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In this answer to the question $$\dfrac{{n+1 \choose x}p^x(1-p)^{n+1-x}}{{n \choose x}p^x(1-p)^{n-x}} = \dfrac{n+1}{n+1-x}(1-p)$$ represents the likelihood ratio $$\frac{L(n+1|x,p)}{L(n|x,p)}$$ If this ratio is larger than one (1), $${L(n+1|x,p)}>{L(n|x,p)}$$ $-$ergo the likelihood increases$-$ and if it is smaller than one (1) $${L(n+1|x,p)}<{L(n|x,p)}...


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If you are really interested in determining how the conditional mean value of the dependent variable varies with the independent variables, then you would address this question by using: Ordinary least squares regression in the absence of heteroskedasticity; Generalized least squares regression or weighted least squares regression in the presence of ...


4

This seems a problem easily solved by additive risk models. Create the regressor 1/n and estimate the additive risk model relating 1/n to the outcome $y$, you will obtain maximum likelihood estimates of $a$ and $b$. This is one of a general class of "binomial models" which are among a general class of "generalized linear models". The most famous example in ...


4

You used a one-sided $t$-test to test against the null that $$\text{H}_0: \mu_\text{women} \leq \mu_\text{men},$$ but you observed $\hat{y}_\text{men} \approx 57.8$ and $\hat{y}_\text{women} \approx 60.3$. Therefore, you have very weak evidence against the null (in fact, your evidence points to the contrary), so you cannot reject it ($p \approx 0.993$).


3

Actually, this problem is a classic example of the Lehmann–Scheffé theorem. The theorem states If a statistic that is unbiased, complete and sufficient for some parameter $\theta$, then it is the UMVUE for $\theta$. Here $\theta$ is $p^3$, and $T = \sum_{i=1}^n X_i$ is a sufficient and complete statistics for $p^3$, so we simply need to construct a ...


3

If $X \sim \mathcal{Pois}(\lambda)$ then $X^* = \mathcal{1}\{X>0\}$ have a Bernoulli distribution with parameter $p=1-e^{-\lambda}$. If you have observations for $n$ boxes then $\sum_i^n X_i^* \sim \mathcal{Bin}(n,p)$ and you can base inference on that. The conjugate prior for the Binomial is the Beta distribution. An example is developed here.


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While the phenomenon is distinct from a type I or II error, I think calling it a type III error is misleading, because it suggests some sort of similarity in interpretation. This phenomenon arises only in null hypothesis significance testing. In the absence of a two-tailed null-hypothesis, (e.g., in binary classification) there exist only false positives ...


2

The traditional approach would be to do a 2-sample test for a difference in proportions: In Minitab, results of this test are as shown below. The warning about the first P-value from a normal approximation causes doubt, roughly for the reasons you mention. However, the result from Fisher's exact test uses an exact hypergeometric probability. It also shows ...


2

Since $X_i \sim \text{Exp}(\lambda)$ you have $\lfloor X_i \rfloor \sim \text{Geom}(1-e^{-\lambda})$, so that: $$C_n = \sum_{i=1}^n \lfloor X_i \rfloor \sim \text{NegBin}(n, e^{-\lambda}).$$ (You have already established this part in your question.) All that you require in your question is an asymptotic size of $\alpha$, so this gives you a broad range of ...


2

As Xi'an correctly points out, this is a maximisation problem over integers, not real numbers. The objective function is quasi-concave, so we can obtain the maximising value by finding the point at which the (forward) likelihood ratio first drops below one. His answer shows you how to do this, and I have nothing to add to that excellent explanation. ...


2

Binomial PMF is a discrete function of $n$, say $f(n)$, given others, i.e. $x,p$. We want to maximize it in terms of $n$. Typically, we would take the derivate and equate it to zero, but in discrete cases we shouldn't do that. This PDF is known to have its peak value(s) around its mean (not exactly but close). Its graph first increases, and then decreases. ...


1

Since $X_i-\beta$ are i.i.d $\mathsf{Exp}(1)$, we have $\min_i(X_i-\beta)=X_{(1)}-\beta\sim \mathsf{Exp}$ with mean $1/n$. Your conclusion is correct, but there is no need for guesswork. Just work out $E\left[X_{(1)}\right]$ from the pdf of $X_{(1)}$, i.e. find $\int xf_{X_{(1)}}(x)\,dx$ directly. If you are starting with $E_{\beta}\left[g(X_{(1)})\right]=\...


1

Based on the analysis you did, the claim is indeed about individual behavior. In your example, you are interested in the probability of being in B1 vs. B2 for an individual in A1. For that individual, their outcome of B1 or B2 is governed by a parameter that applies to them (it may apply to others as well, but the focus here is on the individual). You are ...


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That is exactly what an statistical test is for, but, you must make sure that your sample of 30 individuals has the power to test your hypothesis. The best is to perform an a-priori power analysis, before you sample 30 individuals. Otherwise, how did you decided 30 was enough? It would also depend on the size of the population you are sampling from, if ...


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I did a very rough approximation to $a$ and $b$ via a system of equations. In your data, you sort all of the $n$ values and bin them (in the code below I made bins of same length, it would be probably better to make bins with roughly equal number of $n$'s inside). You count how many $1$'s and $0$'s are in your bins ($\#_1$ and $\#_{1\&0}$), and how many $...


1

Question: Do you have any bounds or relation between $a$ and $b$? If you don't, suppose $n = 10$, you have many answers for $a$ and $b$ that satisfies $p = 0.5$: $b = 0$ and $a = 5$ $b = 1$ and $a = -5$ $b = 10$ and $a = -95$ If you do have bounds or relationship between your parameters, since you have clear Bernoulli experiment with a custom parameter, ...


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Under some circumstances, the paired analysis will have more power and precision than an unpaired analysis. In addition, some research suggests that paired analyses are more robust to unmeasured confounding. We'll assume that both groups have the same sample size and the same variance for simplicity. In an unpaired t-test, the numerator of the t-statistic ...


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This matter is described by Elwert & Winship (2014, p45). I'll try to reproduce their argument but you should probably just read their paper, which is extremely accessible and a must-read for anyone thinking about causal inference. Consider the following causal graph: The important issue is that conditioning on $F$, the friendship tie, induces a non-...


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No, plotting the two variables against each other does not constitute controlling for all other variables. Doing this with two different sets of variables also has no effect on that fact. What you are doing is ignoring all other variables. To better understand the distinction, it may help to read my answer to: Is there a difference between 'controlling ...


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As mentioned in a comment by @tddevlin, it depends on what those objectives are. I give you an example. Suppose I'm solving a vehicle routing problem and I want to minimize both cost and number of vehicles. Then, my $f(x)$ is cost and $g(x)$ is the number of trucks, which both fit the profile that you mentioned. Now, if I prefer costs over the number of ...


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There is a much more general solution to this problem that holds for any distribution whose moments exist. It is a problem known as finding moments of moments. The modus operandi for solving such problems is to work with power sum notation $s_r$, namely: $s_r = \sum_{i=1}^n X_i^r$. We seek $\mathbb{E}[ (\frac{s_1}{n})^3]$ ... which is just the $1^\text{st}$...


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