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The 95% confidence interval $\bar X \pm 1.96\frac{\sigma}{\sqrt{n}}$ for unknown $\mu$ is correct for normal data when the population standard deviation $\sigma$ is known. It is approximately correct for moderately large $n,$ when $\sigma$ is estimated by the sample standard deviation $S.$ However, you are correct to doubt this so-called 'z-interval' when \... 3 Following is what we get based on calculation: \begin{align} t_1&=\frac{\bar x - \mu}{s_1/\sqrt n} \\ \end{align} where,s_1=\sqrt{\frac{\sum\limits_{i=1}^n (x_i-\bar x)^2}{(n-1)}}$For$t_2, \begin{align} s_2 &=\sqrt{\frac{\sum\limits_{i=1}^{4n} (x_i-\bar x)^2}{(4n-1)}} \\ &=\sqrt{\frac{4\sum\limits_{i=1}^{n} (x_i-\bar x)^2}{(4n-1)}} \\ &=... 2 Since the function min(λ, c) meets the Keane–O'Brien theorem, this implies that there are polynomials that converge from above and below to the function. This is discussed, for example, in Thomas and Blanchet (2012) and in Łatuszyński et al. (2009/2011). However, neither paper shows how to automate the task of finding polynomials required for the method to ... 2 A simple example would work best here. Population: Adults aged 20-65 who suffer from heart disease and are patients of a local hospital. Random variable: Heart rate (beats per minute) of a randomly selected adult from this population. The population is a well-defined collection of individuals we are interested in studying. The individuals belonging to this ... 2 Let's denote byT$the number of positive tests in your sample, and by$T_1, T_2$and$T_3$the number of positive tests in categories 1, 2 and 3. Then$T = T_1 + T_2 + T_3$. First, let's consider that$x_1, x_2, x_3$are fixed. Then$T_1, T_2$and$T_3$are independent binomial random variables :$$\begin{array}{ccc} T_1 & \sim & \mathcal{B}(x_1, ... 2 I would be extremely wary of any rule of thumb such as this: a level 2 sample size of 250 organizations and level 1 sample size of 40 individuals per organization will be sufficient. For one thing, sufficient for what ? For another, a lot depends on the correlations within clusters, the higher the correlation, the lower the effective sample size. In my ... 1 Now that I understand your chart better, no, I do not believe you have enough information to write the joint density. For example, if$X$takes a value of$1$, the joint density could say that$Y$is assured of taking a value of$1$...or it could say that$Y$cannot take a value of$1$. EDIT If you are willing to assume independence, there is an answer. ... 1 say this specific sample mean$\bar{x}$follows a sampling distribution$W$, the variance of this$\bar{x}$is$V(\bar{x})$. and the variance of this specific sample is$V(x)$, are$V(x)$and$V(\bar{x})$exactly the same? No$V(\bar{x})=V(x)/n$we know that$V(\bar{x}) = \sigma^2 /n$This part is false, as you stated before that$V(X) = \sigma^2$is the ... 1 You can approach the function but not get equal Suppose we are given a coin with arbitrary (unknown) head probability$p$, I am wondering if there is an easy-to-implement algorithm for generating a$\min\{p, 0.5\}$coin for any$p\in [0,1]$. Function$f(p) = \min\{p, 0.5\} \geq \min\{p, 1-p\}$so the famous paper by Keane and O’Brien guarantees such an ... 1 I don't know how to generate that function, but there is a simple way to generate increasingly good approximations of it by consuming an increasingly large of inputs. Namely, you observe$N$inputs, where$N$is an odd integer. Then: If the majority of those inputs is$0$you "decide" that$p<1/2$. So you take an addtional input an output that. ... 1 Another way, perhaps more indirect, to prove this is to note that the sample quantile is also an M-estimator, specifically when$\rho(y_i,\theta)=\alpha(y_i-\theta)_+ + (1-\alpha)(\theta-y_i)_+\$. Because of that it enjoys all the properties of M-estimators, one of them being that it is consistent in probability (converges in probability) to its mean, which ...