want to determine which variables are crusial and which are almost noisy the scale matters. It is just convinient to standartize variables and forget about their original scales. So regression is "sensetive" in terms of understanding relative impacts.
Why are you calculating sample size if you only have 45 students? Why not use all 45 students and calculate the power of your test?
I wonder about the correctness of these assertions, given that "most of the popular frequentist estimators" are ML and they tend to be biased (depending on the parameterization). Moreover, a good frequentist is deeply concerned about loss and admissibility; a key part of this theory recognizes that admissible procedures come from Bayes procedures, whence--at least in that broad sense--the very heart of frequentist theory relies on Bayes estimators! I might be persuaded to your point of view if you could be clearer about "often," "most," and "rarely," and back that up with evidence.
It is hard to tell whether you are asking about the effects of re-expressing "predictors"--independent or explanatory variables--or response (dependent) variables (or perhaps both?).  Please edit your question so the distinction is made clear.
Hi, as a follow up: The tost test (in R) asks for a magnitude of region of similarity. Any idea what this is?
It would help if you would cite the paper that you're trying to understand.  It would also help to know the form of the likelihood function (which might well be a sum of squares, depending on the details of the model.)
There is every reason not to expect one.  After all, there must be at least one term involving $\\sqrt{p}$ for every prime $p$ less than the maximum value of the variable; in the expectation it will be multiplied by a rational number (the probability) and (as a consequence of Galois theory) those terms will not simplify in the summation.
@brianjd: I don't believe so. Let $g$ be a continuous strictly monotonic function (hence, having a well-defined inverse $g^{-1}$) that satisfies your conditions. Then, it must be that $\\renewcommand{\\Pr}{\\mathbb{P}}\\Phi^{\\alpha}(u) = \\Pr(g(Z) \\leq u) = \\Pr( Z \\leq g^{-1}(u)) = \\Phi(g^{-1}(u))$ and so $g^{-1}(u) = \\Phi^{-1}(\\Phi^{\\alpha}(u))$. So the inverse is identified fairly explicitly, but not $g$ itself. This is what I meant in my previous comment about $g$ being found *implicitly*.
I think it would be helpful if you try to define what "long-term" means for you, or what kind of behaviors you are interested in. For example it is not clear to me if by "long-term" you are referring specifically to something which is related to time or just to asymptotic behavior (where "long-term" would be interpreted as "much more observations").