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9590 | 1 | 9591 | null | 13 | 7756 | I am currently using principal components analysis to select variables to use in modelling. At the moment, I make measurements A, B and C in my experiments -- What I really want to know is: Can I make fewer measurements and stop recording C and or B to save time and effort?
I find that all 3 variables load heavily ... | How to use principal components analysis to select variables for regression? | CC BY-SA 3.0 | null | 2011-04-15T17:57:38.110 | 2015-05-31T07:34:30.303 | 2011-04-16T08:04:39.547 | 930 | 4054 | [
"regression",
"pca",
"model-selection"
] |
9591 | 2 | null | 9590 | 16 | null | You haven't specified what "modeling" you plan on, but it sounds like you're asking about how to select independent variables among $A$, $B$, and $C$ for the purpose of (say) regressing a fourth dependent variable $W$ on them.
To see that this approach can go wrong, consider three independent Normally distributed varia... | null | CC BY-SA 3.0 | null | 2011-04-15T19:38:16.313 | 2011-04-15T20:45:56.973 | 2011-04-15T20:45:56.973 | 919 | 919 | null |
9592 | 1 | null | null | 5 | 5454 | I want to predict a continuous variable like porosity from remote sensing data. Let's say I have a measure of the reflectivity of the surface of the earth, densely sampled over an area. And I have rock porosity at some locations. So I'd like to calibrate the former to the latter, and transform reflectivity to porosity.... | Which parameter on x axis for linear regression? | CC BY-SA 4.0 | null | 2011-04-15T19:58:13.213 | 2018-10-25T09:30:04.200 | 2018-10-25T09:30:04.200 | 128677 | 4187 | [
"regression",
"linear"
] |
9593 | 2 | null | 9592 | 1 | null | As far as I can tell, you have a lot of reflectivity measurements and a few porosity measurements, and you want to estimate porosity in areas where you only have reflectivity information.
In that case, you want to regress porosity ($y$) on reflectivity ($x$) to minimise the sum of squares error in the porosity estimate... | null | CC BY-SA 3.0 | null | 2011-04-15T20:23:14.450 | 2011-04-15T20:23:14.450 | null | null | 2958 | null |
9595 | 1 | null | null | 9 | 635 | I am using a distributed lag model to analyze a time series data. The duration of study period is 18 years, and the observation is yearly data. When including a 1-year lag effect, the first year of the lag variable becomes missing. Then, a 2-year lag effect makes the first two data of lag variable missing, and so on.
... | How to decrease the information loss from lag variables? | CC BY-SA 3.0 | null | 2011-04-15T22:57:39.797 | 2012-08-26T05:00:29.040 | 2011-04-16T08:02:51.610 | 930 | 4083 | [
"time-series",
"missing-data"
] |
9596 | 1 | null | null | 3 | 791 | The health screening for immigration to the US includes a mandatory chest x-ray screening for all adults, with the aim of detecting signs of tuberculosis. If the x-ray indicates the possibility of TB, the applicant then submits sputum samples for culturing. That's a lot of x-rays - and many of them don't result in po... | Using non-random sample to make population estimates | CC BY-SA 3.0 | null | 2011-04-15T23:05:00.143 | 2017-04-29T21:14:14.553 | 2017-04-29T21:14:14.553 | 28666 | 71 | [
"logistic",
"proportion",
"resampling"
] |
9597 | 2 | null | 9592 | 1 | null | It's not a good idea to rearrange an OLS fit equation. If you do, you're changing the assumptions of what variables were assumed to be "independent" versus "dependent".
If you have to rearrange things, one way around this is to use Principal Component Analysis rather than Ordinary Least Squares. Here's an article t... | null | CC BY-SA 3.0 | null | 2011-04-15T23:23:06.797 | 2011-04-17T17:47:38.030 | 2011-04-17T17:47:38.030 | 2775 | 2775 | null |
9598 | 2 | null | 9588 | 1 | null | I did a Google search and bumped into the following title:
"A study of cross-validation and bootstrap for accuracy estimation and model selection"
I would post the link but it's one of those citeseerx.ist.psu.edu links that comes across with an IP number for the URL. If you do the search, you should be able to find i... | null | CC BY-SA 3.0 | null | 2011-04-16T00:01:36.560 | 2011-04-16T00:10:23.233 | 2011-04-16T00:10:23.233 | 2775 | 2775 | null |
9599 | 1 | null | null | 9 | 5684 | For example, say we have three coins. Due to several tests, we have 95% confidence intervals for X (the heads percentage) of each coin. Coin A is 0-5%, coin B is 1-9%, and coin C is 40-70%.
[For the curious, these CI's were created using inverse binomial CDF, based a small sample of ~30 flips of the coin. Not su... | Can confidence intervals be added? | CC BY-SA 3.0 | null | 2011-04-16T00:10:36.150 | 2011-04-16T09:58:53.657 | 2011-04-16T05:37:24.497 | 4191 | 4191 | [
"confidence-interval",
"binomial-distribution",
"frequentist"
] |
9600 | 2 | null | 9596 | 1 | null | The $24,000$ applicant population was divided into two strata
- The $N_1$ applicants who had chest x-rays indicative of TB
- The $N_2=24,000-N_1$ who had chest x-rays not indicative of TB
A stratified random sample was taken. This included $n_1$ participants with x-rays indicative of TB and $n_2=1,500-n_1$ particip... | null | CC BY-SA 3.0 | null | 2011-04-16T00:26:16.457 | 2011-04-16T00:34:17.477 | 2011-04-16T00:34:17.477 | 3874 | 3874 | null |
9601 | 2 | null | 9596 | 1 | null | You have measured 3 variables, X-ray (pos or neg), sputum (pos, neg or not tested), new diagnostic (pos, neg or not tested), and you may have 9 outcome groups:
```
group X-ray sputum new
a 1 + + +
a 2 + + -
a 3 + + nt
b 4 + - +
b 5 + - -
... | null | CC BY-SA 3.0 | null | 2011-04-16T00:43:15.787 | 2011-04-16T01:03:15.690 | 2011-04-16T01:03:15.690 | 3911 | 3911 | null |
9602 | 2 | null | 9599 | 3 | null | There are multiple methods to calculate the [binomial confidence intervals](http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval), and in case of your small (~30) sample size this may mean notable differences.
Can we say anything about the number of heads we would expect to see if we flipped all three? ... | null | CC BY-SA 3.0 | null | 2011-04-16T01:35:00.803 | 2011-04-16T01:54:07.583 | 2011-04-16T01:54:07.583 | 3911 | 3911 | null |
9603 | 2 | null | 9573 | 4 | null | The central limit theorem establishes (under the required conditions) that the numerator of the t-statistic is asymptotically normal. The t-statistic also has a denominator. To have a t-distribution you'd need the denominator to be independent and square-root-of-a-chi-square-on-its-df.
