To find the difference of the means and the quotient of the mean deviations for Team A and Team B, we need to follow these steps:
Step 1: Calculate the means for both teams.
Team A:
Data: \( 1, 2, 3, 3, 3, 3, 4, 5 \)
Number of values (\( n_A \)): 8
\[ \text{Mean of Team A} = \frac{\sum \text{Team A}}{n_A} = \frac{1 + 2 + 3 + 3 + 3 + 3 + 4 + 5}{8} = \frac{20}{8} = 2.5 \]
Team B:
Data: \( 2, 2, 2, 3, 3, 4, 4, 4 \)
Number of values (\( n_B \)): 8
\[ \text{Mean of Team B} = \frac{\sum \text{Team B}}{n_B} = \frac{2 + 2 + 2 + 3 + 3 + 4 + 4 + 4}{8} = \frac{24}{8} = 3 \]
Step 2: Calculate the difference of the means.
\[ \text{Difference of Means} = \text{Mean of Team B} - \text{Mean of Team A} = 3 - 2.5 = 0.5 \]
Step 3: Calculate the mean deviations for both teams.
Mean Deviation for Team A:
-
Calculate absolute deviations:
\[ |1 - 2.5| = 1.5, \quad |2 - 2.5| = 0.5, \quad |3 - 2.5| = 0.5, \quad |3 - 2.5| = 0.5, \quad |3 - 2.5| = 0.5, \quad |3 - 2.5| = 0.5, \quad |4 - 2.5| = 1.5, \quad |5 - 2.5| = 2.5 \] -
Sum of absolute deviations:
\[ 1.5 + 0.5 + 0.5 + 0.5 + 0.5 + 0.5 + 1.5 + 2.5 = 8.0 \] -
Mean deviation:
\[ \text{Mean Deviation of Team A} = \frac{8.0}{8} = 1.0 \]
Mean Deviation for Team B:
-
Calculate absolute deviations:
\[ |2 - 3| = 1, \quad |2 - 3| = 1, \quad |2 - 3| = 1, \quad |3 - 3| = 0, \quad |3 - 3| = 0, \quad |4 - 3| = 1, \quad |4 - 3| = 1, \quad |4 - 3| = 1 \] -
Sum of absolute deviations:
\[ 1 + 1 + 1 + 0 + 0 + 1 + 1 + 1 = 5 \] -
Mean deviation:
\[ \text{Mean Deviation of Team B} = \frac{5.0}{8} = 0.625 \]
Step 4: Calculate the quotient of the mean deviations.
\[ \text{Quotient of Mean Deviations} = \frac{\text{Mean Deviation of Team A}}{\text{Mean Deviation of Team B}} = \frac{1.0}{0.625} = 1.6 \]
Summary of Results
- Difference of Means: \( 0.5 \)
- Quotient of Mean Deviations: \( 1.6 \)