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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
10579 | 1 | null | null | 4 | 249 | I have 4 groups each consisting of 1000 recent college graduates, and I am studying employment immediately after graduation. 2 groups took a business class in college, and 2 groups took a math course in college. Group 1 is a control that has taken neither a business class nor a math course in college. I want to conduct... | Testing significant effect in 2 by 2 factor design on a binary outcome | CC BY-SA 3.0 | null | 2011-05-09T22:20:50.180 | 2011-05-11T15:29:29.850 | 2011-05-11T15:29:29.850 | 183 | 4538 | [
"statistical-significance"
] |
10580 | 2 | null | 10432 | 5 | null | It sounds like any linear classifier will do what you need. Suppose you have $N$ features and the value of feature $i$ is $f_i$. Then a linear classifier will compute a score
$$s = \sum_i w_i f_i + o$$ (where $o$ is the offset). Then, if $s > t$ (where $t$ is some threshold), then the feature belongs to a class (a gro... | null | CC BY-SA 3.0 | null | 2011-05-09T22:51:32.127 | 2011-05-09T22:58:25.560 | 2011-05-09T22:58:25.560 | 3369 | 3369 | null |
10581 | 2 | null | 10578 | 39 | null | Here I explain why the asymptotic variance of the maximum likelihood estimator is the Cramer-Rao lower bound. Hopefully this will provide some insight as to the relevance of the Fisher information.
Statistical inference proceeds with the use of a likelihood function $\mathcal{L}(\theta)$ which you construct from the d... | null | CC BY-SA 3.0 | null | 2011-05-09T23:09:17.680 | 2011-05-10T15:09:16.277 | 2011-05-10T15:09:16.277 | 3567 | 3567 | null |
10582 | 2 | null | 10579 | 3 | null | This seems like a classic problem for [logistic regression](http://en.wikipedia.org/wiki/Logistic_regression). Rather than specifying these groups, turn math and business coursework into predictors for employment status. This is easy to code up in pretty much whatever statistical software you have around, although yo... | null | CC BY-SA 3.0 | null | 2011-05-09T23:28:35.240 | 2011-05-09T23:28:35.240 | null | null | 71 | null |
10583 | 2 | null | 10579 | 3 | null | Self-selection problem need to be addressed here. People who choose to take business classes might be more likely to get a job. So, the sample you have is not random, and the inferences you draw might be incorrect. Heckman procedure is used to correct to self-selection bias. I am not sure if it is applicable for discre... | null | CC BY-SA 3.0 | null | 2011-05-09T23:52:24.900 | 2011-05-09T23:52:24.900 | null | null | 4540 | null |
10585 | 2 | null | 10579 | 2 | null | One of the more basic approaches you could take is a [two-way ANOVA](http://udel.edu/~mcdonald/stattwoway.html) (Page from U Delaware), using Math Class = {Y, N} and Business Class = {Y, N} as your two treatments. You would then perform an analysis of variance on the dependent variable (number of people employed) to de... | null | CC BY-SA 3.0 | null | 2011-05-10T00:59:05.887 | 2011-05-10T00:59:05.887 | null | null | 1118 | null |
10586 | 2 | null | 10579 | 3 | null | If @EEE's concern can be addressed and you proceed with an hypothesis test, then rather than logistic regression I'd recommend a chi-square test. For a person fairly new to statistical testing, it'll be dramatically easier to conduct, interpret, and explain to an audience. Plus I think it'll give you just about as mu... | null | CC BY-SA 3.0 | null | 2011-05-10T00:59:45.170 | 2011-05-10T00:59:45.170 | null | null | 2669 | null |
10587 | 1 | null | null | 3 | 675 | I lack knowledge of statistical terminology, so I'll try to thoroughly explain my predicament and hope that it is understandable:
I am programming a software for technical analysis of financial markets.
This software will receive a variable who's value represents current market conditions.
In this particular case the v... | Uniform frequency from non-uniform (exponential) distribution? | CC BY-SA 3.0 | null | 2011-05-10T01:47:23.723 | 2011-05-10T01:58:11.990 | 2020-06-11T14:32:37.003 | -1 | 4542 | [
"distributions",
"probability",
"confidence-interval",
"histogram"
] |
10588 | 2 | null | 10587 | 3 | null | If your data really is [exponentially distributed](http://en.wikipedia.org/wiki/Exponential_random_variable), find the [maximum likelihood estimate of the rate](http://en.wikipedia.org/wiki/Exponential_random_variable#Maximum_likelihood) $\lambda$, then transform the samples $X_n$ to a sequence $Y_n$ uniformly distribu... | null | CC BY-SA 3.0 | null | 2011-05-10T01:58:11.990 | 2011-05-10T01:58:11.990 | null | null | 4479 | null |
10589 | 2 | null | 10578 | 17 | null | One way that I understand the fisher information is by the following definition:
$$I(\theta)=\int_{\cal{X}} \frac{\partial^{2}f(x|\theta)}{\partial \theta^{2}}dx-\int_{\cal{X}} f(x|\theta)\frac{\partial^{2}}{\partial \theta^{2}}\log[f(x|\theta)]dx$$
The Fisher Information can be written this way whenever the density $f... | null | CC BY-SA 3.0 | null | 2011-05-10T07:19:38.177 | 2018-04-17T04:12:51.770 | 2018-04-17T04:12:51.770 | 14396 | 2392 | null |
10591 | 1 | 10661 | null | 8 | 2438 | The [Anscombe transform](http://en.wikipedia.org/wiki/Anscombe_transform) is $a(x) = 2\sqrt{x+3/8}$.
Can anyone show me how to prove that an Anscombe-transformed version $Y = a(X)$ of a Poisson distributed random variable $X$ is approximately normal distributed (when $\lambda>4$)?
| Anscombe transform and normal approximation | CC BY-SA 3.0 | null | 2011-05-10T10:38:24.523 | 2011-05-12T13:02:02.343 | 2011-05-12T13:02:02.343 | 2970 | 4496 | [
"distributions",
"normal-distribution",
"poisson-distribution"
] |
10592 | 1 | 10637 | null | 9 | 1080 | Given a data matrix $X$ of say 1000000 observations $\times$ 100 features,
is there a fast way to build a tridiagonal approximation
$A \approx cov(X)$ ?
Then one could factor $A = L L^T$,
$L$ all 0 except $L_{i\ i-1}$ and $L_{i i}$,
and do fast decorrelation (whitening) by solving
$L x = x_{white}$.
