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3601 | 2 | null | 3595 | 4 | null | It is hard to give a good answer without knowing what kind of models you have in mind.
For linear regression, I have successfully used the biglm package in R.
| null | CC BY-SA 2.5 | null | 2010-10-14T13:44:36.413 | 2010-10-14T13:44:36.413 | null | null | 439 | null |
3602 | 2 | null | 3589 | 0 | null | Another way to solve such a problem is to impute the missing data for the shorter series using a time series model which may or may not make sense in a particular context.
In your context, imputing the stock prices into the past would mean that you are asking the following counter-factual question: What would be the s... | null | CC BY-SA 2.5 | null | 2010-10-14T13:49:41.147 | 2010-10-14T13:49:41.147 | null | null | null | null |
3603 | 2 | null | 3599 | 1 | null | Your problem is one of multi-collinear regressors (since B and C are correlated). I would suggest that you look at the answers to the question: [Dealing with correlated regressors](https://stats.stackexchange.com/q/3561/28).
The following paper may also be relevant in your context: [Using principal components for esti... | null | CC BY-SA 2.5 | null | 2010-10-14T13:57:04.500 | 2010-10-14T13:57:04.500 | 2017-04-13T12:44:24.947 | -1 | null | null |
3604 | 2 | null | 3592 | 4 | null | I'll concentrate on your example question: does class 1 of the old classification scheme have a better or worse survival than class 1 of the updated classification scheme?
We can form four mutually exclusive groups of patients:
(a) Patients who weren't in class 1 under either scheme. Clearly, they don't help us answer ... | null | CC BY-SA 2.5 | null | 2010-10-14T13:58:37.590 | 2010-10-14T13:58:37.590 | null | null | 449 | null |
3605 | 2 | null | 3589 | 10 | null | No amount of imputation, time series analysis, GARCH models, interpolation, extrapolation, or other fancy algorithms will do anything to create information where it does not exist (although they can create that illusion ;-). The history of Y's price before X went public is useless for assessing their subsequent correl... | null | CC BY-SA 2.5 | null | 2010-10-14T14:07:20.877 | 2010-10-14T14:07:20.877 | null | null | 919 | null |
3606 | 2 | null | 3599 | 2 | null | You have clearly stated a part of your model:
>
C depends on B, in that values of B above a threshold will change C. The change in C will furthermore reduce B in the next measurement.
By "next measurement" I understand you mean next in time. Let's index time as $t = 0, 1, 2, \ldots$. Then the dependence of C on B ... | null | CC BY-SA 2.5 | null | 2010-10-14T14:21:48.683 | 2010-10-14T14:21:48.683 | null | null | 919 | null |
3607 | 2 | null | 3599 | 2 | null | Not easy at all. This starts to sound like the sort of thing that [Jamie Robins](http://en.wikipedia.org/wiki/James_Robins) and colleagues have done a lot of work on. To quote the start of the abstract of one of their papers:
"In observational studies with exposures or treatments that vary over time, standard approache... | null | CC BY-SA 2.5 | null | 2010-10-14T14:23:57.400 | 2010-10-14T14:23:57.400 | null | null | 449 | null |
3609 | 1 | null | null | 4 | 2054 | It seems to me that rather than using non robust
estimation methods with robust standard errors it would be better to use
robust estimation from the outset. I wonder what other people think.
| Robust standard errors for panel data vs robust estimation for panel data | CC BY-SA 2.5 | null | 2010-10-14T17:57:07.050 | 2010-10-15T04:27:41.783 | 2010-10-14T17:59:07.143 | 930 | 1585 | [
"robust",
"panel-data"
] |
3610 | 2 | null | 3595 | 3 | null | We trained 3.5M observations and 44 features using 64-bit R on an EC2 instance with 32GB ram and 4 cores. We used random forests and it worked well. Note that we had to preprocess/manipulate the data before training.
| null | CC BY-SA 2.5 | null | 2010-10-14T18:00:23.600 | 2010-10-14T18:00:23.600 | null | null | 616 | null |
3611 | 1 | 3612 | null | 9 | 5206 | Does X (hazard) variable in Cox proportional hazard regression analysis always have to be time? If not, could you provide an example, please?
Can age of cancer patient be a hazard variable? If so, can it be interpreted as the risk of getting cancer at a certain age? Would Cox regression be a legitimate analysis to stud... | Cox regression and time scale | CC BY-SA 3.0 | null | 2010-10-14T18:40:10.550 | 2014-05-30T12:56:22.647 | 2014-05-30T12:56:22.647 | 28740 | 1586 | [
"regression",
"survival",
"hazard"
] |
3612 | 2 | null | 3611 | 8 | null | Usually, age at baseline is used as a covariate (because it is often associated to disease/death), but it can be used as your time scale as well (I think it is used in some longitudinal studies, because you need to have enough people at risk along the time scale, but I can't remember actually -- just found these slides... | null | CC BY-SA 2.5 | null | 2010-10-14T19:05:25.133 | 2010-10-14T19:05:25.133 | null | null | 930 | null |
3613 | 2 | null | 3611 | 7 | null | No, it doesn't always have to be time. Many censored responses can be modeled with survival analysis techniques. In his book Nondetects and Data Analysis, Dennis Helsel advocates using the negative of a concentration in place of time (in order to cope with nondetects, which when negated become right-censored values).... | null | CC BY-SA 2.5 | null | 2010-10-14T19:10:51.673 | 2010-10-14T19:10:51.673 | null | null | 919 | null |
3614 | 1 | null | null | 27 | 48355 | I want to calculate the probability distribution for the total of a combination of dice.
I remember that the probability of is the number of combinations that total that number over the total number of combinations (assuming the dice have a uniform distribution).
What are the formulas for
- The number of combination... | How to easily determine the results distribution for multiple dice? | CC BY-SA 2.5 | null | 2010-10-14T19:23:39.887 | 2019-10-29T03:58:46.750 | 2010-10-14T22:40:19.860 | null | 1456 | [
"probability",
"dice"
] |
3615 | 2 | null | 3614 | 5 | null | Approximate Solution
I explained the exact solution earlier (see below). I will now offer an approximate solution which may suit your needs better.
Let:
$X_i$ be the outcome of a roll of a $s$ faced dice where $i=1, ... n$.
$S$ be the total of all $n$ dice.
$\bar{X}$ be the sample average.
