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5571 | 1 | 5579 | null | 12 | 6431 | I recently fit 4 multiple regression models for the same predictor/response data. Two of the models I fit with Poisson regression.
```
model.pois <- glm(Response ~ P1 + P2 +...+ P5, family=poisson(), ...)
model.pois.inter <- glm(Response ~ (P1 + P2 +...+ P5)^2, family=poisson(), ...)
```
Two of the models I fit with n... | Comparing regression models on count data | CC BY-SA 3.0 | null | 2010-12-16T18:04:08.740 | 2014-12-10T12:22:50.943 | 2014-12-10T12:22:50.943 | 56216 | 1973 | [
"regression",
"aic",
"count-data",
"likelihood-ratio",
"model-comparison"
] |
5572 | 1 | 5584 | null | 8 | 531 | I am planing a pre-post treatment-control design study with a large number of pre-treatment measurements. I have subjects divided into a control group and a treatment group. For both groups, I will collect hourly data for one year prior to the start of the treatment and then continue collecting data for another year. T... | Taking advantage of many pre-treatment measurements | CC BY-SA 2.5 | null | 2010-12-16T18:16:48.147 | 2010-12-17T21:50:50.030 | 2010-12-17T21:50:50.030 | null | 743 | [
"mixed-model",
"experiment-design",
"repeated-measures",
"panel-data"
] |
5573 | 1 | 5577 | null | 7 | 769 | I would like to create an "equilibrated histogram" with roughly the same number of data points in each bin. The second restriction I have is that I should have only 4 bins. Given the following list of numbers below, how can I achieve that?
-2.153, -1.732, -1.699, -1.559, -1.355, -1.306, -1.151, -1.129, -0.636, 0.4085, ... | How to build an "equilibrated histogram"? | CC BY-SA 2.5 | null | 2010-12-16T20:06:26.387 | 2013-06-18T18:12:49.263 | 2010-12-16T20:49:03.377 | 696 | 64 | [
"r",
"histogram"
] |
5574 | 2 | null | 5563 | 5 | null | As @David mentioned, R is not doing symbolic calculations on its own, and such jobs should be done in other programs or with wrapper packages.
Yet, calculating limit numerically is usually quite easy -- in most cases you just need to find argument $n$ large enough so the difference between $f(n+1)-f(n)$ is smaller than... | null | CC BY-SA 2.5 | null | 2010-12-16T20:10:54.397 | 2010-12-16T20:10:54.397 | null | null | null | null |
5575 | 2 | null | 5573 | 7 | null | You are looking for [quantiles](http://en.wikipedia.org/wiki/Quantile); in R there is a function `quantile` that will calculate them for you; `Hmisc` R package provides `cut2` function which explicitly calculates such "equilibrated bins".
| null | CC BY-SA 2.5 | null | 2010-12-16T20:14:35.177 | 2010-12-16T20:14:35.177 | null | null | null | null |
5577 | 2 | null | 5573 | 8 | null | To follow up on @mbq's suggestion, here's the code to do that with R:
```
require(Hmisc)
x <- c(-2.153, -1.732, -1.699, -1.559, -1.355
, -1.306, -1.151, -1.129, -0.636, 0.4085
, 0.5408, 0.5731, 0.5842, 0.6206, 0.8175
, 0.8274, 0.8710, 1.3214, 1.5552, 2.2342
)
eqBins <- cut2(x, g = 4)
#what are the bins and how many i... | null | CC BY-SA 3.0 | null | 2010-12-16T20:48:42.173 | 2013-06-18T18:12:49.263 | 2013-06-18T18:12:49.263 | 919 | 696 | null |
5578 | 2 | null | 5364 | 2 | null | I discovered a paper by Reiss, et. al.,
[Partial least squares confidence interval calculation for industrial end-of-batch quality prediction](http://dx.doi.org/10.1016/j.chemolab.2009.11.003), in which appears the quote:
>
The PLS prediction should be accompanied by an online confidence
interval to indicate the acc... | null | CC BY-SA 2.5 | null | 2010-12-16T21:07:01.267 | 2010-12-16T22:27:47.410 | 2010-12-16T22:27:47.410 | 795 | 795 | null |
5579 | 2 | null | 5571 | 16 | null | You can compare the negative binomial model to the corresponding Poisson model with a likelihood ratio test. A Poisson model is equivalent to a negative binomial model with an overdispersion parameter of zero. Therefore they are nested models and likelihood ratios are valid. The complication is that the overdispersion ... | null | CC BY-SA 3.0 | null | 2010-12-16T21:47:11.920 | 2013-11-18T16:27:07.357 | 2013-11-18T16:27:07.357 | 17230 | 449 | null |
5580 | 2 | null | 5525 | 5 | null | A more general approach is to use the `logLik()` function. It returns an object with the attribute `df` that gives the fitted models degrees of freedom. The benefit of this approach is that it works with many other model classes (including `glm`). In the case of ordinary linear regression (`lm`) this corresponds to ... | null | CC BY-SA 2.5 | null | 2010-12-16T23:00:58.127 | 2010-12-16T23:16:31.660 | 2010-12-16T23:16:31.660 | 1670 | 1670 | null |
5581 | 2 | null | 4551 | 130 | null | Failing to look at (plot) the data.
| null | CC BY-SA 2.5 | null | 2010-12-16T23:13:28.913 | 2010-12-16T23:13:28.913 | null | null | 1670 | null |
5583 | 2 | null | 5572 | 1 | null | Excuse my previous post. I now see that you are not referring to 9000 different covariates.
What I have written does not apply to your situation.
Sincerest apologies.
Paul
There is a lot of discussion about matching and dimensionality reduction on pre-treatment covariates that may be worthwhile examining - i.e. prope... | null | CC BY-SA 2.5 | null | 2010-12-17T04:14:41.050 | 2010-12-17T05:49:56.120 | 2010-12-17T05:49:56.120 | 2238 | 2238 | null |
5584 | 2 | null | 5572 | 5 | null | This is not a complete answer, but just a few thoughts:
- More pre-treatment measures should increase the reliability of your measurement of baseline differences. Increasing reliability of measuring baseline differences should increase your statistical power in detecting group differences (assuming a real effect exist... | null | CC BY-SA 2.5 | null | 2010-12-17T05:13:59.897 | 2010-12-17T05:13:59.897 | 2017-04-13T12:44:33.310 | -1 | 183 | null |
5585 | 2 | null | 604 | 1 | null | I am having troubling following your reasoning, but here are some things you should consider.
