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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
10797 | 2 | null | 10789 | 12 | null | The issues regarding $\sigma$-algebras are mathematical subtleties, that do not really explain why or if we need a background space. Indeed, I would say that there is no compelling evidence that the background space is a necessity. For any probabilistic setup
$(E, \mathbb{E}, \mu)$ where $E$ is the sample space, $\mat... | null | CC BY-SA 3.0 | null | 2011-05-14T07:07:41.470 | 2011-05-14T17:19:49.897 | 2011-05-14T17:19:49.897 | 4376 | 4376 | null |
10798 | 1 | 10799 | null | 9 | 1836 | Direct Question: Are there any measures of auto-correlation for a sequence of observations of an (unordered) categorical variable?
Background: I'm using MCMC to sample from a categorical variable and I'd like a measure of how well the sampling method I've developed is mixing across the posterior distribution. I'm fam... | Measures of autocorrelation in categorical values of a Markov Chain? | CC BY-SA 3.0 | null | 2011-05-14T07:58:36.470 | 2011-08-10T14:15:42.663 | 2011-05-14T08:04:11.110 | 1499 | 1499 | [
"categorical-data",
"markov-chain-montecarlo",
"autocorrelation"
] |
10799 | 2 | null | 10798 | 6 | null | You can always choose one or several real valued functions of the categorical variables and look at the auto-correlation for the resulting sequence(s). You can, for instance, consider indicators of some subsets of the variables.
However, if I understood your question correctly, your sequence is obtained by an MCMC alg... | null | CC BY-SA 3.0 | null | 2011-05-14T08:26:49.843 | 2011-05-16T05:34:30.250 | 2011-05-16T05:34:30.250 | 4376 | 4376 | null |
10800 | 1 | 10804 | null | 8 | 2998 | I'm just having a look through Hadley's very excellent book about his ggplot2 R package.
He has some code to remove a linear trend in the diamonds dataset, like so:
```
d <- subset(diamonds, carat < 2.5 & rbinom(nrow(diamonds), 1, 0.2) == 1)
d$lcarat <- log10(d$carat)
d$lprice <- log10(d$price)
detrend <- lm(lprice ~... | How do I display detrended data from a linear regression? | CC BY-SA 3.0 | null | 2011-05-14T08:53:36.253 | 2011-08-24T00:34:05.280 | 2011-05-14T23:10:52.190 | 827 | 827 | [
"data-visualization"
] |
10801 | 2 | null | 10795 | 11 | null | Covariance matrix can represent relations between all variables while inverse covariance shows the relations of elements with their neighbors (as wikipedia said partial/pair wise relations).
I borrow the following example from [here](http://videolectures.net/gpip06_mackay_gpb/) in 24:10. Imagine 5 masses are connected ... | null | CC BY-SA 4.0 | null | 2011-05-14T09:07:25.183 | 2021-04-14T15:29:45.470 | 2021-04-14T15:29:45.470 | 102655 | 4581 | null |
10803 | 2 | null | 6883 | 1 | null | simply without using any toolbox box use "polyfit" and in consequence "polyval" function,
here is the example :
X could you you input data, and Y you response value or target value
You can make a n-degree polynomial by :
```
p = polyfit(x,y,n)
```
and fit the polynomial and get the interest output yd, based on fitt... | null | CC BY-SA 3.0 | null | 2011-05-14T11:34:41.720 | 2011-05-14T11:34:41.720 | null | null | 4581 | null |
10804 | 2 | null | 10800 | 2 | null | As for me, it is terribly confusing, especially while you can do much simpler thing -- calculate `price/carat` to get a price of one carat, which would be way easier to interpret.
| null | CC BY-SA 3.0 | null | 2011-05-14T12:49:02.853 | 2011-05-14T16:13:40.960 | 2011-05-14T16:13:40.960 | null | null | null |
10805 | 1 | null | null | 4 | 589 | What is an appropriate econometric model to describe count data with an upper bound?
| Regression model for count data with an upper bound | CC BY-SA 3.0 | null | 2011-05-14T14:05:14.403 | 2015-09-29T20:14:36.727 | 2015-07-06T23:06:35.227 | 22228 | 4496 | [
"regression",
"poisson-distribution",
"count-data",
"bounds"
] |
10806 | 1 | null | null | 10 | 808 | In practical application, I have witnessed often the following practice. One observes a pair $(x_t, y_t)$ over time. Under the assumption that they are linearly related, we regress one against the other using geometric weights rather than uniform ones, i.e., the OLS minimizes
$$\sum_{t=0}^\infty k^{t} (y_{T-t}- a x_{... | Justification for using geometric weights in linear regression | CC BY-SA 3.0 | null | 2011-05-14T15:38:30.307 | 2011-05-16T08:33:33.177 | 2011-05-15T02:21:41.113 | 30 | 30 | [
"regression",
"least-squares"
] |
10807 | 1 | 10808 | null | 5 | 16377 | In R, what is the difference between:
```
if(x>2 & x<3) ...
```
and
```
if(x>2 && x<3) ...
```
Similarly:
```
if(x<2 | x>3) ...
```
and
```
if(x<2 || x>3) ...
```
| Using and/or operators in R | CC BY-SA 3.0 | null | 2011-05-14T19:05:37.203 | 2011-05-15T16:59:09.517 | 2011-05-15T16:59:09.517 | null | 2310 | [
"r"
] |
10808 | 2 | null | 10807 | 10 | null | See `?"&"`: the single version does elementwise comparisons (for when you are doing logical operations on two vectors of the same length, e.g. if in your example x<-c(1.5,3.5). The other one works just like C++'s or java's &&: it only looks at the first element of each vector (this is typically an unexpected downside),... | null | CC BY-SA 3.0 | null | 2011-05-14T19:25:01.863 | 2011-05-14T19:25:01.863 | null | null | 4257 | null |
10809 | 2 | null | 10806 | 7 | null | "Linearly related" usually means
$$y_t = a x_t + b + \varepsilon_t$$
for constant $a$, $b$ and iid random errors $\varepsilon_t$, $t=0,1,\ldots,T$. One reason one would make an exponentially weighted OLS estimate is the suspicion that $a$ and $b$ might themselves be (slowly) varying with time, too. Thus we really th... | null | CC BY-SA 3.0 | null | 2011-05-14T19:26:33.187 | 2011-05-15T14:10:12.053 | 2011-05-15T14:10:12.053 | 919 | 919 | null |
10810 | 1 | null | null | 9 | 914 | Who created sampling distributions? I have looked everywhere and I'm writing a paper, but, so far, all I've been able to come up with is the theory and the definition. Please help me find the who, what, when,and where.
