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3707 | 1 | 3714 | null | 5 | 1532 | I plotted below the standarized results of:
- RAND()
- RAND() * RAND()
- ...
- RAND() * RAND() * RAND() * RAND() * RAND() * RAND()
It seems that the results are getting to zero, is that because you're multiplying a bunch of numbers close to zero together, or is there another explanation?
, RAND()*RAND(), etc | CC BY-SA 2.5 | null | 2010-10-18T15:10:23.190 | 2010-10-19T10:10:46.127 | 2010-10-18T17:43:41.720 | 449 | 59 | [
"probability",
"uniform-distribution",
"monte-carlo",
"excel",
"central-limit-theorem"
] |
3708 | 1 | 3720 | null | 7 | 1494 | I have a data set of 100 geographic regions for which the unemployment rate has been observed during the last 9 years. Now, I want to simulate/forecast from this data the next year unemployment rate for all the 100 regions.
What would be a suitable approach? I was thinking to use an auto-regressive mixed effect model,... | Forecasting unemployment rate | CC BY-SA 2.5 | null | 2010-10-18T15:12:10.027 | 2010-10-19T15:23:18.643 | 2010-10-18T15:36:51.290 | 449 | 1443 | [
"time-series",
"markov-chain-montecarlo",
"mixed-model",
"forecasting"
] |
3710 | 2 | null | 3707 | 5 | null | "In Excel, the Rand function returns a random number that is greater than or equal to 0 and less than 1. The Rand function returns a new random number each time your spreadsheet recalculates." -http://www.techonthenet.com/excel/formulas/rand.php
Because RAND() is always less than one and greater than zero, multiplying ... | null | CC BY-SA 2.5 | null | 2010-10-18T15:18:11.330 | 2010-10-18T15:18:11.330 | null | null | 1118 | null |
3711 | 2 | null | 3707 | 3 | null | I am not sure why your graph has values from -2 to 4 but for what it is worth here is the answer to the text of your question:
Suppose that $U \sim U[0,1]$. Then the cdf of $U$ is given by $F(u) = u$ for $u \in (0,1)$ and 1 otherwise.
When you multiply different iid realizations of the random draws you are essentially ... | null | CC BY-SA 2.5 | null | 2010-10-18T15:29:33.363 | 2010-10-18T19:01:39.267 | 2010-10-18T19:01:39.267 | null | null | null |
3712 | 2 | null | 3704 | 4 | null | Your impact of treatment depends updon quintile status. After running the 'regress' command, type, 'lincom treat + treatXquin2' to get the impact of treatment for those in quintile 2. This gives a confidence interval and standard error for this estimate. Note that, as is typical in Stata, using the names of the variabl... | null | CC BY-SA 2.5 | null | 2010-10-18T15:47:22.743 | 2010-10-18T22:08:54.650 | 2010-10-18T22:08:54.650 | 401 | 401 | null |
3713 | 1 | 3716 | null | 83 | 50779 | When using cluster analysis on a data set to group similar cases, one needs to choose among a large number of clustering methods and measures of distance. Sometimes, one choice might influence the other, but there are many possible combinations of methods.
Does anyone have any recommendations on how to choose among ... | Choosing a clustering method | CC BY-SA 3.0 | null | 2010-10-18T15:58:40.990 | 2017-08-26T18:46:30.470 | 2017-03-21T13:05:38.060 | 7290 | 485 | [
"clustering",
"distance-functions",
"methodology"
] |
3714 | 2 | null | 3707 | 6 | null | Standardization is good, but it's not the right standardization for this situation. It helps to see that multiplying values of RAND() is the same as adding their logarithms (followed by a subsequent exponentiation). Because the different calls to RAND() are supposed to be independent, those logarithms are still indep... | null | CC BY-SA 2.5 | null | 2010-10-18T16:00:46.523 | 2010-10-18T16:06:05.273 | 2010-10-18T16:06:05.273 | 919 | 919 | null |
3715 | 1 | null | null | 6 | 627 | I'd like to use Bayes' Theorem on data obtained through a small random sample, and I want to use Agresti-Coull (or any other alternative technique) to know how big the uncertainty is.
Here is Bayes' Theorem:
$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$
Now, all the data I have on this system is obtained from small random s... | Bayes' Theorem and Agresti-Coull: Will it blend? | CC BY-SA 2.5 | null | 2010-10-18T17:19:09.747 | 2022-05-07T23:15:21.417 | 2010-10-18T17:57:01.283 | 449 | 5793 | [
"bayesian",
"python",
"approximation",
"uncertainty"
] |
3716 | 2 | null | 3713 | 47 | null | There is no definitive answer to your question, as even within the same method the choice of the distance to represent individuals (dis)similarity may yield different result, e.g. when using euclidean vs. squared euclidean in hierarchical clustering. As an other example, for binary data, you can choose the Jaccard inde... | null | CC BY-SA 2.5 | null | 2010-10-18T17:24:05.970 | 2010-11-09T16:38:14.217 | 2017-04-13T12:44:51.060 | -1 | 930 | null |
3717 | 2 | null | 3708 | 3 | null | Given the nature of your data I would suggest you investigate the use of [exponential smoothing](http://en.wikipedia.org/wiki/Exponential_smoothing) as well as fitting ARIMA type models, especially due to the temporal constraints within your data. Although I wouldn't doubt spatial dependencies exist, I would be abit sk... | null | CC BY-SA 2.5 | null | 2010-10-18T18:36:53.263 | 2010-10-19T15:23:18.643 | 2010-10-19T15:23:18.643 | 1036 | 1036 | null |
3718 | 2 | null | 3707 | 1 | null | There is no mysterious reason. If you multiply a bunch of numbers between 0 an 1, the result will forcibly be close to 0. The average result for RAND()*RAND()*RAND()*RAND()*RAND()*RAND() should be something close to (0.5^6), that is, 0.015625.
Be careful using Excel's RAND() function, though. It's not the best random n... | null | CC BY-SA 2.5 | null | 2010-10-18T19:09:55.590 | 2010-10-18T19:18:41.650 | 2010-10-18T19:18:41.650 | 666 | 666 | null |
3719 | 1 | 3721 | null | 2 | 1832 | How is it possible to calculate Pearson/Spearman correlation of more than two judges in MATLAB?
Thanks.
| Is it possible to calculate Pearson/Spearman correlation of more than two judges in MATLAB? | CC BY-SA 2.5 | null | 2010-10-18T19:10:41.067 | 2011-05-08T17:46:41.447 | 2010-10-18T19:13:44.903 | 449 | 1564 | [
"correlation",
"matlab"
] |
3720 | 2 | null | 3708 | 6 | null | The Arellano-Bond estimator has been designed for precisely this type of problems.
