How to set a lower limit @ 95% CL for a measurement I have a result of a measurement, A(1.029) ± s(0.048), where s is one standard deviation. In a theory, A is restricted to be in range of [0,1]. The question is what is the correct way to set a lower limit on A at 95% Confidence level?
A similar discussion: Can one-sided confidence intervals have 95% coverage
Thanks,
MJK
 A: I think what you can do is choose the quantile such that upper limit is $1$ and then back calculate the lower limit. 
Let us first state your problem in a more formal way:
Let you have $n$ independent measurements $a_i$ for $A$. Assume the measurement is done with an error (which has expected value of zero) and therefore $a_i$ is random with some distribution around $A$ (alternatively normal distribution can be assumed but for large $n$ that is unimportant).
Now let $\bar{a}$ and $s$ be the mean and standard deviation of the sample $\{a_i\}_{i=1}^n$.
Using CLT and Slutsky's theorem, we have, for large $n$: (see this for details)
$\dfrac{\sqrt{n}(\bar{a}-A)}s \xrightarrow{d} Z \sim N(0,1)$ 
From this you get, confidence interval on $A$. Now say you objective is to get the most meaningful confidence interval on $A$. From the distribution of the statistic we have created we have:
Probability that the random interval $(\bar{a}-1.96s/\sqrt{n},\bar{a}+1.96s/\sqrt{n})$ contains $A$ be denoted by: $Pr((\bar{a}-1.96s/\sqrt{n},\bar{a}+1.96s/\sqrt{n}) \ni A)$
Therefore, we have from our result that for large $n$:
$Pr((\bar{a}-1.96s/\sqrt{n},\bar{a}+1.96s/\sqrt{n}) \ni A)=0.95$
The number $1.96$ came from $F^{-1}(0.975)$, where $F(.)$ is the cdf of normal distribution. 
Now let $q_2$ s.t. $\bar{a}+q_2s/\sqrt{n}=1$. 
Assume further that $F(q_2)>0.95$  ...................(A1)
Further, calculate $q_1$ s.t. $F(q_2) - F(q_1)=0.95$
This means, that $Pr(Z \in (q_1,q_2]) = 0.95$
Therefore, $Pr((\bar{a}-q_1s/\sqrt{n},\bar{a}+q_2s/\sqrt{n}] \ni A)= 0.95$
$\implies Pr((\bar{a}-q_1s/\sqrt{n},1] \ni A)= 0.95$
So $(\bar{a}-q_1s/\sqrt{n})$ becomes a more meaningful lower limit for $A$.
EDIT (I am retaining the above answer for sake of completeness and clarity)
I realized that your $\bar{a}>1$. In such a case, $q_2$ will come out to be negative to satisfy  $\bar{a}+q_2s/\sqrt{n}=1$. And because of this $F(q_2)<0.95$. In fact, $F(q_2)<0.5$. So A1 is not satisfied. Instead what you can do is the following:
Let $q_1=1.64$. This means that $Pr(Z\leq q_1)=0.95$.
This means, that $Pr((\bar{a}-q_1s/\sqrt{n},\infty) \ni A)= 0.95$
$\implies Pr((\bar{a}-q_1s/\sqrt{n},1] \ni A) + Pr((1,\infty) \ni A)= 0.95$
But we know from theory, $ Pr((1,\infty) \ni A)=0$.
Therefore, $Pr((\bar{a}-q_1s/\sqrt{n},1] \ni A)=0.95$
