Find the expected count and the contribution to the chi-square statistic for the (Treatment, Disagree)...

Solution

The row and column totals are,

$\begin{array}{|ccccccc|}\hline & \text{Strongly Agree}& \text{Agree}& \text{Neutral}& \text{Disagree}& \text{Strongly Disagree}& \text{Total}\\ \text{Control}& 35& 48& 3& 14& 9& 109\\ \text{Treatment}& 56& 43& 11& 5& 0& 115\\ \text{Total}& 91& 91& 14& 19& 9& 224\\ \hline\end{array}$

The expected count for the (Treatment, Disagree) cell is,

$$\begin{gathered} \text{Expected count}=\frac{(\text{Row total})(\text{Column totalt})}{\text{Grand total}} \\ =\frac{115\times 19}{224} \\ \frac{2185}{224} \\ =9.7545 \\ \approx 9.8 \end{gathered}$$

Thus, the expected count for the (Treatment, Disagree) cell is $9.8$. The contribution to the chi-square statistic for the (Treatment, Disagree) cell is,

$$\begin{gathered} \text{Contribution}=\frac{(\text{Observed-Expected}{)}^2}{\text{Expected}} \\ =\frac{(5-9.7545)}{9.7545} \\ \frac{22.60525025}{9.7545} \\ =2.317 \end{gathered}$$

Thus, the contribution to the chi-square statistic for the (Treatment, Disagree) cell is $2.317$.