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Mixed models are good at coping with unbalanced designs. This is one of their advantages compared to other approaches such as ANOVA-type models. So I would not worry about this. You mention small group sizes, but the numbers you mention, are not, in my opinion, small. You will see in my answer to your other question that I was simulating data with maximum ...


3

A) What do you mean by "this situation?" If the issue is calculating sample size for a non-inferiority trial with a time-to-event outcome, I suggest chapter 7 of "Chow SC, Wang H and Shao J, 2007. Sample size calculations in clinical research. CRC press." For a software implementation, dunno what software you use, but if Stata, I suggest ...


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Whether you use the standard "rule of thumb" of 10-20 minority-class cases per predictor or the more refined approaches described in the paper linked by @Chl in a comment, you will find that having only 22 events will severely limit your ability to "find the association" between the event and your approximately 20 candidate predictors. ...


2

Every simulated sample in the sampling distribution is supposed to be generated in the same way as the actual sample. That's the whole point of a sampling distribution: To represent what would have happened if the data were generated the same way as the actual sample but from a hypothetical world (e.g., null hypothesis). Therefore, if the actual data had a ...


2

This is a standard power analysis. To determine how many samples you need in each setting, you need to specify: a baseline value of interest, the difference between values that you want to detect, the chance that you are willing to accept a false-positive difference (Type-I error), and the chance that you are willing to accept that you miss a true-positive ...


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When you're down in the few percent range of rare events you might think in terms of a Poisson distribution, which has a very useful property: the variance of a Poisson rate equals its mean. If you have specified failure rates that you would like to distinguish, Gerald van Belle uses this property to propose a simple rule of thumb (page 40): Suppose the ...


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Your question is a bit vague and it seems the your figure does not quite match the rest of the problem. I think you may have put parts of two similar problems together in your Question. I'll do my best to give most of the information you requested. You say the means of the two normal populations are unknown with $\mu_A \le \mu_B,$ and I will assume the two ...


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The usual approach is to reverse the computation for the confidence interval so we take the half width of the confidence interval and then divide it by the relevant quantile of the normal or Student's $t$ distribution to give the standard error. Multiplying this by the square root of the sample size yields the standard deviation. The only reservation about ...


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Assuming that all samples are from populations with the same variance $\sigma^2,$ you can estimate the variance as $$\hat\sigma^2 = \frac{\sum_{i=1}^n (r_i - 1)S_i^2}{\sum_{i=1}^n r_i\; - n},$$ where you have $n$ samples with $r_i\ge 2, i=1,\dots, n$ replications in each, and $S_i^2$ is the sample variance of the $r_i$ replications in sample $i.$ That is $...


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Your question shows some confusion! A sample is usually represented by a sequence of random variables iid $X_1, X_2, \dotsc, X_n, \dotsc$, where each random variable is a function from some sample space $\omega \mapsto X(\omega)$, the argument $\omega$ often omitted from the notation. So a realization is only one value, always. So $n$ in the CLT refers to ...


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