Question

a b total c x 0.26 0.5 d y 0.60 h e z 0.07 j total 1.0 1.0 1.0 which value for y in the table would be least likely to indicate an association between the variables? 0.06 0.24 0.60 1.0

Explanation:

Step1: Recall Independence in Contingency Tables

For two variables to be independent in a contingency table, the joint probability \( P(A \cap D) \) should equal the product of the marginal probabilities \( P(A) \times P(D) \). The marginal probability of \( A \) is \( 1.0 \) (wait, no, looking at the table, the marginal for \( A \) is \( 1.0 \)? Wait, no, the total for \( A \) column is \( 1.0 \), total for \( D \) row total? Wait, the table is a joint probability table, so the marginal probability of \( A \) is \( P(A) = 1.0 \)? No, wait, the total row and column are \( 1.0 \). Wait, actually, in a joint probability table, the sum of each row is the marginal probability of the row variable, and sum of each column is the marginal probability of the column variable. So for row \( D \) and column \( A \), the joint probability \( P(D \cap A) = Y \), and the marginal probability \( P(A) = 1.0 \)? Wait, no, the total column is \( 1.0 \), so \( P(A) = 1.0 \), and \( P(D) \) is the sum of row \( D \), which is \( Y + 0.60 + \) (wait, no, the row \( D \) has entries \( Y \), \( 0.60 \), and total \( H \). Wait, maybe I misread. Wait, the table has rows \( U \), \( D \), \( E \), Total and columns \( A \), \( B \), Total. So:

- Row \( U \): \( X \), \( 0.26 \), \( U \) (total)
- Row \( D \): \( Y \), \( 0.60 \), \( H \) (total)
- Row \( E \): \( Z \), \( 0.04 \), \( J \) (total)
- Column \( A \): \( X + Y + Z = 1.0 \)
- Column \( B \): \( 0.26 + 0.60 + 0.04 = 0.90 \)? Wait, no, column \( B \) total is \( 1.0 \), so \( 0.26 + 0.60 + 0.04 = 0.90 \), which contradicts the total column \( B \) being \( 1.0 \). Wait, maybe it's a typo, but the key point is: for independence, \( P(D \cap A) = P(D) \times P(A) \).

Since \( P(A) = 1.0 \) (from column \( A \) total), then \( P(D \cap A) = P(D) \times 1.0 = P(D) \). But \( P(D) \) is the sum of row \( D \), which is \( Y + 0.60 + \) (wait, no, row \( D \) total is \( H \), and column \( A \) total is \( 1.0 \), column \( B \) total is \( 1.0 \), total overall is \( 1.0 \).

Wait, maybe a better approach: In a joint probability table, two variables are independent if \( P(\text{row} \cap \text{column}) = P(\text{row}) \times P(\text{column}) \). So for row \( D \) and column \( A \), \( Y = P(D) \times P(A) \).

We know that \( P(A) = 1.0 \) (from column \( A \) total), and \( P(D) \) is the sum of row \( D \), which is \( Y + 0.60 + \) (wait, no, row \( D \) has entries \( Y \) (column \( A \)), \( 0.60 \) (column \( B \)), and total \( H \) (so \( H = Y + 0.60 \)).

But \( P(A) = 1.0 \), so \( P(D \cap A) = Y = P(D) \times P(A) = P(D) \times 1.0 = P(D) \). But \( P(D) = H = Y + 0.60 \). So \( Y = Y + 0.60 \), which is impossible. Wait, maybe \( P(A) \) is not \( 1.0 \). Wait, column \( A \) total is \( 1.0 \), so \( X + Y + Z = 1.0 \), and column \( B \) total is \( 0.26 + 0.60 + 0.04 = 0.90 \), but it's supposed to be \( 1.0 \), so maybe the \( 0.60 \) is a typo, but the question is about independence.

Wait, the key idea is: if two variables are independent, the cell probability should be equal to the product of the row marginal and column marginal. So for row \( D \) and column \( A \), \( Y = P(D) \times P(A) \).

Since the total column \( A \) is \( 1.0 \), \( P(A) = 1.0 \), and the total row \( D \) is \( H = Y + 0.60 \) (assuming column \( B \) entry is \( 0.60 \)). But if \( P(A) = 1.0 \), then \( P(D \cap A) = Y = P(D) \times 1.0 = P(D) \), so \( Y = Y + 0.60 \), which is impossible. So maybe \( P(A) \) is not \( 1.0 \), but the column \( A \) total is \( 1.0 \), so \( P(A) = 1.0 \), which would mean \( P(D \cap A) = Y = P(D) \times 1.0 = P(D) \), so \( Y = P(D) \), and \( P(D) = Y + 0.60 \), so \( Y = Y + 0.60 \), which is a contradiction. Therefore, maybe the table is a frequency table or a relative frequency table, but the key point is: the value of \( Y \) that is closest to \( P(D) \times P(A) \) (for independence) would indicate no association.

Alternatively, since \( P(A) \) is the total of column \( A \), which is \( 1.0 \), and \( P(D) \) is the total of row \( D \), which is \( Y + 0.60 \) (since row \( D \) has \( Y \) in \( A \), \( 0.60 \) in \( B \), and total \( H = Y + 0.60 \)). For independence, \( Y = P(D) \times P(A) = (Y + 0.60) \times 1.0 \), which implies \( Y = Y + 0.60 \), impossible. So maybe the column \( A \) total is not \( 1.0 \), but the total of the table is \( 1.0 \). So the grand total is \( 1.0 \), so:

- Sum of column \( A \): \( X + Y + Z = S \)
- Sum of column \( B \): \( 0.26 + 0.60 + 0.04 = 0.90 \), so \( S + 0.90 = 1.0 \), so \( S = 0.10 \). Therefore, \( X + Y + Z = 0.10 \), and column \( A \) total is \( 0.10 \), column \( B \) total is \( 0.90 \), and grand total \( 1.0 \).

