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A two-way table of sample data is given. If we use this data to conduct a chi-square test for association, fill in the table to give the expected count in each cell.

[tex]\[
\begin{tabular}{|l|lll|l|}
\hline & A & B & C & Total \\
\hline 1 & 34 & 34 & 32 & 100 \\
\hline 2 & 22 & 48 & 30 & 100 \\
\hline 3 & 153 & 74 & 73 & 300 \\
\hline Total & 209 & 156 & 135 & 500 \\
\hline
\end{tabular}
\][/tex]

Enter the exact answers.

[tex]\[
\begin{tabular}{|l|lll|l|}
\hline & A & B & C & Total \\
\hline 1 & $i$ & $j$ & $k$ & 100 \\
\hline 2 & $l$ & $m$ & $n$ & 100 \\
\hline 3 & $o$ & $p$ & $q$ & 300 \\
\hline Total & 209 & 156 & 135 & 500 \\
\hline
\end{tabular}
\][/tex]

Calculate the expected counts using the formula:

[tex]\[ E_{ij} = \frac{(row \ total \times column \ total)}{grand \ total} \][/tex]

Note: Replace [tex]\(i, j, k, l, m, n, o, p, q\)[/tex] with the exact expected counts.

Sagot :

GinnyAnsweridn
Sure! To fill in the expected counts for each cell in the table, we need to use the formula for the expected count in a contingency table. The formula for the expected count for a cell in row [tex]\(i\)[/tex] and column [tex]\(j\)[/tex] is:

[tex]\[ E_{ij} = \frac{( \text{row total for row } i) \times ( \text{column total for column } j )} {\text{grand total}} \][/tex]

Given the table:

[tex]\[ \begin{tabular}{|l|lll|l|} \hline & A & B & C & Total \\ \hline 1 & 34 & 34 & 32 & 100 \\ \hline 2 & 22 & 48 & 30 & 100 \\ \hline 3 & 153 & 74 & 73 & 300 \\ \hline Total & 209 & 156 & 135 & 500 \\ \hline \end{tabular} \][/tex]

We will calculate the expected values for each cell.

### Row 1:
- [tex]\(E_{11}\)[/tex] for Cell (1, A): [tex]\(\frac{(100) \times (209)}{500} = \frac{20900}{500} = 41.8\)[/tex]
- [tex]\(E_{12}\)[/tex] for Cell (1, B): [tex]\(\frac{(100) \times (156)}{500} = \frac{15600}{500} = 31.2\)[/tex]
- [tex]\(E_{13}\)[/tex] for Cell (1, C): [tex]\(\frac{(100) \times (135)}{500} = \frac{13500}{500} = 27.0\)[/tex]

### Row 2:
- [tex]\(E_{21}\)[/tex] for Cell (2, A): [tex]\(\frac{(100) \times (209)}{500} = \frac{20900}{500} = 41.8\)[/tex]
- [tex]\(E_{22}\)[/tex] for Cell (2, B): [tex]\(\frac{(100) \times (156)}{500} = \frac{15600}{500} = 31.2\)[/tex]
- [tex]\(E_{23}\)[/tex] for Cell (2, C): [tex]\(\frac{(100) \times (135)}{500} = \frac{13500}{500} = 27.0\)[/tex]

### Row 3:
- [tex]\(E_{31}\)[/tex] for Cell (3, A): [tex]\(\frac{(300) \times (209)}{500} = \frac{62700}{500} = 125.4\)[/tex]
- [tex]\(E_{32}\)[/tex] for Cell (3, B): [tex]\(\frac{(300) \times (156)}{500} = \frac{46800}{500} = 93.6\)[/tex]
- [tex]\(E_{33}\)[/tex] for Cell (3, C): [tex]\(\frac{(300) \times (135)}{500} = \frac{40500}{500} = 81.0\)[/tex]

So, the expected count table is filled in as follows:

[tex]\[ \begin{tabular}{|l|lll|l|} \hline & A & B & C & Total \\ \hline 1 & 41.8 & 31.2 & 27.0 & 100 \\ \hline 2 & 41.8 & 31.2 & 27.0 & 100 \\ \hline 3 & 125.4 & 93.6 & 81.0 & 300 \\ \hline Total & 209 & 156 & 135 & 500 \\ \hline \end{tabular} \][/tex]

Now, each cell contains the expected count based on the row totals, column totals, and the grand total.
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