You will find a short non-technical paper with a examples [here](http://www.cemmap.ac.uk/wps/cwp0209.pdf). In a nutshell, it combines the information embedded in the large number of cross-section to make up for the small number of points... | null | CC BY-SA 2.5 | null | 2010-10-18T19:19:01.650 | 2010-10-19T15:03:26.300 | 2010-10-19T15:03:26.300 | 603 | 603 | null |
3721 | 2 | null | 3719 | 3 | null | Store the evaluation of the judges in a matrix X where each column represents one judge and the rows are the corresponding evaluations. Then `corr(x)` gives the required pairwise correlation matrix. From Matlab's help:
>
RHO = corr(X) returns a p-by-p matrix containing the pairwise linear correlation coefficient betwe... | null | CC BY-SA 2.5 | null | 2010-10-18T19:37:48.260 | 2010-10-18T19:37:48.260 | null | null | null | null |
3722 | 2 | null | 3715 | 0 | null | [Brown, Cai, and DasGupta, AS, 2002](https://projecteuclid.org/journals/annals-of-statistics/volume-30/issue-1/Confidence-Intervals-for-a-binomial-proportion-and-asymptotic-expansions/10.1214/aos/1015362189.full)
[Brown, Cai, and DasGupta, Stat Sci, 2001](https://projecteuclid.org/journals/statistical-science/volume-16... | null | CC BY-SA 4.0 | null | 2010-10-18T19:53:58.210 | 2022-05-07T23:15:21.417 | 2022-05-07T23:15:21.417 | 79696 | 1307 | null |
3723 | 2 | null | 3715 | 3 | null | [Error propagation](http://en.wikipedia.org/wiki/Error_propagation) won't work in the way handled by the uncertainties package. As you note, they're dependent, so you have to take the covariances into account.
You can obtain the variance of your distribution P(B|A) using the [Delta Method](http://en.wikipedia.org/wiki... | null | CC BY-SA 2.5 | null | 2010-10-18T19:57:08.563 | 2010-10-18T19:57:08.563 | null | null | 251 | null |
3724 | 1 | 3725 | null | 9 | 7424 | According to Microsoft Excel Help:
>
VAR uses the following formula:
where x is the sample mean
AVERAGE(number1,number2,…) and n is
the sample size.
Shouldn't it be n, rather than n - 1, in the denominator?
| Microsoft Excel formula for variance | CC BY-SA 2.5 | null | 2010-10-18T20:54:08.583 | 2010-10-23T06:00:07.847 | 2020-06-11T14:32:37.003 | -1 | 1618 | [
"variance",
"excel",
"unbiased-estimator"
] |
3725 | 2 | null | 3724 | 9 | null | Use VARP for the variance you want ("population variance"). VAR is the unbiased estimator for a normally distributed population.
| null | CC BY-SA 2.5 | null | 2010-10-18T21:01:57.917 | 2010-10-18T21:01:57.917 | null | null | 919 | null |
3726 | 2 | null | 1224 | 3 | null | Are you calculating your chi-square statistic by squaring the difference between the logHRs and dividing by the variance of this diff? If so, that sounds absolutely fine to me. With only two centres, perhaps I wouldn't usually think of this a test for (between-centre) heterogeneity - i might call it a test for interact... | null | CC BY-SA 2.5 | null | 2010-10-18T21:44:50.017 | 2010-10-19T06:44:37.850 | 2017-04-13T12:44:52.660 | -1 | 449 | null |
3727 | 1 | 3789 | null | 11 | 284 | What's the best way to approximate $Pr[n \leq X \leq m]$ for two given integers $m,n$ when you know the mean $\mu$, variance $\sigma^2$, skewness $\gamma_1$ and excess kurtosis $\gamma_2$ of a discrete distribution $X$, and it is clear from the (non-zero) measures of shape $\gamma_1$ and $\gamma_2$ that a normal approx... | Approximating $Pr[n \leq X \leq m]$ for a discrete distribution | CC BY-SA 2.5 | null | 2010-10-18T22:19:27.650 | 2017-06-06T00:56:11.673 | 2017-06-06T00:56:11.673 | 11887 | null | [
"probability",
"distributions",
"moments",
"approximation",
"saddlepoint-approximation"
] |
3728 | 2 | null | 499 | 10 | null | Variable selection techniques, in general (whether stepwise, backward, forward, all subsets, AIC, etc.), capitalize on chance or random patterns in the sample data that do not exist in the population. The technical term for this is over-fitting and it is especially problematic with small datasets, though it is not exc... | null | CC BY-SA 2.5 | null | 2010-10-18T22:26:57.827 | 2010-10-18T22:26:57.827 | null | null | 485 | null |
3729 | 2 | null | 3727 | 2 | null | You could try to use [skew normal distribution](http://en.wikipedia.org/wiki/Skew_normal_distribution) and see if excess kurtosis for your particular data sets is sufficiently close to the excess kurtosis of the distribution for given skewness. If it is, you can use the skew normal distribution cdf to estimate the prob... | null | CC BY-SA 2.5 | null | 2010-10-19T00:21:55.433 | 2010-10-19T00:21:55.433 | null | null | 1448 | null |
3730 | 1 | 3733 | null | 151 | 277079 | I get this question frequently enough in my statistics consulting work, that I thought I'd post it here. I have an answer, which is posted below, but I was keen to hear what others have to say.
Question: If you have two variables that are not normally distributed, should you use Spearman's rho for the correlation?
| Pearson's or Spearman's correlation with non-normal data | CC BY-SA 3.0 | null | 2010-10-19T01:14:16.967 | 2023-05-02T20:06:34.277 | 2017-02-15T01:37:14.343 | 7290 | 183 | [
"correlation",
"normality-assumption",
"pearson-r",
"spearman-rho"
] |
3731 | 2 | null | 3730 | 14 | null | Updated
The question asks us to choose between Pearson's and Spearman's method when normality is questioned. Restricted to this concern, I think the following paper should inform anyone's decision:
- On the Effects of Non-Normality on the Distribution of the Sample Product-Moment
Correlation Coefficient (Kowalski, 19... | null | CC BY-SA 2.5 | null | 2010-10-19T01:27:47.607 | 2010-10-21T02:05:03.130 | 2010-10-21T02:05:03.130 | 251 | 251 | null |
3732 | 2 | null | 499 | 4 | null | Richard Berk has a recent article where he demonstrates through simulation the problems of such data snooping and statistical inference. As Rob [suggested](https://stats.stackexchange.com/questions/499/when-can-you-use-data-based-criteria-to-specify-a-regression-model/531#531) it is more problematic than simply correct... | null | CC BY-SA 2.5 | null | 2010-10-19T01:50:06.413 | 2010-10-19T01:50:06.413 | 2017-04-13T12:44:40.807 | -1 | 1036 | null |
3733 | 2 | null | 3730 | 108 | null | Pearson's correlation is a measure of the linear relationship between two continuous random variables. It does not assume normality although it does assume finite variances and finite covariance. When the variables are bivariate normal, Pearson's correlation provides a complete description of the association.
