tttjjj/elt_bench huggingface dataset:
@statsapprentice - you need to take the square root of 0.867 to get your standard error; and then you also need to multiply that by 1.96 to turn it into the radius of a confidence interval.  And that is for the difference between the two, so you will get 2% +/- 1.8% or so.  The reason zero is particularly interesting is because a difference of zero would mean there is no difference between grant-receiving and no-grant hospitals.  As 2.0+/-1.8 is definitely bigger than zero you can in theory dismiss that - but only just.
"*But does this mean that the means are equal or that we have no evidence to suggest otherwise?*" -- the second. In no way does it imply the first. Indeed, in a very practical sense it can almost never be the case (an exception might be where population parameters may be discrete for some reason).
Not sure what exactly you are looking for. But if your question is answered by "I'm 95% certain that in 95% of the 130 part samples the number of defective parts will be between x and y" then you need to look up *tolerance intervals*.
Good question.  Would this be better on the stats.SE site?  Though it is R/Matlab in nature, it's primarily statistical in nature.  If I were looking up answers to this question, I'd head to CV rather than SO.
@DilipSarwate: I think a lot of the stuff over there is a bit over my head at the moment (although it may just be that I'm not familiar enough with the terminology). If possible, could you point out where I'm going wrong here?
They should be the same. Can you post the full question? Also, if this is homework, please add the self-study tag
The reference is: Cameron and Trivedi (2009), **Microeconometrics** Using Stata.
+1 very nice answer.
@Glen_b: "area" of a region is the Lebesgue measure of the region.
Do you have a rough idea of the size of the proportion $Y$?
@AdamO That's a fair translation of *ad libitum*, which I suggested. I think  *ad liberandum* has quite different overtones, as shown by its Christian liturgical uses.
Please explain more clearly how your situation differs from the one your referenced, which appears *identical* to me.
Yep, that answers it.
@Alexis: There is no special reason, I am only trying to reproduce the results of a paper with R.
(+1) This sort of discrepancies usually appear when a *stopping rule* is involved in one of the models.
@shabbychef: Most of the people grasp it in the _worst_ possible way i.e. probability of making Type I error.
Thank you! A corollary question really relates to how to perform the t-test in this particular situation. I am looking at the agreement between different tasks by group (again, two groups). An example is as follows: for the "select item" task, the horizontal group has 0.36 agreement and the vertical group has 0.33 agreement (that is, there is one observation per group per task). Is there any way to test the difference between the groups for specific tasks or can I only look at differences in the average task agreement by group? I think the latter, but this isn't very interesting.
@DWin: No. Actually I have a number of different data sets which each have different associated constants; 0.6 is merely illustrative.
Hi @Dennis, they give the same result. set.seed(1234); 
x <- rbinom(100,1, 0.2);
y <- rbinom(100,1, 0.2);
chisq.test(table(x, y), correct=FALSE);
summary(glm(y ~ x, family="binomial"));
@whuber:Progression Free Survival. Lot of clinical trials report these statistics. progression free survival rate at 1 year was 60% or something. I have data with tumor assessments at different timepoints. and based on whether subject progrssed or died or stopped treatment their censoring indicator can be programed. It is a typical oncology trial.
@January No, a chi-square test is not best here. Even though weather is categorical, number of goals is not. So, a chi square test is, at best, underpowered. However, in addition, there is a dependent and independent variable, which chisquare does not capture.
Solid. I'd add (1) Variance-component models are explicitly designed to model within cluster correlations (2) as above I've usually used the term for random effects models without covariates.
I don't know why this hasn't been answered yet. My only two suggestions are to take out the g-power bit. It seems like your questions are basically unrelated to that software and it may discourage somebody who is familiar with power analysis and not g-power. Second, it may help to break your questions into two or three separate questions. Just a thought. Good luck.
Note that your distribution for $X_i$ is conditional on only knowing $X_{0}$.  if you know say $X_{i-s}$ then the distribution is $X_{i}|X_{i-s}\\sim N(X_{i-s},(i-s)\\sigma^2)$.  although i'm not sure if they mean parameter is "drift" or the starting point.  If its drift then you need to add $(i-s)\\mu_{0}$ to the mean.  Interesting question though.
