Statistical test for the difference in mean for two normal distributions

Question

I have measured $x$ and $y$ and estimated their uncertainty. They are both distributed normally. I found $x = 5 \pm 0.25$ and $y = 7 \pm 0.5$.

I want to know if these estimates are statistically different. Is it as simple as calculating $\sigma = |5 - 7|/(0.25 + 0.5) = 2.666...$, and hence (using R) $P(z > \sigma) = {\tt 1-pnorm(2.666)}\simeq 0.0038$?

Or should it be: $\sigma = |5 - 7|/\sqrt{0.25^2 + 0.5^2} \simeq 3.57$, and hence $P(z > \sigma) = \simeq 0.000178$?

Answer 1

It sounds like you've measured two samples and want to hypothesis test if they are different from each other. In which case, use chi-squared test if population standard deviation is known (unlikely), in R it's chisq.test. Or if population standard deviation is unknown (very common) then use the t test instead. In R it's just t.test(vector1, vector2). If you go with the T test then your null hypothesis is $H_0:\bar{X_1}=\bar{X_2}$ and the aletnaritve is taht they are not equal. Your formula is very close, but it's

$$\frac{\bar{X_1}-\bar{X_2}}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$$

where

$$S_p=\sqrt{\frac{S^2_1(n_1-1)+S^2_2(n_2-1)}{n_1+n_2-2}}$$

$S_1$ and $S_2$ are just the sample standard deviations which you have already figured out. And then you got to look it up on a T test table with degrees of freedom$=n_1+n_2-2$ If you know the standard deviation then the chi squared test uses $\sigma$ instead of estimating it with $S_p$ and you look it up on the chi squared table. Again, r has all this, but you wrote a lot of maths so i thought I'd write this here just in case.

If you wanted to know how to add and subtract normal distributions from a pure maths perspective then look here. Note that they have to be independent, otherwise there's a different way to work it out (which i can link further, but leave a comment). Sorry if I've missed your point.