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I'm no expert in the bootstrap, but I anticipate that using fewer samples in your bootstrap means that the uncertainty in mean will be inflated. The standard error of the sampling distribution is inversely proportional to the square root of the sample size. A smaller sample size thus means larger uncertainty in your estimates. Its easy to see evidence of ...


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In terms of power against the overall null hypothesis in an ANOVA (all groups same), the most efficient design for $5n$ subjects in 5 groups is to have $n$ subjects per group. The most sensible departure from this balanced design that I have seen is to put slightly more subjects in the control group because, among ad hoc comparisons, the most important are ...


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You don't necessarily need more samples to make multiple inferences, but you may want to keep in mind that as you make more estimates, you're more likely to make a wrong estimate somewhere. Your sample size of 1066 will allow you to estimate proportions that will be within 3 percentage points of the true value 95% of the time (and they'll be even closer if ...


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Yes, $km^2$ can be used as a measurement unit but I would say that 20$km^2$ is not the sample size but rather than these 20 $km^2$ are part of the sample area of the study. For example, if we have 10 areas of 2 $km^2$ each, our sample size is $n$=10 with a total sample area of 20 $km^2$. Usually in cases where the sampling area (or time) is varying among our ...


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The Maths is correct; the interpretation of what is the baseline needs a bit adjusting to get back to the EM result. When comparing two arbitrary proportion coming from equal sized population, indeed the base-line proportion is $\bar{p}$ is $\frac{p_1 + p_2}{2}$. When though making an A/B test the base-line proportion is $p_1$. As such if we use $\bar{p}=p_1$...


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The "worst-case" for the width of a Wald-based CI around a proportion is when p=0.5 (the variance is p(1-p)/n; for a fixed n, p(1-p) is maximised when p=0.5 as you can prove using calculus or show by graphing), so if you want to be safe, this is a sensible choice. If you were confident that most people have a dog (at least 0.8), few people were ...


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I was not able to reproduce the results you got from WebPower using the pilot data you supplied. I was able to reproduce your R code however. You are correct that you can't use the $\eta^2$ for Cohen's f, but $f^2 = \frac{\eta^2}{1-\eta^2}$ "However, how should I compute the effect size from the pilot study" - use the $\eta^2$ from the pilot ...


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