In statistics, why do we sometimes use the symbol $\pm$ to mean interval of numbers, for example in confidence intervals, and sometimes to mean two numbers, for example in the formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$?

  • 6
    $\begingroup$ In both cases the meaning is the same “+ or -“ $\endgroup$
    – Tim
    May 7 '21 at 12:38
  • 1
    $\begingroup$ Hint: An interval is fully defined by its endpoints, which is two numbers. $\endgroup$
    – Ben
    May 7 '21 at 13:45
  • 1
    $\begingroup$ One can say that the confidence interval for an estimate is $(\hat{b}+Z_{\alpha/2} \hat{\sigma}, \hat{b}+Z_{1-\alpha/2} \hat{\sigma})$ xor the confidence limits for an estimate are $\hat{\beta} \pm Z_{1-\alpha/2}\hat{\sigma}$. But the latter is not an interval! That's the only case where I've seen the language abused in probability and statistics. $\endgroup$
    – AdamO
    May 7 '21 at 16:28

It doesn't mean two different things. The meaning is the same, $\pm$ means "the $\pm$ symbol may be used to indicate a symbol that may be replaced by either the plus and minus signs, $+$ or $−$, allowing the formula to represent two values or two equations". "$a \pm b$" means "$a + b$ or $a - b$". Same with $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, you could replace $\pm$ symbol with $+$ or $-$ to get the two possible answers from the formula.

When defining the interval, this is shorthand for defining its bounds, $x \pm c$ means the $[x - c, x + c]$ interval bounds. Obviously, this makes sense only for an interval that is symmetric and centered around the origin, it wouldn't make sense for an asymmetric interval (think of skewed distribution).


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