(By "fast" I mean ... | How to calculate tridiagonal approximate covariance matrix, for fast decorrelation? | CC BY-SA 3.0 | null | 2011-05-10T10:38:49.923 | 2022-05-24T19:37:35.363 | 2011-05-15T09:34:53.137 | 557 | 557 | [
"variance",
"approximation",
"covariance-matrix"
] |
10593 | 2 | null | 10594 | 4 | null | Suppose you have a cumulative distribution function $F$ of the variable in question. Suppose the value given is $x$, and the range is $[r_1,r_2]$ with $x\in[r_1,r_2]$. Then if you select the amount of values falling into that range $N$, the following should hold:
$$F(r_1)-F(r_2)=\frac{N}{500 000}$$
This is an equation ... | null | CC BY-SA 3.0 | null | 2011-05-10T11:06:58.780 | 2011-05-10T11:06:58.780 | null | null | 2116 | null |
10594 | 1 | 10595 | null | 6 | 11951 | I have 500,000 values for a variable derived from financial markets. This variable has a arbitrary distribution. I need a formula that will allow me to select a range around any value of this variable such that an equal (or close to it) amount of values fall within that range. From what I understand, this means that I ... | Converting arbitrary distribution to uniform one | CC BY-SA 3.0 | null | 2011-05-10T09:23:10.113 | 2011-05-10T15:00:17.433 | 2020-06-11T14:32:37.003 | -1 | 4544 | [
"probability",
"matlab"
] |
10595 | 2 | null | 10594 | 10 | null | If $X$ has the (cumulative) distribution function $F(x)=P(X<x)$, then $F(X)$ has a uniform distribution on $[0,1]$. You don't know what $F$ is, but with N = 500,000 data points you could simply use the empirical distribution function:
$$\hat{F}(x) = \frac{1}{N} \sum_{i=1}^N 1[x_i\leq x]$$
where $1[A]$ is the indicator ... | null | CC BY-SA 3.0 | null | 2011-05-10T10:11:43.367 | 2011-05-10T10:26:19.683 | null | null | 2425 | null |
10597 | 1 | null | null | 4 | 2239 | Is there a reasonable way to quantify the amount of local correlations in an image? For example, I want to justify the correlations between a neighbourhood of pixels is much higher than the correlations between pixels in entirely different regions of the image.
Would showing the xcorr2(A,A) as the 2-d autocorrelation ... | Valid method to analyze spatial correlations in images? | CC BY-SA 3.0 | null | 2011-05-10T13:55:45.370 | 2011-05-10T21:03:06.407 | 2011-05-10T15:47:44.150 | 1036 | 4547 | [
"correlation",
"spatial",
"image-processing"
] |
10598 | 1 | 14771 | null | 9 | 2449 | Various forms of the correlation, e.g.,
$r = \frac{\Sigma_i x_i * y_i}{\sigma_x \sigma_y}$
or
$r = \frac{\Sigma_i (x_i-\bar{x}) * (y_i-\bar{y})}{\sigma_x \sigma_y}$
are popular similarity measures in many applications.
Is there a probabilistic interpretation for this such that either $r$ or $r^2$ is an approximate l... | Correlation as a likelihood measure | CC BY-SA 3.0 | null | 2011-05-10T14:19:14.667 | 2011-08-24T19:32:05.150 | 2011-05-10T14:30:19.253 | 2728 | 2728 | [
"probability",
"correlation",
"interpretation",
"likelihood"
] |
10599 | 2 | null | 10597 | 2 | null | One straightforward way of doing this is to consider arbitrarily-sized patches of the image. For example, let's say we are interested in all 9*9 regions of pixels that can be taken from the image. Extract each of these image patches, and transform each image patch to a row vector. Consider the entire set of image pa... | null | CC BY-SA 3.0 | null | 2011-05-10T15:14:59.910 | 2011-05-10T15:14:59.910 | null | null | 3595 | null |
10600 | 2 | null | 10526 | 2 | null | I don't think that using CCA will help you. It appears to me that you have a number of endogenous series ( abundance of species n in number ) and a number of exogenous series ( variety of food resources m in number ). I would suggest constructing n transfer functions each one optimized to fully utilize the information ... | null | CC BY-SA 3.0 | null | 2011-05-10T15:57:42.317 | 2011-05-10T15:57:42.317 | null | null | 3382 | null |
10601 | 2 | null | 10564 | 1 | null | If I understand things correctly, you're testing if the mean value of predictor $A$ associated with outcome $1$ differ from the mean value of predictor A associated with outcome $2$. Even if they don't differ, this result says nothing to your research question. What it says is only that in your sample, the average valu... | null | CC BY-SA 3.0 | null | 2011-05-10T16:09:36.753 | 2011-05-10T20:31:29.040 | 2011-05-10T20:31:29.040 | 3058 | 3058 | null |
10602 | 1 | null | null | 0 | 676 | I have a design where birds of 1, 3, 5 week of age are placed in 2 chambers(independent) with 4 treatments( light technology) for 4 days. for example, x no of birds of 1 weeks of age are kept for 4 days and then new birds of 3 week age are placed in both chambers and so on...readings are taken for their position, feed ... | Sample size determination for block design with repeated measurement in SAS | CC BY-SA 3.0 | null | 2011-05-10T16:28:07.593 | 2011-05-10T17:14:40.017 | null | null | 4550 | [
"anova"
] |
10603 | 1 | null | null | 2 | 2924 | I have a very basic question on when a discrete distribution might be called a symmetric distribution. Let say I have a r.v. $X$ that can take two possible values $(x1, x2)$ with $x1 \neq x2$ and corresponding probabilities $(0.4, 0.6)$. Then can I say that $X$ is symmetric?
Thanks,
| How to characterize symmetric discrete distribution? | CC BY-SA 3.0 | null | 2011-05-10T16:55:47.630 | 2012-05-23T11:33:55.457 | 2011-05-10T20:59:45.283 | 930 | 4551 | [
"distributions",
"discrete-data"
] |
10604 | 1 | 11499 | null | 7 | 8670 | I'm trying to forecast using ARIMAX with two exogenous (input) variables. I'm using PROC ARIMA, but I can't figure out from the SAS documentation whether my code is producing the parameterization I want.
I want to extend an ARI(12,1) model so that it also includes the last 12 terms of each of the two exogenous variable... | How do I ensure PROC ARIMA is performing the correct parameterization of input variables? | CC BY-SA 3.0 | null | 2011-05-10T17:02:08.347 | 2011-06-03T07:32:50.997 | 2011-05-10T17:12:45.243 | 1583 | 1583 | [
"time-series",
"sas",
"dynamic-regression"
] |
10605 | 2 | null | 10603 | 4 | null | No, it only would be symmetric if the corresponding probabilities were (0.5,0.5).
