# Sample size in Confidence Intervals. In repeating confidence interval experiments, are we allowed to take samples of different size every time? Because a confidence interval of 95% means that if the sampling process is repeated infinite times, 95% of all the intervals obtained will have the parameter of interest. So wrt this process of repeating infinite times - do we have to repeat with the same sample size or can we vary it?

Sample size in Confidence Intervals
In repeating confidence interval experiments, are we allowed to take samples of different size every time? Because a confidence interval of 95% means that if the sampling process is repeated infinite times, 95% of all the intervals obtained will have the parameter of interest.
So wrt this process of repeating infinite times - do we have to repeat with the same sample size or can we vary it?
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Bordenauaa
Step 1
As it is always the case, the answer is: it depends. Let me elaborate. Suppose that ${X}_{1},...,{X}_{n}$ are iid $N\left(\mu ,{\sigma }^{2}\right)$. Then, one can show that
$\sqrt{n-1}\frac{{\overline{X}}_{n}-\mu }{\stackrel{^}{\sigma }}\sim {T}_{n-1},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 2.$
where ${\overline{X}}_{n}=\frac{1}{n}\sum _{i=1}^{n}{X}_{i}$, ${\stackrel{^}{\sigma }}^{2}=\frac{1}{n-1}\sum _{i=1}^{n}\left({X}_{i}-{\overline{X}}_{n}\right)$ and ${T}_{n-1}$ has a Students T distribution with $n-1$ degrees of freedom. This say that, for any sample size, the confidence interval
${\overline{X}}_{n}±{t}_{\left(n-1\right),\alpha /2}\frac{\stackrel{^}{\sigma }}{\sqrt{n-1}},$
where $P\left({T}_{n-1}\ge {t}_{\left(n-1\right),\alpha /2}\right)=\alpha /2$, will cover $\mu$ with probability $1-\alpha$ (see "In all likelihood" by Yudi Pawitan). So n is not important because ${X}_{i}\sim N\left(\mu ,{\sigma }^{2}\right)$. (Of course it is important for calculating each term, but the main thing is the probability of covering $\mu$).
Step 2
On the other hand, if you use an asymptotic interval as Walds confidence interval:
${\stackrel{^}{\theta }}_{n}±{z}_{\alpha /2}\sqrt{I\left({\stackrel{^}{\theta }}_{n}\right)},$
where $P\left(N\left(0,1\right)\ge {z}_{\alpha /2}\right)=\alpha /2$, $\stackrel{^}{\theta }$ is the maximum likelihood estimator and $I\left(\theta \right)$ is the Fisher information, then n does matter, since the level of approximation to the normal is improved with n. (See again the book of Pawitan and the Beery-Essen theorem in "Approximation theorems of mathematical statistics" by Robert Serfling). In this case, if you use different n's you will see better or worse intervals depending on n.