The following table lists the number of construction workers who were exposed to carbon monoxide and reported either having a headache or shortness of breath and also their shift (morning vs evening). We want to conduct a test of the claim that a worker's symptom was independent of their shift using a significance level 0.1

Here is the sample data.
\begin{tabular}{|l|l|l|}
\hline & Headache & Breath \\
\hline Morning & 59 & 77 \\
\hline Evening & 21 & 36 \\
\hline
\end{tabular}

The expected observations for this table would be
\begin{tabular}{||l|l|l|}
\hline & Headache & Breath \\
\hline Morning & & \\
\hline & & \\
\hline
\end{tabular}

The resulting $\frac{(O-E)^{2}}{E}$ are:
\begin{tabular}{||l|l|l|}
\hline & Headache & Breath \\
\hline \hline Morning & & \\
\hline Evening & & \\
\hline
\end{tabular}

What is the chi-square test-statistic for this data? $\chi^{2}=$

Step 1 ：We are given a table that lists the number of construction workers who were exposed to carbon monoxide and reported either having a headache or shortness of breath and also their shift (morning vs evening). We want to conduct a test of the claim that a worker's symptom was independent of their shift using a significance level 0.1. The table is as follows: \n\n\begin{tabular}{|l|l|l|}\n\hline & Headache & Breath \\n\hline Morning & 59 & 77 \\n\hline Evening & 21 & 36 \\n\hline\n\end{tabular}

Step 2 ：We first need to calculate the expected data for each category (Headache and Breath, Morning and Evening). The expected data is calculated by multiplying the row total by the column total and dividing by the grand total. The expected observations for this table would be: \n\n\begin{tabular}{||l|l|l|}\n\hline & Headache & Breath \\n\hline Morning & 56.37305699 & 79.62694301 \\n\hline Evening & 23.62694301 & 33.37305699 \\n\hline\n\end{tabular}

Step 3 ：The Chi-square test statistic is calculated by summing the squares of the differences between observed (O) and expected (E) data, divided by the expected data, for all data points. The resulting \(\frac{(O-E)^{2}}{E}\) are: \n\n\begin{tabular}{||l|l|l|}\n\hline & Headache & Breath \\n\hline Morning & & \\n\hline Evening & & \\n\hline\n\end{tabular}

Step 4 ：The chi-square test-statistic for this data is calculated to be 0.707931167634729.

Step 5 ：Final Answer: The chi-square test-statistic for this data is \(\boxed{0.707931167634729}\).