By definition, we have:
$\ba... | null | CC BY-SA 3.0 | null | 2010-10-14T19:58:44.760 | 2017-02-28T01:24:51.983 | 2017-02-28T01:24:51.983 | 805 | null | null |
3616 | 1 | 3617 | null | 16 | 44924 | I have a simple time series with 5-10 data points per data set at regular intervals. I am wondering what is the best way to determine whether two data sets are different. Should i try t-tests on each data point, or look at the area under the curves or is there some kind of multivariate model that would work better?
| What is the best statistical test for a time series? | CC BY-SA 2.5 | null | 2010-10-14T20:17:42.653 | 2010-11-05T14:54:01.720 | 2010-10-15T02:50:22.197 | 159 | 1327 | [
"time-series",
"statistical-significance"
] |
3617 | 2 | null | 3616 | 12 | null | You will need to specify precisely what you mean by "different". You will also need to specify what assumptions you are willing to make about the serial correlation structure within each time series.
With t-tests, you are comparing the mean of each group and you are assuming that the groups consist of independent obse... | null | CC BY-SA 2.5 | null | 2010-10-14T21:14:33.157 | 2010-10-14T21:14:33.157 | null | null | 159 | null |
3618 | 2 | null | 3614 | 19 | null | Exact solutions
The number of combinations in $n$ throws is of course $6^n$.
These calculations are most readily done using the probability generating function for one die,
$$p(x) = x + x^2 + x^3 + x^4 + x^5 + x^6 = x \frac{1-x^6}{1-x}.$$
(Actually this is $6$ times the pgf--I'll take care of the factor of $6$ at the e... | null | CC BY-SA 4.0 | null | 2010-10-14T22:29:58.243 | 2019-10-29T03:58:46.750 | 2019-10-29T03:58:46.750 | 7250 | 919 | null |
3619 | 2 | null | 3595 | 7 | null | There is the [RHIPE](http://www.rhipe.org/) package (R-Hadoop integration). It is can make it very easy (with exceptions) to analyze large amounts of data in R.
| null | CC BY-SA 3.0 | null | 2010-10-15T01:03:14.220 | 2012-03-12T16:11:13.280 | 2012-03-12T16:11:13.280 | 6432 | 1307 | null |
3620 | 2 | null | 3609 | 2 | null | I don't know if I understand you correctly, but still I will give it a shot.
I think robust estimation from the outset would be better in most of the cases.
Reason:
If you estimation is not robust, outliers might severely affect your estimate. Still, in general, you will be far the true value. This may be also looked... | null | CC BY-SA 2.5 | null | 2010-10-15T01:22:06.443 | 2010-10-15T01:22:06.443 | null | null | 1307 | null |
3621 | 2 | null | 3611 | 4 | null | On the age-scale vs. time-scale issue, chl has some good references and captures the essentials -- in particular, the requirement that the at-risk set contain sufficient subjects from all ages as would arise in a longitudinal study.
I would only note that there is no general consensus around this yet, but there is so... | null | CC BY-SA 2.5 | null | 2010-10-15T02:06:12.473 | 2010-10-15T02:06:12.473 | null | null | 251 | null |
3622 | 2 | null | 3609 | 8 | null | They're robust with respect to different things.
If you use robust regression to obtain an estimate of fixed effect in panel data, then you're computing an estimate that's resistant to outliers.
If you use robust standard errors for your OLS estimator, it's because you suspect that the assumption behind your error mode... | null | CC BY-SA 2.5 | null | 2010-10-15T02:40:18.180 | 2010-10-15T04:27:41.783 | 2010-10-15T04:27:41.783 | 251 | 251 | null |
3623 | 1 | 3626 | null | 3 | 421 | I have a logistic regression (in SAS, for reference) with continuous and categorical predictors (with reference coding), and an interaction term between one of each type (assume for now that the categorical variable in question has three response levels, reference coded to $c_1$ and $c_2$):
$logit(p) = a + (continuous ... | Plugging in mean values/proportions to a logistic regression with continuous-discrete interaction | CC BY-SA 2.5 | null | 2010-10-15T04:44:29.807 | 2010-10-15T08:27:19.977 | 2010-10-15T08:27:19.977 | null | 1144 | [
"logistic",
"categorical-data",
"fitting"
] |
3625 | 2 | null | 3614 | 7 | null | There's a very neat way of computing the combinations or probabilities in a spreadsheet (such as excel) that computes the convolutions directly.
I'll do it in terms of probabilities and illustrate it for six sided dice but you can do it for dice with any number of sides (including adding different ones).
(btw it's als... | null | CC BY-SA 3.0 | null | 2010-10-15T06:19:31.640 | 2014-07-28T02:19:47.983 | 2014-07-28T02:19:47.983 | 805 | 805 | null |
3626 | 2 | null | 3623 | 3 | null | Regardless of the interaction term, this procedure isn't going to estimate the average p, because logit is a nonlinear function so the mean of the logit isn't the same as the logit of the mean. If you want to calculate the expected proportion of positive outcomes in a sample, the easiest way is to calculated the predic... | null | CC BY-SA 2.5 | null | 2010-10-15T06:39:37.757 | 2010-10-15T06:39:37.757 | null | null | 449 | null |
3627 | 2 | null | 3623 | 4 | null | The predicted grand average (see onestop's answer) may not be all that informative - after all, you are fitting a model to understand systematic deviations from it.
You can predict your p for any setting of your predictors. Given that you are interested in the effects of the predictors, it would make sense to look at w... | null | CC BY-SA 2.5 | null | 2010-10-15T07:35:14.613 | 2010-10-15T07:35:14.613 | null | null | 1352 | null |
3628 | 1 | 3629 | null | 9 | 38461 | From the output of a logistic regression in JMP, I read about two binary variables:
```
Var1 estimate -0.1007384
Var2 estimate 0.21528927
```
and then
```
Odds ratio for Var1 lev1/lev2 1.2232078 reciprocal 0.8175225
Odds ratio for Var2 lev1/lev2 0.6501329 reciprocal 1.5381471
```
Now I obtain `1.2232078` as `exp(2*0... | Relation between logistic regression coefficient and odds ratio in JMP | CC BY-SA 2.5 | null | 2010-10-15T10:48:41.760 | 2017-06-16T19:24:59.167 | 2010-10-15T14:42:37.287 | 930 | 1219 | [
"logistic",
"odds-ratio",
"jmp"
] |
3629 | 2 | null | 3628 | 15 | null | Ok, I drop a quick response. Your idea is correct in that the regression coefficient is the log of the OR. More precisely, if $b$ is your regression coefficient, $\exp(b)$ is the odds ratio corresponding to a one unit change in your variable. So, to get back to the adjusted odds, you need to know what are the internal ... | null | CC BY-SA 2.5 | null | 2010-10-15T10:54:40.603 | 2010-10-15T11:08:31.683 | 2010-10-15T11:08:31.683 | 930 | 930 | null |
3630 | 1 | null | null | 5 | 435 | I'm trying to evaluate a path dependent function, $f(r_t)$, on a [Cox-Ingersoll-Ross](https://en.wikipedia.org/wiki/Cox%E2%80%93Ingersoll%E2%80%93Ross_model) process:
$$dr_t = \theta (\mu - r_t)dt + \sigma \sqrt r_t dW_t$$
by Monte Carlo simulation.
