Read the introduction to this first. If you don’t, you will be lost, lost, lost.
Logical probability answer to B
The answer to B follows from A. The picture is of what the mean might have been given the assumptions used above. A 1/9 chance the mean was 69.5, 2/9 it was 70, 3/9 for 70.5, etc. No error bars? Well, no: none are needed. This picture is the complete answer.
“Error bars” are classical and come from assuming some sort of parameterized probability model, like a normal (where the probability of seeing any observation is always zero), which we did not use and do not need here. No test statistics, no p-values, no parameters, no priors, no posteriors. Just probability.
Notice that the problem is entirely discrete? It’s not because we only averaged two days, but because the nature of the evidence is discrete (homework: find a non-discrete real-life example; I won’t wait). It would still be discrete no matter what finite number of days on which we took our mean. Our answer is exact given the assumptions; it has been deduced.
What about the month of Maxes, say 30 days? Still discrete. Each daily Max can take three values with equal probability (given our assumptions), and each of these can be combined with each other daily Max so that the mean is comprised of 330 = 2×1014 possibilities. Still discrete, but what a number!
Actually, it’s not as bad as that because not that many unique combinations can occur. It could be that every single time Max was measured, it was low by a degree, or every time high by a degree, or something in between, including the time every measurement was spot on. That makes only 61 possible values the mean of Max can take (every possibility from adding -30 to +30). Quite a reduction!
Start from the left: Assume the average is from every measurement by one degree low, which is equivalent to the sum of the actual temperatures minus 30, all divided by 30. There’s only one way out of the 2×1014 possibilities this can happen, so inverting that gives the probability. This is symmetric with every day being hot by one degree, which has the same probability.
Next: the temp could have been low 29 times and right once. That can happen 30 different ways, with a probability 30/330. This is also symmetric with 29 times high. Next: the temp could have been low 28 times and right twice, or low 29 times and right once, which is also symmetric (they all are).
You get the idea. All we need do is count the number of times each under or over could happen. A cute, eventually tedious, but not overwhelming combinatoric problem. Example: +/- 30 can happen just one way; +/- 29 can happen 30 ways (these are all each); +/- 28 can happen 465 ways; +/- 27 can happen 4930 ways; +/- 26 is 40,020 ways, and so on towards the peak at 0 (where the plus and minus errors balance). Summing all the different ways equals 330 (it must!). (Homework: what are number of ways for +/- 0?)
So I made up 30 days of Max temperatures somewhere around 70. Here’s the picture of what the mean can be, and the probabilities we deduced for these values given our assumptions.
My made-up Maxes were from 60 to 77. The computed average was 70.3666… The most likely value of the true mean is the same. The only values the mean could have been, given these conditions and data, are (rounded) 69.37, 69.4, …, 71.33, 71.37. This distribution is exact. There is an exact (to within roundoff in my calculations) probability of 0.087 for the mean to be 70.37. The others may be drawn from the figure.
For fun, I can report there is a 95.54% chance that the mean is in the set 70.0667, 70.1, …, 70.633 (I can give you all the numbers, but they are beside the point). There is no reason in the world to pick 95.54% except that it is close to the classical magical (magical classical?) value.
Did you notice the language? I did not say that there is a 95.54% chance that the mean is “between” 70.0667 to 70.633, because that is false. For one, those words leave out the endpoints, which are real possibilities. For another, only the discrete values in the set are possible. The mean might have been 70.0667 or 70.1, but it was impossible (given our etc.) that it could have been, say, 70.09, or any other value not in the discrete set.
The red line on the picture, which is cut off and which actually extends from 68.8 to 71.9, is the classical “95% confidence interval” on the parameter of a normal distribution model. Notice that this extends beyond the actual possibilities. The definition of the confidence interval means—ready?—nothing for any particular set of data (except the true mean lies in the interval or not), but even if you took the Bayesian view (same as the frequentist here for a flat prior) the interval still only speaks of a parameter. And even if you integrated the parameters out (let he who readeth understand), you’d still be left with an interval, which gives probabilities for impossible values (actually it gives probability 0 for every value!).
Don’t worry if the last paragraph made little sense. The point is this: the results we have are exact, and not the result of a parameterized probability model. Our results are deduced given the assumptions we used, and not calculated via some ad hoc model.
What happens if we change the assumptions? We change the results! Of course we do. All probability (all logic) is conditional. Change the conditions, change the conclusion.
What this answer isn’t
Because of measurement error, we were not certain of the mean, which is what we wanted to know. But we are certain of what the mean could have been, and its chances.
The results are not a prediction of future values of Max temperature. The are a prediction of what the mean of Max temperatures were during those 30 days, which we don’t know (again) because of measurement error.
There results are not statements about actual past temperatures, which we already knew, up to measurement error.
The results are also not what Kip originally asked for, but the answer to those questions are discovered in just the same was as these.
I’ll do the logical probability example most close to binomial next.