Spearman'... | null | CC BY-SA 2.5 | null | 2010-10-19T01:53:19.697 | 2010-10-19T07:48:02.297 | 2010-10-19T07:48:02.297 | 159 | 159 | null |
3734 | 1 | null | null | 200 | 49234 | In several different contexts we invoke the [central limit theorem](http://en.wikipedia.org/wiki/Central_limit_theorem) to justify whatever statistical method we want to adopt (e.g., approximate the binomial distribution by a normal distribution). I understand the technical details as to why the theorem is true but it ... | What intuitive explanation is there for the central limit theorem? | CC BY-SA 2.5 | null | 2010-10-19T02:14:32.050 | 2022-03-17T02:51:11.573 | null | null | null | [
"intuition",
"central-limit-theorem"
] |
3735 | 2 | null | 3194 | 7 | null | A chi-squared goodness of fit test aims to find all possible kinds of deviations from strict uniformity. This is reasonable with a d4 or a d6, but with a d20, you're probably more interested in checking that the probability that you roll under (or possibly exceed) each outcome is close to what it should be.
What I am g... | null | CC BY-SA 4.0 | null | 2010-10-19T02:53:05.170 | 2018-05-28T23:55:32.270 | 2018-05-28T23:55:32.270 | 805 | 805 | null |
3736 | 2 | null | 3734 | 21 | null | Intuition is a tricky thing. It's even trickier with theory in our hands tied behind our back.
The CLT is all about sums of tiny, independent disturbances. "Sums" in the sense of the sample mean, "tiny" in the sense of finite variance (of the population), and "disturbances" in the sense of plus/minus around a central... | null | CC BY-SA 2.5 | null | 2010-10-19T03:35:38.610 | 2010-10-19T03:55:41.573 | 2010-10-19T03:55:41.573 | null | null | null |
3737 | 2 | null | 3734 | 1 | null | I gave up on trying to come up with an intuitive version and came up with some simulations. I have one that presents a simulation of a Quincunx and some others that do things like show how even a skewed raw reaction time distribution will become normal if you collect enough RT's per subject. I think they help but the... | null | CC BY-SA 4.0 | null | 2010-10-19T04:18:09.647 | 2019-06-26T14:19:09.627 | 2019-06-26T14:19:09.627 | 601 | 601 | null |
3738 | 2 | null | 3730 | 47 | null | From an applied perspective, I am more concerned with choosing an approach that summarises the relationship between two variables in a way that aligns with my research question. I think that determining a method for getting accurate standard errors and p-values is a question that should come second. Even if you chose n... | null | CC BY-SA 2.5 | null | 2010-10-19T06:08:06.590 | 2010-10-19T07:36:34.143 | 2010-10-19T07:36:34.143 | 183 | 183 | null |
3739 | 1 | null | null | 3 | 475 | I have sampled a number of $x$ values from a normal distribution with mean 0 and sd 0.2. I then transformed these $x$ values to $y$ values using the formula $y = e^x/(e^x + 1)$. I know that the $y$ values will have a mean of 0.5 and all lie between 0 and 1.
- Do these y values fit some common distribution?
- Is the... | Does data transformed in a certain way from a normal distribution fit some other common distribution? | CC BY-SA 2.5 | null | 2010-10-19T06:14:17.770 | 2010-10-19T07:48:17.050 | 2010-10-19T07:17:41.907 | 449 | null | [
"distributions",
"data-transformation"
] |
3740 | 2 | null | 3739 | 1 | null | The trivial way is just to realize this transformation and see what happens:
This looks pretty normal, indeed qqnorm confirms it:
| null | CC BY-SA 2.5 | null | 2010-10-19T06:40:30.707 | 2010-10-19T06:40:30.707 | null | null | null | null |
3743 | 2 | null | 3739 | 3 | null | You have:
$X \sim N(\mu,\sigma^2)$
By definition:
$Y = \frac{e^X}{e^X+1}$
Therefore, the cdf of $Y$ is:
$P(Y \le y) = P(\frac{e^X}{e^X+1} \le y)$
Simplifying the RHS, we get:
$P(Y \le y) = P(X \le -log(\frac{1-y}{y}) )$
Therefore,
$P(Y \le y) = \Phi(-log(\frac{1-y}{y}),\mu,\sigma^2)$
Differentiating the above wrt $y$, ... | null | CC BY-SA 2.5 | null | 2010-10-19T06:50:40.420 | 2010-10-19T07:48:17.050 | 2017-04-13T12:44:51.217 | -1 | null | null |
3744 | 2 | null | 3730 | 61 | null | Don't forget Kendall's tau! Roger Newson has argued for the superiority of Kendall's τa over Spearman's correlation rS as a rank-based measure of correlation in a paper whose full text is now freely available online:
Newson R. [Parameters behind "nonparametric" statistics: Kendall's tau,Somers' D and median differences... | null | CC BY-SA 4.0 | null | 2010-10-19T07:07:16.707 | 2023-05-02T20:06:34.277 | 2023-05-02T20:06:34.277 | 344718 | 449 | null |
3745 | 2 | null | 3739 | 2 | null | This is called a [logit-normal distribution](http://en.wikipedia.org/wiki/Logit-normal_distribution) (by analogy with the much more common log-normal distribution). Knowing that doesn't simplify sampling from it, however, or change the parameters used to describe it, which are still the mean and SD (or variance) of the... | null | CC BY-SA 2.5 | null | 2010-10-19T07:14:21.320 | 2010-10-19T07:14:21.320 | null | null | 449 | null |
3746 | 1 | null | null | 11 | 3794 | I need to write a program to find the average GPS point from a population of points.
In practice the following happens:
- Each month a person records a GPS point of the same static asset.
- Because of the nature of GPS, these points differ slightly each month.
- Sometimes the person makes a mistake a records the wro... | Finding the average GPS point | CC BY-SA 2.5 | null | 2010-10-19T08:10:54.190 | 2010-10-19T20:11:33.707 | 2010-10-19T09:28:06.077 | 1352 | 1623 | [
"outliers",
"spatial"
] |
3747 | 2 | null | 3746 | 3 | null | Rob Hyndman recently posed a [question about detecting outliers in multivariate data](https://stats.stackexchange.com/questions/213/what-is-the-best-way-to-identify-outliers-in-multivariate-data). The answers may provide a couple of possible approaches (and otherwise, you may want to put the question of finding 2-d out... | null | CC BY-SA 2.5 | null | 2010-10-19T08:30:31.317 | 2010-10-19T08:30:31.317 | 2017-04-13T12:44:32.747 | -1 | 1352 | null |
3748 | 2 | null | 2214 | 3 | null | The Levy distribution has 4 parameter. Each of them has a quantile-based sample equivalent:
- $\mu$, the location parameter, can be estimated by the median. This is a high efficiency alternative (ARE$\approx 0.85$).