Well, as Hannig himself mentions, "fiducial inference" is now called "generalized inference" and a search for the latter brings up quite a lot of results. Not that I can suggest anything though!
I was trying to make the point that it isn't, or isn't normally thought of as such, but it's late and i'm too tired to word it well...
I don't really have a full answer, but I will mention that I would not usually think of using either method to "choose the number of features".  In general, I interpret Machine Learning and Bayesian Statistics to just include all the features since they all likely have some level of minimal impact.  However, I think the question of relative model complexity is still appropriate.  I will also state I've never actually performed the Bayesian inference you allude to; it just seems to get too messy in practice compared to the simplicity of k-fold or bootstrapping.
I did not say I was confused. If each percent is itself a summary of a sample, then you need the original data. Your summaries do not appear highly skewed to me. Are you measuring proportions or reporting the proportions of a binary variable?
@user3787792 http://stackoverflow.com/questions/6011943/java-normal-distribution
Well - they're asymptotically normal so for small sample sizes it might not quite be normal.
Does each of the 25 patients have 19 electrodes? In that case, I'd imagine you'd have two groups with $19 \\cdot 11$ and $19 \\cdot 25$ points according to the student's strategy. Can you clarify the data structure a bit more?
I now recognize that the answer to my original question depends on algebraic derivation rather than discussion of how a computer program will solve this kind of problem. Whatever answer a computer will arrive to for a given time series, it still might not make sense if one could algebraically prove that the MLE of GARCH(p,0) model yields all sigmas to be equal to the same constant (which means conditional homoskedasticity). Perhaps I should try my luck in solving this algebraic problem after all... :)
could you share how you decided to proceed to run stl decomposition on your weekly data? I was thinking of maybe averaging w52 and w53 if w53 exists?
"*When I simulate normal data in R, I make sure that the sample have the exact mean and sd of the sampling distribution*" --- Except that by doing so, you guarantee it's not actually normally distributed. Is that actually what you want here? And if you do want to absolutely guarantee it's *not* actually a random sample from a binomial, why bring up 'binomial' at all?
To expand a bit on that comment - the error message seems quite clear, so the question then becomes what is `fourier1` doing. You might have made a mistake in programming it.
nice and cool answer.. this will surely help me.. thanks a lot :D
It's hard to imagine a stacked barplot with SE bars (at least for me!). Why not simply use a [dotplot](http://www.statmethods.net/graphs/dot.html)?
+1: Wow, it worked right away! Thank you!
+1 @Glen_b, this is useful! Well, the data aren't implausible, so I believe the mislabeled identification may have something to do with observations closer to the decision boundaries (well separated from its class means). Here is my understanding now, for (1): any model based outlier measures should do the trick. (2) can be worked upon by having multiple models and taking an intersection. Those obs found in (2) can be considered unlabeled and imputed (or predicted).
Could you provide an example? In general your idea seems valid, but  answer with more details would be of great help not only for the asker but for other people who will come to this site with similar problems.
I'm a little confused by the expectation not being 1. I feel I am misinterpreting your answer at http://stats.stackexchange.com/a/96544/9162. If under the null hypothesis the numerator and denominator of the F-ratio are estimators of the population variance, what is the reason the expected value of F is not 1? Is it because they're biased estimators of the population variance?
And would not that rule of not doing any hypothesis testing with fpc be counter to the practice of survey analysis where it seems a fpc is routinely used - for example surveyreg in SAS or the svyglm in R?
@whuber: Thanks for the software suggestion. I got the double exponential to fit all right (you can see it in the plot), but the fit was not good enough in my opinion so I sought a better model. Did you find a better fit with the mixture model?
You've only described the standard deviation's interpretation in the case of the normal distribution. The '68% rule' and (and 95% rule) only apply for normally distributed data. At least state that the two bullet points are only true if haircutting times follow a normal distribution.
Isn't a PCA applied on a correlation matrix equivalent to an MDS with euclidean distances computed on standardized variables?