Could anyone suggest and explain any effective variance reduction tech... | CIR Process-Variance reduction | CC BY-SA 3.0 | null | 2010-10-15T11:00:45.113 | 2016-12-09T08:31:15.880 | 2016-12-09T08:31:15.880 | 113090 | 1443 | [
"r",
"references",
"markov-chain-montecarlo",
"stochastic-processes",
"methodology"
] |
3631 | 2 | null | 3611 | 3 | null | Per the OP's request, heres another application I have seen survival analysis used in a spatial context (although obviously different than measuring environmental substances [mentioned](https://stats.stackexchange.com/questions/3611/cox-regression-and-time-scale/3613#3613) by whuber) is modeling the distance between ev... | null | CC BY-SA 3.0 | null | 2010-10-15T12:01:27.760 | 2011-10-20T19:51:54.067 | 2017-04-13T12:44:33.237 | -1 | 1036 | null |
3632 | 1 | 3665 | null | 2 | 260 | Suppose we need to take an action on a population with income (x) more than $5,000. Income is not observed directly.
Should we use logistic regression to estimate x, or should we use logistic regression to estimate the probability of x>5000 directly? (What are the drawbacks/advantage of the methods?)
Edit: Yes - by log... | Should we regress x or use logistic regression on x>5000 | CC BY-SA 2.5 | null | 2010-10-15T14:14:33.220 | 2010-10-17T04:02:18.357 | 2010-10-16T04:50:04.427 | 994 | 994 | [
"regression",
"logistic"
] |
3633 | 2 | null | 3616 | 6 | null | Maybe repeated measures anova is what you want. It allows you to compare the subjects (inter subject factors) while taking the correlated structure of the "time series" per subject (intra subject factor). It is an easy but dated method and can be found in the context of "general linear models", it needs some additional... | null | CC BY-SA 2.5 | null | 2010-10-15T15:08:03.417 | 2010-10-15T15:08:03.417 | null | null | 1573 | null |
3634 | 1 | null | null | 9 | 6300 | I have to analyze a factorial design with five factors (one of them nested in another one) and numeric responses. I would like to perform a nonparametric ANOVA, but of course I can't use neither Kruskall Wallis nor Friedman test (I have replicated measures). Is there a command or a code in R that could help me?
Thank y... | Multi-way nonparametric anova | CC BY-SA 2.5 | null | 2010-10-15T15:42:56.067 | 2011-03-31T16:14:14.503 | 2010-10-15T16:41:33.940 | null | null | [
"r",
"anova",
"nonparametric"
] |
3635 | 2 | null | 3419 | 4 | null | I might be misunderstanding your goals here, but to me it sounds like a [multi-dimensional scaling](http://en.wikipedia.org/wiki/Multidimensional_scaling) (MDS) problem. I've never used MDS myself, but my sense is that it should allow you to derive a global measure of similarity as well as dimensional measures of simi... | null | CC BY-SA 3.0 | null | 2010-10-15T15:50:12.507 | 2017-11-16T13:21:24.017 | 2017-11-16T13:21:24.017 | 196 | 196 | null |
3636 | 1 | null | null | 1 | 2763 | I have a sample set of values that were taken over a period of time. However, the delta time between each sample is different.
Do I need to account for the different time deltas in the std-dev?
Is std-dev even appropriate for this kind of data?
---
More info...
The data are temperature samples.
The time range is fro... | How to calculate the standard deviation on a sample set with irregular time periods | CC BY-SA 2.5 | 0 | 2010-10-15T15:53:01.893 | 2019-07-21T11:38:26.037 | 2019-07-21T11:38:26.037 | 11887 | 1595 | [
"standard-deviation",
"unevenly-spaced-time-series"
] |
3637 | 2 | null | 3636 | 2 | null | Yes, you do need to account for the irregularity of the time series because volatility scales with time. Depending upon the distribution and independence assumptions, [sometimes a "square root of time" rule can be appropriate](http://ideas.repec.org/p/fmg/fmgdps/dp439.html).
Is this data sampled irregularly intraday... | null | CC BY-SA 2.5 | null | 2010-10-15T16:02:52.387 | 2010-10-15T16:02:52.387 | null | null | 5 | null |
3638 | 2 | null | 3634 | 4 | null | Tukey's Median Polish is implemented in R as medpolish {stats}. See Chapter 6 in [Venables and Ripley](http://www.stats.ox.ac.uk/pub/MASS4/VR4stat.pdf)
| null | CC BY-SA 2.5 | null | 2010-10-15T16:27:19.637 | 2011-03-31T16:14:14.503 | 2011-03-31T16:14:14.503 | 919 | 919 | null |
3639 | 2 | null | 3383 | 2 | null | The one-way ANOVA approach you mention sounds fine to me. Sure the individual change scores aren't going to be the "true change" by any means, but they are better than nothing. If anything the resulting variance in the model should be over estimated as a consequence of this procedure.
In R the easiest way to do ANO... | null | CC BY-SA 2.5 | null | 2010-10-15T16:34:25.457 | 2010-10-15T16:34:25.457 | null | null | 196 | null |
3640 | 1 | 3644 | null | 12 | 8016 | Problem:
I am parameterizing distributions for use as a priors and data in a Bayesian meta-analysis. The data are provided in the literature as summary statistics, almost exclusively assumed to be normally distributed (although none of the variables can be < 0, some are ratios, some are mass, and etc.).