Generally, the harder you fit a model to your training data, the worse the model will perform on independent validation data sets. By over-fitting the model to the training set, you risk capturing predictor-response relationsh... | null | CC BY-SA 2.5 | null | 2010-12-17T05:45:18.537 | 2010-12-17T05:45:18.537 | null | null | 2144 | null |
5586 | 1 | null | null | 3 | 3728 | I have several OLS models with robust s.e.'s that predict an outcome variable Y. For instance:
Model 1:
$Y=B_0 +B_1X_1$
Model 2:
$Y=B_0 + B_1X_1 + B_2X_2$
Model 3:
$Y=B_0 +B_1X_1 + B_2X_2 +B_3X_3$
I am interested in giving an average effect for $B_1$ across Models 1-3 with an accompanied 95% CI.
Can I just take the a... | Average effect of coefficients across multiple linear models? | CC BY-SA 3.0 | null | 2010-12-17T06:31:41.993 | 2021-02-04T22:51:05.250 | 2016-08-15T16:27:59.967 | 22468 | null | [
"regression",
"confidence-interval",
"linear-model",
"regression-coefficients",
"mean"
] |
5588 | 2 | null | 5514 | 3 | null | Conditional logit models for unit $i$ in group $j$ of size $G$, $P(Y_{ij}=1|\sum_{k=1}^GY_{gj}=M)$ for some $0<M<G$. So, suppose you have a group of size 2 in which there is one success. Then, group member number 1's contribution to the likelihood is given by $A^{Y_{1j}}B^{1-Y_{1j}}$, where,
$$
\begin{matrix}
A & = ... | null | CC BY-SA 2.5 | null | 2010-12-17T06:57:01.517 | 2010-12-17T13:15:20.630 | 2010-12-17T13:15:20.630 | 96 | 96 | null |
5589 | 2 | null | 5586 | 4 | null | If these 3 models are estimated from independent samples, then you can assume that $\beta_1$ are independent for these 3 models. Then you can average them. The standard error of the average then will be the square root from the average of the squares of the standard errors.
However you should check if you do not have ... | null | CC BY-SA 2.5 | null | 2010-12-17T07:11:15.503 | 2010-12-17T07:11:15.503 | null | null | 2116 | null |
5590 | 2 | null | 5571 | 5 | null | I believe `anova()` in R can be used for this. Despite its name, it's a likelihood ratio test. Crawley in his [The R Book](http://www.bio.ic.ac.uk/research/mjcraw/therbook/index.htm) has some examples of usage.
| null | CC BY-SA 2.5 | null | 2010-12-17T07:16:03.467 | 2010-12-17T07:16:03.467 | null | null | 144 | null |
5591 | 1 | 5592 | null | 48 | 7219 | Roughly speaking a p-value gives a probability of the observed outcome of an experiment given the hypothesis (model). Having this probability (p-value) we want to judge our hypothesis (how likely it is). But wouldn't it be more natural to calculate the probability of the hypothesis given the observed outcome?
In more d... | Why do people use p-values instead of computing probability of the model given data? | CC BY-SA 3.0 | null | 2010-12-17T10:36:49.853 | 2021-11-05T10:51:01.080 | 2016-08-11T20:26:39.243 | 28666 | 2407 | [
"likelihood",
"p-value"
] |
5592 | 2 | null | 5591 | 34 | null | Computing the probability that the hypothesis is correct doesn't fit well within the frequentist definition of a probability (a long run frequency), which was adopted to avoid the supposed subjectivity of the Bayesian definition of a probability. The truth of a particular hypothesis is not a random variable, it is eit... | null | CC BY-SA 3.0 | null | 2010-12-17T11:06:10.647 | 2012-03-14T10:34:51.520 | 2012-03-14T10:34:51.520 | 887 | 887 | null |
5594 | 2 | null | 490 | 5 | null | If you are only interested in generalization performance, you are probably better off not performing any feature selection and using regularization instead (e.g. ridge regression). There have been several open [challenges](http://clopinet.com/isabelle/Projects/NIPS2003) in the machine learning community on feature sel... | null | CC BY-SA 2.5 | null | 2010-12-17T11:13:27.427 | 2010-12-17T11:13:27.427 | null | null | 887 | null |
5596 | 2 | null | 5115 | 7 | null | [Adolphe Quetelet](http://en.wikipedia.org/wiki/Adolphe_Quetelet) for his work on the "average man", and for pioneering the use of statistics in the social sciences. Before him, statistics were largely confined to the physical sciences (astronomy, in particular).
| null | CC BY-SA 2.5 | null | 2010-12-17T12:20:38.070 | 2010-12-17T12:20:38.070 | null | null | null | null |
5597 | 1 | 5600 | null | 26 | 570 | As a biologist, many of the research projects I work on at some point involve collaboration with a statistician, whether it be for simple advice or for implementing and testing a model for my data. My statistics colleagues admit that they do a significant amount of collaboration, insomuch that the tenure review process... | Statistics collaboration | CC BY-SA 2.5 | null | 2010-12-17T12:27:48.893 | 2017-01-21T09:30:29.850 | 2017-01-21T09:30:29.850 | 28666 | 1973 | [
"academia"
] |
5598 | 2 | null | 5597 | 3 | null | Having no preconceived ideas about the method you should use solely based on papers. Their ideas, logic or methods may be faulty. You want to think about your problem and use the most appropriate set of tools. This reminds me of reproducing cited information without checking the source.
On the other hand, paper with me... | null | CC BY-SA 2.5 | null | 2010-12-17T13:49:05.330 | 2010-12-17T13:49:05.330 | null | null | 144 | null |
5599 | 2 | null | 5597 | 10 | null | I think the concept that few scientists grasp is this: A statistical result can really only be taken at face value when the statistical methods were chosen in advance while the experiment was being planned (or while preliminary data were collected to polish methods).
You are likely to be mislead if you first analyze ... | null | CC BY-SA 2.5 | null | 2010-12-17T14:01:35.943 | 2010-12-18T03:00:48.173 | 2010-12-18T03:00:48.173 | 25 | 25 | null |
5600 | 2 | null | 5597 | 13 | null | My answer is from the point of view of an UK academic statistician. In particular, as an academic that gets judged on advances in statistical methodology.
>
What would make me (or any other
scientist) a better collaborator?