Thanks for a nice answer. I think that what I was wondering about in the original question is why the square-law detector (the term Dilip Sarwate used to describe my test in his comment) works so well.  Besides the test statistic having well-studied distributions under either hypothesis, can we say something about the test itself? While obviously, using the mean of the observation yields no information, why can't we do better by taking, each observation to, say the power of 1.2 (using the magnitudes of complex numbers) or, say power of 4? Why does squaring work so well?
Actually, research exists that supports the ability of software engineers to produce reasonably accurate estimates in groups, hence the creation of Planning Poker, Wideband Delphi, etc. My intuition tells me that coarse rank ordering is probably easy, but fine-grained rank ordering is hard. This is why I am trying to find a measure for "reasonable proficiency." Otherwise, I'll have to do that research too!
It's easy to come up with proportions but I would need confidence intervals? How would I do that? Use the binomial for each proportion and then take a multiple comparison adjustment?
"classification tasks in which you have a number of probabilistic cues which need to be integrated." Per your suggestion, I have been doing some preliminary tests in this area and am getting good results. I am wondering if you could give me some examples of practical problems in this domain, or of some benchmarck datasets.
+1 for Baltic letter, how do you know it? Note that `ū` has a smaller bar compared to $\\bar{u}$, so the end result will depend on the font, it might not look good.
This is probably obvious, but I'd add that for real-world observations, you're testing for lack of a significant difference from zero at some level of confidence. Real populations have a finite size, so chance variations are not guaranteed to average out exactly to zero, even in the total absence of any relationship between the variables.
I'd say it's to give a bit of smoothing, and to avoid artifacts which might come from having the intervals (open, closed] or [closed, open).
In text files, I think that "diff" function that aligns characters, or indicates changes with weights, might be useful preprocessing of "close-together" texts in order to improve meaningful clustering
I'm very sorry I don't know R very well which command would you like to see output from?
I wouldn't evaluate the status of a literature based on a google search. I'm not quite sure small area estimation is what the author would be looking for in this instance. As far as I understand, that literature is focused on either making predictions of characteristics in small areas ([Kriegler & Berk, 2010](http://www-stat.wharton.upenn.edu/~berkr/homeless%20copy.pdf)), or estimating parameters in multi-level models based on a sparse number of samples per aggregate unit.
It is not the same 5 pages for each; there is no difference between (1,1,0,1,0) and (1,1,1,9,0). It's just the distribution of how many views they did for each of the 5 pages they were offered. Neither number of pages nor range is exactly what I want, since neither fully captures the difference between someone who looks at many pages once each and someone who looks at one page several times. Maybe I can report both?
Thank you Alecos! I really appreciate your answer!
What is your goal on generating a probability model for your percentage? Is it possible to generate the necessary descriptives and inference without imposing assumptions about the nature of the data generating process?
I don't think the question is either about the correctness of the algorithm nor about graphical representation. The word "resemble" most likely is intended in some vague sense of "is likely to have all the salient statistical characteristics of." This is the issue @Glen_b is pursuing in comments to the question: it needs clarification from the O.P.
(+1) This is clearly a good approach. But you ought to double-check this work. Starting at the beginning, your expression for the likelihood function causes each observation to appear *twice*: once for $(i,j)$ and again for $(j,i)$. Thus it seems you have written down the square of the likelihood.  The solution will still be the same, but any confidence intervals (from the information matrix) will be wrong. The quadratic program is *very rapidly* solved. Incidentally, I believe the characterization of positive-definiteness may be equivalent to the graph having a unique connected component :-).
No other opinions about this? Or is the question to dumb?
Express the permutation in factorial form, then replace factorial with Gamma function. LnGamma is a common function for stats programs (even Excel) to provide.
If there are any missing values in the categorical predictor then R will eliminate them automatically from the model and they will be treated as though they don't exist - this may be a reason why the null model has a larger log-likelihood (since the likelihood scales with the sample size, and the sample size is smaller). If this is the case, the log-likelihoods are not comparable and you'd have to remove the missing cases before doing the test to make them comparable.