And we know it won't be independe... | null | CC BY-SA 4.0 | null | 2011-04-16T01:44:37.743 | 2022-07-18T22:26:17.110 | 2022-07-18T22:26:17.110 | 805 | 805 | null |
9604 | 1 | null | null | 6 | 2932 | Is it possible have an SES where the component equations are probabilistic, say, logit or probit? I am evaluating a number of quality metrics of services provided by a number of providers. The metrics are binary (pass/fail quality certification). The obvious approach would appear to be to estimate the logit/probit equa... | Simultaneous Equation System for logit/probit? | CC BY-SA 3.0 | null | 2011-04-16T05:18:41.593 | 2014-06-12T17:46:09.247 | 2012-06-20T19:01:37.370 | null | 3671 | [
"r",
"regression",
"logit",
"probit"
] |
9605 | 1 | null | null | 2 | 589 | I am interested in using the MacKinnon product of coefficients test to assess mediation. Is there a quick way to carry out these calculations and determine significance based on the obtained statistic?
| Calculating the MacKinnon empirical distribution test to test mediation | CC BY-SA 3.0 | null | 2011-04-16T06:50:43.337 | 2011-04-16T08:00:23.233 | 2011-04-16T08:00:23.233 | 930 | null | [
"mediation"
] |
9606 | 2 | null | 9605 | 4 | null | Tests for full and partial mediation are well explained on the [Mediation FAQ](http://www.public.asu.edu/~davidpm/ripl/q&a.htm) webpage by David P. MacKinnon, and they can be implemented in any statistical package offering tools for regression modeling. Bootstraping can be used to derive standard errors and confidence ... | null | CC BY-SA 3.0 | null | 2011-04-16T07:59:30.117 | 2011-04-16T07:59:30.117 | null | null | 930 | null |
9607 | 1 | 9683 | null | 2 | 1391 | Background:
I've modeled a project effort prediction as a Google Spreadsheet template. Details of the Model: [http://sites.google.com/site/effortprediction/methodology](http://sites.google.com/site/effortprediction/methodology)
.Google Spreadsheet does not implement beta distribution functions.
In PERT a beta distribut... | Summing normal instead of beta distributions, consequences for the density function of the sum? | CC BY-SA 3.0 | null | 2011-04-16T09:03:50.340 | 2011-06-17T12:59:36.540 | 2011-04-16T11:24:25.440 | 1390 | 778 | [
"normal-distribution",
"central-limit-theorem",
"beta-distribution"
] |
9608 | 2 | null | 9599 | 1 | null | This is a problem which can be relatively easily solved using a Bayesian approach. This is one of the "re-using" the original data answers, rather than one which uses the CIs.
Each coin has some long run frequency of heads $\theta_i$ $(i=1,\dots,50)$. Now you have some partial information about these frequencies, tha... | null | CC BY-SA 3.0 | null | 2011-04-16T09:58:53.657 | 2011-04-16T09:58:53.657 | null | null | 2392 | null |
9609 | 2 | null | 1040 | 2 | null | model misspecification is irrelevant when you are interpreting co-efficients in a model, but any such interpretation is always conditional on the model. This is because a model is about an association, not about a causal relationship - the OLS coefficients just tells you about the linear association between $X$ and $Y... | null | CC BY-SA 3.0 | null | 2011-04-16T10:55:20.060 | 2011-04-16T10:55:20.060 | null | null | 2392 | null |
9610 | 1 | 9613 | null | 15 | 7248 | Is there such a formula? Given a set of data for which the mean, variance, skewness and kurtosis is known, or can be measured, is there a single formula which can be used to calculate the probability density of a value assumed to come from the aforementioned data?
| Closed form formula for distribution function including skewness and kurtosis? | CC BY-SA 3.0 | null | 2011-04-16T14:28:54.517 | 2011-09-26T17:22:30.127 | 2011-09-26T17:22:30.127 | 919 | 226 | [
"distributions",
"density-function",
"kurtosis",
"skewness"
] |
9611 | 2 | null | 9610 | 2 | null | [D’Agostino’s K2 test](http://en.wikipedia.org/wiki/D%27Agostino%27s_K-squared_test) will tell you whether a sample distribution came from a normal distribution based on the sample's skewness and kurtosis.
If you want to do a test assuming a non-normal distribution (perhaps with high skewness or kurtosis), you'll need ... | null | CC BY-SA 3.0 | null | 2011-04-16T14:46:25.153 | 2011-04-16T14:46:25.153 | null | null | 3874 | null |
9612 | 2 | null | 9278 | 0 | null | You might be looking for a goodness-of-fit test to check one of the following
- whether the sample distribution are likely to have come from the population distribution or
- whether multiple sample distributions are likely to have come the same unknown population distribution
Look at chi-square and Kolmogorov-Smirn... | null | CC BY-SA 3.0 | null | 2011-04-16T15:12:30.973 | 2011-04-16T15:12:30.973 | null | null | 3874 | null |
9613 | 2 | null | 9610 | 14 | null | There are many such formulas. The first successful attempt at solving precisely this problem was made by Karl Pearson in 1895, eventually leading to the system of [Pearson distributions](http://en.wikipedia.org/wiki/Pearson_distribution). This family can be parameterized by the mean, variance, skewness, and kurtosis.... | null | CC BY-SA 3.0 | null | 2011-04-16T17:10:33.647 | 2011-04-16T17:10:33.647 | null | null | 919 | null |
9614 | 1 | 9624 | null | 9 | 4062 | I am modeling Diabetes Prediction using Logistic Regression. The dataset used is the [Behavioral Risk Factor Surveillance System (BRFSS)](http://www.cdc.gov/BRFSS/) of the Center for Disease Control (CDC). One of the independent variables is High Blood Pressure. It is categorical with the following levels 'Yes', 'No',... | Treating 'Don't know/Refused' levels of categorical variables | CC BY-SA 3.0 | null | 2011-04-16T17:27:46.990 | 2011-04-16T20:11:08.790 | 2011-04-16T20:01:35.560 | 919 | 3897 | [
"logistic",
"missing-data"
] |
9615 | 1 | null | null | 3 | 537 | I wonder if two identical time series are cointegrated. Can anyone shed some light on this? Thanks
| Are two identical time series cointegrated? | CC BY-SA 3.0 | null | 2011-04-16T17:29:20.930 | 2011-04-16T17:38:37.700 | null | null | 4192 | [
"time-series",
"cointegration"
] |
9616 | 2 | null | 9615 | 3 | null | If I understand your question correctly, you are asking if you take two identical time series (i.e. a time series and a direct copy of it), are these two cointegrated?