To be blunt - money. My time isn't free and I (as an academic) don't get employed to carry... | null | CC BY-SA 2.5 | null | 2010-12-17T15:00:51.533 | 2010-12-18T21:22:28.910 | 2010-12-18T21:22:28.910 | 919 | 8 | null |
5601 | 1 | 5652 | null | 5 | 6067 | I am trying to view the output from the GBM package for boosted trees in R. Below I am fitting a single tree without any sampling in order to compare the tree to the complete dataset. First, create the data set:
```
set.seed(1973)
############## CREATE DATA#############################################
N <- 1000
X1 <... | How to view GBM package trees? | CC BY-SA 2.5 | null | 2010-12-17T15:25:31.870 | 2019-08-20T05:02:05.413 | 2019-08-20T05:02:05.413 | 11887 | 2040 | [
"r",
"boosting"
] |
5602 | 1 | null | null | 2 | 243 | I have data which has several properties (metadata, as key value pairs, where the keyspace is shared over the whole dataset) per object.
I took a sample of objects and divided them in n groups according to an unknown algorithm.
What statistical methods or algorithm exists to find the relevant properties and their weig... | Methods of grouping sets of data | CC BY-SA 2.5 | null | 2010-12-17T17:21:32.020 | 2010-12-21T14:58:31.110 | 2010-12-21T14:58:31.110 | 2423 | 2423 | [
"classification",
"data-mining"
] |
5603 | 1 | 5640 | null | 3 | 775 | How does one approach the problem of modeling a "birth-death process" where the arrivals are dependent on the current state in the following way: if the population is above a certain point, the probability of an arrival decreases.
Basically, I'm interested in complicating (slightly) an existing model of "births" that j... | Modeling a birth-death process that is not memoryless | CC BY-SA 2.5 | null | 2010-12-17T17:24:10.373 | 2010-12-24T02:13:25.680 | 2010-12-23T15:53:46.737 | 446 | 446 | [
"stochastic-processes"
] |
5604 | 1 | null | null | 3 | 2797 | I have a data set that includes the number of visits to a website. Here are some descriptive statistics for my data
Median: 4
Mean: 14.1352
SD: 121.8119
Clearly, there are some huge values (individuals who have visited the site thousands of times.) To remove these outliers I considered simply removing any data that f... | Removing outliers from asymmetric data | CC BY-SA 3.0 | null | 2010-12-17T19:04:04.553 | 2017-04-10T15:45:27.100 | 2017-04-10T15:45:27.100 | 11887 | null | [
"r",
"outliers",
"descriptive-statistics",
"winsorizing"
] |
5605 | 2 | null | 5591 | 6 | null | "Roughly speaking p-value gives a probability of the observed outcome of an experiment given the hypothesis (model)."
but it doesn't. Not even roughly - this fudges an essential distinction.
The model is not specified, as Raskolnikov points out, but let's assume you mean a binomial model (independent coin tosses, fixe... | null | CC BY-SA 2.5 | null | 2010-12-17T19:10:47.387 | 2010-12-17T19:10:47.387 | null | null | 1739 | null |
5606 | 2 | null | 5542 | 2 | null | I propose the following solution to 2), and would appreciate feedback:
- Data include mean, $Y$, sample size $n$, and standard error $\sigma$; calculate precision ($\tau=\frac{1}{\sigma\sqrt{n}}$) because it is required for logN parameterization by BUGS
- data $Y\sim \text{N}(\beta_0,\tau)$
- precision $\tau\sim\tex... | null | CC BY-SA 3.0 | null | 2010-12-17T19:57:51.563 | 2014-04-08T21:43:23.227 | 2014-04-08T21:43:23.227 | 1381 | 1381 | null |
5607 | 2 | null | 5534 | 7 | null | It’s much easier to simultaneously construct $X_i$ and $Y_i$ having the desired properties,
by first letting $Y_i$ be i.i.d. Uniform$[0,1]$ and then taking $X_i = F^{-1}(Y_i)$. This is the basic method for generating random variables with arbitrary distributions.
The other direction, where you are first given $X_i$ an... | null | CC BY-SA 2.5 | null | 2010-12-17T20:13:50.217 | 2010-12-17T20:13:50.217 | null | null | 1670 | null |
5608 | 2 | null | 5591 | 11 | null | As a former academic who moved into practice, I'll take a shot. People use p-values because they are useful. You can't see it in textbooky examples of coin flips. Sure they're not really solid foundationally, but maybe that is not as necessary as we like to think when we're thinking academically.
In the world of data, ... | null | CC BY-SA 2.5 | null | 2010-12-17T20:55:13.000 | 2010-12-17T20:55:13.000 | null | null | 2134 | null |
5609 | 2 | null | 5399 | 5 | null | There are no strong results and it does not depend on Gaussianity. In the case where $x_1$ and $x_2$ are scalars, you are asking if knowing the variance of the variables implies something about their covariance. whuber’s answer is right. The Cauchy-Schwarz Inequality and positive semidefiniteness constrain the possi... | null | CC BY-SA 2.5 | null | 2010-12-17T21:44:43.637 | 2010-12-17T22:05:36.543 | 2010-12-17T22:05:36.543 | 1670 | 1670 | null |
5610 | 2 | null | 4364 | 21 | null | [Stein’s Lemma](http://en.wikipedia.org/wiki/Stein%27s_lemma) provides a very useful characterization. $Z$ is standard Gaussian iff
$$E f’(Z) = E Z f(Z)$$
for all absolutely continuous functions $f$ with $E|f’(Z)| < \infty$.
| null | CC BY-SA 2.5 | null | 2010-12-17T22:00:34.923 | 2010-12-17T23:24:38.617 | 2010-12-17T23:24:38.617 | 1670 | 1670 | null |
5611 | 2 | null | 411 | 4 | null | I think you have to consider the theoretical vs applied advantages of the different notions of distance. Mathematically natural objects don’t necessarily translate well into application.
