Let the "barred" allele be caller B and the "non-white" allele b. Since chickens use chromosomes $Z$ and $W$ to determine sex, hens would have chromosomes $ZW$ , and roosters would have chromosomes $ZZ$ . A Z-linked gene is represented as a superscript on the $Z$ chromosome, $Z^{B}$ for the dominant allele and $Z^{b}$ for the ressesive allele.

A barred hen would have a copy of B on its Z chromosome, a non-barred rooster would have both copies of b on both Z chromosomes. Using Punnet squares to represent the crosses we get the following cases:

$\begin{center}\begin{tabular}{ |c|c|c|c| }\ & Z^{B} & W \\ \ Z^{b} & Z^{B}Z^{b} & Z^{b}W \\ \ Z^{b} & Z^{B}Z^{b} & Z^{b}W \\ \end{tabular}\end{center}$

That is a ratio of two barred heterozygote roosters to two non-barred hens. Crossing them we get:

$\begin{center}\begin{tabular}{ |c|c|c|c| }\ & Z^{b} & W \\ \ Z^{B} & Z^{B}Z^{b} & Z^{B}W \\ \ Z^{b} & Z^{b}Z^{b} & Z^{b}W \\ \end{tabular}\end{center}$