If that is your question, then it really reduces to a simpler form: is a time series stationary? As I understand it, cointegration is the phenomenon wh... | null | CC BY-SA 3.0 | null | 2011-04-16T17:38:37.700 | 2011-04-16T17:38:37.700 | null | null | 781 | null |
9617 | 1 | 9635 | null | 44 | 16316 | I'm hoping that someone can explain, in layman's terms, what a characteristic function is and how it is used in practice. I've read that it is the Fourier transform of the pdf, so I guess I know what it is, but I still don't understand its purpose. If someone could provide an intuitive description of its purpose and pe... | What is the purpose of characteristic functions? | CC BY-SA 3.0 | null | 2011-04-16T18:00:25.273 | 2017-11-27T22:34:54.713 | 2017-11-27T22:34:54.713 | 128677 | 1913 | [
"probability",
"mathematical-statistics",
"characteristic-function"
] |
9618 | 2 | null | 9614 | 0 | null | Do you have any reason to think that study subjects with diabetes were more likely or less likely to end up with the DK/R response? If not (and I'd be pretty surprised to find out you did), including this predictor in the model w/o excluding these cases will result in noise. That is, you'll end up with less precision i... | null | CC BY-SA 3.0 | null | 2011-04-16T18:57:16.713 | 2011-04-16T18:57:16.713 | null | null | 11954 | null |
9619 | 2 | null | 9614 | 2 | null | First you have to think over if the missing data are missing completely at random (MCAR), missing at random (MAR) or missing not at random (MNAR) as deletion (in other words complete-case analysis) may lead to biased results. Alternatives are inverse probability weighting, multiple imputation, the full-likelihood metho... | null | CC BY-SA 3.0 | null | 2011-04-16T18:58:14.193 | 2011-04-16T18:58:14.193 | null | null | 3911 | null |
9620 | 2 | null | 9617 | 4 | null | The purpose of characteristic functions is that they can be used to derive the properties of distributions in probability theory. If you're not interested in such derivations you do not need to learn about characteristic functions.
| null | CC BY-SA 3.0 | null | 2011-04-16T19:24:55.683 | 2011-04-16T19:24:55.683 | null | null | 449 | null |
9621 | 2 | null | 9617 | 2 | null | The characteristic function is the Fourier transform of the density function of the distribution. If you have any intuition regarding Fourier transforms, this fact may be enlightening. The common story about Fourier transforms is that they describe the function 'in frequency space.' Since a probability density is usua... | null | CC BY-SA 3.0 | null | 2011-04-16T19:36:20.890 | 2011-04-16T19:36:20.890 | null | null | 795 | null |
9622 | 2 | null | 9610 | 7 | null | This sounds like a ['moment-matching' approach](http://www.johndcook.com/blog/2010/09/20/skewness-andkurtosis/) to fitting a distribution to data. It is generally regarded as not a great idea (the title of John Cook's blog post is 'a statistical dead end').
| null | CC BY-SA 3.0 | null | 2011-04-16T19:39:11.997 | 2011-04-16T19:39:11.997 | null | null | 795 | null |
9623 | 1 | null | null | 3 | 30095 | I've read the sampling error examples in Wikipedia, but I still don't understand. Could you give me a simple and characteristic example for sampling error?
| Sampling error example | CC BY-SA 3.0 | null | 2011-04-16T19:47:10.053 | 2015-08-06T13:05:40.750 | 2011-05-20T09:04:49.110 | 930 | 4176 | [
"sampling"
] |
9624 | 2 | null | 9614 | 6 | null | I was just wondering about exactly the same question when analyzing the latest [National Hospital Discharge Survey](http://www.cdc.gov/nchs/injury/injury_hospital.htm) data. Several variables have substantial missing values, such as marital status and type of procedure. This issue came to my attention because these c... | null | CC BY-SA 3.0 | null | 2011-04-16T20:11:08.790 | 2011-04-16T20:11:08.790 | null | null | 919 | null |
9625 | 1 | 9630 | null | 7 | 2575 | Suppose you wish to test the hypothesis that the mean of a distribution equals the median, given some samples drawn from the distribution. How would this be done? I am guessing that the test statistic would be (the absolute value of) the sample mean minus the sample median, but am not sure about the standard error of t... | How to test if the mean equals the median? | CC BY-SA 3.0 | null | 2011-04-16T20:11:45.357 | 2011-04-18T03:43:18.880 | null | null | 795 | [
"hypothesis-testing",
"mean",
"median"
] |
9626 | 1 | null | null | 6 | 2046 | I believe that for ANOVA unbalanced means unequal groups. But for randomised controlled trials unbalanced seems to mean something like groups of different characteristics despite randomisation. So in case of RCTs does unbalanced simply mean unlucky randomisation? Nothing to do with group sizes? What's the definition of... | When is a randomised controlled trial (RCT) balanced? | CC BY-SA 3.0 | null | 2011-04-16T20:16:25.510 | 2011-08-15T09:36:58.240 | 2011-04-18T08:36:35.570 | 183 | 4176 | [
"sample-size",
"clinical-trials"
] |
9627 | 1 | 9720 | null | 9 | 4440 | First of all, I realize multiple regression does not really give actually "causal" inferences about the data. Let me explain my current case:
I have four independent variables which I hope (but am not sure) are involved in driving the thing I'm measuring. I wanted to use multiple regression to see how much each of thes... | What should I be aware of when using multiple regression to find "causal" relationships in my data? | CC BY-SA 3.0 | null | 2011-04-16T20:36:49.650 | 2011-04-20T04:35:36.783 | 2011-04-20T04:35:36.783 | 4139 | 4139 | [
"multivariate-analysis",
"multiple-regression"
] |
9628 | 2 | null | 9623 | 1 | null | When you take a sample you do so because you want to know what is happening in the wider population but unless you take a census and ask everyone there will always be a difference between the sample and the population.
There is a good definition below taken from [here](http://www.socialresearchmethods.net/kb/sampstat.... | null | CC BY-SA 3.0 | null | 2011-04-16T20:46:32.990 | 2011-04-16T22:24:59.253 | 2011-04-16T22:24:59.253 | 3597 | 3597 | null |
9629 | 1 | 9641 | null | 14 | 31138 | I've read that the chi square test is useful to see if a sample is significantly different from a set of expected values.