Kolmogorov-Smirnov is the most well-known for application, and is entrenched in testing for goodness of fit. I suppose that one of... | null | CC BY-SA 2.5 | null | 2010-12-17T22:33:50.943 | 2010-12-17T22:42:22.920 | 2010-12-17T22:42:22.920 | 1670 | 1670 | null |
5613 | 2 | null | 5591 | 13 | null | Your question is a great example of frequentist reasoning and is, actually quite natural. I've used this example in my classes to demonstrate the nature of hypothesis tests. I ask for a volunteer to predict the results of a coin flip. No matter what the result, I record a "correct" guess. We do this repeatedly unti... | null | CC BY-SA 2.5 | null | 2010-12-18T05:56:43.220 | 2010-12-18T06:11:01.040 | 2010-12-18T06:11:01.040 | 485 | 485 | null |
5614 | 1 | null | null | 8 | 1746 | While it is easier to use the Pearson chi-square/Cressie-Read type test, I would like to test the equality of proportions in $k$ categories across two groups using a Kolmogorov-Smirnov type test of the form proposed by [Pettitt & Stephens (1977)](http://www.jstor.org/stable/1268631) (see also [here](http://www4.gu.ed... | Two-sample permutation Kolmogorov-Smirnov tests | CC BY-SA 2.5 | null | 2010-12-18T06:56:55.773 | 2010-12-19T06:48:42.843 | null | null | 2399 | [
"hypothesis-testing"
] |
5615 | 2 | null | 5604 | 6 | null | Don't remove any outliers until you explore the data a bit further. I suggest that you should do a log transform on the data and see whether it becomes more nearly symmetrical--the outliers may not be as extreme as you think. (Log values make perfect sense if there is some sort of power law at play.)
| null | CC BY-SA 2.5 | null | 2010-12-18T07:14:17.327 | 2010-12-18T07:14:17.327 | null | null | 1679 | null |
5617 | 1 | 5622 | null | 23 | 73411 | Let's say we have two factors (A and B), each with two levels (A1, A2 and B1, B2) and a response variable (y).
The when performing a two way ANOVA of the type:
```
y~A+B+A*B
```
We are testing three null hypothesis:
- There is no difference in the means
of factor A
- There is no difference
in means of factor B
... | What is the NULL hypothesis for interaction in a two-way ANOVA? | CC BY-SA 2.5 | null | 2010-12-18T13:50:36.403 | 2015-04-24T09:09:14.283 | 2010-12-18T16:12:53.090 | 930 | 253 | [
"hypothesis-testing",
"anova"
] |
5618 | 2 | null | 5617 | 10 | null | An interaction tells us that the levels of factor A have different effects based on what level of factor B you're applying. So we can test this through a linear contrast. Let C = (A1B1 - A1B2) - (A2B1 - A2B2) where A1B1 stands for the mean of the group that received A1 and B1 and so on. So here we're looking at A1B... | null | CC BY-SA 2.5 | null | 2010-12-18T14:14:19.383 | 2010-12-18T15:16:35.160 | 2010-12-18T15:16:35.160 | 930 | 1028 | null |
5619 | 1 | null | null | 5 | 10058 | I know that $r$ is itself a measure of the effect size, but I would like to know if using Spearman's rank test I can argue that the relation between X and Y is significant with $r = 0.33$ and that the effect is medium, as I do with Pearson test.
| Effect size of Spearman's rank test | CC BY-SA 2.5 | null | 2010-12-18T15:33:30.063 | 2010-12-21T17:02:01.640 | 2010-12-18T16:07:11.407 | 930 | null | [
"correlation",
"effect-size"
] |
5620 | 1 | 5621 | null | 9 | 8404 | I'm reading "[The Elements of Statistical Learning](http://statweb.stanford.edu/~tibs/ElemStatLearn/download.html)" and early on there are references to p-vectors (page 10) and K-vectors (page 12).
What exactly is meant by a p-vector and K-vector?
| p-vector and K-vector | CC BY-SA 3.0 | null | 2010-12-18T15:34:36.363 | 2015-10-14T14:50:47.170 | 2015-10-14T14:50:47.170 | 919 | 1212 | [
"mathematical-statistics",
"terminology"
] |
5621 | 2 | null | 5620 | 11 | null | It's merely some generic notation for a vector of $p$ attributes or variables observed on $i=1,\dots, N$ individuals, so that you can define $X^T = (X_1,X_2,\dots,X_p)$ as a vector of inputs, in the feature (or input) space (and each individual will have one such vector of observed inputs).
The $K$ notation seems to b... | null | CC BY-SA 2.5 | null | 2010-12-18T15:53:41.117 | 2010-12-18T15:53:41.117 | null | null | 930 | null |
5622 | 2 | null | 5617 | 19 | null | I think it's important to clearly separate the hypothesis and its corresponding test. For the following, I assume a balanced, between-subjects CRF-$pq$ design (equal cell sizes, Kirk's notation: Completely Randomized Factorial design).
$Y_{ijk}$ is observation $i$ in treatment $j$ of factor $A$ and treatment $k$ of fac... | null | CC BY-SA 2.5 | null | 2010-12-18T17:38:48.130 | 2010-12-20T09:24:11.373 | 2010-12-20T09:24:11.373 | 1909 | 1909 | null |
5623 | 2 | null | 5619 | 3 | null | With increasing sample size $n$, $r_{z} = \sqrt{n-1} r_{S}$ is asymptotically $N(0, 1)$ distributed (standard normal distribution). In R
```
rSz <- sqrt(n-1) * rS
(pVal <- 1-pnorm(rSz)) # one-sided p-value, test for positive rank correlation
```
| null | CC BY-SA 2.5 | null | 2010-12-18T17:55:17.180 | 2010-12-18T17:55:17.180 | null | null | 1909 | null |
5624 | 2 | null | 5604 | 2 | null | The answer you get depends on the question you ask. Dason asked what you are trying to do; I would also ask this question. You say you know some of your data is "tainted" but do you know that this taint applies to the high values? Many sorts of counts have long tails, with no bad data at all.
If you just want to summ... | null | CC BY-SA 2.5 | null | 2010-12-18T20:51:52.010 | 2010-12-18T20:51:52.010 | null | null | 686 | null |
5625 | 2 | null | 5597 | 8 | null | To get a good answer, you must write a good question. Answering a statistics question without context is like boxing blindfolded. You might knock your opponent out, or you might break your hand on the ring post.
What goes into a good question?
- Tell us the PROBLEM you are trying to solve. That is, the substantive pro... | null | CC BY-SA 2.5 | null | 2010-12-18T20:55:08.027 | 2010-12-18T20:55:08.027 | null | null | 686 | null |
5626 | 2 | null | 452 | 7 | null | In Introductory Econometrics (Woolridge, 2009 edition page 268) this question is addressed. Woolridge says that when using robust standard errors, the t-statistics obtained only have distributions which are similar to the exact t-distributions if the sample size is large. If the sample size is small, the t-stats obtai... | null | CC BY-SA 2.5 | null | 2010-12-19T00:59:28.370 | 2010-12-19T00:59:28.370 | null | null | null | null |
5627 | 2 | null | 4364 | 1 | null | Its characteristic function has the same form as its pdf. I am not sure of another distribution which does that.
| null | CC BY-SA 2.5 | null | 2010-12-19T02:56:46.690 | 2010-12-19T02:56:46.690 | null | null | null | null |
5628 | 2 | null | 5517 | 8 | null | Here's (at least most of) a solution with `MCMCglmm`.