- $\gamma$, the scale parameter, can be estimated by the median absolute deviation (or more efficientl... | null | CC BY-SA 2.5 | null | 2010-10-19T09:24:21.443 | 2010-10-19T09:24:21.443 | null | null | 603 | null |
3749 | 1 | 3981 | null | 17 | 4285 | I am discovering the marvellous world of such called "Hidden Markov Models", also called "regime switching models".
I would like to adapt a HMM in R to detect trends and turning points. I would like to build the model as generic as possible so that I can test it on many prices.
Can anyone recommend a paper? I have see... | Usage of HMM in quantitative finance. Examples of HMM that works to detect trend / turning points? | CC BY-SA 2.5 | null | 2010-10-19T10:04:10.973 | 2010-10-26T06:19:55.590 | 2010-10-26T02:19:11.633 | 183 | 1709 | [
"r",
"time-series",
"finance",
"hidden-markov-model"
] |
3751 | 2 | null | 3734 | 32 | null | The nicest animation I know:
[http://www.ms.uky.edu/~mai/java/stat/GaltonMachine.html](http://www.ms.uky.edu/~mai/java/stat/GaltonMachine.html)
[](https://i.stack.imgur.com/IH6cc.gif)
The simplest words I have read: [http://elonen.iki.fi/articles/centrallimit/index.en.html](http://elonen.iki.fi/articles/centrallimit/in... | null | CC BY-SA 3.0 | null | 2010-10-19T10:27:08.170 | 2017-05-12T17:00:53.633 | 2017-05-12T17:00:53.633 | 90687 | 1219 | null |
3752 | 1 | null | null | 19 | 776 | This question is prompted by [discussion elsewhere](https://stats.stackexchange.com/q/3556/159).
Variable kernels are often used in local regression. For example, loess is widely used and works well as a regression smoother, and is based on a kernel of variable width that adapts to data sparsity.
On the other hand, var... | If variable kernel widths are often good for kernel regression, why are they generally not good for kernel density estimation? | CC BY-SA 2.5 | null | 2010-10-19T11:35:15.373 | 2019-04-17T02:20:42.810 | 2017-04-13T12:44:48.803 | -1 | 159 | [
"nonparametric",
"smoothing",
"kernel-smoothing",
"loess"
] |
3753 | 2 | null | 3746 | 8 | null | One of the problems with multivariate data is deciding on, and then interpreting, a suitable metric for calculating distances, hence clever but somewhat hard-to-explain concepts such as Mahalanobis distance. But in this case surely the choice is obvious - [Euclidean distance](http://en.wikipedia.org/wiki/Euclidean_dist... | null | CC BY-SA 2.5 | null | 2010-10-19T11:54:28.387 | 2010-10-19T11:54:28.387 | null | null | 449 | null |
3754 | 1 | 3756 | null | 6 | 11661 | trying to read a large dataset in R the console displayed the follwing errors:
```
data<-read.csv("UserDailyStats.csv", sep=",", header=T, na.strings="-", stringsAsFactors=FALSE)
> data = data[complete.cases(data),]
> dataset<-data.frame(user_id=as.character(data[,1]),event_date= as.character(data[,2]),day_of_week=as.f... | How to read large dataset in R | CC BY-SA 2.5 | null | 2010-10-19T12:00:38.310 | 2013-01-09T11:25:55.163 | null | null | 315 | [
"r",
"large-data"
] |
3755 | 2 | null | 3754 | 2 | null | What platform are you running R on? How much physical and virtual memory does the machine have?
Also, you might find the following relevant: [http://stat.ethz.ch/R-manual/R-devel/library/base/html/Memory-limits.html](http://stat.ethz.ch/R-manual/R-devel/library/base/html/Memory-limits.html)
| null | CC BY-SA 2.5 | null | 2010-10-19T13:02:15.267 | 2010-10-19T13:02:15.267 | null | null | 439 | null |
3756 | 2 | null | 3754 | 9 | null | Two basic things:
- That complaint addresses all of the memory in the R session, not just the one object that you're loading. And unless you're using something like ff, everything in your session is in memory.
- One Windows, you need to specify how much memory can be used by R. Have a look at help(memory.limit). E... | null | CC BY-SA 2.5 | null | 2010-10-19T13:09:51.147 | 2010-10-19T13:16:54.847 | 2010-10-19T13:16:54.847 | 5 | 5 | null |
3757 | 1 | null | null | 6 | 6988 | In my data, the RT (gaze) of individuals (ID) is examined as a function of a visual conditions, the factor size (small, medium, large).
Base model:
```
print(Base <- lmer(RT ~ Size + (1|ID), data=rt), cor=F)
```
Random effect:
```
print(NoCor <- lmer(RT ~ Size + (0+Size|ID) , data=rt))
print(WithCor <- lmer(RT ~ Size... | Random effect slopes in linear mixed models | CC BY-SA 2.5 | null | 2010-10-19T14:00:19.707 | 2010-10-20T21:08:01.323 | 2010-10-20T21:08:01.323 | 8 | 1626 | [
"r",
"mixed-model",
"random-effects-model"
] |
3758 | 1 | null | null | 6 | 1916 | I'm a new R user and had just tried running friedman on non-normal and heteroscedastic data on seagrass. I am testing whether biomass is significantly different between sites across years. R (`friedman` function from `agricolae` package) returned result like this:
```
Friedman's Test
===============
Adjusted for ties... | What is a meaning of "p-value F" from Friedman test? | CC BY-SA 2.5 | null | 2010-10-19T14:27:51.110 | 2014-08-18T07:32:11.940 | 2010-10-20T09:04:39.833 | null | 1627 | [
"r",
"nonparametric"
] |
3759 | 1 | 3761 | null | 5 | 1636 | I'm looking for an implementation of FNN (or better yet, a SOFNN as described by [Forecasting Time Series by SOFNN with Reinforcement Learning](https://www.semanticscholar.org/paper/Forecasting-Time-Series-by-SOFNN-with-Reinforcement-Kuremoto-Obayashi/8a5ce65e52077303b8dcbe39a3953219e910ca3f)). Any language, though p... | Looking for impl of a Fuzzy Neural Network (FNN or SOFNN) | CC BY-SA 4.0 | null | 2010-10-19T14:29:50.193 | 2022-11-21T01:51:40.377 | 2022-11-21T01:51:40.377 | 362671 | 1127 | [
"machine-learning",
"neural-networks"
] |
3760 | 2 | null | 3101 | 4 | null | Way back in 1965, Sir Austin Bradford Hill wrote [a great essay](http://www.edwardtufte.com/tufte/hill) about something very akin to the Pyramid of Evidence, where he discussed how the piling up of evidence can increase our confidence in hypotheses of causality in Medicine.
