# Tag Info

7

The results don't necessarily show that a gamma distribution is a good fit. You might as well as assume that a normal distribution must fit well because you have estimated a mean and a standard deviation. Your graphical assessment is already showing that a good (if not necessarily the best) gamma shape for these data is systematically different from your ...

7

$(X,Y)$ take on value $(0,2)$ and $(2,0)$ with equal probability $\frac{1}{2}$. Then, $$E[X] = 0\times\frac{1}{2} + 2\times\frac{1}{2} = 1,$$ and similarly $E[Y]=1$, while $$E[XY] = 2\times 0\times\frac{1}{2} + 0\times 2\times\frac{1}{2} = 0.$$ Added note: The $X$ and $Y$ in this example are not uncorrelated random variables (cf. the title of the question) ...

7

Indeed. Check out Anderson-Darling, Cramér-von Mises tests, etc. http://en.wikipedia.org/wiki/Anderson%E2%80%93Darling_test is a start. All these tests seem problematic to me for other reasons, not least because they don't tell you directly what is wrong. Increasingly with bigger datasets they reject hypotheses that are practically acceptable on the basis ...

6

Chi is a Greek letter. The canonical modern history references are Karl Pearson's introduction of the chi-square test in 1900 and R.A. Fisher's work in 1924, but there is ancient history too: F.R. Helmert in 1876 deserves more than a nod. http://jeff560.tripod.com/c.html is a good start, especially if other historical bits and pieces are of interest. It ...

6

Let's find out whether this is a good test or not. There's a lot more to it than just claiming it's bad or showing in one instance that it doesn't work well. Most tests work poorly in some circumstances, so often we are faced with identifying the circumstances in which any proposed test might possibly be a good choice. Description of the test Like any ...

5

You could try to directly estimate the CDF via a binomial rate smoother ? Here is an idealized example for x stemming from a normal distribution: ci = seq(from=-3,to=3,length=500) X = rnorm(500) Y = rep(NA, 500) for (i in 1:500) Y[i] = as.numeric(X[i] < ci[i] ) plot(ci,Y, type="s") library(mgcv) library(boot) fit=gam(Y~s(ci), ...

5

this paper might be of interest: http://arxiv.org/pdf/0906.4032v1.pdf It gives a nice summary of some frequentist and Bayesian approaches to the two sample problem, and discusses both the parametric and nonparametric cases. It might add something to the other answers to give a simple example. Say you have two data sets $\mathbf{x}$ and $\mathbf{y}$ where ...

4

Something like that? Dmax occurs at the value max.at. set.seed(12345) x <- rnorm(10000, 5, 5) y <- rnorm(10000, 7, 6.5) # remove any missings from the data x <- x[!is.na(x)] y <- y[!is.na(y)] ecdf.x <- ecdf(x) ecdf.y <- ecdf(y) plot(ecdf.x, xlim=c(min(c(x,y)), max(c(x,y))), verticals=T, cex.lab=1.2, cex.axis=1.3, las=1, ...

4

Welcome to the site. First, I wouldn't use Excel. Whatever its merits as a spreadsheet, it's not a statistics program. On this site, and generally, R is hugely popular (with good reason) but there are also SAS, SPSS, Matlab and other programs that are better suited to this sort of thing. Second, I am not sure why you divided your continuous variable (or ...

4

Is it mathematically correct to say that these scores follow a "truncated normal distribution" rather than a traditional normal distribution? The scores are not truncated normal either. The variable is bounded, but I'm not sure truncation is really the right term for this situation, since it's not like the observations that would otherwise score above ...

3

Think your statement through as a Frequentist and make it more specific first. A Frequentist could not say that "data set A is different from data set B", without any further clarification. First, you'd have to state what you mean by "different". Perhaps you mean "have different mean values". Then again, you might mean "have different variances". Or perhaps ...

3

$X \sim \chi^2_{k}$ is a random variable with mean $k$ and variance $2k$, and so $T$ is a simply a "unitized" version of $X$, meaning $E[T] = 0, \operatorname{var}(T) = 1$. (Parenthetical remark: I would have loved to refer to this process as "normalization" but the risk of being misunderstood is too great!). Now, $X$ is also a Gamma random variable with ...