There are already many answer; so as an aside I would just like to notice that I find the "holding constant" rule very dangerous in practice when people start interpreting their coefficient causally at the same moment. For example, regression earnings over gender and many other factors and concluding that gender has no significant influence; even though gender also may have tremendous influence on other significant predictors. It's not really possible to hold other factors constant in practice.
Can you amplify your last sentence?
I needed to modify the code but the idea was same. Thanks.
@B_Miner I think it's a very minor infraction, but I am not on your thesis committee :P
I thought the Bayesian statistician simply took into account that the chances of the sun exploding are much, much smaller than the chances of the machine lying (so, not necessarily a clueless gambler).
31 models for each day of the month? mmm.... possible, but let me try other ways first...
Could you provide more details of the experimental design and your hypotheses?
As an exercise, constant rate, though it's not.  I have 1000 measured data points, clustered along rivers, so there is an underlying weighted surface.  I'm trying to write a script like spatstat.Gest: compare a CDF of the nearest nbr distances with that of a random set of pts, but my "random" x,y's may not be random in the right (Poisson) way.  I just can't locate the Poisson randomness; if x and y were Poisson distributed, wouldn't they cluster around the centroid?  The distribution must be around the mean intensity at each pixel, but it's a slippery concept; I can't pin it down.
I appreciate that this is something of a "soft-question" but there is no tag on this part of stackexchange to cover this, else I would have used it.
Are you asking if there's an analog of (say) Cohen's kappa for interrater reliability when you have only one rater?
I should point out that this is a slightly abridged cross post of a question that I asked the bioconductor mailing list last week (but didn't get replies to): http://permalink.gmane.org/gmane.science.biology.informatics.conductor/56110.  Not sure on the cross posting etiquette.
Bray-Curtis dissimilarity may do the job... A great ressource is the book 'Numerical Ecology' by Legendre & Legendre (1998).
This is a nice answer.  The quantity $h(m) = \\log 1/p(m)$ can be conceived as the "surprise" of seeing message $m$ (it's a decreasing function of the probability of encountering message $m$ -- if we always observe this message then $p(m)=1$ and surprise $h=0$.).  This quantity averaged over the distribution of all messages $p(m)$ is the Shannon entropy.  So the surprise of seeing $m$ is the contribution of that message to the total entropy, but it's not really an "information" in the classical sense.
$P(X<Y)$ is equivalent to $P(X-Y<0)$; in turn, that's $P(X+(-Y)<0)$ which equals $P(X+(n-Y)<n)$, and $(n-Y)\\sim \\rm{Bin}(n,1-p_2)$; each can be useful. If $n$ is not really large you can do the last by numerical convolution; I've done some decent-sized ones in R on negative binomials in the past; binomials are even better suited to it. If $n$ is not small and the $p$'s are not really close to the boundaries (the two are linked), you can approximate this probability pretty well using normal distributions (I'd suggest using a continuity correction on the difference $X-Y$ and comparing to 0).
And probability.  Shannon-Macmillan-Breiman.
@rolando2, could you show an example of where chi-square / phi correlation would be better than logistic regression for estimating effects of other binaries? I use logistic regression on survey data quite frequently `1=increasing / 0=decreasing ~ personality attributes (0 or 1)...` to understand the effect of personality attributes on likelihood to increase or decrease business with a company.
Actually the variance of the outcome in a random-slopes model is more complicated than the expression that you gave. In particular, the variance is a function of the predictor (i.e., different variances at different values of the predictor). This is discussed by Goldstein et al., which also gives the correct expression for the variance: http://www.bristol.ac.uk/cmm/research/pvmm.pdf
Why do you want to log transform them?
I'd also recommend Alan Pankratz's other book: Forecasting with Dynamic Regression Models. Very similar material, but covers a bit more ground; albeit in less detail on the Box-Jenkins side of things. Great to hear that there's going to be a 2nd edition in Jan 2014!
+1 Scaling the time *up* is tantamount to assuming the rate of 5 deaths per month is going to continue throughout the year and actually produce 60 deaths. That's simply making up 11/12 of one's data, which is why such a ridiculous answer ($10^{-56}$) could be derived.