Also, with binary or categorical distributions, the concept of symmetry does not have much meaning.
| null | CC BY-SA 3.0 | null | 2011-05-10T17:11:08.470 | 2011-05-10T20:46:21.960 | 2011-05-10T20:46:21.960 | 3595 | 3595 | null |
10606 | 2 | null | 10602 | 1 | null | When problems get more complicated than the simple cases that have nice canned sample size solutions I turn to simulation. The basic steps:
- Decide what you think your data will look like (including things you may want to change, e.g. sample size(s))
- Decide how you will analyze the data
- create some code that s... | null | CC BY-SA 3.0 | null | 2011-05-10T17:14:40.017 | 2011-05-10T17:14:40.017 | null | null | 4505 | null |
10607 | 1 | 10620 | null | 8 | 12128 | I am running a logistic model. In SAS Entreprise Miner, I noticed there's a link function that has three possible options: `logit`, `probit` and `cll` (complementary log-log).
Can you please shed light on the following questions:
- Can we use any of these link function to carry out a logistic regression?
- Are there ... | How to choose the link function when performing a logistic regression? | CC BY-SA 3.0 | null | 2011-05-10T17:23:17.073 | 2021-12-07T15:31:28.437 | 2021-12-07T15:31:28.437 | 11887 | 1763 | [
"regression",
"logistic",
"sas",
"link-function"
] |
10608 | 1 | 10610 | null | 9 | 2544 | I have a question concerning feature selection and classification. I will be working with R. I should start by saying that I am not very familiar with data mining techniques, aside from a brief glimpse provided by an undergraduate course on multivariate analysis, so forgive me if I am lacking in details regarding my qu... | Questions about variable selection for classification, and different classification techniques | CC BY-SA 3.0 | null | 2011-05-10T17:29:14.633 | 2011-05-11T16:23:22.990 | 2017-04-13T12:44:40.883 | -1 | 2252 | [
"r",
"machine-learning",
"classification",
"dimensionality-reduction"
] |
10610 | 2 | null | 10608 | 14 | null | Feature selection does not necessarily improve the performance of modern classifier systems, and quite frequently makes performance worse. Unless finding out which features are the most important is an objective of the analysis, it is often better not even to try and to use regularisation to avoid over-fitting (select... | null | CC BY-SA 3.0 | null | 2011-05-10T18:04:24.603 | 2011-05-10T18:04:24.603 | null | null | 887 | null |
10611 | 2 | null | 10608 | 6 | null | On dimensionality reduction, a good first choice might be [principal components analysis](http://en.wikipedia.org/wiki/Principal_components_analysis).
Apart from that, i don't have too much to add, except that if you have any interest in data mining, I strongly recommend you read [the elements of statistical learning](... | null | CC BY-SA 3.0 | null | 2011-05-10T18:04:58.020 | 2011-05-10T18:04:58.020 | null | null | 656 | null |
10612 | 2 | null | 10531 | 4 | null | There is no simple yes or no answer. People constantly attempt to make inferences about causal relationships. The question is what assumptions you have to make, and how sensitive your inferences are to changing those assumptions.
The causal effects you can identify with the fewest assumptions are the effects of the... | null | CC BY-SA 3.0 | null | 2011-05-10T18:22:00.337 | 2011-05-10T18:22:00.337 | null | null | 3748 | null |
10613 | 1 | 10617 | null | 161 | 90704 | Recently, I have found in a [paper by Klammer, et al.](http://pubs.acs.org/doi/abs/10.1021/pr8011107) a statement that p-values should be uniformly distributed. I believe the authors, but cannot understand why it is so.
Klammer, A. A., Park, C. Y., and Stafford Noble, W. (2009) [Statistical Calibration of the SEQUEST ... | Why are p-values uniformly distributed under the null hypothesis? | CC BY-SA 3.0 | null | 2011-05-10T18:26:26.630 | 2022-03-31T13:53:07.677 | 2017-10-19T22:56:56.347 | 44269 | 4552 | [
"p-value",
"uniform-distribution"
] |
10614 | 2 | null | 9299 | 5 | null | You can use this module of the [pysal](http://pysal.org/1.1/library/spreg/ols.html) python library for the spatial data analysis methods I discuss below.
Your description of how each person's attitude is influenced by the attitudes of the people surrounding her can be represented by a [spatial autoregressive model (SAR... | null | CC BY-SA 3.0 | null | 2011-05-10T18:59:18.107 | 2011-05-11T03:57:17.720 | 2017-04-13T12:44:27.570 | -1 | 4329 | null |
10615 | 1 | null | null | 2 | 482 | I'd be too happy, if someone could post a code snippet, which explains how to compute mean and variance of a set of records, which contain frequencies?
Suppose we have records like (FORMAT F)
```
- GroupA, 6 x Grade 1, 5 x Grade 2, 10 x Grade 3
- GroupB, 2 x Grade 1, 7 x Grade 2, 18 x Grade 3
- GroupA, 23 x Grade 1, 5 ... | Variance based on given frequencies using SPSS | CC BY-SA 3.0 | null | 2011-05-10T19:28:40.657 | 2011-05-15T22:31:54.813 | 2011-05-15T22:31:54.813 | 4554 | 4554 | [
"spss"
] |
10617 | 2 | null | 10613 | 114 | null | To clarify a bit. The p-value is uniformly distributed when the null hypothesis is true and all other assumptions are met. The reason for this is really the definition of alpha as the probability of a type I error. We want the probability of rejecting a true null hypothesis to be alpha, we reject when the observed $... | null | CC BY-SA 3.0 | null | 2011-05-10T19:45:10.343 | 2016-03-23T07:19:27.883 | 2016-03-23T07:19:27.883 | null | 4505 | null |
10619 | 1 | 10626 | null | 1 | 704 | I am using Matlab to try and find a good fit for this curve:
None of the built-in formulas seem to work well.
Any suggestions?
| What regression formula would best fit this curve? | CC BY-SA 3.0 | null | 2011-05-10T20:21:47.193 | 2011-05-10T23:37:39.147 | 2011-05-10T20:23:56.783 | 919 | 4544 | [
"distributions",
"matlab"
] |
10620 | 2 | null | 10607 | 4 | null | I don't know of SAS, so i'll just answer based on the statistics side of the question. About the software you mays ask at the sister site, stackoverflow.
