# Tag Info

1

Actually it is very simple. For a one dimensional variable there is a confidence interval, e.g., like $\pm1.5$. For a two dimensional plot, there is a confidence band, which are functions above and below the estimates, e.g., from regression. In the figure above, the 95% CI, and refer to where the linear regression line is located (blue dashes). The 95% "...

0

First, note that you can simplify your problem by setting $a=0$ without loss of generality, which gets rid of $a$ as a parameter. Let $z = nm$, then $m = z/n$ and, substituting, $$F_Z(z) = 1 -\left[1-{z \over nb}\right]^n$$ for $0 < z < nb$. Remembering our limits, specifically, the one relating $(1-x/n)^n$ to $e^{-x}$, we can write: $$\lim_{n \to ... 0 It is not clear what you really want. The Poisson-Binomial distribution on n trials have n probability parameters p_1, \dotsc, p_n so you cannot reasonably expect a confidence interval for each of them! So, for this question to be answerable, you nee to specify some function of the p_i's that you are interested in. So, what is your real inferential ... 4 Let's think about this from first principles. A confidence interval (in this context) is given by a pair of functions l and u. Setting \alpha=95/100, their defining properties are For all numbers x, l(x) \le u(x). This means the interval [l(x),u(x)] is well-defined and non-empty. No matter what \theta\in [0,\infty) might be,$$\alpha \le \Pr(...

0

I think (but am not sure -- for that you may have to ask the authors) that it may be simpler than what you think. Even an identical model has a different AUROC value depending on the dataset used to calculate it. Suppose we have some dataset $\{x_1, \ldots, x_n\}$ and some model $M$. Then, we can calculate its AUROC over this dataset: $$\text{AUROC}(x_1, ... 1 The purpose of the graph appears to be to display the regression results, so it makes sense to use the standard errors from that regression. 0 As an alternative answer, I found that a quadratic logarithmic equation "y = a + b * log(x) + c * pow(log(x), 2.0)" (where log is the natural log) seems to fit the entire range of the posted data fairly well, with parameters a = 3.0496816334683524E-01, b = 2.2339160821508095E+00, and c = -6.8066648613654057E-01 yielding R-squared = 0.9914 and RMSE = 0.01457 1 You would probably use a confidence interval, since it is narrower and more conservative. Inverting null hypothesis significance testing to "confirm a null hypothesis" has innumerable practical and theoretical problems. Case in point, it's hard (impossible?) to state the actual "level" or "power" of the test, whatever you are actually testing, and one can ... 3 This question was asked a long time ago, but I'm posting a response in case anyone discovers it in future. In short, the answer is yes: you can do this in many settings, and you are justified in correcting for the change in sample size by the \sqrt{\frac{M}{N}}. This approach is usually called the M out of N boostrap, and it works in most settings that ... 0 Here's what I understand: a) Distributions with heavy tails may have infinite variance, or mean (Ex: cauchy distribution) b) Heavy tailed means that there are a few outliers that are very different from the most of the samples. And these outliers have non-negligible impact on the future statistic procedures. c) Log-normal or exponential distributions ... 0 If$$ \sqrt{n}[g(\hat{\theta}) - g(\theta)] \rightarrow_{d} N(0, \sigma^2[g'(\theta)]^2) $$(note the typo you made in the variance term) then for a "large" n it is approximately true that$$ \sqrt{n}[g(\hat{\theta}) - g(\theta)] \sim N(0, \sigma^2[g'(\theta)]^2) $$which is true if and only if$$ g(\hat{\theta})\sim N\left(g(\theta), \frac{\sigma^2[g'(\...

3

a) Distributions with heavy tails may have infinite variance, or mean (Ex: Cauchy distribution) True. b) Heavy tailed means that there are a few outliers that are very different from the most of the samples. And these outliers have non-negligible impact on the future statistic procedures. Partly true. This might be how it looks in a realization drawn ...

1

The major issue in this situation ends up being how representative each of the samples is: the 250 households sampled per authority versus all households in the authority, and the 129 authorities in the data set versus the 328 authorities that collect waste (plus any waste of interest generated outside those 328 authorities). To start, however, let's assume ...

4

Judging from its abstract, the JASA article by Weizhen Wang linked in my comment, gives a method to get (nearly) exact hypergeometric confidence confidence intervals. Perhaps a more easily computed style of CI, based on a normal approximation to the hypergeometric distribution, will suffice for your purposes. Main example: Suppose we know there are $T$ ...

1

I would like to add Wilson's method mentioned by Michael M in a comment. From Wikipedia: Binomial proportion confidence interval - Wilson_score_interval. You can get a 95% confidence interval by using the following: $\frac{n_s + \frac{z^2}{2}}{n+z^2} \pm \frac{z}{n+z^2}\sqrt{\frac{n_s n_f}{n}+\frac{z^2}{4}}$ The left term is the center value and the ...

0

A very common statement is that H0 is never true, it's just a matter of sample size. Not among people who know what they're talking about, and are speaking precisely. Traditional hypothesis testing never concludes that the null is true, but whether the null is true or not is separate from whether the null is concluded to be true. This would mean that p-...

1

Main point in my question was that when can we really say that $H_0$ is true, i.e. $\mu_1=\mu_2$ in this case? No, because "absence of evidence is not evidence of absence." Probability can be thought as an extension of logic, with added uncertainties, so imagine for a moment that instead of real numbers on unit interval, the hypothesis test would return ...

7

The CI can have any limits, but it is centered exactly around zero For a two-sample T-test (testing for a difference in the means of two populations), a p-value of exactly one corresponds to the case where the observed sample means are exactly equal.$^\dagger$ (The sample variances can take on any values.) To see this, note that the p-value function for ...

