Interval of confidence (for several tests) Assume I am measuring the glycemic index with a blood glucose meter and its results fall within a $20\%$ range of real lab results.
Example if the real blood glucose (BC) is $100$, the meter can return a value $80 \leq x \leq 120$.
Given several tests (n), I would like to know how can I statistically define my interval of confidence for the actual blood glucose (BC), knowing that each result falls in an interval of $[-20\%, +20\%]$ of the real BC.
A naive approach would create a system on inequalities, but I want to approach this problem from a more rigorous statistic standpoint.
 A: Rather than trying to describe variability of individual meter measurements
as being the meter reading $\pm 20$ or $\pm 20\%,$ it might be better to make a confidence interval based on your $n$ measurements.
If $n$ is small, you might say the the standard deviation of
measurements is $\sigma = 10.$ That amounts to saying that
each individual true glucose value is within $\pm 20$ of the given meter reading 95% of the time. Then using statistical principles, you can get
somewhat shorter intervals based on the average of $n$ measurements. 
One important statistical principle here is that averages are less variable than individual values:

For example, suppose $n = 6$ meter readings on the same subject average $\bar X = 103.2.$ In that case, assuming a normal distribution, a 95%
confidence interval would be of the form $\bar X \pm 1.96\sigma/\sqrt{n}.$ For the numbers assumed above, the confidence interval
would be $(95.2,\, 111.2).$
If $n$ is larger, you might use the sample variance of the
actual meter measurements to estimate $\sigma$ (instead of your rule of $\pm 20$ for individual measurements). For example, here are $n = 15$
hypothetical measurements for which the sample mean is 
$\bar X = 99.54$ and the sample standard deviation 
$S = 7.07.$
 113.8 105.0 104.6  98.1  97.3  96.2 107.1  94.5
  95.2 100.4  99.5 107.8  90.7  96.1  86.8

A 95% confidence interval is of the form $\bar X \pm t^*S/\sqrt{n}.$ This is known as a t confidence interval because it is based on Student's t distribution. In this case with $n = 15$ meter readings,
$t^* = 2.14$ cuts probability $0.025$ from the upper tail of Student's t distribution with $n-1 = 14$ degrees of freedom.
Computation leads to the interval $(95.6, 103.4).$

Notes: (1) Formulas for sample mean and variance: 
$$\bar X = (X_1 + X_2 + \cdots + X_n)/n = \frac 1 n \sum_i X_i.$$
And
$$S = \sqrt{\frac{1}{n-1}\sum_i (X_i - \bar X)^2}.$$ You can
use a statistical calculator or statistical software to compute $\bar X$ and $S.$
(2) You can read about t confidence intervals in an elementary statistics textbook or in Wikipedia under Confidence Intervals.