Most of the factors he discusses can be appli... | null | CC BY-SA 2.5 | null | 2010-10-19T14:34:44.807 | 2010-10-19T14:34:44.807 | null | null | 666 | null |
3761 | 2 | null | 3759 | 4 | null | I'm not an expert on this subject, but I believe that FNN are sometimes referred to as a [neuro-fuzzy system](http://en.wikipedia.org/wiki/Neuro-fuzzy) (also [referenced here](http://www.scholarpedia.org/article/Fuzzy_neural_network)).
There are several implementations that I can find for that subject, including in [... | null | CC BY-SA 2.5 | null | 2010-10-19T14:58:49.700 | 2010-10-19T14:58:49.700 | null | null | 5 | null |
3762 | 1 | null | null | 3 | 550 | I have 6 sets of interval data each of which between 0 and 1. Each set, calculated by a computer program, is related to the degree of similarity between some sounds (pairwise).
What do you think in the best inter-rater reliability measure I can use to see how close the 6 judges are?
If I want to explain the data in eac... | The best measure of reliability for interval data between 0 and 1 | CC BY-SA 2.5 | null | 2010-10-19T15:39:13.843 | 2010-10-20T09:27:39.170 | 2010-10-19T17:11:20.393 | 930 | 1564 | [
"psychometrics",
"reliability",
"agreement-statistics"
] |
3764 | 2 | null | 3758 | 3 | null | I generally used `friedman.test()` which doesn't return any F statistic. If you consider that you have $b$ blocks, for which you assigned ranks to observations belonging to each of them, and that you sum these ranks for each of your $a$ groups (let denote them sum $R_i$), then the Friedman statistic is defined as
$$
F... | null | CC BY-SA 2.5 | null | 2010-10-19T16:00:57.583 | 2010-10-19T16:07:40.880 | 2010-10-19T16:07:40.880 | 930 | 930 | null |
3765 | 2 | null | 3754 | 2 | null | I,m totally agree with Dirk answer. One suggestion. I have found very useful the use of programming languages such as AWK or others when assessing large databases. So, I was able to filter the data I wanted to include in my analysis, reducing the final size of dataset.
Moreover, in your code you are duplicating the sam... | null | CC BY-SA 2.5 | null | 2010-10-19T16:00:58.520 | 2010-10-19T16:00:58.520 | null | null | 221 | null |
3766 | 2 | null | 3762 | 4 | null | Referring to your comments to @Henrik, I'm inclined to think that you rather have continuous measurements on a set of objects (here, your similarity measure) for 6 raters. You can compute an [intraclass correlation](http://en.wikipedia.org/wiki/Intraclass_correlation) coefficient, as described here [Reliability in Elic... | null | CC BY-SA 2.5 | null | 2010-10-19T16:14:39.170 | 2010-10-20T09:27:39.170 | 2017-04-13T12:44:33.310 | -1 | 930 | null |
3767 | 2 | null | 3762 | 1 | null | If you want to compare just two measures, simply take the [correlation coefficient (Pearson's r)](http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient).
| null | CC BY-SA 2.5 | null | 2010-10-19T16:33:04.320 | 2010-10-19T16:33:04.320 | null | null | 442 | null |
3768 | 2 | null | 3754 | 2 | null | Since you're on 64-bit Windows, make sure that you have installed and are running the 64-bit version of R for Windows. Then, follow the instructions on Gary King's page:
- How do I increase the memory for R?
| null | CC BY-SA 2.5 | null | 2010-10-19T16:35:03.540 | 2010-10-19T16:35:03.540 | null | null | 251 | null |
3769 | 2 | null | 3758 | 4 | null | It seems the output is from the `agricolae` package using the method `friedman`. The relevant lines for computing the two statistics in that function are:
```
T1.aj <- (m[2] - 1) * (t(s) %*% s - m[1] * C1)/(A1 - C1)
T2.aj <- (m[1] - 1) * T1.aj/(m[1] * (m[2] - 1) - T1.aj)
```
Comparing this with the formula in chl's a... | null | CC BY-SA 2.5 | null | 2010-10-19T16:43:16.423 | 2010-10-19T16:43:16.423 | null | null | 251 | null |
3770 | 2 | null | 3698 | 3 | null | The author is using a single proportion test. Try the following in R:
```
(p-0.5)/sqrt(0.5*0.5/33)
```
where p = 17/33.
See the wiki for the test statistic where the above test is called the [One-proportion z-test](http://en.wikipedia.org/wiki/Statistical_hypothesis_testing#Common_test_statistics).
| null | CC BY-SA 2.5 | null | 2010-10-19T17:10:57.197 | 2010-10-19T17:10:57.197 | null | null | null | null |
3771 | 2 | null | 3727 | 4 | null | Fitting a distribution to data using the first four moments is exactly what Karl Pearson devised the [Pearson family of continuous probability distributions](http://en.wikipedia.org/wiki/Pearson_distribution) for (maximum likelihood is much more popular these days of course). Should be straightforward to fit the releva... | null | CC BY-SA 2.5 | null | 2010-10-19T17:29:33.383 | 2010-10-19T17:29:33.383 | null | null | 449 | null |
3772 | 1 | 3775 | null | 9 | 4955 | I am evaluating the effectiveness of 5 different methods to predict a particular binary outcome (call them 'Success' and 'Failure'). The data look like so:
```
Method Sample_Size Success Percent_Success
1 28 4 0.14
2 19 4 0.21
3 24 ... | Can I use permutation tests for to avoid the multiple comparison problem in the context of proportions? | CC BY-SA 2.5 | null | 2010-10-19T17:43:46.987 | 2010-11-17T23:55:00.300 | 2010-10-19T18:58:16.153 | null | 1558 | [
"r",
"multiple-comparisons",
"permutation-test"
] |
3773 | 2 | null | 3772 | 1 | null | Your suggested Monte-Carlo permutation test procedure will produce a p-value for a test of the null hypothesis that the probability of success is the same for all methods. But there's little reason for doing a Monte Carlo permutation test here when the corresponding exact permutation test is perfectly feasible. That's ... | null | CC BY-SA 2.5 | null | 2010-10-19T18:17:52.227 | 2010-10-19T18:17:52.227 | null | null | 449 | null |
3774 | 2 | null | 7 | 3 | null | Adding a couple to the list:
- Lots of in-depth financial data on publicly-traded companies, going back many decades: http://www.mergent.com/servius
- Rich information on 16+ million businesses in the US: http://compass.webservius.com
Both available via a REST API and have free trial plans.
| null | CC BY-SA 2.5 | null | 2010-10-19T18:41:58.457 | 2010-10-19T18:41:58.457 | null | null | 1629 | null |
3775 | 2 | null | 3772 | 6 | null | The proposed procedure does not answer your question. It only estimates the frequency, under the null hypothesis, with which your observed order would occur. But under that null, to a good approximation, all orders are equally likely, whence your calculation will produce a value close to 1/5! = about 0.83%. That tel... | null | CC BY-SA 2.5 | null | 2010-10-19T18:58:56.603 | 2010-10-19T19:19:20.613 | 2010-10-19T19:19:20.613 | 919 | 919 | null |
3776 | 2 | null | 3746 | 2 | null | Call the HDOP the independent variable. Use this for weighting later on. So you have sets of co-ordinates - call this (x1,y1); (x2,y2), etc...
