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\begin{tabular}{|c|c|}
\hline
15.9 & 11.5 \\
\hline
11.2 & 13.8 \\
\hline
9.7 & 14.6 \\
\hline
\end{tabular}

\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}

\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}

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 [tex]$A$[/tex] and the mean of group [tex]$B$[/tex] ([tex]$\bar{x}_A - \bar{x}_B$[/tex]). Type the correct answer in each box.

1. After the first randomization, [tex]$\bar{x}_A$[/tex] is [tex]$\square$[/tex], [tex]$\bar{x}_B$[/tex] is [tex]$\square$[/tex], and [tex]$\left(\bar{x}_A - \bar{x}_B\right)$[/tex] is [tex]$\square$[/tex].

2. After the second randomization, [tex]$\bar{x}_A$[/tex] is [tex]$\square$[/tex], [tex]$\bar{x}_B$[/tex] is [tex]$\square$[/tex], and [tex]$\left(\bar{x}_A - \bar{x}_B\right)$[/tex] is [tex]$\square$[/tex].

3. After the third randomization, [tex]$\bar{x}_A$[/tex] is [tex]$\square$[/tex], [tex]$\bar{x}_B$[/tex] is [tex]$\square$[/tex], and [tex]$\left(\bar{x}_A - \bar{x}_B\right)$[/tex] is [tex]$\square$[/tex].



Answer :

GinnyAnswer
To solve the problem, let's calculate the required statistics for each randomization.

### First Randomization
The data for the first randomization is:
- Group A: 15.9, 11.2, 9.7
- Group B: 11.5, 13.8, 14.6

1. Calculate the mean weight for Group A ([tex]\(\bar{x}_A\)[/tex]):
[tex]\[ \bar{x}_A = \frac{15.9 + 11.2 + 9.7}{3} = 12.266666666666666 \][/tex]

2. Calculate the mean weight for Group B ([tex]\(\bar{x}_B\)[/tex]):
[tex]\[ \bar{x}_B = \frac{11.5 + 13.8 + 14.6}{3} = 13.299999999999999 \][/tex]

3. Calculate the difference between the means ([tex]\(\bar{x}_A - \bar{x}_B\)[/tex]):
[tex]\[ \bar{x}_A - \bar{x}_B = 12.266666666666666 - 13.299999999999999 = -1.0333333333333332 \][/tex]

### Second Randomization
The data for the second randomization is:
- Group A: 8.2, 13.8, 15.9, 9.2, 9.7
- Group B: 12.1, 14.6, 13.6, 11.2, 11.5

1. Calculate the mean weight for Group A ([tex]\(\bar{x}_A\)[/tex]):
[tex]\[ \bar{x}_A = \frac{8.2 + 13.8 + 15.9 + 9.2 + 9.7}{5} = 11.36 \][/tex]

2. Calculate the mean weight for Group B ([tex]\(\bar{x}_B\)[/tex]):
[tex]\[ \bar{x}_B = \frac{12.1 + 14.6 + 13.6 + 11.2 + 11.5}{5} = 12.6 \][/tex]

3. Calculate the difference between the means ([tex]\(\bar{x}_A - \bar{x}_B\)[/tex]):
[tex]\[ \bar{x}_A - \bar{x}_B = 11.36 - 12.6 = -1.2400000000000002 \][/tex]

### Third Randomization
The data for the third randomization is:
- Group A: 8.2, 9.7, 11.5, 14.6, 15.9
- Group B: 12.1, 13.8, 13.6, 11.2, 9.2

1. Calculate the mean weight for Group A ([tex]\(\bar{x}_A\)[/tex]):
[tex]\[ \bar{x}_A = \frac{8.2 + 9.7 + 11.5 + 14.6 + 15.9}{5} = 11.98 \][/tex]

2. Calculate the mean weight for Group B ([tex]\(\bar{x}_B\)[/tex]):
[tex]\[ \bar{x}_B = \frac{12.1 + 13.8 + 13.6 + 11.2 + 9.2}{5} = 11.98 \][/tex]

3. Calculate the difference between the means ([tex]\(\bar{x}_A - \bar{x}_B\)[/tex]):
[tex]\[ \bar{x}_A - \bar{x}_B = 11.98 - 11.98 = 0.0 \][/tex]

### Summary
1. After the first randomization:
[tex]\(\bar{x}_{A} = 12.27\)[/tex], [tex]\(\bar{x}_{B} = 13.30\)[/tex], and [tex]\(\bar{x}_{A} - \bar{x}_{B} = -1.03\)[/tex].

2. After the second randomization:
[tex]\(\bar{x}_{A} = 11.36\)[/tex], [tex]\(\bar{x}_{B} = 12.60\)[/tex], and [tex]\(\bar{x}_{A} - \bar{x}_{B} = -1.24\)[/tex].

3. After the third randomization:
[tex]\(\bar{x}_{A} = 11.98\)[/tex], [tex]\(\bar{x}_{B} = 11.98\)[/tex], and [tex]\(\bar{x}_{A} - \bar{x}_{B} = 0.0\)[/tex].

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