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3

Estimating the bias of a statistic was, as I recall, one of the original motivations for developing the bootstrap. Keep in mind the bootstrap principle: The basic idea of bootstrapping is that inference about a population from sample data (sample → population) can be modelled by resampling the sample data and performing inference about a sample from ...


1

Thanks to whuber who provided the answer in the comments. The problem was the expression for the standard error was not correct. It should be $$\text{se}(\hat\beta_1 + \hat\beta_2x_0)= \sqrt{\text{var}(\beta_1) + 2x_0\text{cov}(\beta_1, \beta_2) + \text{var}(\beta_2)x_0^2} $$


3

what is the interpretation of these intervals/regions? Even if you can compute the confidence interval in usual manner, those interval are senseless. The usual story behind the confidence interval rules suppose that you collect a sample by $iid$ sampling on infinite large population. For finite population some ad hoc adjustments exist (read here: http://www....


1

Your argument seems mostly OK. I'm not sure what you mean by "[S]ince it can be assumed that voters think similarly...." and I'm not sure I follow all of your steps toward the end. I think you are describing the Wald confidence interval, which is of the form $$\hat \theta \pm 1.96\sqrt{ \frac{\hat\theta(1-\hat\theta) }{n} }.$$ This kind of interval ...


0

Assuming that you have two independent normal samples of sizes $n_x$ and $n_y,$ you can look at $D = \bar X - \bar Y.$ Also, assume that group population means $\mu_x$ and $\mu_y$ are unknown. If the variances of the two groups are known, then there is no difficulty finding the normal distribution of $D$ which has $E(D) = \mu_x - \mu_y$ and $V(X) = \frac{\...


0

Whether normal distribution approximation is fine depends on how accurate you want. In particular, if Wilson score is used as a normal distribution approximation, compared with the accurate CI, Wilson score will yield a looser lower limit with an error more than 10%, while the upper limit is quite close to the accurate CI.


1

There are a few issues you have going on here that area leading to these differences: In the $t$-test you have run, you do not have the order of the before and after specified the same way as you have it specified for the bootstrap procedure (i.e. you have mixed up which term is subtracted from which). In your $t$-test, you are computing the differences as $...


1

If you choose a significance level $\alpha$, you get a confidence interval with confidence level $1-\alpha$: $$C_{1-\alpha}=\left(X-z_{1-\frac{\alpha}{2}}\text{se},X+z_{1-\frac{\alpha}{2}}\text{se}\right)$$ where $z_{1-\frac{\alpha}{2}}$ is the $(1-\frac{\alpha}{2})$-quantile of a standard normal variable $Z$. In your confidence interval, (0.382-2SE,0.382+...


1

The strategy is to simulate the coefficients from the joint normal distribution of all parameters, including the $\phi$ parameter. The reason is that the coefficients for $\mu$ and $\phi$ are not orthogonal, i.e., depend on each other. After having obtained the full simulated parameter vector, you can than compute any quanity you want from the relevant ...


0

If you are happy with R, you can use the binom.test function. E.g: binom.test(382, 382+578+40, conf.level= 0.9654) Exact binomial test data: 382 and 382 + 578 + 40 number of successes = 382, number of trials = 1000, p-value = 8.291e-14 alternative hypothesis: true probability of success is not equal to 0.5 96.54 percent confidence interval: 0.3494766 ...


1

There is a mistake in your probability result (which should be clear by the fact that it is unbounded). Using the interval $\text{CI}(X) = [X-b, X+c]$ you should have the coverage probability: $$\begin{align} \mathbb{P}(\theta \in \text{CI}(X)) &= \mathbb{P}(X-b \leqslant \theta \leqslant X+c) \\[6pt] &= \mathbb{P}(\theta-c \leqslant X \leqslant \...


1

Since the pdf you provided is a conditional pdf of X under given θ, it is possible to derive the confidence interval (CI) of X under given θ, but not the CI of θ. Contrarily, if the pdf of f(θ|x) is given by the same expression, then the shortest CI of θ can be derived as S(x) = [x+ln(alfa) x-ln(alfa)].


2

One simple trick that avoids any computation if $x_0$ is binary is to get an equivalent model. Let $z_0= 1-x_0$. It corresponds to inverting/recoding of $x_0$. Now the equation $$ y = \mu + \gamma_0 z_0 + \gamma_1 x_1 + \gamma_2 z_0 x_1 + \delta $$ has exactly the same fit as and is in fact equivalent to your equation. But the trick is that when $x_0=1$ i.e. ...