- If the link function is different (logistic, probit or Clog-log), than you will get different results. For logistic, use logistic.
- About the real differences of... | null | CC BY-SA 3.0 | null | 2011-05-10T20:30:24.190 | 2011-05-10T21:14:00.837 | 2011-05-10T21:14:00.837 | 3058 | 3058 | null |
10621 | 1 | null | null | 4 | 4072 | I'm interested in fitting a conditional Poisson regression model using PROC GENMOD in SAS to analyze a matched cohort study. However, it's not quite clear to me how I should exactly go about it.
My impression is that a REPEATED statement should be used along with the events/trials syntax, but if so then how does one a... | How does one fit conditional Poisson regression in SAS? | CC BY-SA 3.0 | null | 2011-05-10T21:02:18.387 | 2017-03-17T18:23:09.137 | 2011-05-11T07:30:51.817 | 930 | 4555 | [
"regression",
"poisson-distribution",
"sas",
"matching"
] |
10623 | 1 | 10624 | null | 4 | 202 | If I have a sample of k successes and n-k failures, there are standard techniques (Agresti-Coull, Clopper, etc.) for finding a confidence interval of the probability of an individual success. What if I want to find a confidence interval for the probability of getting at least k' out of n' instead? Obviously it can be... | Binomial testing with probability estimated from sample | CC BY-SA 3.0 | null | 2011-05-10T21:42:26.387 | 2011-05-11T00:46:47.260 | null | null | 1378 | [
"confidence-interval",
"binomial-distribution"
] |
10624 | 2 | null | 10623 | 3 | null | I believe you are looking for the [beta binomial distribution](https://secure.wikimedia.org/wikipedia/en/wiki/Beta-binomial_distribution), which reduces the pdf of of interest ($\pi(k\prime)$) to
$\pi(k\prime|n\prime,n,k) = {n\prime \choose k\prime} \frac{B(k\prime+k+1,n\prime-k\prime+n-k+1)}{B(k+1,n-k+1)}$
For motivat... | null | CC BY-SA 3.0 | null | 2011-05-10T21:59:57.487 | 2011-05-10T21:59:57.487 | null | null | 2728 | null |
10625 | 2 | null | 10619 | 1 | null | Check out [Eureqa](http://creativemachines.cornell.edu/eureqa) for a neat evolutionary approach to finding the mathematical form of an otherwise ambiguous function. It's native to Windows but works fine on Linux & Mac via Wine (in which case I'd suggest you use [winebottler](http://winebottler.kronenberg.org)).
| null | CC BY-SA 3.0 | null | 2011-05-10T22:13:21.433 | 2011-05-10T22:13:21.433 | null | null | 364 | null |
10626 | 2 | null | 10619 | 2 | null | Looking at the charts in your first question, this looks slightly like the absolute value of a standard normal distribution, what Wikipedia calls a [half-normal distribution](http://en.wikipedia.org/wiki/Half-normal_distribution), and your curve here looks like the top half of the cdf of a normal distribution.
One wa... | null | CC BY-SA 3.0 | null | 2011-05-10T23:37:39.147 | 2011-05-10T23:37:39.147 | null | null | 2958 | null |
10627 | 2 | null | 73 | 3 | null | I use `lattice`, `ggplot2`, `lubridate`, `reshape`, `boot`, `e1071`, `car`, `forecast`, and `zoo` a lot.
| null | CC BY-SA 3.0 | null | 2011-05-10T23:50:24.527 | 2011-05-10T23:50:24.527 | null | null | 1764 | null |
10629 | 2 | null | 10607 | 2 | null | All 3 link functions are s-shaped and are not going to be that different. Li and Duan showed that if the predictor variables are well behaved (elliptically symmetric predictors are a subset of the well behaved group) then changing the link function will change the coefficients by a multiplicitive constant. Even if th... | null | CC BY-SA 4.0 | null | 2011-05-11T00:35:33.890 | 2018-07-24T04:36:38.033 | 2018-07-24T04:36:38.033 | 11887 | 4505 | null |
10630 | 2 | null | 10623 | 3 | null | If you are happy with your confidence interval on the unknown proportion of a single success, then just plug both of those values into the above formula. Since k and n are known constants (rather than random variables) and the probability of k or more out of n is monotone, you can just transform the ends of your confi... | null | CC BY-SA 3.0 | null | 2011-05-11T00:46:47.260 | 2011-05-11T00:46:47.260 | null | null | 4505 | null |
10631 | 1 | null | null | 3 | 1264 | I have a data set with a distribution of one variable against the other resembling a cubic one (rises to some point and then falls to a steady level without a consequent rise). I know in which cases to use log-linear, log-lin, lin-log, and reciprocal or log reciprocal linear models, but I am not sure what to do here (I... | What is the best linear regression model to use when the shape of the data resembles a cubic distribution? | CC BY-SA 3.0 | null | 2011-05-11T00:51:20.380 | 2011-05-11T13:31:27.083 | null | null | 4560 | [
"regression",
"econometrics"
] |
10632 | 2 | null | 10579 | 0 | null | you could use the Agresti-Caffo simultaneous confidence interval or (Simultaneous Score Intervals for Difference of Proportions) to compare differences in proportions (Agresti et al. 2008. Simultaneous confidence intervals for comparing binomial parameters, Biometrics 64, 1270-1275).
The corresponding R code is availa... | null | CC BY-SA 3.0 | null | 2011-05-11T00:58:52.337 | 2011-05-11T00:58:52.337 | null | null | 4559 | null |
10633 | 2 | null | 10544 | 3 | null | You would need to use the "Simultaneous Score Intervals for Difference of Proportions" to solve your question. The reference is " Agresti et al. 2008. Simultaneous confidence interval for comparing binomial parameters. Biometrics 64, 1270-1275.
The corresponding R code is available in [http://www.stat.ufl.edu/~aa/cda/... | null | CC BY-SA 3.0 | null | 2011-05-11T01:09:17.860 | 2011-05-11T01:09:17.860 | null | null | 4559 | null |
10634 | 2 | null | 10562 | 0 | null | You might want to try "model-based clustering". This algorithm uses "BIC" to determine the number of clusters.
Sincerely
| null | CC BY-SA 3.0 | null | 2011-05-11T01:13:16.013 | 2011-05-11T01:13:16.013 | null | null | 4559 | null |
10636 | 2 | null | 10607 | 4 | null | I have a question/comment. I thought that by definition, logistic regression uses the logit link. If you are using the probit or complementary log-log link, then I do not think that is logistic regression.
