Will Osborn

Answered

2022-11-24

When we take ${X}_{1},...,{X}_{n}$ from $F\left(x\right)$. Then the ranks are ${X}_{\left(1\right)}<...<{X}_{\left(n\right)}$ and lets take the test-statistic,
$\begin{array}{r}{t}_{0}\left({X}_{1},...,{X}_{n}\right)=\sum _{i=1}^{n}{X}_{\left(i\right)}\end{array}$
If we look at $t\left(-{X}_{1},...,-{X}_{n}\right)$ to see if $t\left(-{X}_{1},...,-{X}_{n}\right)\stackrel{?}{=}-t\left({X}_{1},...,{X}_{n}\right)$, does the rank change?
For example is then $-{X}_{1},...,-{X}_{n}$ correspond to ${X}_{\left(1\right)}<...<{X}_{\left(n\right)}$?

Answer & Explanation

Jayda King

Expert

2022-11-25Added 8 answers

This test statistic is simply
$t\left({X}_{1},\dots ,{X}_{n}\right)=\sum _{i=1}^{n}{X}_{\left(i\right)}=\sum _{i=1}^{n}{X}_{i}=n{\overline{X}}_{n}$
If you flip the signs of each summand then you get $t\left(-{X}_{1},\dots ,-{X}_{n}\right)=-n{\overline{X}}_{n}$

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