Stated another way: Pearson's correlation coefficient between two random variables $\\zeta$ and $\\eta$ will be +/-1 when you can find an $a$ and $b$ such that $\\zeta = a + b \\eta$. This can be easily done if you only have one sample for each variable by setting $a$ as the difference between the two variables.
@Jonathan check out the packages gbm, mboost, ada, and caTools for various boosting methods in R.
I like ellipses in plots, so this was fun to see these functions in action.
Thanks for adding a real illustration with ppf.
Nice rant. Would've been much easier to read if you'd used the occasional caps though...
A fair evaluation of any analysis involving ratios will assume all quantities are positive (although the OP did not stipulate this): ratios rarely are meaningful otherwise.
In the `nnet` the initial weights are all initialized to a uniform random number between -0.7 and 0.7 if I recall correctly.  And you can control the magnitude in a parameter.  I've honestly had solid luck with the `nnet` package and have never tried any of the other options.  Best of luck!
I'm not sure about the two-sample problem, but as for references for the former, see:
A. P. Dempster, [Generalized $D_n^+$ Statistics](http://projecteuclid.org/euclid.aoms/1177706275), *Ann. Math. Statist.* Volume 30, Number 2 (1959), 593-597. Also, [here](http://www.springerlink.com/content/ln8hj5273837522p/), [here](http://www.springerlink.com/content/mrx551586853v724/) and [here](http://projecteuclid.org/euclid.bsmsp/1200514106) might have some material of interest. The first and last links should be to publicly available articles. Sadly, the middle two aren't.
-1 for increasing the alpha level. I don't think it should always be at 0.05 but doing an analysis, seeing no significance and then just deciding to adhoc increase the alpha level is terrible advice.
By the way, hi Tom!
Have you already decided against using t-tests? What I mean is 1) throw out the data from the control group, 2) ignore the different levels of `treatment`, 3) for each person average over levels of `prePostFup`, 4) for each person average over hours 1,2 (= data group 1) as well as over hours 3,4 (= data group 2), 5) run t-test for 2 dependent groups. Since Maxwell & Delaney (2004) as well as Kirk (1995) discourage doing contrasts with a pooled error term in within-designs, this could be a simple alternative.
Does my edit help clarify? The first tests for differences in risk/change, the second tests for differences in prevalence/levels among matched pairs.
@Tim: I'm not totally certain, but in climate circles they talk about projections. I'd like to say that projections are conditional forecasts (conditioned on a specific scenario), but it's evidently more complex than that. A regular forecast is always conditional on "other things remaining the same", of course.
Could you please be more specific in what context you need to know that, whether this is a homework question and what is wrong with http://en.wikipedia.org/wiki/Affine_transformation ?
@John do you have the time to join me in chat about this: http://chat.stackexchange.com/rooms/9858/meaning-of-insignificance
Perhaps to adjust for distinct unit-level time trends caused by unobserved factors.
Ok, what if instead of comparing the maxima I'll try to compare their quantiles (e.g. 0.99)?
Welcome to the site! You cross-posed it [here](http://math.stackexchange.com/questions/508013/normal-distribution-with-less-probability). Please note that, according to the [help center](http://stats.stackexchange.com/help/on-topic), "cross-posting is not encouraged on SE sites. Choose one best location to post your question."
I think this question is more general that that one, so I am voting to leave this open.
@gung- yeah, but try telling that to a business client :-)
Alex, when you say "For small sample sizes, you should definitely use the unbiased estimator." -- if I care about the invariance properties of MLEs (what's ML for the variance corresponds to ML for the precision and the standard deviation - that is, $\\widehat{g(\\theta)}=g(\\hat \\theta)$, which doesn't hold for unbiasedness nor MMSE), or I prefer smaller MSEs to larger, why should I still 'definitely use the unbiased estimator'? If I am calculating variance in order to obtain the standard deviation, say, or the precision, the $n-1$ divisor is not unbiased for those.
(+1): Thanks @neverknowsbest for your answer. Please my edits too. Thanks