3

Following up on Peter Flom's excellent answer, and its comments. If you have your data in a CSV file with column names at the top, in R you would do something like: messages <- read.csv ("my_data_file", header=TRUE) you could then look at your data to check that it did what you think it did with: str (messages) which will give one line for each ...

3

No, correlation is not a good test of this. x <- 1:100 #Uniform y <- sort(rnorm(100)) #Normal cor(x,y) #.98 I don't know of a good test that compares whether, e.g. two distributions are both normal, but possibly with different mean and s.d. Indirectly, you could test the normality of each, separately, and if both seemed normal, guess they both were. ...

3

The solution is straightforward in Cartesian coordinates and impossibly messy in spherical coordinates. I will describe it for a cone that fits within a hemisphere; larger cones can be treated in a similar manner. Choose units in which the sphere has unit radius. After a suitable rotation by some orthogonal matrix $\mathbb{Q}$, the cone will be vertical ...

2

It is a mistake to suppose that there is a good statistical solution that can be identified regardless of what the science (or economics, or whatever) is here. Why are you calculating a ratio in the first place? Is that essential, standard, conventional, or just a way of combining variables that seemed possible? Would the difference between variables make ...

2

Is this variable going to be an independent variable in some sort of regression? A dependent variable? Or what? Transforming it to normality may not be necessary, depending on how you plan to use the variable. BUT..... If you do need to use the two variables as one variable and you need it to be normally distributed , a better alternative may be the ...

2

This is intended to complement the nice answer from Paul, not contradict it. The simplest theory for estimating correlation (indeed anything) is to postulate that there is one unknown constant value, and each sample provides an estimate of that. There seems a kind of hint in your posting that correlation might vary any way. If that's true the usual kind ...

2

It looks like the problem is that your null hypothesis is on the boundary of your parameter space. You are correct that under $H_0: r = \rho$ and appropriate regularity conditions, the test statistic, $z = \sqrt{df}/2 \left( \log\left(\frac{1+\hat{r}}{1-\hat{r}} \right) - \log\left(\frac{1+\rho}{1-\rho}\right) \right)$ is asymptotically normal. The problem ...

2

I will try to give an answer anyway (I looked at the image and you need the sample standard error of the mean): The standard error is the standard deviation of the sampling distribution of a statistic. Wikipedia So $$SE_\bar{x}\ = \frac{s}{\sqrt{n}}$$ and $$S = \sqrt{S^2} := \sqrt{\frac{1}{n-1} \sum_{i=1}^n{(X_i-\bar{X})^2}}$$ So regarding the ...

2

For any subset of observation $H$ and denoting $$(t,S)=\left(\text{ave}_{i\in H}x_i,\text{cov}_{i\in H}x_i\right)$$ then $$\sum_{i\in H}\;\;\;(x_i-t)'S^{-1}(x_i-t)=p(n-1)$$ where $p$ is the rank of $S$ and $n=\#H$. So when $(t,S)$ have been estimated on the same sample on which the MD's are computed the answer to your first question is 'no'. The ...

1

Survival analysis doesn't necessarily require complete failure - you could model time to any failure (component or complete), time to failure of a sub-component, or time to complete failure. So long as you are model time to some event, any event will work. If you want to model time to failure of a sub-component only, then complete failure would probably be ...

1

If you want to compare the distribution shapes, you might use Kolmogorov-Smirnov. I didn't quite understand: that if one sample size is 20 and the other one is thousands? Please be clear in your question. Explain your study more clearly please. What is timestamping? And what are these duplicates? What are your study goals? What are your variables? How many ...

1

Given data, how strongly do we believe that 2 groups do not come from the same population (H_1: they do not come from the same population vs H_0: they come from the same population). This can be done with a Bayesian t-test. Complexity is used to figure out how much the prior is overlapping with one hypothesis. Fit is used to figure out how much the ...

1

For OLS, which would probably be the BLUE estimator for the problem in your questions and given small sample assumptions (See, for example, Hayashi- Econometrics A1 to A5) we have $$Y=\alpha + \beta X + \epsilon\\ V[\epsilon|X]=\sigma^2 \\ (etc...)\\$$ So that  Y|X \sim U N(\alpha + \beta X, \sigma^2) \\ \hat{\beta}_{OLS} \sim N ( \beta, ...

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