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1. ### Why does bootstraping with 50% samplesize and no replacements always gives appropriate confidence intervals.

@katxt: Thanks for your explanation and I guess you were right. It has something to do with the jackknife method. When d = n/2 then sqrt((n-d)/d) = 1 and I agree that bootstrapping without replacement is a good estimator for the jackknife method.
2. ### Why does bootstraping with 50% samplesize and no replacements always gives appropriate confidence intervals.

@hlsmith: In practise I saw people use different parametrizations of bootstrapping (e. g. 80% sample size with replacement). In papers they use "same sample size with replacement" but I haven't seen an argument why this parametrization is used. I agree with you that "same sample size with...

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4. ### Why does bootstraping with 50% samplesize and no replacements always gives appropriate confidence intervals.

Hello, I tried different kind of distributions. log-normal, chi², alpha, beta, t, uniform and so on. I didn't used smaller sample sizes because I think that this would lead to more variation in the results, so that they are less comparable.
5. ### Why does bootstraping with 50% samplesize and no replacements always gives appropriate confidence intervals.

Hello, I compared different ways to calculate confidence intervals. On the one side I used the direct formular on the other side I used a percentile bootstrap methoden. (Calculate the statistic (e. g. mean) on n subsamples and choose the a/2 and 1-a/2 percentile for the confidence interval.) I...