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怎样计算两个正态分布之间的置信水平 (tension)

怎样计算两个正态分布之间的置信水平 (tension)

作者: lekf123 | 来源:发表于2018-08-15 11:26 被阅读0次

转自:https://physics.stackexchange.com/questions/366367/how-to-calculate-the-sigma-level-between-two-normal-distribution

问题:

Recently, the improved local measurement H_0=73.24 \pm 1.74 km s^{-1}Mpc^{−1} from Riess et al. 2016 (hereafter R16) exhibits a stronger tension with the Planck 2016 release H_0=66.93 \pm 0.62 km s^{−1}Mpc^{−1} (hereafter P15) at the 3.4\sigma level. ... The value of the Hubble constant H_0=68.34^{+0.53}_{−0.92} is consistent with the P15’s result at the 1\sigma confidence level.

How do we figure {3.4\sigma} and 1\sigma level out?

解决:

Trivially, if you have A \pm \delta A and B \pm \delta B, then A−B can be determined (assuming the usual approximations of independent uncertainties that are normally distributed) with a precision of {\sqrt{ (\delta A)^2+(\delta B)^2}}.
In this case, the difference in the two determinations of H_0 is 6.31\pm 1.84 km/s per Mpc. i.e. different from zero by 3.4 error bars. The σ refers to the size of the uncertainty ; or more specifically, it represents the 68% confidence interval in a normal distribution.

计算过程:

  1. 73.24 - 66.93 \simeq 6.31
  2. \sqrt{{1.74^2 + 0.62^2}} \simeq 1.85
  3. 6.31 / 1.85 \simeq 3.4
  1. \frac{{(68.34-66.93)}}{\sqrt{0.62^2 + 0.92^2}} \approx \frac{1.41}{1.11} \approx 1.27

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