| Who first developed the idea of "sampling distributions"? | CC BY-SA 3.0 | null | 2011-05-14T21:23:24.490 | 2011-10-04T23:35:14.370 | 2011-10-04T23:35:14.370 | 183 | 4615 | [
"distributions",
"sampling",
"history"
] |
10811 | 2 | null | 10810 | 8 | null | If you refer to the term "sampling distribution," information is hard to find. But the concept is crucial to Jakob Bernoulli's (Switzerland) recognition that there is a distinction between that distribution and the population distribution itself, leading to his formulation (and proof) of a [Law of Large Numbers](http:... | null | CC BY-SA 3.0 | null | 2011-05-14T22:14:06.040 | 2011-05-14T22:14:06.040 | null | null | 919 | null |
10812 | 2 | null | 10089 | 1 | null | You could use [Kruskal–Wallis test](http://en.wikipedia.org/wiki/Kruskal%E2%80%93Wallis_one-way_analysis_of_variance) to perform a test for
$$
H_0: m_1 = m_2 = m_3 \> ,
$$
where $m_i$ is the median of the $i$th group.
If the $p$-value is $< 0.05$, you could conduct a post hoc test using Bonferroni correction.
Take... | null | CC BY-SA 3.0 | null | 2011-05-14T22:46:19.800 | 2011-05-15T00:18:32.437 | 2011-05-15T00:18:32.437 | 2970 | 4559 | null |
10813 | 2 | null | 10806 | 1 | null | I think in that you actually mean $k^{t}$ as your weight, or that $k>1$. If $0<k<1$ and we take $k^{-t}$ as the weight then $k^{-\infty}=\infty$. So this actually weights the present observation the least. For example, if we take $k=0.5$ then $k^{0}=1,\;k^{-1}=2,\;k^{-2}=4,\dots,k^{-20}\approx 10^{6}$, and so on.
Th... | null | CC BY-SA 3.0 | null | 2011-05-15T00:38:21.330 | 2011-05-16T08:33:33.177 | 2011-05-16T08:33:33.177 | 2392 | 2392 | null |
10814 | 2 | null | 10768 | 11 | null | I like the other answers, but nobody has mentioned the following yet. The event $\{U \leq t,\ V\leq t \}$ occurs if and only if $\{\mathrm{max}(U,V)\leq t\}$, so if $U$ and $V$ are independent and $W = \mathrm{max}(U,V)$, then $F_{W}(t) = F_{U}(t)*F_{V}(t)$ so for $\alpha$ a positive integer (say, $\alpha = n$) take ... | null | CC BY-SA 3.0 | null | 2011-05-15T00:52:55.247 | 2011-05-20T17:59:24.970 | 2011-05-20T17:59:24.970 | null | null | null |
10815 | 2 | null | 9685 | 0 | null | I believe you could use the "simultaneous confidence intervals" for doing multiple comparisons. The reference is Agresti et al. 2008 Simultaneous confidence intervals for comparing binomial parameters. Biometrics 64 1270-1275.
You could find the corresponding R code in [http://www.stat.ufl.edu/~aa/cda/software.html](h... | null | CC BY-SA 3.0 | null | 2011-05-15T01:12:35.080 | 2011-05-15T01:12:35.080 | null | null | 4559 | null |
10816 | 1 | null | null | 6 | 1778 | On page 9 in [http://jenni.uchicago.edu/Oxford2005/four_param_all_2005-08-07_csh.pdf](http://jenni.uchicago.edu/Oxford2005/four_param_all_2005-08-07_csh.pdf)
ATE - the average treatment effect is the expected gain from participation in a program for a random individual. For example, evaluate the impact of going to coll... | Heckman sample selection | CC BY-SA 3.0 | null | 2011-05-15T05:31:38.173 | 2017-11-24T18:01:18.490 | 2017-11-24T18:01:18.490 | 128677 | 4617 | [
"regression",
"econometrics",
"probit",
"heckman"
] |
10817 | 1 | null | null | 4 | 315 | Have anyone tried modeling number of phone calls using OLS?
The dataset is number of calls per months for each customer's account.
The dependent variable is number of calls or average number of calls, and the explanatory variables are customer specific variables including number of purchases, total spend and so on....
... | Modeling number of phone calls with OLS | CC BY-SA 3.0 | null | 2011-05-15T05:58:33.960 | 2011-05-16T05:51:19.287 | 2011-05-16T05:51:19.287 | 2116 | 4617 | [
"regression",
"least-squares"
] |
10818 | 2 | null | 10817 | 3 | null | Just from knowing that there are many zeros in the data, this would suggest to me that you use a zero-inflated poisson model - [a generalised linear model](http://en.wikipedia.org/wiki/Generalized_linear_model).
| null | CC BY-SA 3.0 | null | 2011-05-15T10:35:51.817 | 2011-05-15T10:35:51.817 | null | null | 656 | null |
10819 | 2 | null | 10711 | 4 | null | This is a really, really interesting question (indeed I'm somewhat in love with the idea of modelling the stack exchange sites in general).
On the issue of well-roundedness, one way to assess this would be through the tags that particular users tend to answer, and their distribution across sites. Examples may make this... | null | CC BY-SA 3.0 | null | 2011-05-15T11:44:02.263 | 2011-05-15T11:44:02.263 | null | null | 656 | null |
10820 | 2 | null | 10817 | 1 | null | • Since this is time series data you would be well advised to include some form of "time variable" in the model. This could be accomplished by including seasonal dummies and/or seasonal autoregressive structure. You might also have one or more Level Shifts and/or one or more trends in the data. You might have changes... | null | CC BY-SA 3.0 | null | 2011-05-15T12:26:30.877 | 2011-05-15T14:41:24.497 | 2011-05-15T14:41:24.497 | 3382 | 3382 | null |
10822 | 1 | 10826 | null | 3 | 105 | I am not able to find an existing name for the comparison function I wrote.
I have two lists of values which I want to compare. At first I used jaccard, but then I recognized I need to eliminate the length of the second list as it does not seem to have anything to say in my use-case. So I changed jaccard to:
$$
sim(L1... | Is there a name for the function | L1 sect L2 | / | L1 | | CC BY-SA 3.0 | null | 2011-05-15T13:48:01.293 | 2011-05-15T15:40:04.573 | null | null | 4350 | [
"distance-functions"
] |
10823 | 2 | null | 10407 | 0 | null | I agree with some comments, in that the Poisson approximation sounds nice here (not a 'crude' approximation). It should be asympotically exact, and it seems the most reasonable thing to do, as an exact analytic solutions seems difficult.