First ignore outliers. Calculate the weighted averages of the x co-ordinates as [(x1*h1)+(x2*h2) +....+ (xn*hn)] / [sum(h1,h2,...,hn)] where h1,h2,... is the HDOP value. Do the ... | null | CC BY-SA 2.5 | null | 2010-10-19T20:11:33.707 | 2010-10-19T20:11:33.707 | null | null | null | null |
3778 | 2 | null | 3614 | 9 | null | Yet another way to quickly compute the probability distribution of a dice roll would be to use a specialized calculator designed just for that purpose.
[Torben Mogensen](http://www.diku.dk/hjemmesider/ansatte/torbenm/), a CS professor at [DIKU](http://www.diku.dk/) has an excellent dice roller called [Troll](http://top... | null | CC BY-SA 2.5 | null | 2010-10-19T23:38:01.513 | 2010-10-19T23:38:01.513 | null | null | null | null |
3779 | 1 | 6135 | null | 27 | 4320 | Suppose you had a bag with $n$ tiles, each with a letter on it. There are $n_A$ tiles with letter 'A', $n_B$ with 'B', and so on, and $n_*$ 'wildcard' tiles (we have $n = n_A + n_B + \ldots + n_Z + n_*$). Suppose you had a dictionary with a finite number of words. You pick $k$ tiles from the bag without replacement. Ho... | Probability of not drawing a word from a bag of letters in Scrabble | CC BY-SA 2.5 | null | 2010-10-20T00:28:37.190 | 2011-01-11T08:38:13.353 | 2011-01-11T08:38:13.353 | 2116 | 795 | [
"sampling",
"games",
"probability"
] |
3780 | 2 | null | 3779 | 2 | null | Monte Carlo Approach
The quick and dirty approach is to do a monte carlo study. Draw $k$ tiles $m$ times and for each draw of $k$ tiles see if you can form a word. Denote the number of times you could form a word by $m_w$. The desired probability would be:
$$1 - \frac{m_w}{m}$$
Direct Approach
Let the number of words i... | null | CC BY-SA 2.5 | null | 2010-10-20T00:54:20.850 | 2010-10-20T16:26:06.390 | 2010-10-20T16:26:06.390 | null | null | null |
3781 | 1 | null | null | 4 | 172 | I've got a dataset where someone counted birds in the breeding season over 10 years. For each year (x site), we want to see how reduced sampling might affect our ability to detect a trend. So to that end, I have simulated various datasets from the original where we cut down sampling to once every 45 days (and 60, 90,12... | Is there a way to compute confidence intervals for regression estimates of simulated data? | CC BY-SA 2.5 | null | 2010-10-20T01:04:56.197 | 2010-10-20T12:58:47.653 | null | null | 1451 | [
"confidence-interval",
"negative-binomial-distribution"
] |
3782 | 1 | null | null | 7 | 1403 | I designed a field experiment with 4 independent factors but data is not normal and heteroscedastic. Friedman test (agricolae package) from R only fits for rbd. Can anybody suggest how to analyze my data please?
| How to do factorial analysis for a non-normal and heteroscedastic data? | CC BY-SA 4.0 | null | 2010-10-20T01:11:04.410 | 2020-12-18T17:46:42.550 | 2020-12-18T17:46:03.170 | 11887 | 1627 | [
"r",
"nonparametric",
"experiment-design",
"heteroscedasticity"
] |
3783 | 2 | null | 3782 | 2 | null | Package [vegan](http://cc.oulu.fi/~jarioksa/softhelp/vegan.html) implements some permutation testing procedures using a distance based approach. For factor analysis, you should take a look at section 5 of the [documentation](http://cc.oulu.fi/~jarioksa/opetus/metodi/vegantutor.pdf). There's also more information in t... | null | CC BY-SA 2.5 | null | 2010-10-20T01:52:30.530 | 2010-10-20T02:22:15.280 | 2010-10-20T02:22:15.280 | 251 | 251 | null |
3784 | 2 | null | 3757 | 7 | null | First, you should compare models from lmer after fitting with ML (maximum likelihood) since the default is REML. So something like:
```
fit.nc <- update(NoCor, REML=FALSE)
fit.wc <- udpate(WithCor, REML=FALSE)
anova(fit.nc, fit.wc)
```
It would help to see the output of the random effects variation from your fits. F... | null | CC BY-SA 2.5 | null | 2010-10-20T02:40:28.530 | 2010-10-20T02:40:28.530 | null | null | 251 | null |
3785 | 2 | null | 3757 | 6 | null | In this case it would not be expected to find substantive differences between the 'NoCor' model and the 'WithCor' model that you have specified. This is because 'Size' is a factor, not a numeric covariate, and what changes between the two models is that instead of the random effects being referenced to the intercept (b... | null | CC BY-SA 2.5 | null | 2010-10-20T06:31:31.430 | 2010-10-20T06:31:31.430 | null | null | 966 | null |
3786 | 2 | null | 3782 | 1 | null | The Skillings-Mack test is a general Friedman-type test that can be used in almost any block design with an arbitrary missing-data structure. It's part of the `asbio` package for R, and there's a user-written package `skilmack` for Stata.
Skillings, J. H., and G. A. Mack. 1981. [On the use of a Friedman-type statistic... | null | CC BY-SA 2.5 | null | 2010-10-20T07:22:44.740 | 2010-10-20T07:22:44.740 | null | null | 449 | null |
3787 | 1 | 3790 | null | 46 | 19969 | For a unimodal distribution that is moderately skewed, we have the following empirical relationship between the mean, median and mode:
$$
\text{(Mean - Mode)}\sim 3\,\text{(Mean - Median)}
$$
How was this relationship derived?
