Is $z_{0.025}$ equal to -1.96, or 1.96? I'm unsure about the interpretation of $z_{\alpha}$.
I've seen some source claim that $z_{\alpha}$ is equal to the z value with $\alpha$ of the area to the right of it. Under this view, $z_{0.025}$ is equal to the 97.5th percentile, and therefore equals 1.96.
Alternatively, I've read that $z_{\alpha}$ is the $\alpha$th percentile. Under this view, $z_{0.025}$ is equal to -1.96.
Which is the preferred interpretation? Is $z_{0.025}$ equal to 1.96, or -1.96?
 A: Percentage points. It has been traditional to use $z_{0.025}$ as the value $1.96$ that cuts probability $0.025$ from the upper tail of a standard normal distribution. (This subscript notation is often used in printed tables for normal, t, chi-squared, and other distributions. So you may also see notations such as $t_{10; .025}, \chi^2_{15;.05}, F_{5,15;.05}.$ Sometimes these is called a percentage points.)
From R,  $z_{0.025}, t_{10; .025)}, \chi^2_{15;.05},$ and $F_{5,15;.05}$ respectively, are as follows (to more decimal places than you would see in a printed table).
qnorm(.975)
[1] 1.959964
qt(.975, 10)
[1] 2.228139
qchisq(.95, 15)
[1] 24.99579
qf(.95, 5, 15)
[1] 2.901295

Quantiles. This is to be distinguished from quantile $0.025,$ which cuts probability $0.025$ from the lower tail of a distribution. In R, code qnorm(.025) (without additional parameters, the standard normal quantile function) returns $−1.959964$ and pnorm(-1.96) (CDF) returns $0.0249979.$
qnorm(.025)    # same as: qnorm(.025, 0, 1)
[1] -1.959964
pnorm(-1.96)
[1] 0.0249979