For example, here is a table of results of a survey regarding people's favourite colours (n=15+13+10+17=55 total respondents):
```
red, blue, green, yellow
15, 13, 10, 17
```
A chi square test can... | Can chi square be used to compare proportions? | CC BY-SA 4.0 | null | 2011-04-16T21:08:05.087 | 2022-01-03T17:57:08.743 | 2022-01-02T13:29:49.523 | 11887 | 2830 | [
"chi-squared-test",
"hypothesis-testing",
"proportion"
] |
9630 | 2 | null | 9625 | 4 | null | This is a bootstrap confidence interval for the (median - mean) difference in R:
```
z = function() {s = sample(women$weight, replace=TRUE); median(s)-mean(s)}
k = replicate(10000, z())
c(quantile(k, c(.025, .5, .975)), mean=mean(k), sd=sd(k), qgte0=mean(k>=0))
2.5% 50% 97.5% mean sd qgt... | null | CC BY-SA 3.0 | null | 2011-04-16T21:10:55.900 | 2011-04-16T21:10:55.900 | null | null | 3911 | null |
9631 | 2 | null | 9629 | 3 | null | Yes, you can test the null hypothesis:
```
"H0: prop(red)=prop(blue)=prop(green)=prop(yellow)=1/4"
```
using a chi square test that compares the proportions of the survey (0.273, ...) to the expected proportions (1/4, 1/4, 1/4, 1/4)
| null | CC BY-SA 4.0 | null | 2011-04-16T21:58:42.240 | 2022-01-02T09:47:16.443 | 2022-01-02T09:47:16.443 | 56211 | null | null |
9632 | 2 | null | 9627 | 5 | null | If you want to see if the independent variables are correlated, that's easy - just test the correlations e.g. with PROC CORR in SAS, or cor in R, or whatever in whatever package you use.
You might, though, want to test collinearity instead, or in addition.
But that's only part of the problem for causation. More probl... | null | CC BY-SA 3.0 | null | 2011-04-16T22:01:47.947 | 2011-04-16T22:01:47.947 | null | null | 686 | null |
9633 | 2 | null | 9590 | 4 | null | If you have only 3 IVs, why do you want to reduce them?
That is, is your sample very small (so that 3 IVs risks overfitting)? In this case, consider partial least squares
Or are the measurements very expensive (so, in future, you'd like to measure only one IV)? In this case, I'd consider looking at the different regr... | null | CC BY-SA 3.0 | null | 2011-04-16T22:06:58.427 | 2011-04-16T22:06:58.427 | null | null | 686 | null |
9634 | 2 | null | 9623 | 4 | null | Imagine that you want to know the average height of men on earth. This average height exists but obviously you will never be able to know it (unless you're able to measure several millions men...).
What you can do is measure hundreds or thousands of people and calculate the average height of these people. The average h... | null | CC BY-SA 3.0 | null | 2011-04-16T22:15:11.883 | 2011-04-16T22:15:11.883 | null | null | 3377 | null |
9635 | 2 | null | 9617 | 61 | null | Back in the day, people used logarithm tables to multiply numbers faster. Why is this? Logarithms convert multiplication to addition, since $\log(ab) = \log(a) + \log(b)$. So in order to multiply two large numbers $a$ and $b$, you found their logarithms, added the logarithms, $z = \log(a) + \log(b)$, and then looked... | null | CC BY-SA 3.0 | null | 2011-04-17T00:40:55.120 | 2013-08-19T14:57:12.920 | 2013-08-19T14:57:12.920 | 7290 | 3567 | null |
9636 | 2 | null | 9629 | 5 | null | The chi-square test is good as long as the expected counts are large, usually above 10 is fine. below this the $\frac{1}{E(x_{i})}$ part tends to dominate the test. An exact test statistic is given by:
$$\psi=\sum_{i}x_{i}\log\left(\frac{x_{i}}{np_{i}}\right)$$
Where $x_{i}$ is the observed count in category $i$. $i... | null | CC BY-SA 3.0 | null | 2011-04-17T00:57:53.990 | 2011-04-17T02:04:47.963 | 2011-04-17T02:04:47.963 | 2392 | 2392 | null |
9637 | 1 | null | null | 5 | 2091 | standard deviation (std) can be used when arithmetic mean is the right statistic. What is the equivalent of std when geometric mean is the right statistic?
| Geometric mean and standard deviation | CC BY-SA 3.0 | null | 2011-04-17T03:23:52.840 | 2011-04-17T05:38:56.727 | 2011-04-17T04:42:11.300 | 2116 | null | [
"hypothesis-testing",
"confidence-interval"
] |
9638 | 2 | null | 9627 | 2 | null | The relationship between causation and association is basically in answering the following question:
What else, besides the hypothesised causal relationship, could have caused $X$ and $Y$ to be related to each other?
As long as the answer to this question is not "nothing" then you can only talk definitively about assoc... | null | CC BY-SA 3.0 | null | 2011-04-17T03:46:02.657 | 2011-04-17T03:46:02.657 | null | null | 2392 | null |
9639 | 2 | null | 9637 | 2 | null | The geometric mean can be motivated by noting that it is the exponent of the arithmetic mean of the logarithm:
$$\overline{x}_{geo}=\left(\prod_{i}x_{i}\right)^{\frac{1}{n}}=exp\left[\frac{1}{n}\sum_{i}log(x_{i})\right]=exp\left[\overline{log(x)}\right]$$
So it makes sense to take the exponent of the standard deviation... | null | CC BY-SA 3.0 | null | 2011-04-17T05:38:56.727 | 2011-04-17T05:38:56.727 | null | null | 2392 | null |
9640 | 2 | null | 9627 | 1 | null | I'm not sure what field your work is in, so this may or may not be of any help- but I'm most familiar with using SPSS with psychological constructs. In my experience, if I have a few variables predicting an outcome variable (or dependent variable) in a regression, and I have one or more independent variables show up as... | null | CC BY-SA 3.0 | null | 2011-04-17T06:15:50.957 | 2011-04-17T06:15:50.957 | null | null | 4198 | null |
9641 | 2 | null | 9629 | 7 | null | Correct me if I'm wrong, but I think this can be done in R using this command
```
chisq.test(c(15, 13, 10, 17))
Chi-squared test for given probabilities
data: c(15, 13, 10, 17)
X-squared = 1.9455, df = 3, p-value = 0.5838
```
This assumes proportions of 1/4 each. You can modify expected val... | null | CC BY-SA 4.0 | null | 2011-04-17T08:16:50.087 | 2022-01-02T13:31:55.033 | 2022-01-02T13:31:55.033 | 11887 | 144 | null |
9642 | 2 | null | 9629 | 2 | null | The test statistic for Pearson's chi-square test is
$$\sum_{i=1}^{n} \frac{(O_i - E_i)^2}{E_i}$$
If you write $o_i = \dfrac{O_i}{n}$ and $e_i = \dfrac{E_i}{n}$ to have proportions, where $n=\sum_{i=1}^{n} O_i$ is the sample size and $\sum_{i=1}^{n} e_i =1$, then the test statistic is is equal to
$$n \sum_{i=1}^{n} \f... | null | CC BY-SA 3.0 | null | 2011-04-17T11:21:53.427 | 2011-04-17T11:21:53.427 | null | null | 2958 | null |
9643 | 1 | null | null | 1 | 1566 | I am using R software (R commander) to cluster my data. I have a smaller subset of my data containing 200 rows and about 800 columns. I am getting the following error when trying kmeans cluster and plot on a graph:
>
'princomp' can only be used with more units than variables
I then created a test doc of 10 row and 1... | Clustering can be plotted only with more units than variables? | CC BY-SA 3.0 | null | 2011-04-17T12:42:40.607 | 2011-04-17T13:10:59.397 | 2011-04-17T12:56:15.330 | null | 4020 | [
"r",
"clustering"
] |
9644 | 2 | null | 9643 | 1 | null | The clustering itself has no problems with the p>n situation, however the visualization internally uses `princomp` (which is incapable of handling p>n) to plot the similarity space projection.