Did Karl Pearson plot thousands of these relationships before forming this conclusion, or... | Empirical relationship between mean, median and mode | CC BY-SA 3.0 | null | 2010-10-20T08:22:36.250 | 2018-08-13T03:07:45.363 | 2017-09-18T11:19:07.617 | 60613 | 1636 | [
"distributions",
"mathematical-statistics",
"descriptive-statistics",
"history"
] |
3788 | 1 | null | null | 6 | 10132 | 1- How can I check if a set of data can be assumed as IID data?
I'm not so familiar with statistics, but I guess I should look at the first lag of autocorrelation for independent distribution. Have no idea about identical distribution condition!
2- It seems that I was not clear enough!
I'm trying to detect outliers in... | How can the IID assumption be checked in a given dataset? | CC BY-SA 2.5 | null | 2010-10-20T08:24:46.747 | 2010-10-21T09:21:55.823 | 2010-10-21T09:21:55.823 | 1637 | 1637 | [
"distributions",
"time-series",
"autocorrelation"
] |
3789 | 2 | null | 3727 | 4 | null | This is an interesting question, which doesn't really have a good solution. There a few different ways of tackling this problem.
- Assume an underlying distribution and match moments - as suggested in the answers by @ivant and @onestop. One downside is that the multivariate generalisation may be unclear.
- Saddlepoi... | null | CC BY-SA 2.5 | null | 2010-10-20T08:32:13.313 | 2010-10-20T08:32:13.313 | null | null | 8 | null |
3790 | 2 | null | 3787 | 34 | null | Denote $\mu$ the mean ($\neq$ average), $m$ the median, $\sigma$ the standard deviation and $M$ the mode. Finally, let $X$ be the sample, a realization of a continuous unimodal distribution $F$ for which the first two moments exist.
It's well known that
$$|\mu-m|\leq\sigma\label{d}\tag{1}$$
This is a frequent textbook... | null | CC BY-SA 3.0 | null | 2010-10-20T08:34:16.963 | 2017-10-26T00:38:34.403 | 2017-10-26T00:38:34.403 | 603 | 603 | null |
3791 | 2 | null | 3782 | 0 | null | As you suggest you "designed" an experiment, it would be better if can you give a description of your design and data set. Even if the data is heteroscedastic and non-normal, probably some variable transformations might help and you may be able to take advantage of the design. The t-test is fairly robust to the normali... | null | CC BY-SA 4.0 | null | 2010-10-20T08:53:24.307 | 2020-12-18T17:46:42.550 | 2020-12-18T17:46:42.550 | 11887 | 1307 | null |
3792 | 2 | null | 3788 | 4 | null | You don't frame the two problems the right way.
Given a random dataset, ie a collection of observations $x_{ij}$ lying in general [position](http://en.wikipedia.org/wiki/General_position) you can always make the $n$ $x_{i}\in\mathbb{R}^p$ independent from one another by randomly shuffling the $n$ indexes. The real que... | null | CC BY-SA 2.5 | null | 2010-10-20T08:57:39.543 | 2010-10-20T11:01:37.220 | 2010-10-20T11:01:37.220 | 603 | 603 | null |
3793 | 1 | 3807 | null | 3 | 112 | Is there a term that describes what I'm trying to do below?
Also, how would you do this using something like JMP or Excel? (or do I need to code this in something like perl?)
Given this sort of data:
```
ID| opened | closed | quantity
--------------------------------------
1 | 2010-01-01 | 2010-01-03 | 1
2 | 20... | Is there a term for generating time-line based data from individual points? Also how would I do this? | CC BY-SA 2.5 | null | 2010-10-20T09:55:52.417 | 2010-10-20T15:32:06.020 | 2010-10-20T15:32:06.020 | 919 | 1641 | [
"terminology",
"excel"
] |
3794 | 2 | null | 3793 | 0 | null | One simple algorithm which could be implemented easily would be:
Step 1:
Add variables to your data frame; one for each time you want to graph. In pseudo code
`if time_i >= opened AND time_i < closed then quantity else 0`
Step 2:
sum the rows for each generated time variable to get the quantity at each time
Step 3:
I... | null | CC BY-SA 2.5 | null | 2010-10-20T10:16:28.447 | 2010-10-20T10:16:28.447 | null | null | 183 | null |
3795 | 1 | null | null | 9 | 9335 | I am looking for a software tool (preferably open source) to draw structural equation/mixture models efficiently and prettily.
After looking into xfig and graphviz I now stick to the general vector graphics package [inkscape](http://inkscape.org) because it seems most flexible.
I would like to poll the stat.stackexcha... | How do you draw structural equation/MPLUS models? | CC BY-SA 2.5 | null | 2010-10-20T10:33:59.067 | 2017-03-20T21:11:25.560 | 2017-03-20T21:11:25.560 | 12359 | 767 | [
"data-visualization",
"modeling",
"structural-equation-modeling",
"software"
] |
3796 | 2 | null | 3795 | 4 | null | I use the [psych](http://cran.r-project.org/web/packages/psych/index.html) R package for CFA and John Fox's [sem](http://cran.r-project.org/web/packages/sem/index.html) package with simple SEM. Note that the graphical backend is graphviz. I don't remember if the [lavaan](http://lavaan.ugent.be/) package provides simila... | null | CC BY-SA 2.5 | null | 2010-10-20T10:58:25.770 | 2010-10-20T11:13:51.817 | 2010-10-20T11:13:51.817 | 930 | 930 | null |
3798 | 2 | null | 570 | 13 | null | I met Laura Trinchera who contributed a nice R package for PLS-path modeling, [plspm](http://cran.r-project.org/web/packages/plspm/). It includes several graphical output for various kind of 2- and k-block data structures.
I just discovered the [plotSEMM](http://www.bethanykok.com/plotSEMM.html) R package. It's more re... | null | CC BY-SA 3.0 | null | 2010-10-20T11:31:09.770 | 2012-01-22T19:36:01.517 | 2012-01-22T19:36:01.517 | 930 | 930 | null |
3799 | 1 | 3813 | null | 11 | 312 | I've got following problem:
- We have set of N people
- We have set of K images
- Each person rates some number of images. A person might like or not like an image (these are the only two possibilites) .
- The problem is how to calculate likelihood that some person likes a particular image.
I'll give example present... | Probability that someone will like image | CC BY-SA 2.5 | null | 2010-10-20T11:59:04.173 | 2010-10-21T09:06:04.907 | 2010-10-21T09:06:04.907 | 183 | 1643 | [
"missing-data",
"rating"
] |
3800 | 2 | null | 3799 | 6 | null | This looks like a good problem for machine learning, so I'll concentrate on this group of methods.
First and the most obvious idea is the kNN algorithm. There you first calculate the similarity among viewers and then predict the missing votes with the average vote on this picture cast by similar users. For details see ... | null | CC BY-SA 2.5 | null | 2010-10-20T12:29:20.677 | 2010-10-20T12:49:42.360 | 2010-10-20T12:49:42.360 | null | null | null |
3801 | 2 | null | 3781 | 3 | null | Yes, you have repetitions of each scenario, so you can make a histogram of each coefficient values. Now treat it as a real distribution and just find a band that encloses this 99, 95 or whatever per cent of its area -- this will be the nonparametric approximation of CI.