I have come ac... | How to parameterize the ratio of two normally distributed variables, or the inverse of one? | CC BY-SA 2.5 | null | 2010-10-15T16:45:49.320 | 2010-10-15T19:45:31.940 | 2010-10-15T19:28:25.393 | 1381 | 1381 | [
"distributions",
"bayesian",
"variance",
"random-variable",
"meta-analysis"
] |
3641 | 1 | null | null | 4 | 216 | I would like to project the data in this graph for at least 4 or 5 periods. Unfortunately, that won't be possible with a moving average. A regression will result in negative values after the 3rd period. What are my forecasting options?
Basically, what i'm trying to do, is predict where the boomer hump is gonna be based... | Forecasting Age distribution | CC BY-SA 2.5 | null | 2010-10-15T16:50:38.817 | 2010-10-15T19:43:31.437 | 2010-10-15T17:25:32.207 | 59 | 59 | [
"time-series",
"forecasting"
] |
3642 | 2 | null | 3634 | 1 | null | You might check out the ezBoot() function in the ez package for bootstrapping confidence intervals on your effects of interest.
| null | CC BY-SA 2.5 | null | 2010-10-15T16:52:13.963 | 2010-10-15T16:52:13.963 | null | null | 364 | null |
3643 | 2 | null | 3640 | 0 | null | Could you not assume that $y^{-1} \sim N(.,.)$ for the inverse of a normal random variable and do the necessary bayesian computation after identifying the appropriate parameters for the normal distribution.
My suggestion below to use the Cauchy does not work as pointed out in the comments by ars and John.
The ratio of ... | null | CC BY-SA 2.5 | null | 2010-10-15T17:05:22.430 | 2010-10-15T18:42:03.223 | 2010-10-15T18:42:03.223 | null | null | null |
3644 | 2 | null | 3640 | 6 | null | You might want to look at some of the references under the Wikipedia article on [Ratio Distribution](http://en.wikipedia.org/wiki/Ratio_distribution). It's possible you'll find better approximations or distributions to use. Otherwise, your approach seems sound.
Update I think a better reference might be:
- Ratios of... | null | CC BY-SA 2.5 | null | 2010-10-15T17:17:18.427 | 2010-10-15T19:45:31.940 | 2010-10-15T19:45:31.940 | 251 | 251 | null |
3645 | 2 | null | 3634 | 4 | null | The [vegan](http://cc.oulu.fi/~jarioksa/softhelp/vegan.html) package implements permutation testing for distance based ANOVA, which should work with multi-way, repeated measures data.
| null | CC BY-SA 2.5 | null | 2010-10-15T17:43:27.083 | 2010-10-15T17:43:27.083 | null | null | 251 | null |
3646 | 1 | 3648 | null | 2 | 536 | I have a univariate data set that's approximately normally distributed. I am happy to assume that the population is normally distributed, and I'd like to estimate the mean and variance of the population.
My textbook suggests (as I understand it) that since my sample size is large (1000's of data points), it is reasonab... | Simple Estimates vs Model for calculating mean and variance of population | CC BY-SA 2.5 | null | 2010-10-15T19:27:02.353 | 2010-10-15T20:20:29.600 | null | null | 1598 | [
"normal-distribution"
] |
3647 | 2 | null | 3641 | 6 | null | For demographic forecasting of any quality whatsoever you need to account for birth and death rates and, if possible, migration, breaking them down by gender (at a minimum) and, if possible, by race. These rates have all changed substantially during your time period and are likely to continue changing in the future. ... | null | CC BY-SA 2.5 | null | 2010-10-15T19:43:31.437 | 2010-10-15T19:43:31.437 | null | null | 919 | null |
3648 | 2 | null | 3646 | 7 | null | If I understand your question, and you mean using a least squares model of the form $Y=\beta + \epsilon$ where $\epsilon\sim N(0,\sigma^2)$ these two approaches are equivalent.
A simple R example will demonstrate this:
```
#generate pseudo-data
set.seed(0)
n <- 1000
x <- rnorm(n)
# approach 1: calculation
sum(x)/... | null | CC BY-SA 2.5 | null | 2010-10-15T19:48:34.417 | 2010-10-15T20:20:29.600 | 2010-10-15T20:20:29.600 | 1381 | 1381 | null |
3649 | 2 | null | 3646 | 3 | null | I was about to make the same point as David, except illustrating using Stata rather than R:
. summarize length
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------------------
length | 74 187.9324 22.26634 142 ... | null | CC BY-SA 2.5 | null | 2010-10-15T19:58:01.210 | 2010-10-15T19:58:01.210 | null | null | 449 | null |
3650 | 1 | 3658 | null | 5 | 802 | I have a two part question;
First Part:
I have an urn with 20 balls, 2 of those balls are purple, and I pull out 6 balls at random. I witness 100 realizations of this process.
Given the observed frequency at which I drew purple balls, how do I determine if I am really pulling balls out at random? Also given that there... | Expected distribution of random draws | CC BY-SA 2.5 | null | 2010-10-15T20:17:13.240 | 2010-10-22T16:38:05.127 | 2017-04-13T12:44:33.310 | -1 | 1036 | [
"distributions",
"probability"
] |
3651 | 2 | null | 3650 | 2 | null | First Part: The draws from the urn follow a [hypergeometric distribution](http://en.wikipedia.org/wiki/Hypergeometric_distribution) assuming random draws. Any deviation from the theoretical probabilities vis-a-vis the observed frequencies can be evaluated using chi-square tests.
Second Part:
Let:
$n \sim U(20,30)$ be ... | null | CC BY-SA 2.5 | null | 2010-10-15T20:23:22.837 | 2010-10-16T00:13:23.233 | 2010-10-16T00:13:23.233 | null | null | null |
3652 | 1 | null | null | 20 | 21738 | Suppose one has two independent samples from the same population, and different methods were used on the two samples to derive point estimate and confidence intervals. In trivial cases a sensible person would just pool the two samples and use one method to do the analysis, but let's suppose for the moment that differen... | Combining two confidence intervals/point estimates | CC BY-SA 2.5 | null | 2010-10-15T20:24:37.687 | 2015-01-09T15:52:22.550 | 2010-10-15T23:18:35.950 | 449 | 1600 | [
"confidence-interval",
"meta-analysis"
] |
3653 | 1 | 3657 | null | 15 | 9699 | I have a multivariate regression, which includes interactions. For example, to get the estimate of the treatment effect for the poorest quintile I need to add the coefficients from the treatment regressor to the coefficient from the interaction variable (which interacts treatment and quintile 1). When adding two coeffi... | Adding coefficients to obtain interaction effects - what to do with SEs? | CC BY-SA 2.5 | null | 2010-10-15T20:30:05.690 | 2012-01-24T17:10:44.107 | 2010-10-15T20:31:52.317 | 930 | 834 | [
"regression",
"standard-deviation",
"standard-error"
] |
3654 | 2 | null | 3652 | 1 | null | This is not unlike a stratified sample. So, pooling the samples for a point estimate and standard error seems like a reasonable approach. The two samples would be weighted by sample proportion.