You can't fix that, at most try to reproduce similar graph by obtaining similarity space projection with `cmdscale(dist(...))` ... | null | CC BY-SA 3.0 | null | 2011-04-17T13:03:21.863 | 2011-04-17T13:10:59.397 | 2011-04-17T13:10:59.397 | null | null | null |
9645 | 1 | 9646 | null | 3 | 2301 | I'm kind of new doing data mining, so sorry if my question is not very clear.
I'm working in a project that is aiming to do data mining over the interactions of the students with a e-learning platform. So, I'm trying to generate decision trees with some data that I collected (number of times that student use a resourc... | Minimum number of instances to create a decision tree | CC BY-SA 3.0 | 0 | 2011-04-17T14:01:40.170 | 2011-05-17T07:51:12.127 | 2011-04-17T14:50:36.583 | null | 4203 | [
"cart",
"weka"
] |
9646 | 2 | null | 9645 | 2 | null | {I'm guessing that for each person and each variable you have a single value covering the 2 months--i.e., you don't have repeated measures.} 44 sounds like an awfully small number. Is it a random sample of a larger population? The answer to that, for one thing, would affect your confidence in the findings.
I won't ... | null | CC BY-SA 3.0 | null | 2011-04-17T14:45:36.953 | 2011-04-17T14:45:36.953 | null | null | 2669 | null |
9647 | 1 | null | null | 3 | 247 | Sometimes they ask questions in different orders, or use different prompts.
Or datasets with instruments (with at least one variable randomized)?
I would like to use at least one of them for my causal modelling course (Stat 566), whose syllabus is at [https://www.stat.washington.edu/tsr/s566/syllabus566.pdf](https://ww... | What are some examples of public datasets that have randomized instruments? | CC BY-SA 3.0 | null | 2011-04-17T17:16:26.730 | 2011-04-17T18:41:11.000 | 2011-04-17T18:11:52.157 | 919 | null | [
"dataset"
] |
9650 | 2 | null | 9647 | 3 | null | Donald Green in Yale's political science department has posted the replication data for many of his experiments. You can find them here:
- Donald Green's Replication Data
Green is also the director for the Yale Institution for Social and Policy Studies. They have a replication data archive here:
- ISPS Data Archive... | null | CC BY-SA 3.0 | null | 2011-04-17T18:41:11.000 | 2011-04-17T18:41:11.000 | null | null | 3265 | null |
9653 | 1 | 9655 | null | 23 | 46636 | I've learnt that small sample size may lead to insufficient power and type 2 error. However, I have the feeling that small samples just may be generally unreliable and may lead to any kind of result by chance. Is that true?
| Can a small sample size cause type 1 error? | CC BY-SA 3.0 | null | 2011-04-17T21:55:55.793 | 2013-10-15T11:43:49.583 | 2011-04-18T07:45:15.140 | null | 4176 | [
"hypothesis-testing",
"small-sample"
] |
9654 | 1 | 9660 | null | 11 | 344 | Many randomised controlled trial (RCT) papers report significance tests on baseline parameters just after/before randomisation to show that the groups are indeed similar. This is often part of a "baseline characteristics" table. However, significance tests measure the probability of getting the observed (or a stronger)... | Does significance test make sense to compare randomised groups at baseline? | CC BY-SA 3.0 | null | 2011-04-17T22:31:33.500 | 2011-04-18T10:52:12.323 | 2011-04-18T10:52:12.323 | 183 | 4176 | [
"statistical-significance",
"clinical-trials"
] |
9655 | 2 | null | 9653 | 27 | null | As a general principle, small sample size will not increase the Type I error rate for the simple reason that the test is arranged to control the Type I rate. (There are minor technical exceptions associated with discrete outcomes, which can cause the nominal Type I rate not to be achieved exactly especially with small... | null | CC BY-SA 3.0 | null | 2011-04-17T22:41:17.137 | 2011-04-17T22:41:17.137 | null | null | 919 | null |
9656 | 1 | null | null | 3 | 2309 | am writing a simple R script to test the spectral clustering algorithm but for the eigenvalues I don't get them all positive and lambda0 is different from 0. here is my script
```
Dsqrt<-matrix(0,length(V(g)),length(V(g))) # The D^-0.5 matrix
for(i in 1:(length(V(g))-1))
Dsqrt[i,i]<-1/sqrt(degree(g,V(g)[i-... | Problem with R code for spectral clustering | CC BY-SA 3.0 | null | 2011-04-18T00:43:44.917 | 2011-04-18T07:01:02.667 | 2011-04-18T07:01:02.667 | 2116 | 2325 | [
"r",
"clustering"
] |
9657 | 2 | null | 8371 | 7 | null | As whuber says, the short answer is that quota samples are the "poster child for outmoded, known-bad sampling methods" and "have long been discredited." The longer answer is that there may be conditions under which "quota-like" samples can work reasonably well.
Exhibit A here is recent work on reconstructing represent... | null | CC BY-SA 4.0 | null | 2011-04-18T01:45:03.810 | 2019-07-17T12:03:09.373 | 2019-07-17T12:03:09.373 | 142322 | 4110 | null |
9659 | 1 | null | null | 6 | 1050 | I wish to infer the posterior distribution on the probability of success $\theta$ in some binomial process, the twist being that I know that $\theta$ lies in the interval [0.5, 1].
