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Answer :
We wish to “fill in” the missing expected counts by verifying that the large counts condition is met. Under the null hypothesis of independence, the expected count in any cell is given by
[tex]$$
\text{Expected count} = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}.
$$[/tex]
Although we do not have the full marginal totals explicitly provided, we can use the following idea: when the expected counts are computed, the ratio between the “Yes” and “No” entries in each column equals the overall ratio of Yes responses to No responses. That is, if the overall proportion of Yes responses is
[tex]$$
r = \frac{\text{Yes row total}}{\text{grand total}},
$$[/tex]
then for a given column with total [tex]$T$[/tex] we have
[tex]$$
\text{Yes expected count} = r \,T \quad \text{and} \quad \text{No expected count} = (1-r) \,T.
$$[/tex]
In the table below, some cells are given and the missing expected counts are labeled as [tex]$A$[/tex], [tex]$B$[/tex], [tex]$C$[/tex], [tex]$D$[/tex], and [tex]$E$[/tex]:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
\textbf{Education} & \text{Less than HS} & \text{HS Grad} & \text{Some College} & \text{College Grad} & \text{Postgrad Degree} \\
\hline
\text{Yes} & 76.5 & 109.4 & A & B & 150.4 \\
\hline
\text{No} & C & D & 90.8 & 163.5 & E \\
\hline
\end{array}
\][/tex]
Notice that in three columns one of the two cells is provided and in two columns the “Yes” value is provided. We can use these “paired” cells to solve for the missing counts as follows.
1. In any column, the ratio of the Yes to No expected counts equals the overall ratio of Yes to No:
[tex]$$
\frac{\text{Yes cell}}{\text{No cell}} = \frac{r}{1-r}.
$$[/tex]
2. Although the full marginal totals are not given, we can use the known cells to approximate the overall ratio. In the Yes row, three cells are given: for Less than HS, HS Grad, and Postgrad Degree. Their sum is
[tex]$$
76.5 + 109.4 + 150.4 = 336.3.
$$[/tex]
In the No row, two cells are given (Some College and College Grad):
[tex]$$
90.8 + 163.5 = 254.3.
$$[/tex]
Taking these as partial totals, the overall proportion of Yes responses is approximated by
[tex]$$
r \approx \frac{336.3}{336.3 + 254.3} = \frac{336.3}{590.6} \approx 0.56942.
$$[/tex]
Then, the proportion of No responses is
[tex]$$
1 - r \approx 0.43058.
$$[/tex]
3. We now compute the missing expected counts using the formulas
- For a column where the No cell is known and the Yes cell is missing (like “Some College”):
The expected Yes count (labeled [tex]$A$[/tex]) satisfies
[tex]$$
A = \frac{r}{1-r} \times 90.8.
$$[/tex]
Substituting the values we have:
[tex]$$
A \approx \frac{0.56942}{0.43058} \times 90.8 \approx 120.0788.
$$[/tex]
- For the column “College Grad,” where the No cell is known (163.5) and the Yes cell is missing (labeled [tex]$B$[/tex]):
[tex]$$
B = \frac{r}{1-r} \times 163.5 \approx \frac{0.56942}{0.43058} \times 163.5 \approx 216.2212.
$$[/tex]
- For a column where the Yes cell is known and the No cell is missing, the formula is rearranged. For “Less than HS” where the Yes cell is 76.5 and the No cell is missing (labeled [tex]$C$[/tex]):
[tex]$$
76.5 = r \, (76.5 + C) \quad\Longrightarrow\quad C = \frac{76.5(1-r)}{r}.
$$[/tex]
Hence,
[tex]$$
C \approx \frac{76.5 \times 0.43058}{0.56942} \approx 57.8470.
$$[/tex]
- Similarly, for “HS Grad” where the Yes cell is 109.4 and the No cell is missing (labeled [tex]$D$[/tex]):
[tex]$$
D = \frac{109.4(1-r)}{r} \approx \frac{109.4 \times 0.43058}{0.56942} \approx 82.7250.
$$[/tex]
- Finally, for “Postgrad Degree” where the Yes cell is 150.4 and the No cell is missing (labeled [tex]$E$[/tex]):
[tex]$$
E = \frac{150.4(1-r)}{r} \approx \frac{150.4 \times 0.43058}{0.56942} \approx 113.7280.
$$[/tex]
4. The computed missing expected counts are:
[tex]$$
A \approx 120.0788,\quad B \approx 216.2212,\quad C \approx 57.8470,\quad D \approx 82.7250,\quad E \approx 113.7280.
$$[/tex]
These values indicate that each cell of the contingency table has been filled so that the large counts condition for conducting the chi-square test is met.
Thus, the answers are:
[tex]$$
\boxed{A \approx 120.08,\quad B \approx 216.22,\quad C \approx 57.85,\quad D \approx 82.73,\quad E \approx 113.73.}
$$[/tex]
All calculations are rounded appropriately for clarity.
