Search Results

Someone made this statement: 1 out of 1000 corresponds to a proportion of 0,1%. In this case you have a chance of 37% of having 0 outcome, significance level 63%. 3 out of 1000 translating to a pro …

The probability density function I have is as following: Now, I want to rewrite the cumulative probability function bellow into R. The variables ($S$, $\mu$, $\sigma$) are constant variables. … For $t$ i.e. t <- seq(0.1,1,0.1) How the cumulative probability would look like in R? Non of my solutions yields desired output! …

<- data.frame(group=c(1,2,3), o=c(695301,154100, 224140), e=c(930785, 192893, 273400)) e <- x$e # expected frequency under Null Hypothesis o <- x$o # obseved frequency r … lsim <- lapply(x[ ,"o", drop=FALSE], function(x) replicate(100, jitter(x))) o <- data.frame(lsim) e <- x$e apply(o, 2, function(x) { r <- sum( (x - e) / sqrt(e)); chq <- sum(r^2); p.value <- 1 - …

Here are some data: (R language) set.seed(1234) dat <- data.frame( ins=sample(c(1,2,3,4,5,6,7,8), 100, replace=T), outs=sample(c(1,2,3,4,5,6,7,8), 100, replace=T) ) df <- ifelse(with … This is the chi-squared test from R: with(tab, chisq.test(Freq, p=exp_freq, rescale.p=TRUE)) Well, I'm indifferent, would this be correct approach given setup hypothesis? …

I run this example in r and it shows already impact with adding only 10 (representing ony 1% of the total dataset) such opposing variables. …

Actual example (fitted model) within R would be desired result. For nice answer I'm offering double the current bounty. … EDIT: I'm going to disappoint in term of data and provide data from the forecast R package, since I think (for my purpose) it is going to be sufficient to answer this question. …

I'm using r statistical software so suggesting a r package would be also helpful. …

Sample data and the attempt (in r): The data are percentages per group and represent the proportion of observations per group of the overall dataset per year (each column should sum to 100% or 1, but because …

I'm using software R. EDIT: I'm thinking to model each of the independent (explanatory) variable one by one, using survival analysis. Would that be good approach? …

I have as target variable probability (observed/realised p over time) so its p~(0,1). I'm looking for the right model to fit the observed probability over time and "predict" n steps ahead. I'm not su …

These have been calculated in R based on this formula for $A$: A = obs / mean(obs.window) The values of $B$ and $C$ in R are based on the formulas: B = obs / min(obs.window) and C = obs / max(obs.window …

I'm not sure but here is the best solution I can provide: I feel sort of optimization should be used to solve this issue, or definitely better model (rather then linear OLS model) but nonlinear... d …

I have past 12 month data (a ratio of occurence: ranges between 0%-100%) and I have most recent data (January, so 13th observatios). I would calculate the average of the first 12 months (first 12 ob …

In this case the upper CI is, using R: qbeta(.95, 0.15*1000, 1000-(0.15*1000)) # [1] 0.1689547 Since 16% < then 16.90% we fail to reject the $H_{0}$ hypothesis. …

Here are two examples of binomial model fitting. In the second example, the independent variable is modeled using poly() as a second order polynomial. How do I interpret these 2 results? Why someone …