As an intermediate alternative, (if you really need it) I suggest I fisrt order co... | null | CC BY-SA 3.0 | null | 2011-05-15T14:13:09.697 | 2011-05-15T14:13:09.697 | null | null | 2546 | null |
10824 | 2 | null | 10774 | 4 | null | 1) What is your spatial explaining variable? Looks like the x*y plane would be a poor model for the spatial effect.
>
i=c(1,3,5,7,8,11,14,15,16,17,18,22,23,25,28,30,31,32,35,36,39,39,41,42)
l=rep(NA,42)[i];l[i]=level
r=rep(NA,42)[i];r[i]=response... | null | CC BY-SA 3.0 | null | 2011-05-15T14:33:44.393 | 2011-05-15T20:09:18.353 | 2011-05-15T20:09:18.353 | 2456 | 2456 | null |
10825 | 2 | null | 10407 | 0 | null | Just a word of explanation: Part out of curiosity, part for lack of a better more theoretical method, i approached the problem in a completely empirical/inductive way. I'm aware that there is the risk of getting stuck in a dead end without gaining much insight, but i thought, i'll just present what i got so far anyway,... | null | CC BY-SA 3.0 | null | 2011-05-15T15:05:05.490 | 2011-05-15T15:05:05.490 | null | null | 4360 | null |
10826 | 2 | null | 10822 | 2 | null | From where I stand it looks like the conditional relative frequency of $L2$ given $L1$.
| null | CC BY-SA 3.0 | null | 2011-05-15T15:40:04.573 | 2011-05-15T15:40:04.573 | null | null | 4376 | null |
10827 | 1 | null | null | 10 | 4178 | I'm investigating interplay between two variables ($x_1$ and $x_2$). There is a great deal of linear correlation between these variables with $r>0.9$. From the nature of the problem I cannot say anything about the causation (whether $x_1$ causes $x_2$ or the other way around). I would like to study deviations from the ... | Does the variable order matter in linear regression | CC BY-SA 3.0 | null | 2011-05-15T15:48:13.567 | 2011-05-16T15:13:08.160 | null | null | 4622 | [
"regression",
"outliers",
"linear-model"
] |
10828 | 2 | null | 10827 | 3 | null | It surely can (actually, it even matters with regard to the assumptions on your data - you only make assumptions about the distribution of the outcome given the covariate). In this light, you might look up a term like "inverse prediction variance". Either way, linear regression says nothing about causation! At best, yo... | null | CC BY-SA 3.0 | null | 2011-05-15T15:53:39.430 | 2011-05-15T15:53:39.430 | null | null | 4257 | null |
10829 | 2 | null | 10827 | 3 | null | To make the case symmetrical, one may regress the difference between the two variables ($\Delta x$) vs their average value.
| null | CC BY-SA 3.0 | null | 2011-05-15T16:53:06.540 | 2011-05-15T17:01:17.343 | 2011-05-15T17:01:17.343 | 1496 | 1496 | null |
10830 | 2 | null | 10407 | 5 | null | Partition the outcomes by the frequency of occurrences $x$ of the "double outcome", $0 \le x \le t$. Conditional on this number, the distribution of the remaining $t-x$ outcomes is multinomial across $n-1$ equiprobable bins. Let $p(t-x, n-1, x)$ be the chance that no bin out of $n-1$ equally likely ones receives more... | null | CC BY-SA 3.0 | null | 2011-05-15T19:07:36.967 | 2011-05-15T19:07:36.967 | null | null | 919 | null |
10831 | 2 | null | 10827 | 0 | null | Your x1 and x2 variables are collinear. In the presence of multicollinearity, your parameter estimates are still unbiased, but their variance is large, i.e., your inference on the significance of the parameter estimates is not valid, and your prediction will have large confidence intervals.
Interpretation of the param... | null | CC BY-SA 3.0 | null | 2011-05-15T19:50:27.713 | 2011-05-15T19:50:27.713 | null | null | 4617 | null |
10832 | 1 | 10834 | null | 6 | 4657 | I've written a small hierarchical clustering algorithm (for better or for worse). I'd like a quick way of visualizing it, any tooling ideas?
| Anyone know of a simple dendrogram visualizer? | CC BY-SA 3.0 | null | 2011-05-15T19:51:12.777 | 2011-12-16T06:06:24.933 | 2011-05-15T20:18:15.013 | null | 4623 | [
"data-visualization",
"dendrogram"
] |
10834 | 2 | null | 10832 | 5 | null | [TreeView](http://taxonomy.zoology.gla.ac.uk/rod/treeview.html) -- it is not a statistical tool, but it is very light and I have a great sentiment to it; and it is easy to make output to [Newick format](http://en.wikipedia.org/wiki/Newick_format), which TV eats without problems.
More powerful solution is to use R, but ... | null | CC BY-SA 3.0 | null | 2011-05-15T20:17:01.317 | 2011-05-15T20:17:01.317 | null | null | null | null |
10835 | 2 | null | 10554 | 3 | null | Homogeneity Definition: To start, let's define homogeneity as the degree to which households grouped in same area are like one another for some attribute.
## MAUP Approach
Paraphrasing the stated problem: We are uncertain how homogeneity changes as we decrease the spatial resolution of the design of how we group hou... | null | CC BY-SA 3.0 | null | 2011-05-15T20:20:23.437 | 2011-05-15T20:31:46.160 | 2011-05-15T20:31:46.160 | 4329 | 4329 | null |
10838 | 1 | null | null | 32 | 221823 | I wonder if there is a simple way to produce a list of variables using a for loop, and give its value.
```
for(i in 1:3)
{
noquote(paste("a",i,sep=""))=i
}
```
In the above code, I try to create `a1`, `a2`, `a3`, which assign to the values of 1, 2, 3. However, R gives an error message. Thanks for your help.
| Produce a list of variable name in a for loop, then assign values to them | CC BY-SA 3.0 | null | 2011-05-16T00:17:52.007 | 2013-08-15T13:26:48.480 | 2011-05-16T04:18:00.890 | 159 | 4625 | [
"r"
] |
10839 | 2 | null | 10838 | 46 | null | Your are looking for `assign()`.
```
for(i in 1:3){
assign(paste("a", i, sep = ""), i)
}
```
gives
```
> ls()
[1] "a1" "a2" "a3"
```
and
```
> a1
[1] 1
> a2
[1] 2
> a3
[1] 3
```
Update
I agree that using loops is (very often) bad R coding style (see discussion above). Using `list2env()` (tha... | null | CC BY-SA 3.0 | null | 2011-05-16T00:51:17.090 | 2011-05-16T11:53:29.887 | 2011-05-16T11:53:29.887 | 307 | 307 | null |
10840 | 2 | null | 206 | 1 | null | In the case of database, we would always store the data in discrete even the nature of the data is continuous. Why should I emphasize the nature of data? We should take the distribution of data that could help us to analyze the data. IF the nature of data is continuous, I suggest you to use them by continuous analysis.... | null | CC BY-SA 3.0 | null | 2011-05-16T02:19:18.120 | 2011-05-16T02:19:18.120 | null | null | 4588 | null |
10841 | 1 | 10843 | null | 7 | 462 | If you're modelling a proportion response against numerous predictors that are also proportions, is it necessary to transform the response if the standard OLS model is seemingly well behaved?
