Teachers are being trained to standardize the scores they give to students' essays. The same essay was scored by 10 different teachers at the beginning and at the end of their training. The results are shown in the tables.

\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline
\multicolumn{7}{|c|}{Scores for Essay at Beginning of Teachers' Training} \\
\hline
76 & 81 & 85 & 79 & 89 & 86 & 84 & 80 & 88 & 79 \\
\hline
\end{tabular}

\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline
\multicolumn{7}{|c|}{Scores for Essay at End of Teachers' Training} \\
\hline
79 & 82 & 84 & 81 & 77 & 85 & 82 & 80 & 78 & 83 \\
\hline
\end{tabular}

7. Calculate the MADs for the teachers' scores. Did the teachers make progress in standardizing their scores?

8. What would it mean if the teachers had a MAD of 0?



Answer :

Sure, let's work through this step-by-step.

### 7. Calculate the Mean Absolute Deviations (MADs) for the teachers' scores. Did the teachers make progress in standardizing their scores?

1. Mean Absolute Deviation (MAD): MAD is a measure of the average absolute deviation from the mean of a dataset.

#### Scores at the Beginning:
The scores at the beginning of the training are:
[tex]\[76, 81, 85, 79, 89, 86, 84, 80, 88, 79\][/tex]

- Step 1: Calculate the mean (average) of the scores.
[tex]\[ \text{Mean at the beginning} = 82.7 \][/tex]

- Step 2: Calculate the absolute deviations from the mean for each score.
[tex]\[ \begin{align*} |76 - 82.7| & = 6.7 \\ |81 - 82.7| & = 1.7 \\ |85 - 82.7| & = 2.3 \\ |79 - 82.7| & = 3.7 \\ |89 - 82.7| & = 6.3 \\ |86 - 82.7| & = 3.3 \\ |84 - 82.7| & = 1.3 \\ |80 - 82.7| & = 2.7 \\ |88 - 82.7| & = 5.3 \\ |79 - 82.7| & = 3.7 \\ \end{align} \][/tex]

- Step 3: Find the average of these absolute deviations.
[tex]\[ \text{MAD at the beginning} = \frac{6.7 + 1.7 + 2.3 + 3.7 + 6.3 + 3.3 + 1.3 + 2.7 + 5.3 + 3.7}{10} = 3.7 \][/tex]

#### Scores at the End:
The scores at the end of the training are:
[tex]\[79, 82, 84, 81, 77, 85, 82, 80, 78, 83\][/tex]

- Step 1: Calculate the mean (average) of the scores.
[tex]\[ \text{Mean at the end} = 81.1 \][/tex]

- Step 2: Calculate the absolute deviations from the mean for each score.
[tex]\[ \begin{align
} |79 - 81.1| & = 2.1 \\ |82 - 81.1| & = 0.9 \\ |84 - 81.1| & = 2.9 \\ |81 - 81.1| & = 0.1 \\ |77 - 81.1| & = 4.1 \\ |85 - 81.1| & = 3.9 \\ |82 - 81.1| & = 0.9 \\ |80 - 81.1| & = 1.1 \\ |78 - 81.1| & = 3.1 \\ |83 - 81.1| & = 1.9 \\ \end{align*} \][/tex]

- Step 3: Find the average of these absolute deviations.
[tex]\[ \text{MAD at the end} = \frac{2.1 + 0.9 + 2.9 + 0.1 + 4.1 + 3.9 + 0.9 + 1.1 + 3.1 + 1.9}{10} = 2.1 \][/tex]

#### Progress in Standardizing Scores:
The MAD at the beginning is [tex]\(3.7\)[/tex] and the MAD at the end is [tex]\(2.1\)[/tex]. Since the MAD decreased from [tex]\(3.7\)[/tex] to [tex]\(2.1\)[/tex], it indicates that the teachers have made progress in standardizing their scores. Lower MAD indicates that the scores are more consistent and have less deviation from the mean.

### 8. What If? What would it mean if the teachers had a MAD of 0?

If the teachers had a MAD of 0, it would mean that all the teachers gave the exact same score to the essay. This implies perfect standardization, as there would be no deviation whatsoever from the mean score. In simpler terms, every teacher would have agreed completely and assigned the same score.