| null | CC BY-SA 2.5 | null | 2010-10-15T20:32:35.200 | 2010-10-15T20:32:35.200 | null | null | 485 | null |
3655 | 2 | null | 3652 | 8 | null | You could do a pooled estimate as follows. You can then use the pooled estimates to generate a combined confidence interval. Specifically, let:
$\bar{x_1} \sim N(\mu,\frac{\sigma^2}{n_1})$
$\bar{x_2} \sim N(\mu,\frac{\sigma^2}{n_2})$
Using the confidence intervals for the two cases, you can re-construct the standard er... | null | CC BY-SA 3.0 | null | 2010-10-15T20:33:13.343 | 2015-01-09T15:52:22.550 | 2015-01-09T15:52:22.550 | -1 | null | null |
3657 | 2 | null | 3653 | 10 | null | I think this the expression for $SE_{b_{new}}$:
$$\sqrt{SE_1^2 + SE_2^2+2Cov(b_1,b_2)}$$
You can work with this new standard error to find your new test statistic for testing $H_o: \beta=0$
| null | CC BY-SA 2.5 | null | 2010-10-15T20:44:47.467 | 2010-10-15T22:09:51.290 | 2010-10-15T22:09:51.290 | 1307 | 1307 | null |
3658 | 2 | null | 3650 | 3 | null | The expected frequency of observing $k$ purple balls in $d$ draws (without replacement) from an urn of $p$ purple balls and $n-p$ other balls is obtained by counting and equals
$$\frac{{p \choose k} {n-p \choose d-k} }{{n \choose d}}.$$
Test a sample (of say $100$) such experiments with a chi-squared statistic using th... | null | CC BY-SA 2.5 | null | 2010-10-15T22:00:06.187 | 2010-10-18T18:21:25.497 | 2010-10-18T18:21:25.497 | 919 | 919 | null |
3659 | 2 | null | 3652 | 4 | null | Sounds a lot like [meta-analysis](http://en.wikipedia.org/wiki/Meta-analysis) to me. Your assumption that the samples are from the same population means you can use fixed-effect meta-analysis (rather than random-effects meta-analysis). The generic inverse-variance method takes a set of independent estimates and their v... | null | CC BY-SA 2.5 | null | 2010-10-15T23:18:18.057 | 2010-10-15T23:18:18.057 | null | null | 449 | null |
3660 | 2 | null | 3595 | 8 | null | Most of the algorithms on [Apache Mahout](http://mahout.apache.org/) scale way beyond 20M records, even with high-dimensional data. If you only need to build a prediction model, there are specific tools like Vowpal Wabbit (http://hunch.net/~vw/) that can easily scale to billions of records on a single machine.
| null | CC BY-SA 2.5 | null | 2010-10-16T01:09:31.757 | 2010-10-16T01:09:31.757 | null | null | 635 | null |
3661 | 1 | null | null | 6 | 2904 | I have 3 observers that each take 2 measurements (length and weight) on 100 individuals; these procedures are repeated once (i.e., the same measurements are taken on the same 100 individuals by the same 3 observers), so that the data set is duplicated (i.e., early reading and late reading).
- What is the best way to ... | Repeatability and measurement error from and between observers | CC BY-SA 2.5 | null | 2010-10-16T01:44:55.913 | 2010-10-18T07:17:51.227 | 2010-10-16T08:43:26.653 | null | 1603 | [
"anova",
"error",
"measurement",
"reliability",
"agreement-statistics"
] |
3663 | 2 | null | 3661 | 8 | null | What you describe is a reliability study where each subject is going to be assessed by the same three raters on two occasions. Analysis can be done separately on the two outcomes (length and weight, though I assume they will be highly correlated and you're not interested in how this correlation is reflected in raters' ... | null | CC BY-SA 2.5 | null | 2010-10-16T08:43:54.767 | 2010-10-18T07:17:51.227 | 2017-04-13T12:44:52.277 | -1 | 930 | null |
3664 | 2 | null | 3661 | 3 | null | You need to repeat the same process separately for length and weight, as these are completely separate outcomes with different units and methods of measurement.
I'd start, as so often, by plotting some exploratory graphs. In this case a set of [Bland–Altman](http://en.wikipedia.org/wiki/Bland-Altman_plot) (diffference ... | null | CC BY-SA 2.5 | null | 2010-10-16T08:50:42.457 | 2010-10-16T08:50:42.457 | null | null | 449 | null |
3665 | 2 | null | 3632 | 4 | null | If your only interest is whether their income is over $\$$5,000 or not, and it doesn't make a difference how far from that threshold their income actually is, then I would recommend using a classification technique (not necessarily logistic regression, try a range of methods and use whatever gives best out-of-sample pe... | null | CC BY-SA 2.5 | null | 2010-10-16T14:58:21.727 | 2010-10-16T15:08:36.650 | 2010-10-16T15:08:36.650 | 919 | 887 | null |
3666 | 2 | null | 2730 | 10 | null | [Applied Linear Statistical Models](http://rads.stackoverflow.com/amzn/click/0256117365) by Neter, Kutner, Wasserman, and Nachtscheim, has a very exhaustive (and exhausting!) treatment of ANOVA and ANCOVA.
It also covers power analysis, linear regression, multilinear regression, and introduces some MANOVA. It's a very ... | null | CC BY-SA 2.5 | null | 2010-10-16T15:47:52.163 | 2010-10-16T15:47:52.163 | null | null | 1118 | null |
3667 | 2 | null | 1015 | 5 | null | Maybe I misunderstood the question, but what you are describing sounds like a test-retest reliability study on your Q scores. You have a series of experts each going to assess a number of items or questions, at two occasions (presumably fixed in time). So, basically you can assess the temporal stability of the judgment... | null | CC BY-SA 2.5 | null | 2010-10-16T18:41:49.013 | 2010-11-01T16:32:27.973 | 2017-04-13T12:44:37.420 | -1 | 930 | null |
3668 | 2 | null | 3632 | 3 | null | What are you trying to predict? Is your outcome just an indicator of whether income is above 5000 dollars or not? If so, that is the best that you can predict; that is, you can't predict anyone's income, only whether he has high (above 5000) income or not.
