 # 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? obojeneqk 2022-09-11 Answered
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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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.