# Test $Z$ and uncertainty

I am having some problems when I need to calculate the test $$Z$$. Suppose I got a value, by a measurement, of $$x = 3.44433$$ with uncertainty $$0.00003$$. And, for example, it is given to me a "theoretical/standard" value of this $$x$$, let's say it is $$3.44$$

How can I see, if both are compatible? I mean, the test $$Z$$ says that if $$|Z| = \frac{x-\mu}{\delta} < 3$$, then they are compatible.

The problem is, here, in the formula, should I use $$x-\mu = 0.00433$$, as a normal subtraction, or should I use $$x-\mu = 0.00$$, since the rules of significant figures claim that in the subtraction we just consider the resultant number until the last decimal of the number which has lesser number of significant figures?

Now, using the first method gives an incompatibility, and the second method, compatibility. So what is right?

• Where do the values and uncertainty come from? Are they emasured directly or calculated as averages of multiple measurements (in which case Student-t test may be more relevant)? Sep 20, 2021 at 11:39
• @RogerVadim Hello. It is just an example. I am aware of the concept of Student t test, but suppose here that there is a lot of measurements in such way that the test Z is allowed.
– LSS
Sep 20, 2021 at 11:43
• The question is essentially whether your number is really $3.44$ or $3.44000$ - you have to check this in the source where you take it from. Sep 20, 2021 at 11:53
• @RogerVadim noo. I think my question is not clear. My question is: SUPPOSE someone gives to you a lot of data, you need to use this data to ger the value of a parameter, let's call it x. Ok? Now, suppose we obtain the value of x as 3.44433 with the uncertainty said above. So we return to the person that give us the data, and he say that the value, for example (suppose we achieved our value manipulating the data, using propagation of error, adjust of functions, LSQ etc, matematically), for x is 3.44. A standard value of 3.44. How do we know if they are compatible? I use which method cited
– LSS
Sep 20, 2021 at 12:05
• @RogerVadim i Know the question is a little confused, since the value given by "the person" should have uncertainty. But that's literally my situation, the discipline relatory gives me a value without uncertainty. So i suppose we need to use the equation "given above, subtract the values and divide by the uncertainty i obtained. Again, the problem is about the subtraction, which results is right?
– LSS
Sep 20, 2021 at 12:07