The trouble is that a Beta distribution that is supported on [0.5, 1] is no longer conjugate. In particular, the prior looks like this ($p... | Fast integration of a posterior distribution | CC BY-SA 3.0 | null | 2011-04-18T02:52:39.853 | 2014-03-02T17:40:49.110 | 2011-04-18T07:40:42.710 | null | 4209 | [
"bayesian",
"beta-binomial-distribution"
] |
9660 | 2 | null | 9654 | 6 | null | A hypothesis test would be nonsensical, but a significance test may be useful.
The hypothesis test would be testing a null hypothesis that is already known to be true, as your question makes clear. It is silly to apply a statistical test to any hypothesis that has a truth value already known via completely reliable inf... | null | CC BY-SA 3.0 | null | 2011-04-18T03:25:38.537 | 2011-04-18T03:25:38.537 | null | null | 1679 | null |
9661 | 2 | null | 9625 | 0 | null | A permutations test can easily be set up to use the (mean - mode difference) as its test statistic. That would give you an exact P value for the difference.
| null | CC BY-SA 3.0 | null | 2011-04-18T03:43:18.880 | 2011-04-18T03:43:18.880 | null | null | 1679 | null |
9662 | 1 | 9703 | null | 4 | 257 | I have 3 different groups: A, B, and C:
- A (has a medical condition) has 30 entries
- B (another medical condition) has 31 entries
- C (control group) has 55 entries
All participants have information on the same set of variables.
I want to assess whether there is a statistically significant difference in risk o... | How to assess differential risk of disease across three groups after adjusting for other risk factors? | CC BY-SA 3.0 | null | 2011-04-18T04:37:50.350 | 2011-04-18T19:55:39.293 | 2011-04-18T12:19:04.447 | 4210 | 4210 | [
"correlation",
"statistical-significance",
"spss"
] |
9663 | 1 | 9665 | null | 12 | 1162 | If an item follows normal distribution, average also follows normal distribution. What about minimum and maximum?
| Distribution of extremal values | CC BY-SA 3.0 | null | 2011-04-18T05:01:08.790 | 2011-10-30T22:02:55.253 | 2011-10-30T22:02:55.253 | null | 4211 | [
"normal-distribution",
"order-statistics"
] |
9664 | 1 | 9722 | null | 106 | 19334 | Suppose I have a set of sample data from an unknown or complex distribution, and I want to perform some inference on a statistic $T$ of the data. My default inclination is to just generate a bunch of bootstrap samples with replacement, and calculate my statistic $T$ on each bootstrap sample to create an estimated distr... | What are examples where a "naive bootstrap" fails? | CC BY-SA 3.0 | null | 2011-04-18T05:44:33.810 | 2013-12-30T13:46:24.387 | 2011-04-19T13:03:21.823 | 2970 | 1106 | [
"hypothesis-testing",
"confidence-interval",
"bootstrap"
] |
9665 | 2 | null | 9663 | 13 | null | You should have a look at the [order statistics](http://mathworld.wolfram.com/OrderStatistic.html).
Here is a very brief overview.
Let $X_{1}, \ldots X_{n}$ be an i.i.d. sample of size $n$ drawn from a population with distribution function $F$ and probability density function $f$. Define $Y_{1}=X_{(1)}, \ldots, Y_{r} ... | null | CC BY-SA 3.0 | null | 2011-04-18T05:46:01.393 | 2011-04-18T06:59:14.667 | 2011-04-18T06:59:14.667 | 2116 | 3019 | null |
9666 | 1 | null | null | 5 | 1423 | My new computer clock runs at a rate that changes in odd steps over time, even after I tuned it via the Linux adjtimex software. Here is a plot of the change in the cumulative clock drift for each of about 1700 samples taken every 10000 seconds, with a bunch of missed points and outliers when I was off the network and... | Identifying modes in floating point data | CC BY-SA 4.0 | null | 2011-04-18T06:10:33.577 | 2019-01-16T10:28:23.557 | 2019-01-16T10:28:23.557 | 79696 | 2434 | [
"mode"
] |
9667 | 1 | null | null | 12 | 2691 | >
Possible Duplicate:
Books for self-studying time series analysis?
I am new to time series modelling altogether. But I am aware about regression modelling and some data mining algorithms like decision trees.
I want to learn time series from scratch. My background is Mathematics. Please suggest me any good book/mat... | Learning material about time series | CC BY-SA 3.0 | null | 2011-04-18T07:44:31.833 | 2011-04-19T15:57:14.973 | 2017-04-13T12:44:27.570 | -1 | 861 | [
"time-series",
"references"
] |
9668 | 2 | null | 9617 | 7 | null | @charles.y.zheng and @cardinal gave very good answers, I will add my two cents. Yes the characteristic function might look like unnecessary complication, but it is a powerful tool which can get you results. If you are trying to prove something with cumulative distribution function it is always advisable to check whethe... | null | CC BY-SA 3.0 | null | 2011-04-18T07:47:28.053 | 2011-04-18T07:47:28.053 | null | null | 2116 | null |
9669 | 2 | null | 9280 | 4 | null | Note that @whuber adresses all the points raised in the question, I will provide only the mathematical details. The regression with spatial error assumes that model is the following:
$$y=X\beta+u$$
where $Euu'=\Omega$. The efficient estimate of $\beta$ then is
$$\hat\beta=(X'\Omega X)^{-1}X'\Omega y$$
When you have ne... | null | CC BY-SA 3.0 | null | 2011-04-18T09:08:31.047 | 2011-04-18T13:52:57.690 | 2011-04-18T13:52:57.690 | 2116 | 2116 | null |
9670 | 2 | null | 6989 | 20 | null | No need to leave the lme4 package to account for overdispersion; just include a random effect for observation number. The BUGS/JAGS solutions mentioned are probably overkill for you, and if they aren't, you should have the easy to fit lme4 results for comparison.
```
data$obs_effect<-1:nrow(data)
overdisp.fit<-lmer(y~1... | null | CC BY-SA 3.0 | null | 2011-04-18T09:09:04.673 | 2015-06-22T11:38:13.537 | 2015-06-22T11:38:13.537 | 179 | 1893 | null |
9671 | 1 | 11196 | null | 4 | 2925 | I am assessing how good different features are for unsupervised classification of a set of objects. For each different feature I test, I have computed a feature vector that describes the object. I then want to get a metric out for how 'good' this vector is at separating the objects into their respective classes.