By well behaved I mean:
- None of the fitted values are outside the range [0,1] (In fact they are fairly accurate)
- Residual... | Is it necessary to perform a transformation on proportion data if it's reasonably well behaved? | CC BY-SA 3.0 | null | 2011-05-16T02:23:47.057 | 2011-05-16T04:33:27.030 | null | null | 845 | [
"modeling",
"data-transformation",
"proportion"
] |
10842 | 1 | 10844 | null | 5 | 4982 | A simple question.
If $Y=\frac{1}{X}$ and I know $f_X(x)$, is it true that $E(Y) = E(1/X) = \int_{-\infty}^\infty \frac{1}{x}f_X(x) dx$?
| Expected value of a transformation | CC BY-SA 3.0 | null | 2011-05-16T04:32:57.500 | 2011-05-16T15:28:28.480 | 2011-05-16T09:37:28.880 | null | 4627 | [
"data-transformation",
"expected-value"
] |
10843 | 2 | null | 10841 | 4 | null | It depends. If your goal is prediction, then you may not need to do any gymnastics to get a more theoretically sound model if the one in hand does well. But of course you should be always be aware that a good-fitting model to present data may not perform well on new data. You can try to get a feel for that using cross ... | null | CC BY-SA 3.0 | null | 2011-05-16T04:33:27.030 | 2011-05-16T04:33:27.030 | null | null | 26 | null |
10844 | 2 | null | 10842 | 6 | null | Yes. In general if $X\sim f(x)$ then for a function $g(x)$ you have $E(g(X)) = \int g(x)f(x)dx$. You can verify this for simple cases by deriving the distribution of the transformed variable. The completely general result takes some more advanced math which you can probably safely avoid :)
| null | CC BY-SA 3.0 | null | 2011-05-16T04:37:24.950 | 2011-05-16T04:37:24.950 | null | null | 26 | null |
10845 | 2 | null | 10838 | 23 | null | If the values are in vector, the loop is not necessary:
```
vals <- rnorm(3)
n <- length(vals)
lhs <- paste("a", 1:n, sep="")
rhs <- paste("vals[",1:n,"]", sep="")
eq <- paste(paste(lhs, rhs, sep="<-"), collapse=";")
eval(parse(text=eq))
```
As a side note, this is the reason why I love R.
| null | CC BY-SA 3.0 | null | 2011-05-16T05:46:43.297 | 2013-08-15T13:26:48.480 | 2013-08-15T13:26:48.480 | 7290 | 2116 | null |
10846 | 2 | null | 10816 | 5 | null |
- The answer is yes, you do not need to use the parameters of inverse Mills ratios. But you must include them in the regression nevertheless, or your other parameters will be biased.
- According to the article yes. Although if different variables are statistically significant in different regression there is no probl... | null | CC BY-SA 3.0 | null | 2011-05-16T06:11:15.550 | 2011-05-16T06:11:15.550 | null | null | 2116 | null |
10847 | 1 | 10848 | null | 8 | 920 | What kind of function is:
$f_X(x) = 2 \lambda \pi x e^{-\lambda \pi x ^2}$
Is this a common distribution? I am trying to find a confidence interval of $\lambda$ using the estimator $\hat{\lambda}=\frac{n}{\pi \sum^n_{i=1} X^2_i}$ and I am struggling to prove if this estimator has Asymptotic Normality.
Thanks
| What kind of distribution is $f_X(x) = 2 \lambda \pi x e^{-\lambda \pi x ^2}$? | CC BY-SA 3.0 | null | 2011-05-16T07:22:55.747 | 2012-12-11T16:09:46.907 | 2011-05-17T03:53:29.227 | 2116 | 4627 | [
"distributions",
"confidence-interval",
"estimation",
"self-study"
] |
10848 | 2 | null | 10847 | 10 | null | It is a square root of [exponential distribution](http://en.wikipedia.org/wiki/Exponential_distribution) with rate $\pi\lambda$. This means that if $Y\sim\exp(\pi\lambda)$, then $\sqrt{Y}\sim f_X$.
Since your estimate is [maximum likelihood estimate](http://en.wikipedia.org/wiki/Maximum_likelihood) it should be asympt... | null | CC BY-SA 3.0 | null | 2011-05-16T08:17:54.307 | 2011-05-16T08:24:30.760 | 2011-05-16T08:24:30.760 | 2116 | 2116 | null |
10849 | 2 | null | 10847 | 10 | null | Why do you care about asymptotics when the exact answer is just as simple (and exact)? I am assuming that you want asymptotic normality so that you can use the $\mathrm{Est}\pm z_{\alpha}\mathrm{StdErr}$ type of confidence interval
If you make the probability transformation $Y_{i}=X_{i}^{2}$ then you have an exponenti... | null | CC BY-SA 3.0 | null | 2011-05-16T10:22:52.400 | 2012-12-11T16:09:46.907 | 2012-12-11T16:09:46.907 | 17230 | 2392 | null |
10850 | 1 | 10853 | null | 6 | 114 | My current dataset has three conditions, and we've measured the activity levels of 10,000 genes in each condition. Replicated 8 times.
Using 10,000 linear models, we determine for each pair of conditions (ie for each of three contrasts) the number of genes with significantly different activity levels. This is [standard... | Comparing numbers of p-values from many linear models | CC BY-SA 3.0 | null | 2011-05-16T11:28:46.337 | 2011-05-16T12:12:57.250 | 2011-05-16T11:51:00.613 | 930 | 3773 | [
"genetics",
"microarray"
] |
10851 | 1 | null | null | 2 | 124 | I am trying to devise a simple algorithm to create a Bayesian prior from measurements obtained from time series data.
