# Tag Info

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I suggest reading this explanatory post on Bayes's experiment. The general idea, however, it that the landing spot of the ball on the table is random, and therefore the position of the ball measured from the left side (or any, as long as it's chosen before throwing) of the table can be described through a Uniform distribution.

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Frequentist statistics assumes parameters are fixed quantities. As in, there is a "true" parameter out there that we are trying to estimate. Bayesians instead assume parameters are random, that is, they have a distribution. Hence the $f(\theta)$ in $1.2$. Lastly, if a ball lands $N$ times on a $1 \times 1$ table, you can record whether the ball landed ...

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All coins are biased. It's a question of how biased they are, but no coin has exactly equal chances of being heads or tails. In the frequentist approach, I think this calls for a test of equivalence: Decide how close to 0.50000 you will accept as unbiased. Do a power analysis to determine how many flips you will need to have a good (0.80? 0.90?) chance of ...

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Presumably, this code is meant to show that the maximum likelihood estimator is 12/20. To answer your question, the term ${n \choose k}$ does not affect the maximization of the function since it only modifies the likelihood by a constant of proportionality that does not depend on the unknown parameter. Thus, the point on the x-axis where the maximum occurs ...

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In the case of a mixed-effects models in general you would want to specify the fixed- and random-effects structures appropriately. The former concerns the mean of your outcome and the latter the correlation structure. There can be different considerations on how you would go about to build each part, affected by the design of your study and how much ...

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For a binomial RV, we have $\operatorname{var}(X)=np(1-p)$, $E[X]=np$. To estimate $p$, we need to get rid of $n$. If you take the ratio: $${\operatorname{var}(X) \over E[X]}=1-p$$ Put $E[X]\approx \bar{x}$, and $\operatorname{var}(X)\approx \frac{1}{n}\sum x_i^2 - \frac{1}{n}\bar{x}^2$: p\approx1-\frac{\sum x_i^2}{n\bar{x}}+\bar{x}=1+\bar{x}-\frac{\sum ...

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To give some quick answers to the points raised: The additional "$+4$"observed when calculated the "adjusted version of recall". This comes from the viewing the occurrence of a True Positive as a success and the occurrence of a False Negative as a failure. Using this rationale, we follow the general recommendation from Agresti & Coull (1998) "...

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I'd like to point out that the "exact" solution to this problem in the frequentist sense is the Clopper-Pearson interval. As the author noted, there are a number of nice approximate intervals with different characteristics, and the Wilson score interval is often a good choice. However I thought I would also point out that Bayesian inference with a uniform ...

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I think what you want to do is to compare two proportions. R's prop.test function does that. Edited after discussion: Apparently this uses an asymptotic approximation that gives a warning for these data, probably because of too small success numbers. Thinking about this more, Fisher's exact test should do what you want. This is a permutation test for ...

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I would recommend the binomial test as the assumptions are less restrictive than the chi-squared test. Search on binom.test function in R.

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The likelihood ratio test supposes nested hypothesis, that is, the smaller model is a special case of the larger model. And, that is the case for the binomial and beta-binomial models. See parametrizations for the beta distribution, you could reparametrize the beta binomial distribution using the mean/variance parametrization for the beta distribution, ...

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For the case $g(p)=1/p$, a more rigorous proof for nonexistence of unbiased estimators is For the binomial distribution, why does no unbiased estimator exist for $1/p$?. With $g(p)=\log\frac{p}{1-p}$, you could mimick your argument using limits when $p \to 0$, since $\log\frac{p}{1-p} = \log p - \log (1-p)$ which goes to $-\infty$ when p goes to zero. But ...

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Plotting a seaborn distplot needs an adjustment, as it is primarily meant for continuous distributions. The distplot will put the data in 16 equally size bins, that don't align with the integer numbers. For discrete distributions, distplot would need explicit bins, e.g. range(30). However, with that many bins, the default calculated kde will not be as ...

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You're asked to calculate $P(7\leq X \leq 13)$, but you only find $P(X=7)$, where $X\sim \text{Bin(n,p)}$ denotes the number of low-birth weight babies. So, your script should be: sum((binomial>=7) & (binomial <= 13))/runs And, it's better to increase your number of runs to decrease your estimate's variance. You can also calculate it ...

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This is a table for a sample of either $n=500$ or $501$ subjects, not $100.$ Perhaps the best confidence interval method for binomial data of this kind (where each observation is independent of the others and all are modeled as random binary outcomes with a common probability) is the Clopper-Pearson interval. Given a sample size $n$ and an observed ...

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It's a good start with the hypotheses and models. The choice of model really boils down to what we know a priori about the distributions. But two important things to keep in mind: We don't choose GLM specifications deterministically based on the probability model of the response. For instance, if a response is binary, we can fit any model we choose. If I ...

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What is the most robust statistical method (test) to compare each coin flip observation distribution with the other coin observation distribution to determine whether both of distributions are the same or different? This sounds a lot like a binomial test of proportions. Comparing binomial proportions has been analyzed several ways, including Frequentist ...

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