My cur... | How can I assess how descriptive feature vectors are? | CC BY-SA 3.0 | null | 2011-04-18T09:15:14.207 | 2011-05-24T15:05:40.330 | null | null | 4213 | [
"clustering",
"classification",
"feature-selection",
"unsupervised-learning"
] |
9672 | 2 | null | 9667 | 4 | null | I have yet still not found the time series book which I like. Here is the list of books which I found very useful:
- Time Series Analysis by J. D. Hamilton. This is the book which contains practically everything.
- Applied econometric time series by W. Enders. Classical reference
- Introductory econometrics for fina... | null | CC BY-SA 3.0 | null | 2011-04-18T09:17:26.670 | 2011-04-18T09:17:26.670 | null | null | 2116 | null |
9673 | 2 | null | 9587 | 2 | null | No, that isn't really it. One side point is you should consider looking into multi-modal inference a la Anderson and Burnham. However, there are plenty of resources, many of which also refer to the P+B book; such as this one, with code: [http://glmm.wikidot.com/basic-glmm-simulation](http://glmm.wikidot.com/basic-glmm-... | null | CC BY-SA 3.0 | null | 2011-04-18T09:28:42.100 | 2011-04-18T09:28:42.100 | null | null | 1893 | null |
9674 | 1 | null | null | 2 | 2861 | I have a bit of a problem with the [tm](http://cran.r-project.org/web/packages/tm/index.html) R package for cleaning text documents.
Here is how my code looks like:
```
library("tm")
# import text files in corpus text – ok
c_txt <- Corpus((DirSource(directory = ".", pattern ="txt",
encoding = "UTF-8")... | How to remove stopwords with Russian documents? | CC BY-SA 3.0 | null | 2011-04-18T09:37:37.580 | 2018-05-21T12:37:46.633 | 2011-04-18T09:46:25.323 | 930 | 4212 | [
"r",
"text-mining"
] |
9675 | 2 | null | 9508 | 1 | null | A Likelihood ratio test on an lme object is probably fine to test a certain hypothesis; but that is not an appropriate way to build or choose a model. You would do that with REMLoff and:
```
anova(fit1, fit2)
```
AICc is a well developed criterion for model selection that directly implies and estimates from the data... | null | CC BY-SA 3.0 | null | 2011-04-18T09:38:45.867 | 2011-04-18T09:38:45.867 | null | null | 1893 | null |
9676 | 2 | null | 9327 | 1 | null | Try the following packages on cran: Rcapture and mra. The easiest method to use in R for a log linear model, open closed or robust, M0, Mh, Mt, or Mth is well detailed here: [http://www.jstor.org/pss/3695557](http://www.jstor.org/pss/3695557) and that is what Rcapture is based on. You forgot to specify in your lmer cal... | null | CC BY-SA 3.0 | null | 2011-04-18T09:50:16.247 | 2011-04-18T09:50:16.247 | null | null | 1893 | null |
9677 | 2 | null | 8501 | 1 | null | You could always just calculate try to calculate R^2 from the squared covariance of Ypred and Ynew over the product of their variances: [http://www.stator-afm.com/image-files/r-squared.gif](http://www.stator-afm.com/image-files/r-squared.gif)
This works for any continuous model, regardless of how the parameters are est... | null | CC BY-SA 3.0 | null | 2011-04-18T10:01:20.853 | 2011-04-18T10:01:20.853 | null | null | 1893 | null |
9678 | 2 | null | 9664 | 2 | null | The naive bootstrap depends on the sample size being large, so that the empirical CDF for the data are a good approximation to the "true" CDF. This ensures that sampling from the empirical CDF is very much like sampling from the "true" CDF. The extreme case is when you have only sampled one data point - bootstrapping... | null | CC BY-SA 3.0 | null | 2011-04-18T10:14:17.253 | 2011-04-19T08:24:30.707 | 2011-04-19T08:24:30.707 | 2116 | 2392 | null |
9679 | 2 | null | 7675 | 3 | null | I disagree with the other responder, you shouldn't just fit varying intercept, fixed slope by default.
y~1+condition*distractor*direction+(1+condition*distractor*direction|subject) should at least be fit for model testing. Otherwise, you are just assuming the subject effect is only on the intercept and you are assuming... | null | CC BY-SA 3.0 | null | 2011-04-18T10:22:26.577 | 2011-04-18T10:22:26.577 | null | null | 1893 | null |
9680 | 2 | null | 5517 | 4 | null | It looks like your error message isn't about varying slopes, it is about correlated random effects. You can fit the uncorrelated as well; that is, a mixed-effects model with independent random effects:
```
Linear mixed model fit by REML
Formula: Reaction ~ Days + (1 | Subject) + (0 + Days | Subject)
Data: sleepstudy
... | null | CC BY-SA 3.0 | null | 2011-04-18T10:35:40.950 | 2013-08-24T14:58:15.637 | 2013-08-24T14:58:15.637 | 7290 | 1893 | null |
9681 | 2 | null | 9659 | 0 | null | You can always use Monte Carlo Integration or the Midpoint method. With Monte Carlo, you simply generate a bunch of points in your parameter space and see if they are in the area or volume or hyper-dimensional space you are trying to integrate.
From:
[http://farside.ph.utexas.edu/teaching/329/lectures/node109.html](htt... | null | CC BY-SA 3.0 | null | 2011-04-18T11:41:02.280 | 2011-04-18T11:41:02.280 | null | null | 1893 | null |
9682 | 2 | null | 9674 | 5 | null | I've just downloaded the Windows binary version of `tm` from CRAN: [http://cran.r-project.org/web/packages/tm/index.html](http://cran.r-project.org/web/packages/tm/index.html)
The `russian.dat` file that's included there isn't UTF-8, but is encoded using [KOI-8](http://en.wikipedia.org/wiki/KOI8-R). I am unfamiliar wit... | null | CC BY-SA 3.0 | null | 2011-04-18T11:46:17.800 | 2011-04-18T12:04:01.667 | 2011-04-18T12:04:01.667 | 439 | 439 | null |
9683 | 2 | null | 9607 | 2 | null | Just go ahead and calculate a random uniform, then use the inverse cdf method (http://en.wikipedia.org/wiki/Inverse_transform_sampling) to get a random Beta. Here is the formula for the inverse beta cdf: [http://www.mathworks.com/help/toolbox/stats/betainv.html](http://www.mathworks.com/help/toolbox/stats/betainv.html)... | null | CC BY-SA 3.0 | null | 2011-04-18T11:47:34.240 | 2011-04-18T11:54:48.510 | 2011-04-18T11:54:48.510 | 1893 | 1893 | null |
9684 | 2 | null | 7236 | 3 | null | Yeah, the sampling package handles this, you can do cluster sampling or stratified or a few others: [http://cran.r-project.org/web/packages/sampling/sampling.pdf](http://cran.r-project.org/web/packages/sampling/sampling.pdf)
It can then also handle a lot of the special variance estimation techniques you'll have to do ... | null | CC BY-SA 3.0 | null | 2011-04-18T12:02:13.153 | 2011-04-18T12:02:13.153 | null | null | 1893 | null |
9685 | 1 | null | null | 9 | 12640 | My data set is comprised of either total mortality or survival of an organism at three site types, inshore, midchannel and offshore. The numbers in the table below represent the number of sites.