Firstly, I presuppose that the data can take on one of five possible "shapes" or "patterns" and then measure the Euclidean distance of each time series data point from its equivalent point in time in e... | Algorithm to create Bayesian priors from measurements | CC BY-SA 3.0 | null | 2011-05-16T11:42:56.027 | 2017-08-01T10:41:07.663 | 2017-08-01T10:41:07.663 | 11887 | 226 | [
"bayesian",
"modeling",
"prior"
] |
10852 | 1 | 10864 | null | 3 | 182 | I have annual returns and standard deviations for two funds, $r_{a}$, $r_{b}$, $\mathrm{SD}_{a}$ and $\mathrm{SD}_{b}$ but I do not have individual data, just the annual data. The annual correlation between the prices of the funds is 0.7. If I had the individual data I could use
$\mathrm{Cov}(A,B) =\sum_i (\bar{r}_{a} ... | Half annual covariance $\mathrm{Cov}(X,Y)_{\text{6 months}}$ from annual covariance? | CC BY-SA 3.0 | null | 2011-05-16T12:02:12.167 | 2011-05-16T15:59:16.300 | 2011-05-16T14:47:44.590 | 2970 | 2914 | [
"covariance"
] |
10853 | 2 | null | 10850 | 3 | null | Prior to answering your question - is the distribution of effect of the genes justify using a linear model (e.g: are they distributed more or less normally?)
Now to your question - I might offer to go a different way about it. It sounds like what you are asking for is to measure the correlation (e.g similarity of beh... | null | CC BY-SA 3.0 | null | 2011-05-16T12:12:57.250 | 2011-05-16T12:12:57.250 | null | null | 253 | null |
10855 | 1 | null | null | 5 | 414 | I have this long running experiment. Each time I run it I get a new goodness value, since the algorithm has random variables in it. So I need to report the mean and the std of some n runs. What should n be?
I need to be able to defend n based on some statistical ideas. Some kind of scientific reference (a book, a paper... | Number of times to run a lengthy experiment | CC BY-SA 3.0 | null | 2011-05-16T12:59:12.657 | 2011-05-18T03:09:31.003 | 2011-05-17T12:27:13.450 | 4629 | 4629 | [
"experiment-design"
] |
10856 | 1 | 10857 | null | 24 | 14633 | ... and why ?
Assuming $X_1$,$X_2$ are independent random-variables with mean $\mu_1,\mu_2$ and variance $\sigma^2_1,\sigma^2_2$ respectively. My basic statistics book tells me that the distribution of the $X_1-X_2$ has the following properties:
- $E(X_1-X_2)=\mu_1-\mu_2$
- $Var(X_1-X_2)=\sigma^2_1 +\sigma^2_2$
No... | What is the distribution of the difference of two-t-distributions | CC BY-SA 3.0 | null | 2011-05-16T13:52:03.117 | 2022-01-25T23:30:49.427 | 2015-11-28T00:05:10.073 | 805 | 264 | [
"distributions",
"degrees-of-freedom",
"t-distribution"
] |
10857 | 2 | null | 10856 | 16 | null | The sum of two independent t-distributed random variables is not t-distributed. Hence you cannot talk about degrees of freedom of this distribution, since the resulting distribution does not have any degrees of freedom in a sense that t-distribution has.
| null | CC BY-SA 3.0 | null | 2011-05-16T13:58:53.843 | 2011-05-16T13:58:53.843 | null | null | 2116 | null |
10858 | 1 | 10860 | null | 7 | 6335 | This is more of a general statistics question, though if it matters I'm writing PHP code.
Let's say I'm trying to compute the average value of a toy that is commonly bought and sold on the secondary market, and I have a set of price values culled both from auctions and from user-entered "price paid" data. The data poin... | Computing average value ignoring outliers | CC BY-SA 3.0 | null | 2011-05-16T14:01:19.187 | 2011-07-08T14:55:05.550 | 2011-05-17T17:02:35.693 | null | 4631 | [
"standard-deviation",
"outliers"
] |
10859 | 2 | null | 10858 | 2 | null | I originally posted this on SO before it was deleted:
[https://stats.stackexchange.com/](https://stats.stackexchange.com/) will probably help you better with this, and give a more comprehensive answer. I'm not a mathematician, but I suspect there are multiple ways to solve this issue.
As a programmer this is how I woul... | null | CC BY-SA 3.0 | null | 2011-05-16T14:13:37.913 | 2011-05-16T14:26:03.333 | 2017-04-13T12:44:39.283 | -1 | 4632 | null |
10860 | 2 | null | 10858 | 10 | null | In boxplots, values that are more than 1.5 times the IQR (interquartile range, difference between quartile 1 and 3) away from (as in: in the direction away from the median) the quartiles are typically considered outliers.
I cannot say whether this is an appropriate measure for your data, though...
| null | CC BY-SA 3.0 | null | 2011-05-16T14:22:43.363 | 2011-05-16T14:22:43.363 | null | null | 4257 | null |
10861 | 2 | null | 10827 | 3 | null | Standard regression minimizes the vertical distance between the points and the line, so switching the 2 variables will now minimize the horizontal distance (given the same scatterplot). Another option (which goes by several names) is to minimize the perpendicular distance, this can be done using principle components.
... | null | CC BY-SA 3.0 | null | 2011-05-16T15:13:08.160 | 2011-05-16T15:13:08.160 | null | null | 4505 | null |
10862 | 2 | null | 10842 | 1 | null | Another approach if you are happy with a numerical estimate (as opposed to the theorectical exact value) is to generate a bunch of data from the distribution, do the transformation, then take the mean of the transformed data as the estimate of the expected value. This avoids integration which can be nice in ugly cases... | null | CC BY-SA 3.0 | null | 2011-05-16T15:28:28.480 | 2011-05-16T15:28:28.480 | null | null | 4505 | null |
10864 | 2 | null | 10852 | 2 | null | The [Wikipedia entry for covariance](http://en.wikipedia.org/wiki/Covariance) is good for learning the rules for manipulating and calculating covariance. The [Wikipedia entry on variance](http://en.wikipedia.org/wiki/Variance) is good for that special case.
In your particular case, you can solve your problem fairly ea... | null | CC BY-SA 3.0 | null | 2011-05-16T15:59:16.300 | 2011-05-16T15:59:16.300 | null | null | 1146 | null |
10865 | 2 | null | 10774 | 10 | null | 1) You can model spatial correlation with the `nlme` library; there are several possible models you might choose. See pages 260-266 of Pinheiro/Bates.
A good first step is to make a variogram to see how the correlation depends on distance.
```
library(nlme)
m0 <- gls(response ~ level, data = layout)
plot(Variogram(m... | null | CC BY-SA 3.0 | null | 2011-05-16T16:09:29.857 | 2011-05-16T19:56:57.893 | 2011-05-16T19:56:57.893 | 3601 | 3601 | null |
10867 | 1 | null | null | 4 | 1342 | I have 3 experiments, where some quantity was measured 3 times. Thus, 3 biological replicas, 3 technical replicas in each biological replica, 9 measurements in total. I need to answer the following questions:
- For each biological replica: are my 3 measurements consistent? To do this, I'm using Dixon's Q test.