If this is your outcome and what you'd like to predict, the que... | null | CC BY-SA 2.5 | null | 2010-10-17T04:02:18.357 | 2010-10-17T04:02:18.357 | null | null | 401 | null |
3669 | 2 | null | 3653 | 2 | null | To be more general, if you create a (row) vector for the estimate that you care about $R$ such that your estimator is equal to $R\beta$, then the variance of that estimator is $R\hat{V}R^\prime$, where $\hat{V}$ is the estimated variance-covariance matrix of your regression. Your estimate is distributed Normal or t, de... | null | CC BY-SA 2.5 | null | 2010-10-17T04:09:30.837 | 2010-10-17T04:24:19.630 | 2010-10-17T04:24:19.630 | 401 | 401 | null |
3670 | 2 | null | 3542 | 11 | null | You might check out the documentation for the LaTeX package [booktabs](http://www.ctan.org/tex-archive/macros/latex/contrib/booktabs/booktabs.pdf); it gives general guidance and implements its design suggestions in LaTeX tables.
| null | CC BY-SA 2.5 | null | 2010-10-17T04:16:04.893 | 2010-10-17T04:16:04.893 | null | null | 401 | null |
3671 | 2 | null | 3542 | 6 | null | I hope this answer is not too off topic, but a couple of days ago I have seen this link on visualizing tables at StackExchange:
[Visual Representation of Tabular Information – How to Fix the Uncommunicative Table](http://flowingdata.com/2009/04/21/visual-representation-of-tabular-information-how-to-fix-the-uncommunicat... | null | CC BY-SA 2.5 | null | 2010-10-17T07:36:38.073 | 2010-10-17T07:36:38.073 | null | null | 1607 | null |
3672 | 1 | 3674 | null | 7 | 591 | Could anyone suggest some statistical measures to describe the distribution of a dendrogram? If I have two dendrograms, how could can I quantify their structural differences?
| A measure to describe the distribution of a dendrogram | CC BY-SA 2.5 | null | 2010-10-17T11:24:28.837 | 2010-10-18T08:07:08.010 | 2010-10-18T08:07:08.010 | 449 | 1250 | [
"distributions",
"time-series",
"clustering",
"dendrogram"
] |
3673 | 2 | null | 3419 | 7 | null | You asked a difficult question, but I'm a little bit surprised that the various clues that were suggested to you received so little attention. I upvoted all of them because I think they basically are useful responses, though in their actual form they call for further bibliographic work.
Disclaimer: I never had to deal ... | null | CC BY-SA 2.5 | null | 2010-10-17T12:06:38.520 | 2010-10-17T19:06:12.167 | 2010-10-17T19:06:12.167 | 930 | 930 | null |
3674 | 2 | null | 3672 | 5 | null | See this SO question: [https://stackoverflow.com/questions/2218395/how-do-you-compare-the-similarity-between-two-dendrograms-in-r](https://stackoverflow.com/questions/2218395/how-do-you-compare-the-similarity-between-two-dendrograms-in-r)
| null | CC BY-SA 2.5 | null | 2010-10-17T14:37:56.007 | 2010-10-17T14:37:56.007 | 2017-05-23T12:39:26.150 | -1 | null | null |
3675 | 2 | null | 3307 | 3 | null | I agree with @ars that you are unlikely to get one answer for this (you may also have more success on [http://mathoverflow.net](http://mathoverflow.net), since our community tends to be more applied, while this technique would have very little real-world usage). The Abrahao/Barbosa paper is a good reference. Just to ... | null | CC BY-SA 2.5 | null | 2010-10-17T15:45:28.087 | 2010-10-17T15:59:56.937 | 2010-10-17T15:59:56.937 | 5 | 5 | null |
3676 | 1 | null | null | 5 | 896 | Background:
I'm a junior researcher at an institute dealing with regional issues, particularly involving drug policy. Almost two years ago, one of our senior researchers began collecting arrest data about a nearby large city. He had been transcribing newspaper police blotter by hand until I joined a year and change ago... | "Multiple response" analysis of arrest records | CC BY-SA 2.5 | null | 2010-10-17T16:13:28.657 | 2017-02-01T23:42:59.623 | null | null | 1609 | [
"dataset",
"spss"
] |
3677 | 2 | null | 633 | 4 | null | [Ingo Ruczinski](http://biostat.jhsph.edu/~iruczins/) has contributed to promote the use of [Logic regression](http://kooperberg.fhcrc.org/logic/documents/intro.html) for data set consisting of binary variables, with an emphasis on higher-order interaction terms. The main advantage compared to usual or penalized GLMs i... | null | CC BY-SA 2.5 | null | 2010-10-17T18:21:12.987 | 2010-11-02T18:47:07.033 | 2010-11-02T18:47:07.033 | 930 | 930 | null |
3678 | 2 | null | 3676 | 4 | null | I can't particularly comment on how to handle multiple response categories, but you need to further refine your question for people on this forum to be able to give useful advice.
You mention various interests, such as some sort of drug policy intervention, and differential charges according to race, arrest location, a... | null | CC BY-SA 2.5 | null | 2010-10-17T18:37:01.210 | 2010-10-17T19:27:17.977 | 2010-10-17T19:27:17.977 | 1036 | 1036 | null |
3679 | 2 | null | 3676 | 2 | null | It is not clear what you questions you are trying to answer but here are are several ways to deal with the multiple-response data:
- Arresting Officer
Convert the two columns into a single count variable (1 or 2) which indicates the no of arresting officers. You will lose the arresting officer's identities but perhaps... | null | CC BY-SA 2.5 | null | 2010-10-17T18:41:41.327 | 2010-10-17T18:41:41.327 | 2020-06-11T14:32:37.003 | -1 | null | null |
3680 | 2 | null | 1621 | 5 | null | There was a read paper last week at the Royal Statistical Society on MCMC techniques over Riemann manifolds, primarily using the Fisher information metric:
[http://www.rss.org.uk/main.asp?page=1836#Oct_13_2010_Meeting](http://www.rss.org.uk/main.asp?page=1836#Oct_13_2010_Meeting)
The results seem promising, though as ... | null | CC BY-SA 2.5 | null | 2010-10-17T20:07:00.620 | 2010-10-17T20:12:38.687 | 2010-10-17T20:12:38.687 | 495 | 495 | null |
3681 | 2 | null | 3676 | 2 | null | I've examined associations between multiple response categorical variables in the past basically following the log-linear approach for marginal data outlined in the following:
- Strategies for Modeling Two Categorical Variables with Multiple Category Choices (Bilder, Loughlin, 2003)
Your case may be more complicated... | null | CC BY-SA 3.0 | null | 2010-10-17T20:10:30.190 | 2017-02-01T23:42:59.623 | 2017-02-01T23:42:59.623 | 11887 | 251 | null |
3682 | 1 | 3683 | null | 10 | 505 | What is the meaning of $\|a\|_p=\left(\sum _{i=1}^n \left|a_i(t)\right|{}^p\right){}^{\frac{1}{p}}$?
