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The lower and upper bounds of the 95% confidence interval of the mean of a random variable with normal distribution of known variance $\sigma$ are $\bar{x}-1.96\frac{\sigma}{\sqrt{n}}$ and $\bar{x}+1.96\frac{\sigma}{\sqrt{n}}$.

$n$ is the size of the sample and $\bar{x}$ is the mean value of the sample.

One assumption is that the sample of size is obtained through a simple random sampling but what does it happen when the sample is not obtained through a simple random sampling but for example through a systematic sampling?

So, the alternative to simple random sampling is the systematic sampling where (quoting from wikipedia) the sampling starts by selecting an element from the list at random and then every $k$-th element in the frame is selected, where $k$, the sampling interval (sometimes known as the skip) is $k=\frac{N}{n}$ where $n$ is the sample size, and $N$ is the population size.

Is it meaningful to compute a confidence interval in this case?

What are the errors in using the above formula when the sampling is not a simple random sampling?

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  • $\begingroup$ Maybe take a look at highest denstity intervals. $\endgroup$ – Drey Feb 22 '17 at 10:18
  • $\begingroup$ @Drey are you meaning en.wikipedia.org/wiki/Credible_interval ? $\endgroup$ – Alessandro Jacopson Feb 22 '17 at 11:21
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    $\begingroup$ Yes. On another note: what you trying to compute? Systematic sampling is usefull when you specify group weights - I don't see any in hint about this in your question. If you have groups - why do you want to compute an overall mean? Why not group means? Why not, if you assume normality an ANOVA ... and so on. The answer to you question depends on the information which you want to extract, not on the method per se. $\endgroup$ – Drey Feb 22 '17 at 11:41
  • $\begingroup$ @Drey I have no "group weights", in my application a simple random sampling is not feasible while a systematic sampling is feasible. $\endgroup$ – Alessandro Jacopson Feb 22 '17 at 12:33
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    $\begingroup$ It would help alot if you describe what your application is. As for CIs, from frequentist appraoch, they rely on the underlying distribution (normal in your example above). However, if you can assume that your sample is i.i.d (which is probably not, based on your question) and you compute a statistic (on a subset) that involves sums you can apply CLT. Otherwise, it may help to open another question describing your data and what approaches to use. $\endgroup$ – Drey Feb 26 '17 at 17:14
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You're not specific as to which alternative to simple random sampling you have in mind. One could deliberately pick the $100$ largest observations in a simple random sample of $200$. Or the $100$ smallest. If one assumes a normally distributed population, then application of usual formula that you quote to the $100$ largest observations would hardly ever yield an interval that includes the population mean.

Or one could make the same of $200$ the $200$ largest in the whole population. That would be even worse: The validity of any conclusion you could draw would depend on the size of the population.

In any of these cases, you would not get a valid confidence interval from the usual formulas.

If you were to specify a particular alternative way of sampling then one might be able to work out a valid formula from a confidence interval when that kind of sampling is used. For example, suppose the observations are jointly normal and identically distributed and the correlation between any two of them is $1/2$. I expect one could work out a formula for use in that case. (Although just how such a distribution for the sample would arise I don't know.)

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  • $\begingroup$ The alternative to simple random sampling, as I wrote in the question, is the systematic sampling where (quoting from wikipedia) the sampling starts by selecting an element from the list at random and then every $k$-th element in the frame is selected, where $k$, the sampling interval (sometimes known as the skip) is $k=\frac {N}{n}$ where $n$ is the sample size, and $N$ is the population size. $\endgroup$ – Alessandro Jacopson Feb 18 '17 at 10:58
  • $\begingroup$ I've edited the question. $\endgroup$ – Alessandro Jacopson Feb 22 '17 at 10:13
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    $\begingroup$ If you sample from a finite population of size $N$, then the standard formula does not apply, because of the finite population size (although if $N$ is really really large and $n<<N$, the formula is about right). You will find plenty of information on that, if you google for survey sampling, Alessandro. However, it certainly makes sense to do a CI derived in a sensible way. $\endgroup$ – Björn Feb 22 '17 at 10:49
  • $\begingroup$ @Björn : You don't need to write $n<<N;$ you can write $n\ll N. \qquad$ $\endgroup$ – Michael Hardy Feb 23 '17 at 2:24

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