First fit the equivalent intercept-variance-only model with `MCMCglmm`:
```
library(MCMCglmm)
primingHeid.MCMCglmm = MCMCglmm(fixed=RT ~ RTtoPrime * ResponseToPrime + Condition,
random=~Subject+Word, data = primingHeid)
```
Comparin... | null | CC BY-SA 3.0 | null | 2010-12-19T03:07:09.180 | 2013-08-24T15:05:00.227 | 2013-08-24T15:05:00.227 | 7290 | 2126 | null |
5629 | 1 | 5745 | null | 2 | 1283 | Is there a software package that supports, or could support, through operator overloading or extensions, code such as the following;
```
x = rand_arr(10) ; array 10 elements long
y = rand_arr(10)
z = x + y ; elemental addition (z[0]=x[0]+y[0];z[1]=x[1]+y[1],...)
print z
```
If the above, or something similar, is valid... | Implementing error propagation | CC BY-SA 2.5 | null | 2010-12-19T04:20:06.560 | 2016-09-07T11:45:57.263 | 2010-12-21T20:02:06.103 | null | 957 | [
"error-propagation"
] |
5630 | 2 | null | 5629 | 4 | null | It's difficult to determine what you're asking for because you haven't specified the semantics of "x.err" etc., but it sounds like you might be interested in [interval arithmetic](http://en.wikipedia.org/wiki/Interval_arithmetic#Implementations). [Implementations](http://www.cs.utep.edu/interval-comp/intlang.html) are... | null | CC BY-SA 2.5 | null | 2010-12-19T05:08:59.810 | 2010-12-21T19:22:49.363 | 2010-12-21T19:22:49.363 | 919 | 919 | null |
5631 | 2 | null | 5629 | 3 | null | I recommend you look into [R](http://www.r-project.org/), the open source software for statistical computing. R supports many vectorized functions to deal with element by element computation of objects. Chapter 3 of [The R Inferno](http://lib.stat.cmu.edu/S/Spoetry/Tutor/R_inferno.pdf) provides a good outline on vector... | null | CC BY-SA 2.5 | null | 2010-12-19T05:12:26.403 | 2010-12-19T05:12:26.403 | null | null | 696 | null |
5632 | 2 | null | 5620 | 8 | null | In mathematics and physics, the "x" in "x-vector" stands for the dimension of the vector. The meanings of $K$ and $p$ were previously established. Typically a "p-vector" is written as a column vector and a "p-covector" would be written as a row vector.
| null | CC BY-SA 2.5 | null | 2010-12-19T05:18:34.387 | 2010-12-19T05:18:34.387 | null | null | 919 | null |
5633 | 2 | null | 5614 | 3 | null | The answer depends on the nature of the data generation process and on the alternative hypothesis you have in mind.
Your test is a kind of unweighted chi-square. Because of this lack of weighting, changes that principally affect the less-populated categories will be difficult to detect. For example, your test is goin... | null | CC BY-SA 2.5 | null | 2010-12-19T06:15:26.400 | 2010-12-19T06:48:42.843 | 2010-12-19T06:48:42.843 | 919 | 919 | null |
5634 | 1 | 5635 | null | 3 | 31346 | So I have data like:
```
Cost 20 30 10 5
Rating 5 3 2 5
```
I want to make a chart of rating vs. cost, so the points would be
```
[(5,20), (3,30), (2,10), (5,5)]
```
I can't seem to get excel to do anything other than put the two rows as independent series. Am I missing something, or I do have to pivo... | In Excel, how do I plot two rows against each other? | CC BY-SA 2.5 | null | 2010-12-19T15:53:17.357 | 2014-07-29T14:28:00.070 | 2010-12-19T18:14:33.343 | 930 | 1531 | [
"data-visualization",
"excel"
] |
5635 | 2 | null | 5634 | 5 | null | Select the two rows and do a scatterplot which I think is called an XY plot in Excel (sorry, I run a Linux machine, so I do not have Excel installed).
| null | CC BY-SA 2.5 | null | 2010-12-19T15:56:51.113 | 2010-12-19T15:56:51.113 | null | null | 582 | null |
5636 | 2 | null | 3497 | 4 | null | The answer to your original question is yes, because the classical theory applies under your sampling scheme. You don’t need any assumptions on the original data matrix. All of the randomness (implicitly behind standard errors and consistency) comes from your scheme for sampling $N$ rows from the data matrix.
Think o... | null | CC BY-SA 2.5 | null | 2010-12-19T18:04:59.073 | 2010-12-19T21:49:54.970 | 2010-12-19T21:49:54.970 | 1670 | 1670 | null |
5637 | 2 | null | 5418 | 2 | null | Another piece of advise might be to look at packages yours will be depending on or interacting with, especially if these implement some [items](https://stats.stackexchange.com/questions/5418/first-r-packages-source-code-to-study-in-preparation-for-writing-own-package/5433#5433) [Joshua Ulrich](https://stats.stackexchan... | null | CC BY-SA 2.5 | null | 2010-12-19T18:29:21.010 | 2010-12-19T18:29:21.010 | 2017-04-13T12:44:31.577 | -1 | 1355 | null |
5640 | 2 | null | 5603 | 3 | null | If this is for a particular application, I would first consider the question: "do you need to fit such a complex model?" which is code for, what other, more simple methods have you tried prior to this one? Have you looked at a plot of the "births & deaths" data that you have?
I would advise that you look up some of th... | null | CC BY-SA 2.5 | null | 2010-12-20T03:39:18.017 | 2010-12-20T03:39:18.017 | null | null | 2392 | null |
5641 | 2 | null | 5090 | 7 | null | Here's the solution I came up with: The trick is to add NAs to the end of the observation data. When seeing NA as a response variable the Kalman filter algorithm will simply predict the next value and not update the state vector. This is exactly what we want to make our forecast.
```
nAhead <- 12
mod <- dlmModSeas(4)+d... | null | CC BY-SA 2.5 | null | 2010-12-20T03:50:36.193 | 2010-12-20T03:50:36.193 | null | null | 2451 | null |
5642 | 2 | null | 2182 | 6 | null | It’s easier to explain in terms of standard deviations, rather than confidence intervals.
