# 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.

• Are these $n$ tests from the same person (i.e. the same assumed underlying true BC)? Apr 20 '19 at 7:27
• @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.
– Alex
Apr 21 '19 at 1:32
• Great, then the answer of @BruceET is correct, in my view. Apr 21 '19 at 6:07

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.

• (+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. Apr 20 '19 at 7:21
• Got that. But whatever the the $\pm 20\%$ turns out to be, taking the average of several will give a shorter confidence interval. Apr 20 '19 at 7:24
• 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. Apr 20 '19 at 7:25
• 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?
– Alex
Apr 21 '19 at 1:36
• 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 $\sigma.$ Apr 21 '19 at 1:52