Ah, that makes sense. So:

- Column \( A \) total: \( P(A) = 0.10 \)
- Column \( B \) total: \( P(B) = 0.90 \)
- Row \( U \) total: \( P(U) = X + 0.26 \)
- Row \( D \) total: \( P(D) = Y + 0.60 \)
- Row \( E \) total: \( P(E) = Z + 0.04 \)
- Grand total: \( P(U) + P(D) + P(E) = 1.0 \), and \( P(A) + P(B) = 1.0 \)

For independence, \( P(D \cap A) = Y = P(D) \times P(A) \)

We know \( P(A) = 0.10 \), so \( Y = 0.10 \times P(D) \)

But \( P(D) = Y + 0.60 \), so substituting:

\( Y = 0.10 \times (Y + 0.60) \)

\( Y = 0.10Y + 0.06 \)

\( Y - 0.10Y = 0.06 \)

\( 0.90Y = 0.06 \)

\( Y = \frac{0.06}{0.90} = 0.066... \approx 0.06 \)? Wait, no, solving \( Y = 0.10 \times (Y + 0.60) \):

\( Y = 0.10Y + 0.06 \)

\( Y - 0.10Y = 0.06 \)

\( 0.9Y = 0.06 \)

\( Y = 0.06 / 0.9 = 0.066... \approx 0.07 \), but the options are 0.06, 0.24, 0.60, 1.0.

Wait, maybe the table is a contingency table where the expected frequency (for independence) is \( E = \frac{\text{row total} \times \text{column total}}{\text{grand total}} \). Since grand total is 1.0 (probability), the expected value for \( Y \) (which is the observed frequency in cell \( D,A \)) under independence is \( E = P(D) \times P(A) \).

To have no association, \( Y \) should be equal to \( E = P(D) \times P(A) \).

Since \( P(A) = 1.0 \) (if column \( A \) total is 1.0), but that leads to a contradiction, so maybe the table is a 2x2 table? No, it's 3x2.

Alternatively, the question is about which \( Y \) is closest to the value that would be expected under independence, i.e., \( Y \approx P(D) \times P(A) \).

Since the total of column \( A \) is 1.0 (from the table), and total of row \( D \) is \( Y + 0.60 \), then for independence, \( Y = (Y + 0.60) \times 1.0 \), which is impossible, so maybe the table is misprinted, and the column \( A \) total is \( X + Y + Z = 1.0 \), and column \( B \) total is \( 0.26 + 0.60 + 0.04 = 0.90 \), which is wrong, so the key is: the value of \( Y \) that is closest to \( 0.60 \times \frac{1.0 - 0.26 - 0.04}{1.0} \)? No, maybe simpler: the least association is when \( Y \) is equal to the product of the row total and column total. Since the total of column \( A \) is 1.0, and total of row \( D \) is \( Y + 0.60 \), then \( Y = (Y + 0.60) \times 1.0 \) is impossible, so the next approach: in a contingency table, the closer the observed frequency is to the expected frequency (under independence), the less association. The expected frequency for cell \( (D,A) \) is \( E = \frac{\text{row } D \text{ total} \times \text{column } A \text{ total}}{\text{grand total}} \). Since grand total is 1.0, \( E = \text{row } D \text{ total} \times \text{column } A \text{ total} \).

Column \( A \) total is \( 1.0 - 0.26 - 0.60 - 0.04 = 0.10 \)? Wait, no, column \( B \) has 0.26, 0.60, 0.04, sum 0.90, so column \( A \) sum is 0.10. So column \( A \) total is 0.10, column \( B \) total is 0.90, grand total 1.0.

Row \( D \) total is \( Y + 0.60 \), column \( A \) total is 0.10, so expected \( Y \) (under independence) is \( (Y + 0.60) \times 0.10 \).

We want \( Y \) to be equal to this expected value, so:

\( Y = 0.10(Y + 0.60) \)

\( Y = 0.10Y + 0.06 \)

\( 0.9Y = 0.06 \)

\( Y = 0.06 / 0.9 = 0.066... \approx 0.06 \) (since 0.06 is one of the options). So the value closest to 0.066 is 0.06, so the answer is 0.06.

Step2: Evaluate the Options

Now, check the options: 0.06, 0.24, 0.60, 1.0.

- If \( Y = 0.06 \), then \( E = 0.10 \times (0.06 + 0.60) = 0.10 \times 0.66 = 0.066 \), which is close to 0.06.
- If \( Y = 0.24 \), \( E = 0.10 \times (0.24 + 0.60) = 0.10 \times 0.84 = 0.084 \), which is far from 0.24.
- If \( Y = 0.60 \), \( E = 0.10 \times (0.60 + 0.60) = 0.10 \times 1.20 = 0.12 \), far from 0.60.
- If \( Y = 1.0 \), \( E = 0.10 \times (1.0 + 0.60) = 0.16 \), far from 1.0.

So the value closest to the expected value (under independence) is 0.06, so it's least likely to indicate an association.

Answer:

0.06