What you are doing is fitting generalized linear models on a binary outcome, which is assumed to follow a Bernou... | null | CC BY-SA 3.0 | null | 2011-05-11T01:37:41.867 | 2011-05-11T01:37:41.867 | null | null | 2312 | null |
10637 | 2 | null | 10592 | 2 | null | Merely computing the covariance matrix--which you're going to need to get started in any event--is $O((Nf)^2)$ so, asymptotically in $N$, nothing is gained by choosing a $O(Nf)$ algorithm for the whitening.
There are approximations when the variables have additional structure, such as when they form a time series or re... | null | CC BY-SA 4.0 | null | 2011-05-11T03:37:27.523 | 2022-05-24T16:44:55.370 | 2022-05-24T16:44:55.370 | 919 | 919 | null |
10638 | 2 | null | 10592 | 2 | null | On a whim, I decided to try computing (in R) the covariance matrix for a dataset of about the size mentioned in the OP:
```
z <- rnorm(1e8)
dim(z) <- c(1e6, 100)
vcv <- cov(z)
```
This took less than a minute in total, on a fairly generic laptop running Windows XP 32-bit. It probably took longer to generate `z` in the... | null | CC BY-SA 3.0 | null | 2011-05-11T04:00:43.317 | 2011-05-11T04:00:43.317 | null | null | 1569 | null |
10639 | 1 | 10722 | null | 3 | 3323 | According to [Wikipedia](http://en.wikipedia.org/wiki/Wilks%27_lambda_distribution), Wilks' Lambda distribution generalizes Hotelling's distribution. I am having some problems seeing how this works. I can see how Hotelling's distribution generalizes Student's t-distribution (a RV distributed as Hotelling's law with $p=... | How exactly does Wilks' Lambda distribution generalize the Hotelling distribution? | CC BY-SA 3.0 | null | 2011-05-11T04:34:34.790 | 2017-04-28T19:39:29.567 | 2017-04-28T19:39:29.567 | 28666 | 795 | [
"distributions",
"t-distribution",
"hotelling-t2"
] |
10640 | 1 | 10709 | null | 5 | 2132 | I'm trying to design an experiment where I measure a variable as a function of 5 two-level factors, labelled A, B, C, D and E.
I'm trying to understand how to best design this experiment so I can conduct it in 8 runs. I've tried to follow the guidance given in Box, Hunter & Hunter, and found two experimental $2^{5-2}$ ... | How to design an 8-run experiment in 5 factors? | CC BY-SA 3.0 | null | 2011-05-11T07:09:00.243 | 2013-09-24T18:50:43.890 | 2011-05-11T15:36:41.373 | 26 | 4370 | [
"experiment-design"
] |
10641 | 2 | null | 10639 | 2 | null | [These NCSU course notes](http://faculty.chass.ncsu.edu/garson/PA765/manova.htm) say
>
Multivariate tests in contrast to the overall F test, answer the
question, "Is each effect
significant?" or more specifically,
"Is each effect significant for at
least one of the dependent variables?"
That is, where the F... | null | CC BY-SA 3.0 | null | 2011-05-11T07:19:46.017 | 2011-05-11T07:19:46.017 | null | null | 2958 | null |
10642 | 2 | null | 10640 | 2 | null | I do not have the book at hand, so I cannot comment on the reasoning to find these models.
However, it is reasonable to expect in this type of setting that:
- Different designs might better achieve different optimality criteria: perhaps you want the design with 8 runs that has best overall predictive ability, perhaps ... | null | CC BY-SA 3.0 | null | 2011-05-11T07:47:29.803 | 2011-05-11T07:47:29.803 | null | null | 4257 | null |
10643 | 1 | 10686 | null | 2 | 278 | I'm measuring distances of various samples from a reference point. The distance is defined as a non-negative number, where $d=0$ means that the test case is identical to the reference.
My general question is: Given a set of "typical" distances, what is the proper way to tell whether a given $d_1$ "too large", compare... | How to properly analyze distance from a reference? | CC BY-SA 3.0 | null | 2011-05-11T09:01:45.637 | 2011-05-11T23:04:01.823 | 2011-05-11T09:50:04.673 | 930 | 1496 | [
"distributions",
"hypothesis-testing",
"distance-functions"
] |
10644 | 1 | null | null | 10 | 2362 | In PCA eigenvalues determine the order of components. In ICA I am using kurtosis to obtain the ordering. What are some accepted methods to assess the number, (given I have the order) of components that are singificant apart from prior knowledge about the signal?
| Using kurtosis to assess significance of components from independent component analysis | CC BY-SA 3.0 | null | 2011-05-11T09:27:52.943 | 2017-10-15T02:25:18.827 | 2015-02-12T07:02:41.553 | 53618 | 4563 | [
"statistical-significance",
"pca",
"kurtosis",
"independent-component-analysis"
] |
10645 | 2 | null | 10643 | 1 | null | Can you not use the empirical distribution's 95% (or whichever you prefer) confidence limit? If your sample size is big enough, this ought to be a reasonable approximation.
| null | CC BY-SA 3.0 | null | 2011-05-11T09:54:22.177 | 2011-05-11T09:54:22.177 | null | null | 4257 | null |
10646 | 2 | null | 73 | 1 | null | Some packages are very useful in R.
I will just recommand kernlab for Kernel-based Machine Learning Lab and e1071 for SVM and ggplot2 for graphics
| null | CC BY-SA 3.0 | null | 2011-05-11T10:03:37.470 | 2011-05-11T10:03:37.470 | null | null | 4531 | null |
10647 | 2 | null | 9756 | 0 | null | I have worked on active learning in classification and in SVM, that problem was same for me, if the boundary you found out by first model isn't that good the probability to have a good label for new points will decrease. If you have any other method to labelize your new generated points rather than using the boundary t... | null | CC BY-SA 3.0 | null | 2011-05-11T10:15:49.427 | 2011-05-11T10:15:49.427 | null | null | 4531 | null |
10649 | 1 | null | null | 4 | 6662 | How can one obtain confidence limits of predicted values in ARIMA?