- For ... | Tests for consistent measurements and outliers | CC BY-SA 3.0 | null | 2011-05-16T17:36:32.373 | 2011-05-16T20:12:36.277 | null | null | 4337 | [
"repeated-measures",
"outliers",
"reproducible-research"
] |
10868 | 2 | null | 10858 | 0 | null | Perhaps a robust estimator like [RANSAC](http://en.wikipedia.org/wiki/RANSAC) could be used here.
| null | CC BY-SA 3.0 | null | 2011-05-16T18:31:08.690 | 2011-05-16T18:31:08.690 | null | null | 4360 | null |
10869 | 2 | null | 10858 | 0 | null | hope this helps
Simplistic approaches , as suggested here , often fail to their lack of generality. In general you may have a series that has multiple trends and/or multiple levels thus to detect anomalies one has to "control" for these effects. Additionally there may be a seasonal effect that may have started in the l... | null | CC BY-SA 3.0 | null | 2011-05-16T18:58:16.157 | 2011-05-16T20:39:41.933 | 2011-05-16T20:39:41.933 | 3382 | 3382 | null |
10870 | 2 | null | 10858 | 5 | null | You could consider using a [trimmed mean](http://en.wikipedia.org/wiki/Truncated_mean). This would involve discarding, say, the highest 10% of values and the lowest 10% of values, regardless of whether you consider them to be bad.
| null | CC BY-SA 3.0 | null | 2011-05-16T19:10:27.770 | 2011-05-16T19:10:27.770 | null | null | 3835 | null |
10871 | 1 | 10879 | null | 10 | 394 | I am interested to know what level of statistics kids are learning in different countries around the world. Could you please suggest data/links that shed light on what is happening in this regards?
I'll start. Israel: The students who are taking advanced math study more or less - mean, sd, histogram, normal distributio... | Children's statistical education in different countries? | CC BY-SA 3.0 | null | 2011-05-16T19:35:27.120 | 2017-02-13T16:29:58.350 | 2017-02-13T16:29:58.350 | 22468 | 253 | [
"dataset",
"teaching"
] |
10872 | 2 | null | 10867 | 1 | null | It seems highly unlikely that there would be a test that, based upon three observations, decides whether one is an outlier! The fact that 'the other two' are closer could just as well be an anomaly.
At best, you could use Dixon's Q test to find out whether the largest/smallest value of your 9 observations is an outlier... | null | CC BY-SA 3.0 | null | 2011-05-16T20:12:36.277 | 2011-05-16T20:12:36.277 | null | null | 4257 | null |
10873 | 2 | null | 10795 | 40 | null | There are basically two things to be said. The first is that if you look at the density for the multivariate normal distribution (with mean 0 here) it is proportional to
$$\exp\left(-\frac{1}{2}x^T P x\right)$$
where $P = \Sigma^{-1}$ is the inverse of the covariance matrix, also called the precision. This matrix is p... | null | CC BY-SA 3.0 | null | 2011-05-16T20:24:14.100 | 2011-05-16T20:24:14.100 | null | null | 4376 | null |
10874 | 1 | null | null | 8 | 472 | What are important/notable publishing houses for books in statistics?
When I come across a book published at O'Reilly or Springer I imagine its quality will be high. What other notable publishing houses are out there (for statistics books)? Any recommendation on a way to find out? (I'd imagine we could check it if we... | High quality publishing house for books in the field of statistics | CC BY-SA 3.0 | null | 2011-05-16T20:41:15.513 | 2011-10-30T03:14:35.967 | 2011-10-30T03:14:35.967 | 5256 | 253 | [
"references"
] |
10875 | 2 | null | 10871 | 5 | null | Good question.
For my answer, I'll talk about Ireland.
In Senior Cycle (16-18 years) students study very basic statistics, mean, histograms, standard deviation. Basic probability is covered (completely seperately). Calculus, up to the level of integration by parts. Matrices (only 2*2) are an option on the Higher level... | null | CC BY-SA 3.0 | null | 2011-05-16T20:51:58.830 | 2011-05-16T20:51:58.830 | null | null | 656 | null |
10879 | 2 | null | 10871 | 9 | null | Statistics eduction in the US is in flux, in no small part because we now expect even grade school students (ages 5-12) to become proficient not only with fundamental concepts of statistical thinking, but also with techniques of data summary and presentation that many of their teachers do not even know!
For an authorit... | null | CC BY-SA 3.0 | null | 2011-05-16T22:32:27.220 | 2011-05-16T22:32:27.220 | null | null | 919 | null |
10880 | 1 | null | null | 3 | 155 | I hope this is a stats question :) There's a game (RftG) with cards, each of which has various attributes (generally bonuses of some kind) and a price. I'm wondering if there's some kind of technique for determining the average value of each attribute, and therefore calculating the expected cost of each card as a whole... | Calculating the value of attributes of cards in a card game | CC BY-SA 3.0 | null | 2011-05-16T23:24:57.510 | 2011-06-16T01:59:02.513 | null | null | 4636 | [
"bayesian",
"games",
"nonparametric",
"multiple-comparisons"
] |
10881 | 2 | null | 10880 | 1 | null | Regression analysis could be used for this sort of thing. But it wouldn't be perfect. For one reason, regression analysis is meant for random data, and the prices fixed to playing cards is definitely not random.
There's not really a standard method here. One reason is that certain abilities, in different contexts becom... | null | CC BY-SA 3.0 | null | 2011-05-17T00:25:11.637 | 2011-05-17T01:03:57.033 | 2011-05-17T01:03:57.033 | 4637 | 4637 | null |
10882 | 1 | null | null | 3 | 856 | I'm not a statician/mathematician, so please be gentle with me
This is a cross-post [from Stack Overflow](https://stackoverflow.com/questions/6021170/formula-to-discard-items-by-votes-lower-bound-of-wilson-score-confidence-interva). I'm working with a site that lets the users create 'sections'. Each section have multip... | Formula to discard items by votes (Lower bound of Wilson score confidence interval) | CC BY-SA 3.0 | null | 2011-05-17T01:32:50.910 | 2011-05-20T17:39:28.577 | 2017-05-23T12:39:26.143 | -1 | 4638 | [
"confidence-interval"
] |
10883 | 2 | null | 10394 | -3 | null | I guess Gaussian efficiency is something related to computation cost.
