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\begin{tabular}{|c|c|c|c|c|}
\cline{2-5} \multicolumn{1}{c|}{} & [tex]$A$[/tex] & [tex]$B$[/tex] & [tex]$C$[/tex] & Total \\
\hline [tex]$D$[/tex] & 0.12 & 0.78 & 0.10 & 1.0 \\
\hline [tex]$E$[/tex] & [tex]$R$[/tex] & [tex]$S$[/tex] & [tex]$T$[/tex] & 1.0 \\
\hline Total & [tex]$U$[/tex] & [tex]$X$[/tex] & [tex]$Y$[/tex] & 1.0 \\
\hline
\end{tabular}

Which value for [tex]$R$[/tex] in the table would most likely indicate an association between the conditional variables?

A. 0.09
B. 0.10
C. 0.13
D. 0.79

Sagot :

GinnyAnswer
To determine the value of \( R \) that most likely indicates an association between the conditional variables, let's perform a detailed analysis.

We are given the following table:

[tex]\[ \begin{tabular}{|c|c|c|c|c|} \cline { 2 - 5 } \multicolumn{1}{c|}{} & A & B & C & Total \\ \hline D & 0.12 & 0.78 & 0.10 & 1.0 \\ \hline E & R & S & T & 1.0 \\ \hline Total & U & X & Y & 1.0 \\ \hline \end{tabular} \][/tex]

From the data provided:
- The totals for each column (A, B, and C) add up to the overall total of 1.0.
- The total of row D is 1.0.
- The total of row E is also 1.0, which means:

[tex]\[ R + S + T = 1.0 \][/tex]

### Step-by-Step Solution

#### Step 1: Identify values for A, B, and C
Each entry in the D row must sum up to 1.0:
- \( A_D = 0.12 \)
- \( A_B = 0.78 \)
- \( A_C = 0.10 \)

#### Step 2: Calculate \( R \)
We need to calculate the left-over value \( R \).

[tex]\[ A_D + A_B + A_C = 0.12 + 0.78 + 0.10 = 1.0 \][/tex]

Since \( A_D, A_B, \) and \( A_C \) together already sum to 1.0 under row D, the sum of row E \( (R, S, T) \) should individually make up 1.0:

Let’s look at the possible values of \( R \):

- If \( R = 0.09 \):
[tex]\[ R + S + T = 0.09 + S + T = 1.0 \][/tex]
[tex]\[ S + T = 0.91 \][/tex]
- If \( R = 0.10 \):
[tex]\[ R + S + T = 0.10 + S + T = 1.0 \][/tex]
[tex]\[ S + T = 0.90 \][/tex]
- If \( R = 0.13 \):
[tex]\[ R + S + T = 0.13 + S + T = 1.0 \][/tex]
[tex]\[ S + T = 0.87 \][/tex]
- If \( R = 0.79 \):
[tex]\[ R + S + T = 0.79 + S + T = 1.0 \][/tex]
[tex]\[ S + T = 0.21 \][/tex]

#### Step 3: Choose the value of \( R \) that shows a likely natural distribution
Since values for \( S \) and \( T \) need to sum to a much more naturally balanced distribution, let’s analyze according to the mixing of higher interdependencies:

Considering our odd distributions for the plausible values:
- \( R = 0.79 \) leaves \( S \) and \( T \) very low combined \( 0.21 \).

Detailly, \( R = 0.09, 0.10, 0.13 \) leaves the remainder of \( S \) and \( T \):
- Hence \( R = 0.09, 0.13 \) is more equally probable.

Thus, a normalized association highly advises \( R = 0.13 \).

### Conclusion

[tex]\( R = 0.13 \)[/tex] is the valuable distributional figure that seems explainably natural, showing an association amongst the variables.
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