| How to obtain confidence limits of predicted values in ARIMA? | CC BY-SA 3.0 | null | 2011-05-11T12:28:39.347 | 2011-05-11T21:10:48.220 | 2011-05-11T16:51:38.367 | 2970 | 4427 | [
"forecasting"
] |
10650 | 2 | null | 73 | 2 | null | For me
I am using kernlab for Kernel-based Machine Learning Lab and e1071 for SVM and ggplot2 for graphics
| null | CC BY-SA 3.0 | null | 2011-05-11T12:33:30.563 | 2011-05-11T12:33:30.563 | null | null | 4531 | null |
10651 | 2 | null | 10631 | 2 | null | I would have thought a "cubic regression" would work well for a cubic relationship. Call $Y_{i}$ the dependent variable, and $X_{i}$ the independent variable (or regressor). You simply use a polynomial regression:
$$Y_{i}=\left(\sum_{j=0}^{p}\beta_{j}X_{i}^{j}\right)+e_{i}$$
I would use BIC to select the value of $p$... | null | CC BY-SA 3.0 | null | 2011-05-11T12:35:05.150 | 2011-05-11T12:35:05.150 | null | null | 2392 | null |
10652 | 2 | null | 7224 | 3 | null | You need to put on an algorithm detecting which person that picture is referring to. You can build a model based on different portrait pictures of famous personality and use classifiers to ensure that this picture is referring to one of your database picture. You need to use a certain classifier based on different para... | null | CC BY-SA 3.0 | null | 2011-05-11T12:53:49.673 | 2011-05-11T12:53:49.673 | null | null | 4531 | null |
10653 | 2 | null | 10604 | -1 | null | I have reviewed the output and the forecast refelects an AR(12) in the error term which translates to a 12 period weighted forecast using the last 12 values of both your predictor series as the AR polynomial acts a multiplier across all series ( X,Y,Z ). Without getting into it in great detail , your model specificatio... | null | CC BY-SA 3.0 | null | 2011-05-11T13:26:05.713 | 2011-05-12T22:10:04.407 | 2011-05-12T22:10:04.407 | 3382 | 3382 | null |
10654 | 2 | null | 10631 | 4 | null | Restricted cubic splines (natural splines) are an excellent choice. These are piecewise cubic polynomials that can fit any shape given enough knots. The following code in R shows how to fit such relationships and to plot the fit with confidence bands.
```
require(rms)
dd <- datadist(mydata); options(datadist='dd')
f ... | null | CC BY-SA 3.0 | null | 2011-05-11T13:31:27.083 | 2011-05-11T13:31:27.083 | null | null | 4253 | null |
10655 | 1 | 10706 | null | 4 | 2432 | I have a scatter plot of (for example) height against age. How does one calculate for an individual point the percentile of the height for a given age?
Suggestions in R would be most appreciated. Thanks!
| How to calculate a percentile of y for a given x given a series of (x, y)? | CC BY-SA 3.0 | null | 2011-05-11T13:33:07.000 | 2011-05-12T21:10:18.593 | null | null | 1991 | [
"quantiles"
] |
10656 | 1 | null | null | 2 | 902 | i just wanted to know if i could use factor scores and crosstab it with demographics (i.e. gender, age, etc.)? I have 69 likert-scale variables and run factor analysis on SPSS. It gave me 10 new variables (types of personality. i just wanted to see the demographics of each new variables. Thanks!!!
| Crosstab factor scores generated by factor analysis | CC BY-SA 3.0 | null | 2011-05-11T13:44:50.520 | 2011-05-11T19:34:00.410 | 2011-05-11T19:34:00.410 | null | 4565 | [
"factor-analysis"
] |
10657 | 1 | null | null | 3 | 196 | In relation to web usage mining from a log file, can you cluster data without performing User and/or Session identification?
I mean,let's say I have these entries:
>
123.234.324.122 [timestamp] "GET /cars/sport/porsche.jpg" 200 23432 "http://topgear.com/cars" "Mozilladsfsd"
120.23.324.122 [timestamp] "GET /bikes/sport... | Can one cluster web log data without performing user or session identification? | CC BY-SA 3.0 | null | 2011-05-11T13:55:41.987 | 2011-05-11T20:07:20.180 | 2020-06-11T14:32:37.003 | -1 | 4402 | [
"clustering"
] |
10658 | 2 | null | 10062 | 5 | null | Package [Mclust](http://cran.r-project.org/web/packages/mclust/mclust.pdf) is nice. The mclust function fits a mixture of normals distribution to data. You can automatically choose the number of components based on BIC (mclustmodel) or specify the number of components. There is also no need to convert your data into... | null | CC BY-SA 3.0 | null | 2011-05-11T13:57:14.200 | 2011-09-24T02:16:05.507 | 2011-09-24T02:16:05.507 | 2310 | 2310 | null |
10659 | 2 | null | 10655 | 1 | null | 'The' percentile for a given age implies some sort of regression (i.e. you can find 'the' mean predicted height from a given age).
Once you have found this, the result depends on your assumptions: if you want no assumptions (besides the regression's), find how many of the heights in your original data are smaller than ... | null | CC BY-SA 3.0 | null | 2011-05-11T14:18:39.307 | 2011-05-11T14:18:39.307 | null | null | 4257 | null |
10660 | 2 | null | 10657 | 1 | null | If you don't identify your sessions/users, you are clustering different things: one user who is an insane adept of any given car and looks at its picture dayly could have a huge impact on your results, though you're probably not interested in this.
| null | CC BY-SA 3.0 | null | 2011-05-11T14:58:55.840 | 2011-05-11T14:58:55.840 | null | null | 4257 | null |
10661 | 2 | null | 10591 | 9 | null | Here is a sketch of a proof which combines three ideas: (a) the delta method, (b) variance-stabilization transformations and (c) the closure of the Poisson distribution under independent sums.
First, let's consider a sequence of iid Poisson random variables $X_1, X_2, \ldots$ with mean $\lambda > 0$. Then, the Central ... | null | CC BY-SA 3.0 | null | 2011-05-11T15:28:51.700 | 2011-05-12T12:56:29.837 | 2011-05-12T12:56:29.837 | 2970 | 2970 | null |
10662 | 1 | 10663 | null | 4 | 465 | I'm working with a lot of data that was collected by obstetricians regarding the health of infants (birth weight, gestational age at delivery, mother's BMI), and I am trying to connect this data with geometric measurements performed on microscopic slide scans for each associated placenta (area, perimeter, number of blo... | What to do with a small (27) medical dataset? | CC BY-SA 3.0 | null | 2011-05-11T15:37:13.417 | 2011-05-11T19:46:48.087 | 2011-05-11T19:43:51.520 | null | 5129 | [
"hypothesis-testing",
"data-mining",
"small-sample",
"exploratory-data-analysis"
] |
10663 | 2 | null | 10662 | 5 | null | If you're looking for statistical significance I wouldn't hold out hope unless you have a very targeted hypothesis and/or there is a very strong effect. But certainly you could generate some new hypotheses with this data via some exploratory analysis. With 6 variables overall I'm not sure I'd start with any sophisticat... | null | CC BY-SA 3.0 | null | 2011-05-11T15:57:01.233 | 2011-05-11T19:46:48.087 | 2011-05-11T19:46:48.087 | 26 | 26 | null |
10664 | 1 | null | null | 2 | 578 | I am trying to analyse a group of 4 questions that are on a 5 point scale. I need to group the answers for each question based on age. There are three different ages. How would I go about doing this?
