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For each table, calculate the mean weight for each group, [tex]$\bar{x}_A$[/tex] and [tex]$\bar{x}_B$[/tex], and find the difference of the mean of group A and the mean of group B ([tex]$\bar{x}_A - \bar{x}_B$[/tex]).

Type the correct answer in each box.

[tex]\[
\begin{tabular}{|c|c|}
\hline 13.6 & 9.2 \\
\hline 12.1 & 8.2 \\
\hline 15.9 & 11.5 \\
\hline 11.2 & 13.8 \\
\hline 9.7 & 14.6 \\
\hline
\end{tabular}
\][/tex]

[tex]\[
\begin{tabular}{|c|c|}
\hline \multicolumn{2}{|c|}{Second Randomization} \\
\hline Group A & Group B \\
\hline 8.2 & 12.1 \\
\hline 13.8 & 14.6 \\
\hline 15.9 & 13.6 \\
\hline 9.2 & 11.2 \\
\hline 9.7 & 11.5 \\
\hline
\end{tabular}
\][/tex]

[tex]\[
\begin{tabular}{|c|c|}
\hline \multicolumn{2}{|c|}{Third Randomization} \\
\hline Group A & Group B \\
\hline 8.2 & 12.1 \\
\hline 9.7 & 13.8 \\
\hline 11.5 & 13.6 \\
\hline 14.6 & 11.2 \\
\hline 15.9 & 9.2 \\
\hline
\end{tabular}
\][/tex]

Sagot :

GinnyAnsweridn
Certainly! Let's go through each of the randomizations and compute the required values step-by-step.

### First Randomization:

| Group A | Group B |
|---------|---------|
| 13.6 | 9.2 |
| 12.1 | 8.2 |
| 15.9 | 11.5 |
| 11.2 | 13.8 |
| 9.7 | 14.6 |

Mean of Group A ([tex]\(\bar{x}_A\)[/tex]):

[tex]\[ \bar{x}_A = \frac{13.6 + 12.1 + 15.9 + 11.2 + 9.7}{5} = \frac{62.5}{5} = 12.5 \][/tex]

Mean of Group B ([tex]\(\bar{x}_B\)[/tex]):

[tex]\[ \bar{x}_B = \frac{9.2 + 8.2 + 11.5 + 13.8 + 14.6}{5} = \frac{57.3}{5} = 11.46 \][/tex]

Difference of means:

[tex]\[ \bar{x}_A - \bar{x}_B = 12.5 - 11.46 = 1.04 \][/tex]

### Second Randomization:

| Group A | Group B |
|---------|---------|
| 8.2 | 12.1 |
| 13.8 | 14.6 |
| 15.9 | 13.6 |
| 9.2 | 11.2 |
| 9.7 | 11.5 |

Mean of Group A ([tex]\(\bar{x}_A\)[/tex]):

[tex]\[ \bar{x}_A = \frac{8.2 + 13.8 + 15.9 + 9.2 + 9.7}{5} = \frac{56.8}{5} = 11.36 \][/tex]

Mean of Group B ([tex]\(\bar{x}_B\)[/tex]):

[tex]\[ \bar{x}_B = \frac{12.1 + 14.6 + 13.6 + 11.2 + 11.5}{5} = \frac{63.0}{5} = 12.6 \][/tex]

Difference of means:

[tex]\[ \bar{x}_A - \bar{x}_B = 11.36 - 12.6 = -1.24 \][/tex]

### Third Randomization:

| Group A | Group B |
|---------|---------|
| 8.2 | 12.1 |
| 9.7 | 13.8 |
| 11.5 | 13.6 |
| 14.6 | 11.2 |
| 15.9 | 9.2 |

Mean of Group A ([tex]\(\bar{x}_A\)[/tex]):

[tex]\[ \bar{x}_A = \frac{8.2 + 9.7 + 11.5 + 14.6 + 15.9}{5} = \frac{59.9}{5} = 11.98 \][/tex]

Mean of Group B ([tex]\(\bar{x}_B\)[/tex]):

[tex]\[ \bar{x}_B = \frac{12.1 + 13.8 + 13.6 + 11.2 + 9.2}{5} = \frac{59.9}{5} = 11.98 \][/tex]

Difference of means:

[tex]\[ \bar{x}_A - \bar{x}_B = 11.98 - 11.98 = 0.0 \][/tex]

### Summary:

For each table, we have the following results:

First Randomization:

[tex]\[ \bar{x}_A = 12.5, \quad \bar{x}_B = 11.46, \quad \bar{x}_A - \bar{x}_B = 1.04 \][/tex]

Second Randomization:

[tex]\[ \bar{x}_A = 11.36, \quad \bar{x}_B = 12.6, \quad \bar{x}_A - \bar{x}_B = -1.24 \][/tex]

Third Randomization:

[tex]\[ \bar{x}_A = 11.98, \quad \bar{x}_B = 11.98, \quad \bar{x}_A - \bar{x}_B = 0.0 \][/tex]
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