```
100% Mortality 100% Survival
Inshore 30 31
Midchannel ... | How to carry out multiple post-hoc chi-square tests on a 2 X 3 table? | CC BY-SA 3.0 | null | 2011-04-18T13:50:07.170 | 2022-01-02T09:40:58.107 | 2011-04-18T19:26:00.267 | 930 | null | [
"logistic",
"multiple-comparisons",
"chi-squared-test"
] |
9686 | 2 | null | 9685 | 7 | null | A contingency table should contain all the mutually exclusive categories on both axes. Inshore/Midchannel/Offshore look fine, however unless "less than 100% mortality" means "100% survival" in this biological setting you may need to construct tables that account for all the cases observed or explain why you restrict yo... | null | CC BY-SA 3.0 | null | 2011-04-18T14:20:42.363 | 2011-04-19T15:42:39.223 | 2011-04-19T15:42:39.223 | 3911 | 3911 | null |
9687 | 2 | null | 9663 | 5 | null | You might also want to read up on the [generalized extreme value (GEV) distribution](http://en.wikipedia.org/wiki/Generalized_extreme_value_distribution). It turns out that as $n\rightarrow\infty$, the (shifted and scaled) distribution of the maximal value of the sample converges to one of the three special cases of th... | null | CC BY-SA 3.0 | null | 2011-04-18T14:37:04.670 | 2011-04-18T14:37:04.670 | null | null | 279 | null |
9688 | 1 | 9689 | null | 1 | 461 | Even simple models in probability theory can be quite confusing (e.g. drawing something with or without returning it, conditional probabilities, one and only one outcome, at least one outcome asoasf...)
In other areas of mathematical modelling (e.g. differential equations) you often have powerful and easy to use modell... | Generator and/or interpreter of probability models | CC BY-SA 3.0 | null | 2011-04-18T14:56:54.450 | 2011-05-18T18:44:21.447 | 2011-05-18T18:44:21.447 | null | 230 | [
"probability",
"modeling"
] |
9689 | 2 | null | 9688 | 0 | null | I would recommend downloading R and then following a good R book on Statistics
R download:
[http://cran.r-project.org/bin/windows/base/](http://cran.r-project.org/bin/windows/base/)
Here's a reasonable R/Stats book:
[http://www.amazon.com/Introductory-Statistics-R-Peter-Dalgaard/dp/0387954759](http://rads.stackoverflow... | null | CC BY-SA 3.0 | null | 2011-04-18T15:42:57.210 | 2011-04-18T15:42:57.210 | null | null | 2775 | null |
9690 | 2 | null | 8371 | 2 | null |
- In most non-compulsory survey contexts, there is a substantial problem with nonresponse. This from 2002: "the recently reported estimate of survey cooperation rates from CMOR, the Council for Market and Opinion Research [USA], averaged only 14.7 percent." and from Paul Gerhold, "I believe that it is still possible t... | null | CC BY-SA 3.0 | null | 2011-04-18T16:21:07.000 | 2011-04-18T16:21:07.000 | null | null | 3919 | null |
9691 | 1 | null | null | 5 | 3473 | As a follow up to my [previous question](https://stats.stackexchange.com/questions/9629/can-chi-square-be-used-to-compare-proportions), I learned that the chi square test can be used to compare expected and observed frequency distributions. But ...
Suppose I have the following data table in CSV, which surveys people on... | Equivalent of Tukey test for chi-square? | CC BY-SA 4.0 | null | 2011-04-18T16:56:18.330 | 2023-02-12T02:44:22.600 | 2021-02-06T18:09:59.713 | 11887 | 2830 | [
"distributions",
"chi-squared-test",
"multiple-comparisons",
"tukey-hsd-test"
] |
9692 | 1 | 9698 | null | 4 | 7286 | In a (one or multi) way anova model, once a new individual is assigned to a treatment, the predicted value for him is calculated using the coefficients of the ANOVA model (simply assigning the treatment mean value to the individual).
How should I construct a confidence (or prediction) interval for that predicted value... | Prediction in simple and multiple ANOVA | CC BY-SA 3.0 | null | 2011-04-18T17:01:07.550 | 2011-04-19T03:30:54.897 | 2011-04-19T03:30:54.897 | 3582 | 632 | [
"r",
"anova",
"confidence-interval"
] |
9693 | 1 | 14379 | null | 7 | 3485 | What are some hot topics that mathematical statistics researchers are studying now?
| Hot topics in mathematical statistics | CC BY-SA 3.0 | null | 2011-04-18T17:20:16.697 | 2016-10-28T09:51:48.300 | null | null | 4111 | [
"mathematical-statistics"
] |
9695 | 1 | null | null | 23 | 768 | In [Statistical Modeling: The Two Cultures](https://projecteuclid.org/journals/statistical-science/volume-16/issue-3/Statistical-Modeling--The-Two-Cultures-with-comments-and-a/10.1214/ss/1009213726.full) Leo Breiman writes
>
Current applied practice is to check the data model fit using goodness-of-fit tests and residu... | Difficulty of testing linearity in regression | CC BY-SA 4.0 | null | 2011-04-18T18:24:55.707 | 2022-07-04T10:57:01.637 | 2022-07-04T10:57:01.637 | 79696 | 319 | [
"regression",
"goodness-of-fit"
] |
9696 | 2 | null | 9691 | 5 | null | With this many questions to ask of the data, you are approaching the border between a "test" and a "model" (not that such a sharp distinction exists except in the conceptualization of the problem).
A [Poisson regression model](https://secure.wikimedia.org/wikipedia/en/wiki/Poisson_regression), aka log-linear model, can... | null | CC BY-SA 4.0 | null | 2011-04-18T19:00:36.163 | 2023-02-12T02:44:22.600 | 2023-02-12T02:44:22.600 | 345611 | 2975 | null |
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