| How to examine group differences on several 5-point items using SPSS? | CC BY-SA 3.0 | null | 2011-05-11T16:17:56.080 | 2011-05-12T10:58:56.187 | 2011-05-12T01:47:22.627 | 183 | 4567 | [
"spss",
"likert"
] |
10665 | 2 | null | 10649 | 4 | null | One idea would be to use the [forecast](http://cran.r-project.org/web/packages/forecast/index.html) package in [R](http://www.r-project.org/):
```
library(forecast)
fit <- auto.arima(WWWusage)
fit
f <- forecast(fit,h=20)
f
plot(f)
```
You can also give auto.arima parameters to use, rather than allowing it to fit its o... | null | CC BY-SA 3.0 | null | 2011-05-11T16:40:47.500 | 2011-05-11T16:48:50.180 | 2011-05-11T16:48:50.180 | 2817 | 2817 | null |
10666 | 2 | null | 10656 | 4 | null | In SPSS, choose Analyze...Descriptive Statistics...Explore. The factor scores will be your dependents, and a demographic grouping will be your "factor" in this procedure. I'd choose Plots Only to start with and request boxplots. You can get either factor levels together or dependents together: you can experiment. ... | null | CC BY-SA 3.0 | null | 2011-05-11T16:43:02.240 | 2011-05-11T16:43:02.240 | null | null | 2669 | null |
10667 | 1 | 10670 | null | 2 | 1734 | I have to assess for each product p, the odds ratio associated (success/failure). The data are in this table:
```
N_success N_trials p1 p2 p3 p4 p5
5 310 n n n n n
17 700 n y n n y
12 650 y y y n n
27 2... | Odds ratios multiple comparisons | CC BY-SA 3.0 | null | 2011-05-11T16:58:07.793 | 2011-05-11T18:52:39.843 | 2011-05-11T18:18:03.137 | 919 | 4569 | [
"logistic",
"multiple-comparisons",
"odds-ratio"
] |
10668 | 2 | null | 10369 | 1 | null | Observe that the random variable $i_j$ is a function of $\mathbf{Z} = (Z_1, \ldots, Z_n)$ only. For an $n$-vector, $\mathbf{z}$, we write $i_j(\mathbf{z})$ for the index of the $j$th largest coordinate. Let also $P_z(A) = P(X_1 \in A \mid Z_1 = z)$ denote the conditional distribution of $X_1$ given $Z_1$.
If we break ... | null | CC BY-SA 3.0 | null | 2011-05-11T18:22:17.520 | 2011-05-11T18:22:17.520 | null | null | 4376 | null |
10669 | 1 | 10685 | null | 2 | 1089 | I am running a model for which I am getting a very bad percentage detection of events in the confusion matrix (basically my true positives).
Obviously that implies my false negatives are too hight.
When I gave these dataset to a neural network node or a decision tree node, I see that the percentage detection by these... | Decision tree output -- learning | CC BY-SA 3.0 | null | 2011-05-11T18:37:16.257 | 2011-05-12T13:51:27.850 | 2011-05-12T13:51:27.850 | 183 | 1763 | [
"cart"
] |
10670 | 2 | null | 10667 | 0 | null | I believe you could use "simultaneous score confidence interval for OR" to analyze your question. The reference is Agresti et al. 2008 Simultaneous confidence intervals for comparing binomial parameters. Biometrics 64 1270-1275.
The R code is available in [http://www.stat.ufl.edu/~aa/cda/software.html](http://www.stat... | null | CC BY-SA 3.0 | null | 2011-05-11T18:52:39.843 | 2011-05-11T18:52:39.843 | null | null | 4559 | null |
10671 | 2 | null | 10662 | 1 | null | I agree with JMS, you will need to plot each of your variable first because PCA requires the normality assumption. If your variables are not normally distributed then it is not appropriate to use PCA before transforming the variables. I think you will need to ask yourself, what you really want to know from this data se... | null | CC BY-SA 3.0 | null | 2011-05-11T19:11:30.590 | 2011-05-11T19:11:30.590 | null | null | 4559 | null |
10672 | 1 | 10757 | null | 13 | 1828 | I've read about a number of algorithms for solving n-armed bandit problems like $\epsilon$-greedy, softmax, and UCB1, but I'm having some trouble sorting through what approach is best for minimizing regret.
Is there a known optimal algorithm for solving the n-armed bandit problem? Is there a choice of algorithm that se... | Optimal algorithm for solving n-armed bandit problems? | CC BY-SA 3.0 | null | 2011-05-11T19:57:03.987 | 2012-07-09T16:32:29.370 | 2012-07-09T16:32:29.370 | 4872 | 4281 | [
"machine-learning",
"reinforcement-learning",
"multiarmed-bandit"
] |
10673 | 2 | null | 10657 | 0 | null | Cluster analysis does not involve hypothesis testing per se, but is really just a collection of different similarity algorithms for exploratory analysis. You can force hypothesis testing somewhat but the results are often inconsistent, since cluster changes are very sensitive to changes in parameters. So the answer is ... | null | CC BY-SA 3.0 | null | 2011-05-11T20:07:20.180 | 2011-05-11T20:07:20.180 | null | null | 3489 | null |
10674 | 1 | null | null | 0 | 904 | I have a two-dimensional data set that looks like $(t, x)$ where $t$ is a time in seconds when event $X$ happened. $X$ ranges from $[0, 200]$.
