Statistics - suitable test for value to belong to a list of values

Question

I have a group of lists of numerical values representing a measurement of distance. In each group, the values are close to each other. Sometimes I have to pick one between two lists to add a new measurement to or I have to determine that the measurement doesn't statistically belong to the list it's been assigned to:

List A: $\{1.0, 1.0, 1.0, 0.998, 1.0,...\}$ Test Value $X = 0.96.$ Question: Does $X$ belong to List A?

The test should take into account how many elements are there in List A, such that if there is only one element {1.0} and the added value is close (within a given threshold), say 0.98 with threshold 0.05, then the new value is accepted. But if List A has 100 elements and all of them are 1.0 then the new value is rejected.

I'm somewhat familiar with Z-test, but I'm not sure it would work well because the $\sqrt{N}$ would be always $\sqrt{1}$ in the formula.

Answer 1

If you have $n$ observations from a normal population with sample mean $\bar X$ and sample standard deviation $S$ then a 95% prediction interval for an additional observation from the same population is $$\bar X \pm t^*S\sqrt{1 + \frac{1}{n}},$$ where $t^*$ cuts 2.5% of the probability from the upper tail of Student's t distribution with $n-1$ degrees of freedom. [If $n$ is as large as 100, then $t^* \approx 2.0.$] This interval is intended to predict how far from $\bar X$ a new observation from the same population might fall. That is not exactly what you have in mind.

I have to say that the kind of ad hoc tinkering with data you have described seems inappropriate. One might have hoped that the original list of $n$ is a random sample from a particular population. To append other values after the fact, possibly when conditions have changed, makes the new sample of size $n+1$ seem possibly inauthentic. But if you feel you must do this, then appending a value only if it lies within the prediction interval given above might do the least harm. Better, of course, would be to get the sample right on the first go.

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Example: If your original sample of 100 observations had $\bar X = 9.999$ and $S = 0.093,$ min = 9.77, max = 10.22, then the 95% PI was $(8.03, 11.97).$ The new observation 10.53 falls into the PI. If you append it to the original sample of 100, then the new sample mean is $\bar X_+ = 10.004$ and the new sample standard deviation is $S_+ = 0.107.$ The sample mean and SD have not been changed much by the tinkering. However, as a cautionary note, a boxplot of he 'enlarged sample' shows 10.53 as an outlier (see below), while the original sample showed no outliers.