Your friend’s conclusion is basically correct under the simplest model where you have simple random sampling and two candidates. Now the sample proportions satisfy $p_A + p_B = 1$ so that $p_B = 1 - p_A$. Thus,
$$Var(p_A - p_B)... | null | CC BY-SA 2.5 | null | 2010-12-20T05:45:35.210 | 2010-12-22T06:32:25.093 | 2010-12-22T06:32:25.093 | 1670 | 1670 | null |
5644 | 2 | null | 5382 | 0 | null | I worked on it and thought this Bayesian equation will be useful.
RatingTM = SRTM/(1 + AWD/WDTM) + MSRT/(1 + WDTM/AWD)
The variables are:
SR = Team member's self rating
AWD = Average work done by team
WD = Work done by team member
MSRT = Mean self rating of team
(TM: TeamMember)
Please comment if you think this is not ... | null | CC BY-SA 2.5 | null | 2010-12-20T07:25:21.537 | 2010-12-20T07:25:21.537 | null | null | 2344 | null |
5645 | 2 | null | 2982 | 6 | null | $L_1$ penalization is part of an optimization problem. Soft-thresholding is part of an algorithm. Sometimes $L_1$ penalization leads to soft-thresholding.
For regression, $L_1$ penalized least squares (Lasso) results in soft-thresholding when the columns of the $X$ matrix are orthogonal (assuming the rows correspond ... | null | CC BY-SA 2.5 | null | 2010-12-20T07:31:41.610 | 2010-12-21T01:03:16.813 | 2010-12-21T01:03:16.813 | 1670 | 1670 | null |
5646 | 2 | null | 2854 | 3 | null | First I would like to point out that to get the object `ks3` you do not need to do that much of the book-keeping code. Use the features of `ddply`:
```
ks3 <- ddply(o,.(ai,Gs),function(d){
temp <- ols(value~as.numeric(bc)+as.numeric(age),data=d,x=T,y=T)
t2 <- bootcov(temp,B=1000,coef.reps=T)
data.frame(... | null | CC BY-SA 2.5 | null | 2010-12-20T07:59:32.343 | 2010-12-24T11:45:34.523 | 2010-12-24T11:45:34.523 | 2116 | 2116 | null |
5647 | 1 | null | null | 2 | 2467 | I am fitting a conditional logistic regression model with 1:4 controls using `R`. I wish to obtain `AIC` from the model. How can I extract the appropriate parameters based on the object `m`?
```
library(survival)
m<-clogit(cc~exp+ factor1+ factor2 + strata(stratum),data=data1)
```
| How to obtain AIC with conditional logistic regression using R? | CC BY-SA 3.0 | null | 2010-12-20T12:30:03.800 | 2016-03-19T09:50:14.343 | 2013-09-03T11:43:32.840 | 22047 | null | [
"r",
"logistic",
"clogit"
] |
5648 | 2 | null | 5647 | 3 | null | It seems you must do it manually, so something like this:
```
2*length(m$coefficients)-2*(m$loglik[2])
```
| null | CC BY-SA 2.5 | null | 2010-12-20T12:40:11.920 | 2010-12-20T12:40:11.920 | null | null | null | null |
5649 | 2 | null | 5602 | 2 | null | Since your metadata is probably discrete, I suggest using [classification trees](http://en.wikipedia.org/wiki/Classification_tree). Note that from your example it is highly likely that your unknown algorithm is random, i.e. there is no algorithm.
| null | CC BY-SA 2.5 | null | 2010-12-20T14:10:31.630 | 2010-12-20T14:10:31.630 | null | null | 2116 | null |
5650 | 1 | null | null | 9 | 976 | I am currently working on some time series data, I know I can use
LOESS/ARIMA model.
The data is written to a vector whose length is 1000, which is a queue,
updating every 15 minutes,
Thus the old data will pop out while the new data push in the vector.
I can rerun the whole model on a scheduler, e.g. retrain the m... | Incremental learning for LOESS time series model | CC BY-SA 2.5 | null | 2010-12-20T14:49:58.113 | 2013-05-07T05:59:47.533 | 2013-03-31T20:37:54.133 | 919 | 2454 | [
"time-series",
"model-evaluation"
] |
5651 | 2 | null | 5602 | 1 | null | This isn't a clustering, it's (supervised) classification.
There are a lot of methods to do classification, such as naïve bayes for binary features and linear discriminants for continuous, or neural networks for some opaque combination of both.
For the example you mention, the feasibility of what you suggest depends a ... | null | CC BY-SA 2.5 | null | 2010-12-20T16:55:46.273 | 2010-12-20T16:55:46.273 | null | null | 2456 | null |
5652 | 2 | null | 5601 | 8 | null | This is not a bug. The model is stored using a 0-based index. So SplitVar=0 is X1, SplitVar=1 is X2, and SplitVar=2 is X3. So this split corresponds to a split on X3. Since X3 is an ordinal factor and the split is at 1.5, this corresponds to splitting levels 0&1 from 2&3.
> sum(data$X3<="c")
[1] 522
> sum(data$X3>="b")... | null | CC BY-SA 2.5 | null | 2010-12-20T18:12:43.150 | 2010-12-20T18:12:43.150 | null | null | null | null |
5653 | 2 | null | 5111 | 2 | null | Under one interpretation of your situation there is no need to modify the p values at all.
For example, let's posit that a sequence of (unknown) bivariate distributions $p_i(x,y)$ govern $A$ and $B$ for each organism $i$. That is, $\Pr(A=x, B=y) = p_i(x,y)$ for all possible outcomes $(x,y)$ of $(A,B)$. To test whethe... | null | CC BY-SA 2.5 | null | 2010-12-20T19:26:32.303 | 2010-12-20T19:26:32.303 | null | null | 919 | null |
5654 | 1 | null | null | 9 | 1855 | I'm trying to help a scientist design a study for the occurrence of salmonella microbes. He would like to compare an experimental antimicrobial formulation against a chlorine (bleach) at poultry farms. Because background rates of salmonella differ over time, he plans to measure % poultry w/salmonella before treatment... | Sample size for proportions in repeated measures | CC BY-SA 2.5 | null | 2010-12-20T22:24:03.600 | 2010-12-22T11:04:25.630 | 2010-12-22T11:04:25.630 | null | 2473 | [
"sample-size",
"repeated-measures",
"proportion"
] |
5655 | 2 | null | 1875 | 0 | null | How about just binning the given predictions and taking the observed fractions as your estimate for each bin?