1

The straightforward answer (+1 to Noah) will explain that the confidence interval for the mean difference may still be of nonzero length because it depends on the observed variation in the sample in a different way than the p-value does. However you might still wonder why it is like that. Since it is not soo strange to imagine that a high p-value also means ...

10

Being super-lazy, using R to solve the problem numerically rather than doing the calculations by hand: Define a function that will give normally distributed values with a mean of (almost!) exactly zero and a SD of exactly 1: rn2 <- function(n) {r <- rnorm(n); c(scale(r)) } Run a t-test: t.test(rn2(16),rn2(16)) Welch Two Sample t-test data: ...

3

It is difficult to have a cogent philosophical discussion about things that have 0 probability of happening. So I will show you some examples that relate to your question. If you have two enormous independent samples from the same distribution, then both samples will still have some variability, the pooled 2-sample t statistic will be near, but not exactly ...

15

A confidence interval for a t-test is of the form $\bar{x}_1 - \bar{x}_2 \pm t_{\text{crit}, \alpha}s_{\bar{x}_1 - \bar{x}_2}$, where $\bar{x}_1$ and $\bar{x}_2$ are the sample means, $t_{\text{crit}, \alpha}$ is the critical $t$ value at the given $\alpha$, and $s_{\bar{x}_1 - \bar{x}_2}$ is the standard error of the difference in means. If $p=1.0$, then $\... 0 Nothing stops you from using standard t- or Gauss-formulae for computing the confidence interval - all informations needed are given in your question. p=1 doesn't mean that there's anything wrong with that. Note that p=1 does not mean that you can be particularly sure that the H0 is true. Random variation is still present and if u0=u1 can happen under the H0,... 5 It's a bit nuanced. You could pull out the big guns and use a poisson regression # example sample of counts (n = 50, 'true' mean = 5) set.seed(25) d <- rpois(50, 5) model = glm(d~1, family = poisson) exp(confint(model)) >>> 2.5 % 97.5 % 4.143126 5.348048 Or you could use the normal approximation since, as$\lambda$gets big ... 1 To add to @mkt 's answer, which does capture all the most critical mathematical aspects, a few observations: The intercept CI spans from -70 to -33. Assuming that body fat is a percentage, then this means that the baseline amount of fat in the cohort is very variable. If the distribution of BodyFat is left or right tailed then the mean will be skewed ... 5 You are comparing two very different things. In the first case, you are making pairwise comparisons when calculating the correlation coefficient between BodyFat and Weight. In the second, you are doing a multiple regression that also accounts for the variation in BodyFat that is explained by all your other variables. To oversimplify a bit: after accounting ... 1 The standard approach for a question like this is to calculate a confidence-interval for the mean accident rate in your two groups and check whether they overlap. Typically, one would use 95% CIs. Now, things that can happen or not in$n$independent samples, like accidents, are binomially distributed. So we need to calculate CIs for the binomial parameter$...

0

I think I figured out an answer my self. Regarding this page https://rpubs.com/aaronsc32/regression-confidence-prediction-intervals they call it "Confidence (Mean) Intervals" and decribe the calulation there as: \hat{y}h \pm t{\alpha / 2, n - 2} \sqrt{MSE \left(\frac{1}{n} + \frac{(x_k - \bar{x})^2}{\sum(x_i - \bar{x}^2)} \right)} Did I get this right? ...

1

The "base" level (often called the 'reference level') should not have been left out of the model. It is represented by the intercept. The other levels are typically specified in your output, but those coefficients are not actually the values for those levels, instead they are the differences between the values for the indicated level and the base level. ...

5

$\color{white}{^\text{Likelihood contour showing the usual case two boundary points for}\\^\text{a given choice of one parameter when the selected value is within the region}}$

2

1&2) You are working on a prediction problem. You need to solve $$\pi(k\ge450|4\text{ success of 10 tries})=1516927277253024\sum_{k=450}^{1000}\int_0^1\binom{1000}{k}p^k(1-p)^{1000-k}(1-p)^{25}p^{23}\mathrm{d}p$$ That is your posterior probability that the next 1000 tosses will result in 450 or more successes. You should gamble no more than the prize ...

1

A pdf is a pdf. Just as you could integrate the binomial pdf from 450 to 1000 based on 1000 coin flips, you can integrate the Beta(24,26) from p = 0.45 to p = 1, based on your posterior, to get the probability of 450 or more out of 1000. I don't know that there is any single "correct notation" for the probability that you seek. You specified quite nicely in ...

2

This answer is for those readers who could not fully understand the previous answers. Let's discuss a specific example. Suppose you try to predict the people's weight from their height, sex (male, female) and diet (standard, low carb, vegetarian). Currently, there are more than 8 billion people on Earth. Of course, you can find many thousands of people ...

1

My first attempt to analyze such data would be to use a two-sample Wilcoxon test. Here are two nonnormal samples of size $n = 15$ to use for illustration. set.seed(2019) x1 = round(rgamma(15, 6, 1/.5),2) x2 = round(rgamma(15, 5, 1/.4),2) x1 [1] 3.68 2.18 5.99 1.91 3.17 3.40 2.66 4.98 1.60 [10] 4.99 2.62 3.88 1.07 2.59 3.10 summary(x1) Min. 1st Qu. ...

0

If you have the standard errors of the kappa values and are willing to assume they are normally distributed you could summarise them using techniques from meta-analysis. Each would be inversely weighted by its variance to form the summary and the weights used to form the standard error of the summary and hence to form confidence intervals. Meta-analysis is ...

0

It's okay to report the mean of Cohen's kappa. However instead of reporting mean upper/lower confidence limits, report the mean standard error.

Top 50 recent answers are included