This formula is called out on the fifth page of An Improved Data Stream Summary: The Count-Min Sketch and its Applications (which can be found [here](http://www.madalgo.au.dk/img/SumSchoo2007_Lecture%20slides/Bibliograp... | What is the meaning of $\|a\|_p=\left(\sum _{i=1}^n \left|a_i(t)\right|{}^p\right){}^{\frac{1}{p}}$? | CC BY-SA 2.5 | null | 2010-10-17T20:29:40.313 | 2020-04-05T00:05:45.660 | 2020-04-05T00:05:45.660 | 11887 | 1515 | [
"descriptive-statistics",
"notation"
] |
3683 | 2 | null | 3682 | 11 | null | It's the $L^p$ norm. See for example the Wikipedia articles:
- $L^p$ space
- Minkowski distance
If you use $p = 2$, you'll find it resolves to the more familiar Euclidean norm -- i.e. the most familiar measure used as length of the vector $a$. Other values of p give others ways of measuring length as outlined in ... | null | CC BY-SA 2.5 | null | 2010-10-17T20:36:56.287 | 2010-10-17T20:42:02.297 | 2010-10-17T20:42:02.297 | 251 | 251 | null |
3684 | 2 | null | 3614 | 5 | null | [Characteristic functions](http://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29) can make computations involving the sums and differences of [random variables](http://en.wikipedia.org/wiki/Random_variable) really easy. [Mathematica](http://www.wolfram.com/) has lots of functions to work with st... | null | CC BY-SA 2.5 | null | 2010-10-17T20:44:19.483 | 2010-10-17T20:44:19.483 | null | null | null | null |
3685 | 1 | 3692 | null | 76 | 97745 | Hierarchical clustering can be represented by a dendrogram. Cutting a dendrogram at a certain level gives a set of clusters. Cutting at another level gives another set of clusters. How would you pick where to cut the dendrogram? Is there something we could consider an optimal point? If I look at a dendrogram across tim... | Where to cut a dendrogram? | CC BY-SA 2.5 | null | 2010-10-17T21:57:55.460 | 2020-03-28T07:00:29.263 | 2010-10-17T22:01:28.690 | 930 | 1250 | [
"clustering",
"dendrogram"
] |
3686 | 2 | null | 3685 | 6 | null | Perhaps one of the simplest methods would be a graphical representation in which the x-axis is the number of groups and the y-axis any evaluation metric as the distance or the similarity. In that plot you usually can observe two differentiated regions, being the x-axis value at the 'knee' of the line the 'optimal' numb... | null | CC BY-SA 2.5 | null | 2010-10-17T22:30:44.930 | 2010-10-18T06:13:33.683 | 2010-10-18T06:13:33.683 | 221 | 221 | null |
3687 | 2 | null | 3682 | 6 | null | This paper does not appear to use $L^p$ norms in any essential way--every one of the results references the $L^1$ norm explicitly. The problem itself determines which norm to use. In this case interest focuses on the cardinality of multisets. A multiset is represented as a vector of counts of its elements, whence it... | null | CC BY-SA 2.5 | null | 2010-10-18T00:16:28.227 | 2010-10-18T00:16:28.227 | null | null | 919 | null |
3688 | 2 | null | 3595 | 3 | null | SAS Enterprise Miner version 6.2 would have no problem handling 20 million observations, and a variety of models which can be adapted to your situation. The issue with SAS is usually the cost however. Here's a summary of what SAS EM can do:
[SAS EM 6.2: What's New ](http://support.sas.com/documentation/cdl/en/whatsnew/... | null | CC BY-SA 2.5 | null | 2010-10-18T03:05:26.233 | 2010-10-18T03:12:18.987 | 2010-10-18T03:12:18.987 | null | null | null |
3689 | 1 | 3690 | null | 6 | 268 | I'm interested in exploring statistical models (or modifications thereof) designed to handle a specific type of problem. Due to my ignorance of statistical terminology, I can only describe this type of problem by (contrived) examples:
Suppose we're interested in estimating the likelihood that a given cell phone custome... | Coefficient / model averaging to control for exogenous circumstances in prediction | CC BY-SA 3.0 | null | 2010-10-18T03:08:15.450 | 2011-08-08T02:26:47.610 | 2011-08-08T02:26:47.610 | 919 | 1611 | [
"time-series",
"logistic",
"classification",
"forecasting"
] |
3690 | 2 | null | 3689 | 5 | null | I am not sure you need any special tricks as long as the relevant factors are captured by the model. To keep things simple I will discuss the issue in the context of linear regression. The same intuition carries over to the time series setting.
Suppose, that you want to predict the monthly sales of cell phones for bran... | null | CC BY-SA 2.5 | null | 2010-10-18T03:53:26.107 | 2010-10-18T14:21:03.917 | 2010-10-18T14:21:03.917 | null | null | null |
3691 | 2 | null | 3685 | 12 | null | There isn't really an answer. It's somewhere between 1 and N.
However, you can think about it from a profit perspective.
For example, in marketing one uses segmentation, which is much like clustering.
A message (an advertisement or letter, say) that is tailored for each individual will have the highest response rate. ... | null | CC BY-SA 3.0 | null | 2010-10-18T04:49:46.067 | 2011-12-24T20:01:26.503 | 2011-12-24T20:01:26.503 | 74 | 74 | null |
3692 | 2 | null | 3685 | 58 | null | There is no definitive answer since cluster analysis is essentially an exploratory approach; the interpretation of the resulting hierarchical structure is context-dependent and often several solutions are equally good from a theoretical point of view.
Several clues were given in a related question, [What stop-criteria ... | null | CC BY-SA 3.0 | null | 2010-10-18T05:56:44.473 | 2013-12-05T19:21:11.280 | 2017-04-13T12:44:28.813 | -1 | 930 | null |
3694 | 2 | null | 3614 | 4 | null | Here's another way to calculate the probability distribution of the sum of two dice by hand using convolutions.
