Question 3 You are conducting a multinomial hypothesis test for the claim that all 5 categories are equally likely to be selected. Complete the table. Round to 3 decimal places. \begin{tabular}{|c|c|c|c|} \hline Category & \begin{tabular}{c} Observed \\ Frequency \end{tabular} & \begin{tabular}{c} Expected \\ Frequency \end{tabular} & $\frac{(O-E)^{2}}{E}$ \\ \hline A & 22 & & \\ \hline B & 6 & & \\ \hline C & 12 & \\ \hline D & 23 & \\ \hline E & 22 & \\ \hline \end{tabular} What is the chi-square test-statistic for this data? \[ \chi^{2}= \] Report all answers accurate to three decimal places.

Step 1 ：Calculate the total observed frequency, which is the sum of the observed frequencies for all categories: \(22 (A) + 6 (B) + 12 (C) + 23 (D) + 22 (E) = 85\).

Step 2 ：Since we are testing the claim that all 5 categories are equally likely to be selected, the expected frequency for each category is the total observed frequency divided by the number of categories: \(Expected frequency = \frac{Total observed frequency}{Number of categories} = \frac{85}{5} = 17\).

Step 3 ：Complete the table with the observed and expected frequencies, and the calculation of \(\frac{(O-E)^{2}}{E}\) for each category.

Step 4 ：Calculate the chi-square test-statistic, which is the sum of the values in the last column of the table: \(\chi^{2} = \frac{(22-17)^{2}}{17} + \frac{(6-17)^{2}}{17} + \frac{(12-17)^{2}}{17} + \frac{(23-17)^{2}}{17} + \frac{(22-17)^{2}}{17}\).

Step 5 ：Simplify the chi-square test-statistic: \(\chi^{2} = \frac{25}{17} + \frac{121}{17} + \frac{25}{17} + \frac{36}{17} + \frac{25}{17} = 1.471 + 7.118 + 1.471 + 2.118 + 1.471 = 13.649\).

Step 6 ：So, the chi-square test-statistic for this data is \(\boxed{13.649}\) (rounded to three decimal places).