What is the tension of the Hubble constant in standard deviations? Depending on the data, the tension in the measurement of the Hubble constant $H_0$ is up to 9 percent. This corresponds to about 5 sigma. I am interested how this standard deviation is calculated.
 A: You can compute the difference between the two measurements and propagate the error linearly. This yields (neglecting the units, which are always $\mathrm{km / s / Mpc}$):
$$
H_0^{\text{late}} - H_0^{\text{early}} \approx 5.9 \pm 0.9
$$
where $H_0^{\text{late}} = 73.3 \pm 0.8$ and $H_0^{\text{early}} = 67.4 \pm 0.5$ (Verde et al, 2019, Planck versus all late measurements combined, figure 1), while the error is computed by linear propagation: $\sigma_{\text{tot}} = \sqrt{\sigma_{\text{late}}^2 + \sigma_{\text{early}}^2}$.
If the measurements were compatible the difference should be around zero; it is not so by $5.9 / 0.9 \gtrsim 6$ standard deviations (the number is not exact since the decimals are truncated).
Linear error propagation means we approximate the distributions of the parameters with Gaussians, if they significantly differ from Gaussians (for example, if they are very asymmetric) you need more sophisticated ways of testing for compatibility; fortunately when averaging several measurements the result tends to resemble a Gaussian.