4

The confidence interval for $\hat\beta_1$ is: $$\hat{\beta}_1 \pm t_{n-4,1-\alpha/2}\sqrt{\hat{\text{var}}(\hat\beta_1)}$$ The confidence interval for $\hat\beta_1+\hat\beta_2$, when $x_1$ is binary (0,1), is: $$(\hat\beta_1+\hat\beta_2)\pm t_{n-4,1-\alpha/2} \sqrt{\hat{\text{var}}(\hat\beta_1)+\hat{\text{var}}(\hat\beta_2)+2\hat{\text{cov}}(\hat\beta_1,\hat\...


1

estimate the variance of the estimator using the usual CLT-based approach. ... Can I use this information to produce estimates with smaller confidence intervals? Yes, you can. (This is true in general. In many cases, you can do better than a normal approximation, especially when the distribution is not really a normal distribution but just approximately) ...


3

An estimator that is obviously better in some ways is $$\hat\mu= \frac{\sum_{\textrm{observed }k} n_kx_k}{\sum_{\textrm{observed }k} n_k}$$ In particular, if $|J|$ is large enough that all $K$ distinct items will be observed at least once (with probability going to 1) and the error of $\hat\mu$ will be exactly zero, where your estimator (call it $\bar x$) ...


1

You are asked to find out how many survey plots are needed to make the confidence interval (CI) equals 56.5 ~ 71.5 with the confidence level (CL) equals 80%. In other words, how many survey plots are needed in order to be able to claim with 80% confidence that there are 56.5~71.5 trees per hectare on average.


6

Let's break this down into easier problems. To keep the post reasonably short, I will only sketch a good confidence interval procedure without going into all the details. What is interesting about this situation is that because $Y$ varies in such a complex, nonlinear fashion with the distribution parameters, a careful analysis and special solution are ...


0

From the description you provided, your first question is about the distribution of people's age. Normal (i.e. Gaussian) distribution applies to such kind of applications. It will be helpful if you know how the confidence interval (CI) was calculated, because there are many different possible ways that the CI was calculated. For instance, if the distribution ...


1

Here is a fairly straightforward and intuitive argument for both estimators being asymptotically normal. Call $$T(X|y_1)=\frac{1}{n} \sum_{i=1}^{n} I(x_i>y_1)$$ As with $S$, central limit theorem implies this is asymptotically normal. Furthermore, take $$T = \frac{1}{n}\sum_{i=1}^{n} T(X|y_i)$$ Since each $T(X|y_i)$ is asymptotically normal, and any ...


1

There is a median test, based on a chi-square test. It breaks each group up into cases above and below the individual medians, and determines whether the frequencies of those groups differ from what would be predicted if all the data came from the same distribution. "[I]t only considers the position of each observation relative to the overall median,&...


1

I would try something completely different, a logistic mixed effects model. In R that can be estimated by lme4::glmer, search this site for examples, there are plenty. You seem to have no covariates, so that would be a random intercepts model. Another, maybe more direct way: If all the success probabilities where equal, you had binomial data with the same $p$...


0

What you are plotting is the distribution of your statistic among the bootstrapped samples, the basis for the percentile bootstrap estimate of CI. Oddly enough, the percentile bootstrap method doesn't actually follow the bootstrap principle, which is that the relationship of your bootstrapped re-samples to your data sample represents the relationship of your ...


3

I have tried running the loocv experiment multiple times with different random seeds [...] I always get the same results for each run so the variance is zero. Of course you get the same results, for LOO the random seed cannot change anything but the order in which the different surrogate models are evaluated: one run of LOO consists of n surrogate models ...


3

One way would be to take the mean and standard deviation and apply the central limit theorem to justify the old mean + 2 standard error formula. Because each fold is very highly correlated there may or may not be some objections to doing this. I think the best way is to actually bootstrap the entire process and then correct for optimism in the training ...


1

If we understand the expression "the confidence interval of my evaluation" to mean "a credible range of values for each parameter I infer when I perform logistic regression on training data using the LOO cross validation technique." For a training set with $n$ values of $p$-dimensional predictors $x_i$ and response $y_i, i=1\ldots n$, you ...


1

It is not unusual at all to get negative confidence limits with small datasets. They are due to using naive normal-theory intervals: $\hat p \pm z_{\alpha/2}SE(\hat p)$. When $\hat p$ is small and the sample size is small, the $\pm$ part can easily exceed $\hat p$, leading to a negative bound. There is kind of a workaround for this. The "prob" mode ...


2

Here is an approach using the maximum observation (the sufficient statistic for $\theta),$ rather than the sample mean and standard deviation. (Of course $\mu = \theta/2$ can also be estimated by $\bar X,$ but with more variability; see Notes at end.) Let $W$ be the maximum of $n=10$ observations from $\mathsf{Unif}(0, \theta).$ Then it is not difficult to ...


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