[tex]$$
\text{Expected count} = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}.
$$[/tex]
Although we do not have the full marginal totals explicitly provided, we can use the following idea: when the expected counts are computed, the ratio between the “Yes” and “No” entries in each column equals the overall ratio of Yes responses to No responses. That is, if the overall proportion of Yes responses is
[tex]$$
r = \frac{\text{Yes row total}}{\text{grand total}},
$$[/tex]
then for a given column with total [tex]$T$[/tex] we have
[tex]$$
\text{Yes expected count} = r \,T \quad \text{and} \quad \text{No expected count} = (1-r) \,T.
$$[/tex]
In the table below, some cells are given and the missing expected counts are labeled as [tex]$A$[/tex], [tex]$B$[/tex], [tex]$C$[/tex], [tex]$D$[/tex], and [tex]$E$[/tex]:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
\textbf{Education} & \text{Less than HS} & \text{HS Grad} & \text{Some College} & \text{College Grad} & \text{Postgrad Degree} \\
\hline
\text{Yes} & 76.5 & 109.4 & A & B & 150.4 \\
\hline
\text{No} & C & D & 90.8 & 163.5 & E \\
\hline
\end{array}
\][/tex]
Notice that in three columns one of the two cells is provided and in two columns the “Yes” value is provided. We can use these “paired” cells to solve for the missing counts as follows.
1. In any column, the ratio of the Yes to No expected counts equals the overall ratio of Yes to No:
[tex]$$
\frac{\text{Yes cell}}{\text{No cell}} = \frac{r}{1-r}.
$$[/tex]
2. Although the full marginal totals are not given, we can use the known cells to approximate the overall ratio. In the Yes row, three cells are given: for Less than HS, HS Grad, and Postgrad Degree. Their sum is
[tex]$$
76.5 + 109.4 + 150.4 = 336.3.
$$[/tex]
In the No row, two cells are given (Some College and College Grad):
[tex]$$
90.8 + 163.5 = 254.3.
$$[/tex]
Taking these as partial totals, the overall proportion of Yes responses is approximated by
[tex]$$
r \approx \frac{336.3}{336.3 + 254.3} = \frac{336.3}{590.6} \approx 0.56942.
$$[/tex]
Then, the proportion of No responses is
[tex]$$
1 - r \approx 0.43058.
$$[/tex]
3. We now compute the missing expected counts using the formulas
- For a column where the No cell is known and the Yes cell is missing (like “Some College”):
The expected Yes count (labeled [tex]$A$[/tex]) satisfies
[tex]$$
A = \frac{r}{1-r} \times 90.8.
$$[/tex]
Substituting the values we have:
[tex]$$
A \approx \frac{0.56942}{0.43058} \times 90.8 \approx 120.0788.
$$[/tex]
- For the column “College Grad,” where the No cell is known (163.5) and the Yes cell is missing (labeled [tex]$B$[/tex]):
[tex]$$
B = \frac{r}{1-r} \times 163.5 \approx \frac{0.56942}{0.43058} \times 163.5 \approx 216.2212.
$$[/tex]
- For a column where the Yes cell is known and the No cell is missing, the formula is rearranged. For “Less than HS” where the Yes cell is 76.5 and the No cell is missing (labeled [tex]$C$[/tex]):
[tex]$$
76.5 = r \, (76.5 + C) \quad\Longrightarrow\quad C = \frac{76.5(1-r)}{r}.
$$[/tex]
Hence,
[tex]$$
C \approx \frac{76.5 \times 0.43058}{0.56942} \approx 57.8470.
$$[/tex]
- Similarly, for “HS Grad” where the Yes cell is 109.4 and the No cell is missing (labeled [tex]$D$[/tex]):
[tex]$$
D = \frac{109.4(1-r)}{r} \approx \frac{109.4 \times 0.43058}{0.56942} \approx 82.7250.
$$[/tex]
- Finally, for “Postgrad Degree” where the Yes cell is 150.4 and the No cell is missing (labeled [tex]$E$[/tex]):
[tex]$$
E = \frac{150.4(1-r)}{r} \approx \frac{150.4 \times 0.43058}{0.56942} \approx 113.7280.
$$[/tex]
4. The computed missing expected counts are:
[tex]$$
A \approx 120.0788,\quad B \approx 216.2212,\quad C \approx 57.8470,\quad D \approx 82.7250,\quad E \approx 113.7280.
$$[/tex]
These values indicate that each cell of the contingency table has been filled so that the large counts condition for conducting the chi-square test is met.
Thus, the answers are:
[tex]$$
\boxed{A \approx 120.08,\quad B \approx 216.22,\quad C \approx 57.85,\quad D \approx 82.73,\quad E \approx 113.73.}
$$[/tex]
All calculations are rounded appropriately for clarity.
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