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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.

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  • $\begingroup$ Are these $n$ tests from the same person (i.e. the same assumed underlying true BC)? $\endgroup$ Apr 20 '19 at 7:27
  • $\begingroup$ @COOLSerdash yes, these n tests are from the same person and taken in a short amount of time - hence we can assume the true BC stays constant. $\endgroup$
    – Alex
    Apr 21 '19 at 1:32
  • $\begingroup$ Great, then the answer of @BruceET is correct, in my view. $\endgroup$ Apr 21 '19 at 6:07
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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:

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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.

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  • $\begingroup$ (+1) The problem is that the range changes according to the real BC: For a real BC of 120, the $\pm$20% range is not $\pm$20 anymore, but $\pm$24 (from 96 to 144). As we don't know the real BC, assuming a standard deviation of 10 is assuming the real BC is 100, if I'm not mistaken. For a real BC of 120, the corresponding standard deviation would be 12.25. $\endgroup$ Apr 20 '19 at 7:21
  • $\begingroup$ Got that. But whatever the the $\pm 20\%$ turns out to be, taking the average of several will give a shorter confidence interval. $\endgroup$
    – BruceET
    Apr 20 '19 at 7:24
  • $\begingroup$ Are you talking about repeated measurements of the same underlying true BC, as in repeated measurements of the same person? If yes, your approach works fine. $\endgroup$ Apr 20 '19 at 7:25
  • $\begingroup$ The problem is indeed that I do not know what's the real BC, therefore I don't know the standard deviation to use for computing the interval of confidence from a normal distribution. How do I get the standard deviation when I have, let's say, n=6? $\endgroup$
    – Alex
    Apr 21 '19 at 1:36
  • $\begingroup$ Not sure exactly what you're asking. If you can guess the half width of the CI for normal data that's about $2\sigma/\sqrt{n}.$ If you know $n,$ you might get a good guess at $\sigma.$ $\endgroup$
    – BruceET
    Apr 21 '19 at 1:52

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