The simpler way is to assume normality and just c... | null | CC BY-SA 2.5 | null | 2010-10-20T12:58:47.653 | 2010-10-20T12:58:47.653 | null | null | null | null |
3802 | 2 | null | 3795 | 11 | null | I use [OpenMx](http://openmx.psyc.virginia.edu/) for SEM modeling where I simply use the [omxGraphViz](http://openmx.psyc.virginia.edu/docs/OpenMx/latest/_static/Rdoc/omxGraphviz.html) function to return a dotfile. I haven't found it too inflexible -- the default output looks pretty good and though I've rarely needed ... | null | CC BY-SA 2.5 | null | 2010-10-20T13:33:51.450 | 2010-10-20T14:55:41.667 | 2010-10-20T14:55:41.667 | 251 | 251 | null |
3803 | 2 | null | 3779 | 7 | null | Srikant is right: a Monte Carlo study is the way to go. There are two reasons. First, the answer depends strongly on the structure of the dictionary. Two extremes are (1) the dictionary contains every possible single-letter word. In this case, the chance of not making a word in a draw of $1$ or more letters is zero... | null | CC BY-SA 2.5 | null | 2010-10-20T14:29:00.610 | 2010-10-20T14:29:00.610 | null | null | 919 | null |
3804 | 1 | 3832 | null | 4 | 1689 | Suppose I'm modeling a set of processes using a beta-binomial prior. I can build parameterized beta-binomial models that average over large groups of the processes to give reasonable, although coarse, priors.
$p_i \sim \beta B(n, \alpha_i, \beta_i)$ (roughly)
I know how to update those priors using observed partial da... | Updating a beta-binomial | CC BY-SA 3.0 | null | 2010-10-20T14:43:05.913 | 2017-08-27T12:45:40.327 | 2017-08-27T12:45:40.327 | 11887 | 6 | [
"bayesian",
"prior",
"beta-binomial-distribution"
] |
3805 | 1 | 3806 | null | 5 | 681 | I have a very simple model. This model uses data that are not given as continuous distributions, but are described by percentiles. What is the best way to sample these percentile bins, when the bins are of unequal size?
So, for example, to select the body weight for a given individual, I pick a random number between ... | Sampling with unequal bins? | CC BY-SA 2.5 | null | 2010-10-20T14:47:38.207 | 2010-10-20T15:06:32.823 | 2010-10-20T15:06:32.823 | 919 | 1645 | [
"distributions",
"modeling",
"sampling",
"discrete-data",
"quantiles"
] |
3806 | 2 | null | 3805 | 2 | null | In the spirit of simplicity, while aiming to attain some realism (which is not possible without both interpolation and extrapolation), consider fitting a distribution to the percentiles and sampling from it.
For body weights we can expect a power between 1/2 and 1/3 to be normally distributed. By trial and error you c... | null | CC BY-SA 2.5 | null | 2010-10-20T15:05:41.183 | 2010-10-20T15:05:41.183 | null | null | 919 | null |
3807 | 2 | null | 3793 | 2 | null | Use SUMIF() to compute the total open to date and the total closed to date. The difference at any time is the total currently open.
Let's suppose the data you show are in the range A1:D4 in Excel. Reserve four columns for your output: the two shown plus two for intermediate calculations. Let's suppose they are colum... | null | CC BY-SA 2.5 | null | 2010-10-20T15:30:33.467 | 2010-10-20T15:30:33.467 | null | null | 919 | null |
3808 | 2 | null | 3752 | 2 | null | There seem to be two different questions here, which I'll try to split:
1) how is KS, kernel smoothing, different from KDE, kernel density estimation ?
Well, say I have an estimator / smoother / interpolator
```
est( xi, fi -> gridj, estj )
```
and also happen to know the "real" densityf() at the xi. Then running
`es... | null | CC BY-SA 2.5 | null | 2010-10-20T15:51:18.260 | 2010-10-20T15:51:18.260 | null | null | 557 | null |
3809 | 2 | null | 3804 | 1 | null | Assume that prior2 is a beta random variable and set $\alpha$ and $\beta$ as needed subject to the constraint that $\frac{\alpha-1}{\alpha + \beta -2} = 6$.
In response to your comment:
- Getting to prior2:
Fix either $\alpha$ or $\beta$ at the same value as prior1 and tweak the other to match the desired mode.
If ... | null | CC BY-SA 2.5 | null | 2010-10-20T16:09:43.653 | 2010-10-20T18:39:15.817 | 2010-10-20T18:39:15.817 | null | null | null |
3810 | 1 | null | null | 4 | 3374 | I have data from a survey comprised of several measures that used different Likert-type scaling (4-, 5-, and 6-point scales). I would like to run a principal components analysis using the data from these measures. It seems to me that I need to transform this data in some way so that the power of all items is equivale... | Data transformation for Principal Components Analysis from different Likert scales | CC BY-SA 2.5 | null | 2010-10-20T16:38:35.493 | 2020-11-15T08:13:18.257 | 2020-11-15T08:13:18.257 | 930 | 1647 | [
"pca",
"data-transformation",
"likert",
"scales",
"psychometrics"
] |
3812 | 2 | null | 3810 | 9 | null | As suggested by @whuber, you can "abstract" the scale effect by working with a standardized version of your data. If you're willing to accept that an interval scale is the support of each of your item (i.e. the distance between every two response categories would have the same meaning for every respondents), then linea... | null | CC BY-SA 2.5 | null | 2010-10-20T17:20:23.490 | 2010-10-20T18:18:35.790 | 2017-04-13T12:44:33.310 | -1 | 930 | null |
3813 | 2 | null | 3799 | 6 | null | I believe this is a standard problem of [Collaborative Filtering](http://en.wikipedia.org/wiki/Collaborative_filtering). A google search gives thousands of results.
| null | CC BY-SA 2.5 | null | 2010-10-20T19:01:44.367 | 2010-10-20T19:01:44.367 | null | null | 795 | null |
3814 | 1 | null | null | 105 | 7058 | I recently asked a question regarding general principles around [reviewing statistics in papers](https://stats.stackexchange.com/questions/3460/reviewing-statistics-in-papers). What I would now like to ask, is what particularly irritates you when reviewing a paper, i.e. what's the best way to really annoy a statistical... | How to annoy a statistical referee? | CC BY-SA 3.0 | null | 2010-10-20T19:09:31.910 | 2019-06-25T12:56:38.810 | 2017-04-13T12:44:45.783 | -1 | 8 | [
"references",
"referee"
] |
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