To keep the example really simple, we're going to calculate the probability distribution of the sum of a three-sided die (d3) whose random variable we will call X and a two-sided die (d2) whose random variabl... | null | CC BY-SA 2.5 | null | 2010-10-18T07:19:52.593 | 2010-10-18T07:19:52.593 | null | null | null | null |
3695 | 1 | 3697 | null | 15 | 25795 | When plotting a boxplot with python matplotblib, the lines halfway the plot is the median of the distribution.
Is there a possibility to instead have the line at the average. Or to plot it next to it in a different style.
Also, because it is common for the line to be the median, will it really confuse my readers if I m... | Show average instead of median in boxplot | CC BY-SA 2.5 | null | 2010-10-18T07:51:12.160 | 2010-10-18T12:02:05.757 | 2010-10-18T12:02:05.757 | 8 | 190 | [
"data-visualization",
"python",
"matplotlib",
"boxplot"
] |
3696 | 2 | null | 3695 | 20 | null | To answer your second question: Yes, I think it will be confusing to put the line at the mean instead of the median. The precise rules controlling the length of the 'whiskers' (if any) and treatment of outliers vary, but everyone keeps to Tukey's use of the box as displaying the median and lower and upper quartiles. Fo... | null | CC BY-SA 2.5 | null | 2010-10-18T08:05:49.140 | 2010-10-18T08:28:14.397 | 2010-10-18T08:28:14.397 | 449 | 449 | null |
3697 | 2 | null | 3695 | 26 | null | This code makes the boxplots then places a circle marking the mean for each box. You can use a different symbol by specifying the [marker](http://matplotlib.sourceforge.net/api/pyplot_api.html?highlight=scatter#matplotlib.pyplot.scatter) argument in the call to `scatter`.
```
import numpy as np
import pylab
# 3 boxes... | null | CC BY-SA 2.5 | null | 2010-10-18T08:40:10.110 | 2010-10-18T08:40:10.110 | null | null | 251 | null |
3698 | 1 | 3770 | null | 2 | 1002 | I am trying to understand a published analysis. This is the data of interest:
D1>0 D1<0
D2>0 7 2 9
D2<0 9 15 24
total 16 17 33
The author notes that 17/33 is 51.5% and states:
>
"we expect about 50% of the D1's to be negative, and that is what we actually observe here (z=.08,... | Understanding published data: z-ratio for proportions | CC BY-SA 2.5 | null | 2010-10-18T09:08:51.310 | 2010-10-28T10:40:02.167 | 2010-10-28T10:40:02.167 | null | 1614 | [
"contingency-tables"
] |
3700 | 2 | null | 3636 | 0 | null | Based on Shane's answer I entered some data into Excel. What I came up with is that I need to multiple the square of the different from the mean by the number of seconds until the next sample. This assumes the reading was steady just until the next reading.
| null | CC BY-SA 2.5 | null | 2010-10-18T12:24:04.177 | 2010-10-18T12:24:04.177 | null | null | 1595 | null |
3701 | 2 | null | 3698 | 2 | null | I'm not sure the discrepancy is worth worrying about. The exact p-value is clearly 1 for a 2-sided test of p=0.5 given 17 positive responses out of 33, as there's no integer closer to 33/2 than 17. With small or moderate N as here there's no good reason for not doing the exact test (even without a PC, as the cdf of the... | null | CC BY-SA 2.5 | null | 2010-10-18T12:27:47.530 | 2010-10-18T12:27:47.530 | null | null | 449 | null |
3702 | 1 | null | null | 1 | 1132 | I am newbee and i am trying with functional data analysis. I have a 8x11 matrix data, how can i input into R as an object in this form:
```
$hgtm
boy01 boy02 boy03 boy04 boy05
1 81.3 76.2 76.8 74.1 74.2
1.25 84.2 80.4 79.8 78.4 76.3
1.5 86.4 83.2 82.6 82.6 78.3
1.75 88.9 85.4 84.7 85.4 ... | Data manipulation in R for functional data analysis | CC BY-SA 2.5 | null | 2010-10-18T13:48:33.180 | 2019-12-03T03:17:04.480 | 2019-12-03T03:17:04.480 | 11887 | 1615 | [
"r",
"panel-data",
"functional-data-analysis"
] |
3703 | 2 | null | 3702 | 3 | null | It looks like you want to use a named list, since each object is of different dimensions above. Here is an example with some dummy data:
```
hgtm <- matrix(1:100, ncol=5)
hgtf <- matrix(1:100, ncol=5)
age <- 1:10
namvan <- list(hgtm=hgtm, hgtf=hgtf, age=age)
```
Now you can reference each object by name:
```
> str(na... | null | CC BY-SA 2.5 | null | 2010-10-18T13:56:05.490 | 2010-10-18T13:56:05.490 | null | null | 5 | null |
3704 | 1 | 3712 | null | 3 | 5393 | I posted this question earlier and am rewriting it in hopes of getting some guidance. I am using a weighted regression (after propensity score matching) to obtain estimates of the effects of a treatment (treat) on an outcome for different income quintiles. I include interactions of the quintiles and the treatment in th... | Adding coefficients to obtain interaction effects - can I add standard errors? | CC BY-SA 2.5 | null | 2010-10-18T14:21:06.687 | 2010-10-18T22:08:54.650 | 2010-10-18T17:48:59.000 | 449 | 834 | [
"regression",
"standard-error",
"multivariate-analysis",
"stata"
] |
3705 | 2 | null | 3702 | 4 | null | As to how you would get the data into R and into one of these formats, we'd need to know more about what your input file looks like and the format that it is in. However, here are some general tips on formatting the type of data that you have for analysis in any system.
Singer (Applied Longitudinal Data Analysis) sugg... | null | CC BY-SA 2.5 | null | 2010-10-18T14:59:44.807 | 2010-10-18T18:24:25.100 | 2010-10-18T18:24:25.100 | 485 | 485 | null |
3706 | 2 | null | 3704 | -1 | null | I am quite sure it's fine to just add the standard errors on the two coefficients together. I can then obtain the t-stats by dividing the sum of the coefficients by the sum of the standard errors. If anyone disagrees please let me know.
| null | CC BY-SA 2.5 | null | 2010-10-18T15:09:48.413 | 2010-10-18T15:09:48.413 | null | null | 834 | null |
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