If your sample size is 5 the chance is large that all 5 units have a value very close to the large top (chance to ad randomly draw a unit there is the largest). Suppose you have a bimodal population distribution and one top is a lot larger than the other one.
Smartpls bootstrapping sample mean large how to#
This is merely an idea on how to determine how large your original sample size needs to be in order to be reasonably certain that the sample distribution corresponds with the population distribution. Your original population sample needs to be large enough to be moderately certain that the distribution of your population sample corresponds (equals) with the 'real' distribution of the population.
I know that the bootstrap samples take over the distribution of the population sample. Thanks for replying, Your answer was very useful to me especially the book links.īut I am afraid that in my attempt to give information I completely clouded my question. What do you guys think? If my premise makes sense, do I need to chose an $x$ larger than $10$, or smaller? Depending on your answers (do I need to feel embarrassed or not? :-) I'll be posting some more discussion ideas. But since I am not a statistician but an engineer whose statistics lessons took place in the days of yonder I cannot exclude the possibility I just generated a lot of rubbish :-). (Indirectly, the $n=45$ determines the size of my timespan: time needed to finish $45$ demands.) Since I know my population of interest (all the demands that are finished between present day and a day in the past) has less variance I can safely use a sample size of $n=45$ to bootstrap. If so I repeat the same steps but with a sample size of $40$, if not repeat with a sample size of $60$ (etc.).Īfter a while I conclude that $n=45$ is the absolute minimum sample size to get a more or less good representation of my 2011 population. I take an $n=50$, and see if my sample mean population is normally distributed by using Kolmogorov-Smirnov.
Smartpls bootstrapping sample mean large trial#
Now I use trial and error to determine a good sample size. I choose a value of $x$, suppose $10$ ( $x=10$). My 2011 population exists out of enough units to make $x$ samples of a sample size $n$. In my case (administrative processes: time needed to finish a demand vs amount of demands) I have a population with a multi-modal distribution (all the demands that are finished in 2011) of which I am 99% certain that it is even less normally distributed than the population (all the demands that are finished between present day and a day in the past, ideally this timespan is as small as possible) I want to research. In other words, your samples need to represent your population (distribution) well enough. Only when your sample size is large enough you can be certain that the population of your sample means is normally distributed (around the population mean). If my premise is correct you have the same problem when using the central limit theorem to determine the population mean. In order to be certain this is the case you need to make your sample size large enough. The bootstrap method is only useful if your sample follows more or less (read exactly) the same distribution as the original population. Nevertheless I am wondering if the following approach couldn’t be useful.
However, I'm not sure about how to handle this.I know this is a rather hot topic where no one really can give a simple answer for.
sequence of random variables with the common distribution function $F(\cdot) := \mathbb P(X_1 \leq \cdot)$ and finite first moment $\mathbb E <\infty$.
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