You can generalise this to a continuous model by weighing all the observations around your value of interest (say the prediction by tomorrow) by a Gaussian and seeing what the weighted average is.
You can guess... | null | CC BY-SA 2.5 | null | 2010-12-21T00:33:47.953 | 2010-12-21T00:33:47.953 | null | null | 2067 | null |
5656 | 1 | 5662 | null | 10 | 11800 | Wondering if anyone has run across a package/function in R that will combine levels of a factor whose proportion of all the levels in a factor is less than some threshold? Specifically, one of the first steps in data preparation I conduct is to collapse sparse levels of factors together (say into a level called 'Other'... | R package for combining factor levels for datamining? | CC BY-SA 3.0 | null | 2010-12-21T01:35:25.853 | 2017-05-16T23:32:03.043 | 2017-05-16T23:32:03.043 | 11887 | 2040 | [
"r",
"many-categories"
] |
5657 | 2 | null | 5656 | 5 | null | I wrote a quick function that will accomplish this goal. I'm a novice R user, so it may be slow with large tables.
```
Merge.factors <- function(x, p) {
#Combines factor levels in x that are less than a specified proportion, p.
t <- table(x)
y <- subset(t, prop.table(t) < p)
z <- subset(t, prop.table(t... | null | CC BY-SA 2.5 | null | 2010-12-21T02:33:01.607 | 2010-12-21T06:58:11.473 | 2010-12-21T06:58:11.473 | 1118 | 1118 | null |
5658 | 2 | null | 5656 | 5 | null | The only problem with Christopher answer is that it will mix up the original ordering of the factor. Here is my fix:
```
Merge.factors <- function(x, p) {
t <- table(x)
levt <- cbind(names(t), names(t))
levt[t/sum(t)<p, 2] <- "Other"
change.levels(x, levt)
}
```
where `change.levels` is the foll... | null | CC BY-SA 2.5 | null | 2010-12-21T04:51:40.117 | 2010-12-21T04:51:40.117 | null | null | 2116 | null |
5659 | 2 | null | 5399 | 0 | null | suppose $(X,Y)$ is bivariate normal with zero means and correlation $\rho$. then
${\mathrm E} XY= cov(X,Y)= \rho\sigma_X\sigma_Y$.
all of the entries in the matrix $x_1x_2^T$ are of the form $XY$.
| null | CC BY-SA 2.5 | null | 2010-12-21T05:30:02.967 | 2010-12-21T05:30:02.967 | null | null | 1112 | null |
5660 | 2 | null | 5602 | 0 | null | You can use SOM which is a kind of supervised clustering. Or as sesqu said their are tons of other algorithms such as support vector machines, regression or logistic regression. Neural Networks, etc ..
| null | CC BY-SA 2.5 | null | 2010-12-21T08:36:12.760 | 2010-12-21T08:36:12.760 | null | null | 1808 | null |
5661 | 2 | null | 1875 | 2 | null | The [Brier Score](http://docs.lib.noaa.gov/rescue/mwr/078/mwr-078-01-0001.pdf) approach is very simple and the most directly applicable way verify accuracy of a predicted outcome versus binary event.
Don't rely on just formulas ...plot the scores for different periods of time, data, errors, [weighted] rolling average... | null | CC BY-SA 2.5 | null | 2010-12-21T08:54:45.653 | 2010-12-21T22:21:01.123 | 2010-12-21T22:21:01.123 | 2342 | 2342 | null |
5662 | 2 | null | 5656 | 11 | null | It seems it's just a matter of "releveling" the factor; no need to compute partial sums or make a copy of the original vector. E.g.,
```
set.seed(101)
a <- factor(LETTERS[sample(5, 150, replace=TRUE,
prob=c(.1, .15, rep(.75/3,3)))])
p <- 1/5
lf <- names(which(prop.table(table(a)) < p))
level... | null | CC BY-SA 2.5 | null | 2010-12-21T10:16:38.273 | 2010-12-21T10:16:38.273 | null | null | 930 | null |
5663 | 2 | null | 5115 | 23 | null | [Florence Nightingale](http://en.wikipedia.org/wiki/Florence_Nightingale) for being "a true pioneer in the graphical representation of statistics" and developing the polar area diagram. Yes, that Florence Nightingale!
| null | CC BY-SA 3.0 | null | 2010-12-21T11:49:21.513 | 2011-12-14T06:47:23.300 | 2011-12-14T06:47:23.300 | 183 | null | null |
5664 | 1 | 5665 | null | 6 | 409 | I'm a complete newbie to statistics (although I find it really interesting!), and I have taken the task of distributing feedback to speakers of a conference I'm co-organizing. Each speaker was given a grade on the scale 1-5 from the participants, and we combine feedback from all participants into a mean score, for inst... | Is there a name for 10% best individual grades? | CC BY-SA 3.0 | null | 2010-12-21T13:18:15.667 | 2015-12-19T16:39:46.913 | 2015-12-19T16:39:46.913 | 28666 | 2470 | [
"terminology",
"quantiles"
] |
5665 | 2 | null | 5664 | 9 | null | If I understand you correctly, you may refer to [Percentiles](http://en.wikipedia.org/wiki/Percentile), perhaps espacially Quartiles.
Perhaps you can elaborate a little more on which percentages should be enclosed in each bin, to get a more accurate answer.
UPDATE: Based on the comments below decile seems to be the ter... | null | CC BY-SA 2.5 | null | 2010-12-21T13:37:09.843 | 2010-12-21T14:34:48.360 | 2010-12-21T14:34:48.360 | 442 | 442 | null |
5666 | 2 | null | 5654 | 2 | null | Let's take a stab at a first-order approximation assuming simple random sampling and a constant proportion of infection for any treatment. Assume the sample size is large enough that a normal approximation can be used in a hypothesis test on proportions so we can calculate a z statistic like so
$z = \frac{p_t - p_0}{\... | null | CC BY-SA 2.5 | null | 2010-12-21T14:10:44.480 | 2010-12-21T14:10:44.480 | null | null | 5792 | null |
5667 | 2 | null | 726 | 20 | null | >
The primary product of a research
inquiry is one or more measures of
effect size, not p values.
