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Suppose you have the observation $X$ from an exponential distribution with mean $\mu$ and rate $\lambda=\frac{1}{\mu}$ Then $\mathbb P(X>x_1) = e^{-\lambda x_1}$ and $\mathbb P(X<x_2) = 1-e^{-\lambda x_2}$. Suppose we want both of these probabilities to be less than or equal to $\alpha$. This can happen in the first case when $\lambda \le -\log_e(\...


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This is a vague question without key specifics. Here is a correspondingly rough and vague outline that may be of some use. Output from your Markov process may be very far from a sequence of independent observations from its limiting distribution. A thousand observations is a very small run, but perhaps useful. (If not, run your simulation program again to ...


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I have known what exactly I have done wrongly. The exact solution in R is as follows: dif=(New-Old) m=mean(dif) st=sd(dif) CI=m+c(-1,1)*qt(0.975,4)*st *sqrt(1/5) ExpCI=exp(CI) # equals [0.09 2.49] MEAN=mean(ExpCI) # equals 1.29


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$\newcommand{\convp}{\stackrel{\text p}{\to}}$By assumption we have $\sqrt n(G_n - \theta) \implies Y$ for some random variable $Y$. If $\gamma \neq 0$ and we have a consistent estimator $\hat\gamma_n$ of $\gamma$, so $\hat\gamma_n \convp \gamma$, then $$ \frac{\sqrt n (G_n - \theta)}{\sqrt{\hat\gamma_n}} \implies \frac{Y}{ \sqrt \gamma} $$ by a combination ...


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They are both exact. Exact is a weaker property than you probably think it is. A confidence interval method is exact if the probability of a $1-\alpha$ confidence interval covering the true value is at least $1-\alpha$. Here's an exact 95% confidence interval method for any real-valued parameter: roll a twenty-sided die. If you get a 20, the interval is the ...


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You calculated the mean of logarithms of ratios, not the mean of ratios themselves. To obtain ratios from their logarithms, you have to raise $e$ to power of them. In Python and NumPy: import numpy as np logs = [1.304, -0.768, -0.473, 1.237, 2.405] # Logarithms of ratios ratios = np.exp(logs) # You omitted this! np.mean(...


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If you have datapoints $x_1, \ldots, x_n$ and $y_1, \ldots, y_n$ and you want to find the mean ratio $\frac{1}{n} \sum_{i=1}^n \frac{x_i}{y_i}$, this is not equal to the exponentiated mean of the log ratios. In other words, this is not the same as computing $\exp\left(\frac{1}{n}\sum_{i=1}^n \log \frac{x_i}{y_i}\right)$, which is what it sounds like you are ...


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What you're asking is essentially the general form of how do I get a confidence interval for the mean of a function broadcast over a sample. This should help you out. You're already 95% of the way there. Thus, the inference here is (usually) about a nonlinear function of the parameter vector $\beta$. Presumably this means that forming a confidence interval ...


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It is not exactly correct to say the blue curves represent the 95% CIs. For a given sample size, the 95% CI would be different for every different observed mean. But, it is correct that any mean will fall within the blue curves with 95% probability.


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You could try a non-parametric bootstrap: set.seed(1976) N <- 86 sample <- c(rexp(floor(N/2), 5), rexp(ceiling(N/2), 1/10)) hist(sample) length(sample) mean(sample) # expected (floor(N/2)*1/5 + ceiling(N/2)*10) / N me <- qt(.975, length(sample) - 1) * sd(sample)/sqrt(length(sample)) lower <- mean(sample) - me upper <- mean(sample) + me c(...


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To begin, I confess I'm not familiar with either the SAS/binomial method or (just from your bit of R code) the exact type of bootstrap CI you are using. I notice from your plot that the median is not near the middle of either of these types of CIs. There are various correct methods for making CIs, based on different assumptions and criteria, so both kinds of ...


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You have some true interest-parameter $\theta$ and have computed a confidence interval for $\theta$, $(l,u)$. You also have an estimator of $\theta$, $\hat\theta$, which is between $l$ and $u$. That $(l,u)$ is a 95% CI means that $$ \DeclareMathOperator{\P}{\mathbb{P}} \P(l \le \theta \le u)=0.95 $$ I suppose the function $f$ is monotone increasing (it is ...


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As I understand it, you want to know when to use a certain quantile (qnorm). This will depend on the level of significance set for your test. For example, consider a hypothesis test on which you want to test, $H_0: \mu = \mu_0 \: \times \: \mu \neq \mu_0$. It is known that $$Z_{test} = \frac {X- \mu_0}{\sigma} \sim N(0,1)$$ As it is a bilateral test (see the ...


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Bootstrap is not used to estimate mean. It can be used to estimate the uncertainty of the estimate for mean. You can use it in same way to estimate the uncertainty around any statistic, by following the same procedure: sampling with replacement from the data and evaluating the statistic on the sample, to eventually learn the distribution of the estimates by ...


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By specifying ci='sd', you are indicating at what range you could be 95% sure that a sample of the population data contains the population standard deviation. From graphpad: Interpreting the CI of the SD is straightforward. If you assume that your data were randomly and independently sampled from a Gaussian distribution, you can be 95% sure that the CI ...


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The Confidence Interval (CI) definition the OP has provided is more or less the working definition - perhaps not the most rigourous one if we were to involve samples, but this form is used in general. However, one should note that The CI is random, not the parameter The population parameter $\theta$ is a fixed, unknown constant. It is by definition the true ...


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Here's the most degenerate example. If I want to build a 95% confidence interval $I$ for a real-valued parameter $\mu$, I can use the following distribution: $$ \mathbb P(I = (-\infty, +\infty)) = 0.95 \\ \mathbb P(I = \emptyset) = 0.05 $$ (Some definitions of confidence interval may not technically include infinite or empty intervals, but this doesn't ...


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The lm function does its inferential work under a fairly standard assumption that the error term has constant variance. That appears to be a poor assumption for your data, but lm does not know that, so it just treats the groups as having equal variance and calculates the error term variance thinking it is the same for each group. Weighted least squares is ...


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The bootstrap reuses samples from the data to simulate the sampling distribution of the statistic you're interested in. Your samples is comprised entirely of ones. So resampling from your data will always yield a bootstrapped dataset comprised of ones. Thus, the bootstrapped statistics would be homogeneous and there would be no variation. No variation ...


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100% proportion is an extreme and illustrative case of the limitations of the bootstrap assumption. Consider the case of having only a single observation of a coin flip. The bootstrap CI of the coin's bias will always be 0% or 100% since the bootstrap assumes that all possible flips are like the one you've observed. This is obviously wrong. Similarly, also ...


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The point of the Bonferroni correction is to limit the probability of reporting at least one false positive to $\le \alpha$, regardless of which null hypotheses are true. I think that doing the correction only for the confidence intervals that you have found to be significant does not achieve this limit, so is hard to justify. Suppose your 5 null hypotheses ...


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