@memester,
memester;113577 wrote:the sets used are ( 80, 50) and (50, 80).
I haven't been ignoring your question, by the way. I just discovered that my copy of SPSS is not compatible with Windows 7, which I upgraded to last week, so a few hundred bucks out of my pocket and I'll be up and running (which I need for more pressing things anyway).
At any rate, if you take your data set and use a statistical comparison that compares means, like the Student's T-test, you'll get a P-value of 1 because the means are identical. (This is using a 2-tailed test, which is necessary because a temperature can be below 50 or above 80, i.e. you can have variance in either direction -- a 1-tailed test would give you a P-value of 0.5).
It's sort of a strange question, though, because embedded in most of these formulae is the assumption of a sort of distribution (i.e. normal or not normal), and it's mathematically impossible to make that determination with only two data points per set.
Given that we cannot "assume" normal distribution or pairing for these samples, I would normally use a Mann-Whitney U test. This I cannot do in Excel, which I incidentally have found rather limited for stats -- better for organizing data than for analyzing it. If I get SPSS up and running next week when I can make it to the bookstore, I'd be happy to run it.
As for the Chi-squared analysis, this was perhaps not the best test to choose because it's really a measurement of distribution and not means (it's also properly speaking a Chi value and not a p-value).
That's the best way I have to answer the question without just throwing out numbers. Obviously having greater sample sizes allows for more robust statistics because the distribution of data becomes more precise and the effect of outliers is minimized. The statistical test you choose depends in part on the question you're asking, the assumptions you make about your data, and parameters like whether they're paired or not and unidirectional or bidirectional.