Hi I was trying to understand the hubble tension. I know that the value infered for $H_{0}$ with supernovae is $H_{0}=74.03\pm 1.5 Km s^{-1} Mpc^{-1}$ and the value infered with the CMB using the standar model of cosmology is $H_{0}=67.39\pm 0.54 Km s^{-1} Mpc^{-1}$ and this values are in a $4.2\sigma$ tensión.
Statistically what is the meaning of a $4.2\sigma$ tension? How do I know when the tension is low and when the tension is high?
The figure 4.2 sigma means that the discrepancy between the two numbers is 4.2 times the estimated standard deviation of the difference between the two numbers, under the assumptions that the measurements are independent and the uncertainties have an approximately normal distribution.
If I have two measurements $a \pm b$ and $c \pm d$, where $b$ and $d$ are the standard deviations corresponding to the uncertainties in $a$ and $c$, then standard error propagation theory tells me that the standard deviation of $a-c$ is $\sigma =\sqrt{b^2 + d^2}$.
In this case $a-c$ is different from zero by 4.2 times that standard deviation.
Qualitatively, the larger this multiple is then the more significantly the difference between $a$ and $c$ is from zero (i.e. that $a$ and $c$ are different).
Quantitatively, we look at the properties of the normal distribution function and ask: if the distribution is centred at $4.2\sigma$ and has a standard deviation of $\sigma$, what fraction of the probability of the distribution function is at zero or below? Equivalently and symmetrically, we could say, if the distribution is centered at zero, what is the fraction of the probability contained beyond $4.2\sigma$.
You can look those values up in probability tables that give the integral under a normal distribution between $-\infty$ and $z$, where in this case $z=4.2$. This is the probability of rejecting the hypothesis that the two measurements are the same and is 0.99998665.
EDIT: A.V.S makes the point (as did I), that the above analysis assumes the probability distributions of the measurements follow a normal distribution. In particular, if the wings of the probability distribution are bigger - e.g. a Student's t-distribution - then this means the "overlap" between the measurement probability distributions is larger and so the significance of a $4.2\sigma$ discrepancy is lower than a calculation based on the normal distribution.
You have two data sets $A+\delta A$ and $B+\delta B$. To see that these two measurements agree or not we can calculate their difference. Let us call this value as $D$ such that $D=A-B$ and $\delta D = \sqrt{\delta A^2 + \delta B^2}$.
For instance, in some measurement, we find that
$D\pm\delta D \equiv 0.03\pm0.05$
In this case, as you can see D ranges between $(-0.02,0.08)$ and most importantly contains $0$. So we can say that, the measurement $A$ can be equal to $B$.
For the Hubble constant measurements, we have $A = 74.03\pm 1.5$ and $B = 67.39\pm 0.54$ which yields $D = 6.64 \pm 1.59$ In this case $D$ ranges from $(5.05, 8.23)$ and clearly does not contain $0$.
Well how far $D$ is to $0$ ? We can calculate it by taking $D / \delta D = 6.64 / 1.59 = 4.18$ or in rounded form $4.2$.
So $D$ is $4.2\sigma$ away from the $0$.