The efficiency of Gaussian adaptation relies on the theory of information due to Claude E. Shannon. When an event occurs with probability P, then the information −log(P) may be achieved. For instance, if the mean fitness is P, the information gained ... | null | CC BY-SA 3.0 | null | 2011-05-17T01:53:56.790 | 2011-05-17T01:53:56.790 | null | null | 4286 | null |
10884 | 1 | null | null | 8 | 34906 | I received this elementary question by email:
>
In a regression equation am I correct
in thinking that if the beta value is
positive the dependent variable has
increased in response to greater use
of the independent variable, and if
negative the dependent variable has
decreased in response to an increase
... | Interpretation of positive and negative beta weights in regression equation | CC BY-SA 3.0 | null | 2011-05-17T05:26:40.100 | 2021-06-24T05:16:22.860 | null | null | 183 | [
"regression",
"causality"
] |
10885 | 2 | null | 10884 | 7 | null | In explaining the meaning of regression coefficient I found that the following explanation very useful. Suppose we have the regression
$$Y=a+bX$$
Say $X$ changes by $\Delta X$ and $Y$ changes by $\Delta Y$. Since we have the linear relationship we have
$$Y+\Delta Y= a+ b(X+\Delta X)$$
Since $Y=a+bX$ we get that
$$\... | null | CC BY-SA 3.0 | null | 2011-05-17T05:36:19.520 | 2011-05-17T06:03:25.247 | 2011-05-17T06:03:25.247 | 2116 | 2116 | null |
10886 | 1 | null | null | 5 | 1662 | I'm wondering how to proceed to perform Canonical Correspondence Analysis and Multiple Correspondence Analysis in R on the following three-way contingency table
```
Eco_region3 Eco_region4 Species Freq
A A1 S1 10
A A1 S2 12
A A1 S3 8... | Correspondence analysis for a three-way contingency table | CC BY-SA 3.0 | null | 2011-05-17T06:24:48.690 | 2013-11-16T07:49:56.723 | 2011-05-17T17:10:47.323 | 3903 | 3903 | [
"r",
"contingency-tables",
"correspondence-analysis"
] |
10887 | 2 | null | 10884 | 7 | null | As @gung notes, there are varying conventions regarding the meaning of ($\beta$, i.e., "beta"). In the broader statistical literature, beta is often used to represent unstandardised coefficients. However, in psychology (and perhaps other areas), there is often a distinction between b for unstandardised and beta for sta... | null | CC BY-SA 3.0 | null | 2011-05-17T06:33:30.290 | 2013-07-07T04:23:19.687 | 2013-07-07T04:23:19.687 | 183 | 183 | null |
10889 | 1 | null | null | 3 | 59 | Say I want to classify my data into two categories. I am pretty sure that my data has been generated by two mixtures of Gaussians -- on has a bimodal and one a trimodal form.
I then train the generating models $p(x|c_1)$ and $p(x|c_2)$ and combine them via Bayes' theorem into a classifier of the form $p(c_i|x)$.
Since ... | Generative modelling: what if the generating models have very different "quality of fit" | CC BY-SA 3.0 | null | 2011-05-17T08:42:53.303 | 2011-05-17T08:42:53.303 | null | null | 2860 | [
"classification",
"generative-models"
] |
10890 | 1 | 11133 | null | 14 | 33811 | What is the difference between [endogeneity](http://en.wikipedia.org/wiki/Endogeneity_%28economics%29) and unobserved heterogeneity? I know that endogeneity comes for example from omitted variables? But as far as I understand, unobserved heterogeneity causes the same problem. But where exactly lays the difference betwe... | Endogeneity versus unobserved heterogeneity | CC BY-SA 3.0 | null | 2011-05-17T09:38:43.160 | 2018-04-22T17:18:49.153 | 2011-05-17T10:16:50.513 | 2116 | 4496 | [
"regression",
"assumptions"
] |
10891 | 2 | null | 10271 | -1 | null | After a friend of mine pointed me into the direction of clustering algorithms, I stumbled across [DBSCAN](http://en.wikipedia.org/wiki/DBSCAN) which builds clusters in n-dimensional space according to two predefined parameters. The basic idea is density-based clustering, i.e. dense regions form clusters. Outliers are r... | null | CC BY-SA 3.0 | null | 2011-05-17T14:09:26.937 | 2011-05-18T08:32:24.623 | 2011-05-18T08:32:24.623 | 4446 | 4446 | null |
10892 | 1 | null | null | 7 | 1110 | I'm trying to run R's multidimensional scaling algorithm, `cmdscale`, on roughly 2,200 variables, i.e. a 2,200x2,200 distance matrix. It's taking forever (about a day so far). I don't know much about the algorithm used under the hood. What's its big-O notation? Is a more efficient, even if only approximate, algorit... | Big-O Scaling of R's cmdscale() | CC BY-SA 3.0 | null | 2011-05-17T14:16:14.383 | 2011-05-17T18:24:11.260 | null | null | 1347 | [
"r",
"algorithms",
"multidimensional-scaling"
] |
10893 | 1 | null | null | 2 | 329 | I'm developing an application in which users can create 'sections' (à la subreddit in reddit), in which items/posts can be created and voted with a thumbs-up/down system.
[A great article](http://www.evanmiller.org/how-not-to-sort-by-average-rating.html) guided me on how to sort these votes so that an item with a 100% ... | Formula to discard items by votes (Lower bound of Wilson score confidence interval) | CC BY-SA 3.0 | 0 | 2011-05-16T17:39:58.170 | 2011-06-02T22:10:16.550 | 2011-06-02T22:10:16.550 | null | 4638 | [
"confidence-interval"
] |
10894 | 2 | null | 10882 | 4 | null | Perhaps you could extend the idea of using the lower bound of the confidence interval for sorting: you could throw away items that have a low upper bound. The items with only a few votes will have pretty high upper bounds; the lowest upper bounds will correspond to the lowest "quality" items.
| null | CC BY-SA 3.0 | null | 2011-05-16T18:27:42.933 | 2011-05-16T18:27:42.933 | null | null | 279 | null |
10897 | 1 | null | null | 6 | 398 | I would like to compute the probability distribution for the length of the fragments which I would obtain by fragmenting a linear rod of length $L$ in the following way:
- I choose at random (uniformly) $n$ breakpoints
- I cut the rod at those breakpoints, creating $(n+1)$ fragments.
Now, while it is easy to see t... | Probability distribution of fragment lengths | CC BY-SA 3.0 | null | 2011-05-17T16:06:19.143 | 2011-05-18T16:41:17.197 | 2011-05-17T16:38:47.773 | 2970 | 4642 | [
"probability"
] |
10899 | 2 | null | 10897 | 0 | null | Let $\{X_i\}$ be the locations of the cuts.
