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For reference, here is the $\chi^2$ distribution table for degrees of freedom = 7:

p     1%        5%        95%     99%
v=7   1.239     2.167     14.07   18.48

It means that 99% of the time $V \leq 18.48$, where $V$ is the chi-squared statistic. And 1% of the time we expect $V \leq 1.239$.

Can we combine these so that there is a range of expected values $1.239 < V \leq 18.48$ ?

And do we expect it to be outside this range 2% of the time?

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up vote 1 down vote accepted

If $V\sim \chi^2(7)$ and $F$ denotes its distribution function, i.e. $F(v)=P(V\leq v)$, then these values express that $F(1.239)=0.01, F(2.167)=0.05,\ldots, F(18.58)=0.99$. Using this we can obtain $$ P(1.239<V\leq 18.48)=P(V\leq 18.48)-P(V\leq 1.239)=F(18.58)-F(1.239)=0.98. $$ And then $$ P(V\notin (1.239,18.48])=1-0.98=0.02 $$ as you claimed.

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