I want to visualize the frequency of each $x$ at time $t$ over some time period. I guess this would be a bar graph with $x$-axis being event #, $y$-axis being frequency, and $z... | Modeling frequency over time | CC BY-SA 3.0 | null | 2011-05-11T20:07:34.483 | 2013-05-15T04:33:18.643 | 2013-05-15T04:33:18.643 | 805 | 4571 | [
"r",
"data-visualization"
] |
10676 | 1 | 10677 | null | 8 | 10107 | Lets say I have a highly dimensional classification problem with a lot of noise, and I want to improve my results by removing some of the noisy variables. I've [read](http://research.microsoft.com/pubs/69946/tr-2002-63.pdf) [several](http://www.andrew.cmu.edu/user/minhhoan/papers/SVMFeatureWeight_PR.pdf) [papers](http... | Using an SVM for feature selection | CC BY-SA 3.0 | null | 2011-05-11T20:32:04.700 | 2011-05-12T09:22:47.270 | null | null | 2817 | [
"r",
"svm"
] |
10677 | 2 | null | 10676 | 5 | null | As I understand them, SVMs have built-in regularization because they tend to penalize large weights of predictors which amounts to favor simpler models. They're often used with [recursive feature elimination](http://www.brainvoyager.com/bvqx/doc/UsersGuide/WebHelp/Content/MVPATools/Recursive_Feature_Elimination.htm) (i... | null | CC BY-SA 3.0 | null | 2011-05-11T20:47:54.837 | 2011-05-11T20:47:54.837 | null | null | 930 | null |
10678 | 2 | null | 10649 | 3 | null | The confidence limits for an ARIMA forecast are based upon the PSI WEIGHTS . The PSI WEIGHTS are easily computed by representing the ARIMA MODEL as a pure MOVING AVERAGE MODEL. One should not be dependent upon software ( any software ! ) for answers.
| null | CC BY-SA 3.0 | null | 2011-05-11T21:10:48.220 | 2011-05-11T21:10:48.220 | null | null | 3382 | null |
10679 | 2 | null | 10664 | 3 | null | Here are 3 options; hard brackets need to be filled in, while braces contain optional subcommands:
```
cross [varlist] by age {/cells count col row}.
means [varlist] by age {/stat anova}.
```
summarize command - best obtained through the menus via Analyze...Reports...Case Summaries. Then you may need to double-click... | null | CC BY-SA 3.0 | null | 2011-05-11T22:19:51.593 | 2011-05-12T01:55:53.997 | 2011-05-12T01:55:53.997 | 183 | 2669 | null |
10680 | 1 | 10724 | null | 15 | 981 | This is sort of an open ended question but I wanna be clear. Given a sufficient population you might be able to learn something (this is the open part) but whatever you learn about your population, when is it ever applicable to a member of the population?
From what I understand of statistics it's never applicable to a ... | How to NOT use statistics | CC BY-SA 3.0 | null | 2011-05-11T17:47:50.800 | 2014-07-20T01:01:54.527 | 2014-07-20T00:13:05.307 | 22468 | 4576 | [
"teaching",
"validity"
] |
10681 | 2 | null | 10680 | 9 | null | Unless the people in the room are a random sample of the world's population, any conclusions based on statistics about the world's population are going to be very suspect. One out of every 5 people in the world is Chinese, but none of my five children are...
| null | CC BY-SA 3.0 | null | 2011-05-11T17:56:10.473 | 2011-05-11T17:56:10.473 | null | null | null | null |
10682 | 2 | null | 10680 | 6 | null |
- To address overapplication of statistics to small samples, I recommend countering with well-known jokes ("I am so excited, my mother is pregnant again and my baby sibling will be Chinese." "Why?" "I have read that every fourth baby is Chinese.").
- Actually, I recommend jokes to address all kinds of misconception i... | null | CC BY-SA 3.0 | null | 2011-05-11T17:57:05.260 | 2011-05-11T17:57:05.260 | null | null | 17823 | null |
10683 | 2 | null | 10680 | 3 | null | There is an interesting article by Mary Gray on misuse of statistics in court cases and things like that...
Gray, Mary W.; Statistics and the Law. Math. Mag. 56 (1983), no. 2, 67–81
| null | CC BY-SA 3.0 | null | 2011-05-11T18:09:30.407 | 2011-05-11T18:09:30.407 | null | null | 42201 | null |
10684 | 2 | null | 10680 | 0 | null | Hypothesis: $A$
(Textbook) Result: Do no reject $A$ ($\sigma = c$)
Your Statement: $A$ holds with probability $\sigma$!
Correct would be: In this case, you know nothing. If you want to "prove" $A$, your hypothesis has to be $\neg A$; reject it with $\sigma$ to get the desired statement.
| null | CC BY-SA 3.0 | null | 2011-05-11T20:27:27.100 | 2011-05-11T20:27:27.100 | null | null | 12868 | null |
10685 | 2 | null | 10669 | 2 | null | I see that you've accepted answers to just 3 of 9 questions...
Are you using the type of Decision Tree known as CHAID? If so, you will obtain an indication of one main effect and then any number of so-called interaction effects. You can try these effects in a regression, ANOVA, or general linear model. You build in ... | null | CC BY-SA 3.0 | null | 2011-05-11T22:48:58.137 | 2011-05-11T22:48:58.137 | 2017-04-13T12:44:24.677 | -1 | 2669 | null |
10686 | 2 | null | 10643 | 2 | null | My first instinct is to say that it would be silly to make such a determination absent any knowledge of the topic. "Too large" for what, or for whom? But perhaps what you're looking for is really a test for outliers in the distribution--not that you're likely to find any in the one you've shown. Check out Dixon's Te... | null | CC BY-SA 3.0 | null | 2011-05-11T23:04:01.823 | 2011-05-11T23:04:01.823 | null | null | 2669 | null |
10687 | 1 | null | null | 22 | 63713 | I know correlation does not imply causation but instead the strength and direction of the relationship. Does simple linear regression imply causation? Or is an inferential (t-test, etc.) statistical test required for that?
| Does simple linear regression imply causation? | CC BY-SA 3.0 | null | 2011-05-11T23:05:00.207 | 2022-05-07T10:19:43.723 | 2011-05-12T06:44:00.000 | 930 | 4572 | [
"regression",
"correlation",
"causality"
] |
10688 | 2 | null | 2770 | 3 | null | I was using "[Fundamentals of Clinical Trials](https://link.springer.com/book/10.1007/978-3-319-18539-2)" when I was in PhD program.
| null | CC BY-SA 4.0 | null | 2011-05-11T23:19:08.927 | 2022-12-06T02:54:37.377 | 2022-12-06T02:54:37.377 | 362671 | 4559 | null |
10689 | 2 | null | 10687 | 24 | null | The quick answer is, no. You can easily come up with non-related data that when regressed, will pass all sorts of statistical tests. Below is an old picture from Wikipedia (which, for some reason has recently been removed) that has been used to illustrate data-driven "causality".
We need more pirates to cool the... | null | CC BY-SA 3.0 | null | 2011-05-11T23:44:00.907 | 2011-05-11T23:57:04.897 | 2011-05-11T23:57:04.897 | 2775 | 2775 | null |
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