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]
