Can temperature data be averaged to a greater resolution than the raw data?

[statdad]

If I understand #35 correctly, your program is generating a pair of random numbers, A and B. A has a uniform distribution on [42.45, 42.55), so the mean $$\mu_A = 42.5$$. B has a uniform distribution on [42.95, 43.05), so its mean is $$\mu_B = 43$$. If you calculate $$\frac{A+B} 2$$ a large number of times (I assume you mean this instead of (AB)/2), you should expect most of the results to be close to

$$\mu_{\frac{A+B}2} = \frac{\mu_A + \mu_B}2 = 42.75$$

I don't know what you mean by this:

" I show that 85.44% of the time, 42.75 is closer to the actual average M than 42.8?"

 

[zgozvrm]

If I understand #35 correctly, your program is generating a pair of random numbers, A and B. A has a uniform distribution on [42.45, 42.55), so the mean $$\mu_A = 42.5$$. B has a uniform distribution on [42.95, 43.05), so its mean is $$\mu_B = 43$$.

Yes, but the point wasn't so much that the mean values were 42.5 and 43 as that the values, when rounded to the nearest 0.1, were 42.5 and 43 (basically the same difference).

And, for instance, A might be the value 42.45032157926 and B might be 43.049000783 in one iteration of the program.

If you calculate $$\frac{A+B} 2$$ a large number of times (I assume you mean this instead of (AB)/2), ...

Yes, that is what I meant

... you should expect most of the results to be close to

$$\mu_{\frac{A+B}2} = \frac{\mu_A + \mu_B}2 = 42.75$$

and they are, as shown by my program.

I don't know what you mean by this:

" I show that 85.44% of the time, 42.75 is closer to the actual average M than 42.8?"

I counted the number of times 42.75 was closer to the actual average and compared that sum to the number of times 42.8 was closer to the actual average. 42.75 was closer 85.44% of the time.

Thus making my point that 42.75 is a better representation of the average between measurements of 42.5 and 43, given that these measurements may actually be any value in the ranges given above (to any number of decimal places).

 

[statdad]

Then this shouldn't be a surprise - this is exactly what what SHOULD happen. I'm still not sure why you're mentioning 42.8.

 

[zgozvrm]

Then this shouldn't be a surprise - this is exactly what what SHOULD happen. I'm still not sure why you're mentioning 42.8.

It's not a surprise to me! According to Mark44, in post #24, there is no justification for "the 5" in 42.75, therefore he wants to say that the average is either 42.7 or 42.8. I chose 42.8 since 42.75 rounds to 42.8 (obviously, the result would've been the same, had I chosen 42.7).

 

[Mark44]

It's not a surprise to me! According to Mark44, in post #24, there is no justification for "the 5" in 42.75, therefore he wants to say that the average is either 42.7 or 42.8. I chose 42.8 since 42.75 rounds to 42.8 (obviously, the result would've been the same, had I chosen 42.7).

If you're going to cite what I said, please get it right. Here's what I said at the end of post #24 (complete with two typos I made).

If you take the average of the two rod lengths, you get 42.75 +/- 0.05, which represents a number somewhere between 42.7 and 42.8. As it happens, 42.75 is right smack in the middle of that interval, by there is no justifaction whatsoever for the 5 in the hundredths' place.

I DID NOT say that the average was either 42.7 or 42.8, as the quote above shows. Since we are dealing with only two measurements, and the measurements are of two different things, we can't invoke the increased precision that would come about from numerous (i.e., much more than two) measurements of the same thing (the rods are different).

 

[zgozvrm]

I DID NOT say that the average was either 42.7 or 42.8, as the quote above shows.

No, you didn't say that the average was either 42.7 or 42.8, but you DID say that there was no justification for the 5 in the hundredths' place. I'm saying that there IS.

Since we are dealing with only two measurements

All my posts refer to multiple measurements, which pertains to the OP.

 

[Mark44]

But you were responding to my post, and my example had two rods, with each one measured only once.