Cohen, J. (1990). [Things I have learned (so far)](http://www.cps.nova.edu/marker/whatIhavelearnedsofarcohen.pdf). American Psychologist, 45, 1304-1312.
| null | CC BY-SA 2.5 | null | 2010-12-21T15:48:06.727 | 2010-12-21T15:48:06.727 | null | null | 930 | null |
5668 | 2 | null | 5664 | 3 | null | You have the answer you asked for, but along with how to communicate this information you might also want to think about how to asses the reliability & precision of the scores. If the evaluators aren't really using the same standards, the scores will furnish a misleading measure of the quality of the speakers no matter... | null | CC BY-SA 2.5 | null | 2010-12-21T16:02:19.503 | 2010-12-21T18:29:28.220 | 2010-12-21T18:29:28.220 | 11954 | 11954 | null |
5669 | 2 | null | 859 | 4 | null | Look into models with spatially correlated errors (and spatially correlated covariates). A brief introduction, with references to [GeoDa](http://geodacenter.asu.edu/), is available [here](http://www.s4.brown.edu/s4/courses/SO261-John/lab9.pdf). There are plenty of texts; good ones are by [Noel Cressie](http://rads.st... | null | CC BY-SA 2.5 | null | 2010-12-21T16:08:11.367 | 2010-12-21T16:08:11.367 | null | null | 919 | null |
5670 | 2 | null | 5619 | 5 | null | I see no obvious reason not to do so. As far as I know, we usually make a distinction between two kind of effect size (ES) measures for qualifying the strength of an observed association: ES based on $d$ (difference of means) and ES based on $r$ (correlation). The latter includes Pearson's $r$, but also Spearman's $\rh... | null | CC BY-SA 2.5 | null | 2010-12-21T16:44:46.750 | 2010-12-21T17:02:01.640 | 2010-12-21T17:02:01.640 | 930 | 930 | null |
5671 | 2 | null | 5207 | 3 | null | In fact, you should not do MCMC, since your problem is so much simpler. Try this algorithm:
Step 1: Generate a X from Log Normal
Step 2: Keeping this X fixed, generate a Y from the Singh Maddala.
Voilà! Sample Ready!!!
| null | CC BY-SA 3.0 | null | 2010-12-21T17:16:52.633 | 2018-02-24T13:37:18.813 | 2018-02-24T13:37:18.813 | 7224 | 2472 | null |
5672 | 2 | null | 5054 | 1 | null | Maybe I'm missing something here, but if you plot the number of times these hashtags are mentioned over time, shouldn't that tell you something?
Of course, maybe you need automated processing. In that case fit splines to these series, and take the derivative. (They are easy : just look up what the Function Data Analyst... | null | CC BY-SA 2.5 | null | 2010-12-21T17:23:16.220 | 2010-12-21T17:23:16.220 | null | null | 2472 | null |
5675 | 1 | 5676 | null | 4 | 1001 | When computing a confidence interval of slope in linear regression, should you use the z- or t-statistic?
| Confidence interval of slope in linear regression | CC BY-SA 2.5 | null | 2010-12-21T19:37:50.770 | 2010-12-22T07:04:10.727 | 2010-12-21T20:59:47.620 | null | 1395 | [
"regression",
"confidence-interval"
] |
5676 | 2 | null | 5675 | 5 | null | If you're doing linear regression using least squares, you should use base confidence intervals on Student's t-distribution.
| null | CC BY-SA 2.5 | null | 2010-12-21T19:43:52.090 | 2010-12-21T19:43:52.090 | null | null | 449 | null |
5677 | 2 | null | 5675 | 5 | null | Rule of thumb: Use Student's t distribution if you must estimate the variance.
Since the distribution's variance is estimated (not known), you should use Student's t distribution rather than the standard normal distribution (z), which requires a known variance.
Although the t distribution becomes almost exactly the sam... | null | CC BY-SA 2.5 | null | 2010-12-21T20:10:22.910 | 2010-12-21T21:18:12.697 | 2010-12-21T21:18:12.697 | 1583 | 1583 | null |
5678 | 2 | null | 5054 | 1 | null | As far as I can tell from my reading, a common method for determining a time series' trend is to smooth the series, perhaps in an iterated fashion, as in:
[A Pakistan SBP paper](http://www.sbp.org.pk/departments/stats/sam.pdf). In the Seasonal Adjustment Methodology section, it describes how X-12 ARIMA does it, though ... | null | CC BY-SA 2.5 | null | 2010-12-21T20:21:54.227 | 2010-12-21T20:55:04.573 | 2010-12-21T20:55:04.573 | 1764 | 1764 | null |
5679 | 2 | null | 5675 | 3 | null | Depends on assumptions on your disturbances. If they are normal and homoscedastic, then yes use t-statistic. In economic applications though these assumptions rarely hold, so in that case I would suggest using z-statistic with [robust standard errors](http://en.wikipedia.org/wiki/White_standard_errors).
| null | CC BY-SA 2.5 | null | 2010-12-21T21:10:43.467 | 2010-12-22T07:04:10.727 | 2010-12-22T07:04:10.727 | 2116 | 2116 | null |
5680 | 1 | 5946 | null | 32 | 151636 | I have analyzed an experiment with a repeated measures ANOVA. The ANOVA is a 3x2x2x2x3 with 2 between-subject factors and 3 within (N = 189). Error rate is the dependent variable. The distribution of error rates has a skew of 3.64 and a kurtosis of 15.75. The skew and kurtosis are the result of 90% of the error rate me... | Can I trust ANOVA results for a non-normally distributed DV? | CC BY-SA 2.5 | null | 2010-12-21T21:38:00.003 | 2019-05-08T13:47:21.173 | null | null | 2322 | [
"anova",
"normality-assumption"
] |
5681 | 1 | null | null | 1 | 125 | Measurement outcome is a continuous number measured in $p$ locations.
This yields a vector $X_1$ to $X_p$ per subject. These vectors are obtained on $N$ subjects.
In addition these $N$ subjects come from $k$ groups.
What method should I use to compare the average vector across the entire study population to an hypothes... | Comparing a vector of averages to a null one | CC BY-SA 2.5 | null | 2010-12-22T00:04:01.290 | 2010-12-22T11:47:26.107 | 2010-12-22T11:31:41.020 | null | null | [
"multiple-comparisons",
"multivariate-analysis"
] |
5682 | 1 | 6236 | null | 7 | 3016 | I was wondering if anyone had experience with SAS/IML and R, and can give some pointers as to the relative advantages/disadvantages of the two. I've used R extensively for programming and statistical analyses, but haven't had much experience with IML. However, as this company is a SAS shop, I'm likely to have to use it... | SAS/IML compared to R | CC BY-SA 2.5 | null | 2010-12-22T01:19:57.423 | 2019-02-14T17:43:29.250 | null | null | 1569 | [
"r",
"sas"
] |
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