I'd approach this problem by finding the order statistics $\{Y_i\}$ so that $Y_1$ would be the location of the leftmost cut. Then I'd calculate the probability distributions of the differences between the variables $Y_i-Y_{i-1}$. Don't forget to also calculate $Y_1-0$ and $L-... | null | CC BY-SA 3.0 | null | 2011-05-17T16:17:19.443 | 2011-05-17T16:17:19.443 | null | null | 4637 | null |
10900 | 1 | null | null | 4 | 3195 | I have to generate random numbers for my algorithm based on probability distributions. I want a distribution which has heavy tails and is unskewed, which can produce numbers far away from location parameter. There should be a parameter to control the tail heaviness (e.g., like levy distribution where alpha determines t... | Long tailed distributions for generating random numbers with parameters to control tail heaviness | CC BY-SA 3.0 | null | 2011-05-17T17:13:30.487 | 2012-07-13T07:12:05.503 | 2011-06-08T16:11:41.350 | 183 | 4319 | [
"distributions",
"randomness"
] |
10901 | 2 | null | 10892 | 4 | null | You should be able to test this on your system via Monte Carlo. For example:
```
try.cmdscale <- function(n) {
x <- matrix(rnorm(n*n) ** 2,nrow=n,ncol=n);
took.time <- system.time(cmdscale(x))
}
#repeat multiple times and take the median of the user.self + sys.self
#probably a better way to do timing?
multi.cm... | null | CC BY-SA 3.0 | null | 2011-05-17T17:33:50.177 | 2011-05-17T17:33:50.177 | null | null | 795 | null |
10902 | 2 | null | 10855 | 4 | null | If we assume an underlying success probability $\theta$ for a specific training set and number of points used for image analysis, we can expect a binomial distribution for the number of successful object recognitions $k$ out of $n$ runs:
$P(k|\theta,n)=\left(n\atop k\right)\theta^k (1-\theta)^{n-k}$
What we're actually... | null | CC BY-SA 3.0 | null | 2011-05-17T17:50:45.363 | 2011-05-18T03:09:31.003 | 2011-05-18T03:09:31.003 | 4360 | 4360 | null |
10903 | 2 | null | 10892 | 3 | null | Ironically, the MDS actually wasn't my problem. The preprocessing I was doing was the issue. I'm used to coding in lower-level languages and forgot how slow looping is in R. I rewrote the preprocessing code using vector ops and the MDS actually only takes a few seconds.
| null | CC BY-SA 3.0 | null | 2011-05-17T18:24:11.260 | 2011-05-17T18:24:11.260 | null | null | 1347 | null |
10904 | 1 | 10922 | null | 15 | 1780 | What are the pros and cons of learning about a distribution's properties algorithmically (via computer simulations) versus mathematically?
It seems like computer simulations can be an alternative learning method, especially for those new students who do not feel strong in calculus.
Also it seems that coding simulations... | What are the pros and cons of learning about a distribution algorithmically (simulations) versus mathematically? | CC BY-SA 3.0 | null | 2011-05-17T18:24:27.733 | 2017-02-20T10:35:22.387 | 2017-02-20T10:35:22.387 | 9964 | 4329 | [
"distributions",
"algorithms",
"teaching"
] |
10905 | 1 | 10909 | null | 10 | 44835 | I have searched for this online for hours but none of online posts is what I am looking for. My question is very easy to implement in SAS Proc mixed procedure but I am not sure how to do it in lme and/or lmer packages. Assume, I have a model,
$y = \mu + \alpha + \beta +\alpha\beta + e$,
where $\alpha$ is fixed but $\b... | How to specify random effects in lme? | CC BY-SA 3.0 | null | 2011-05-17T18:39:01.290 | 2011-05-17T20:15:56.673 | 2011-05-17T18:40:30.710 | 2116 | 4559 | [
"r",
"mixed-model",
"cross-validation",
"biostatistics"
] |
10906 | 2 | null | 10905 | 2 | null | It would help a lot if you provided a data.frame. Now it is not clear what is a grouping factor. I judge that it is $\beta$. Then in `lme` notation your model should be written as follows:
```
lme(y~a,random=~a|b, data=mydata)
```
| null | CC BY-SA 3.0 | null | 2011-05-17T19:01:07.637 | 2011-05-17T19:01:07.637 | null | null | 2116 | null |
10907 | 1 | 10944 | null | 5 | 2632 | I have data with standard error, included below for clarity,
```
X Y Error in Y
0.0105574 -28.831027 0.04422
0.0070382 -27.800385 0.04225
0.0052787 -27.314088 0.04209
0.0042229 -27.054207 0.04185
0.0035191 -27.000188 0.04143
0.0030164 -26.891275 0.04108
... | Obtaining standard error on a data point obtained from linear regression | CC BY-SA 3.0 | 0 | 2011-05-17T19:14:12.547 | 2017-11-12T21:25:20.757 | 2017-11-12T21:25:20.757 | 11887 | 4643 | [
"regression",
"least-squares",
"error"
] |
10909 | 2 | null | 10905 | 13 | null | Try this, it's a standard way to do a split plot. The notation `/` means that method is nested in day.
```
lme(level~method, random=~1|day/method, data=d)
```
| null | CC BY-SA 3.0 | null | 2011-05-17T20:15:56.673 | 2011-05-17T20:15:56.673 | null | null | 3601 | null |
10910 | 1 | null | null | 1 | 21770 | I have a 37-question scale, where each item belongs to one of four scales.
The items for each scale are interspersed.
How can I compute a scale score for each of the four scales?
I need to be able to analyse each subscale independently and then compare scale responses on and independent variable.
| How do I create subscales based on sets of items in SPSS | CC BY-SA 3.0 | null | 2011-05-17T20:50:11.410 | 2015-07-07T15:15:10.360 | 2011-05-18T04:59:26.770 | 183 | 4644 | [
"spss",
"scales"
] |
10912 | 2 | null | 10910 | 1 | null | This code divides the sum of scores (or of weighted scores) by
the number of valid values a given case has for that set of variables.
In this example, if a case has values only for verbal and quant, then
the score will be simply the sum of those divided by 2.
```
Compute SCORE = sum (.05* jobs, .15* exper, verbal, q... | null | CC BY-SA 3.0 | null | 2011-05-17T21:07:35.910 | 2011-05-17T21:07:35.910 | null | null | 2669 | null |
10913 | 2 | null | 10832 | 0 | null | While it's not a tool per-se, ascii art is a fairly safe option actually, and not as hard as it seems at first. It's not pretty, but it gets the point accross.
| null | CC BY-SA 3.0 | null | 2011-05-17T21:13:53.457 | 2011-05-17T21:13:53.457 | null | null | 4623 | null |
Subsets and Splits
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