Answer :
Sure, let's analyze the data to determine if there is a linear relationship between the number of absences and the students' grades for the semester. The data provided are as follows:
[tex]\[ \begin{array}{|l|c|c|} \hline \text{Student} & \text{Number of Absences} & \text{Algebra Grade (\%)} \\ \hline \text{Albert} & 2 & 87 \\ \hline \text{Brianna} & 7 & 68 \\ \hline \text{Carlos} & 3 & 90 \\ \hline \text{Diana} & 1 & 92 \\ \hline \text{Elaine} & 2 & 88 \\ \hline \text{Freddy} & 3 & 73 \\ \hline \text{Giana} & 10 & 60 \\ \hline \text{Helen} & 0 & 93 \\ \hline \text{Irene} & 6 & 70 \\ \hline \text{Jacob} & 1 & 88 \\ \hline \end{array} \][/tex]
### Step 1: Calculate the Mean of Absences and Grades
First, we need to calculate the mean (average) for the number of absences and the students' grades.
The data for absences are: [tex]\(2, 7, 3, 1, 2, 3, 10, 0, 6, 1\)[/tex]
The data for grades are: [tex]\(87, 68, 90, 92, 88, 73, 60, 93, 70, 88\)[/tex]
To find the mean:
[tex]\[ \text{Mean of Absences} = \frac{2+7+3+1+2+3+10+0+6+1}{10} = \frac{35}{10} = 3.5 \][/tex]
[tex]\[ \text{Mean of Grades} = \frac{87+68+90+92+88+73+60+93+70+88}{10} = \frac{809}{10} = 80.9 \][/tex]
### Step 2: Calculate the Terms Needed for the Correlation Coefficient
Next, we will compute the individual deviations from the mean for both absences and grades, then determine the following terms needed for the correlation coefficient: the numerator, and the sum of the squared deviations for absences and grades.
#### Numerator
[tex]\[ \text{Numerator} = \sum (a_i - \text{mean of absences}) \times (g_i - \text{mean of grades}) \][/tex]
[tex]\[ = (2-3.5)(87-80.9) + (7-3.5)(68-80.9) + (3-3.5)(90-80.9) + (1-3.5)(92-80.9) + (2-3.5)(88-80.9) + \][/tex]
[tex]\[ (3-3.5)(73-80.9) + (10-3.5)(60-80.9) + (0-3.5)(93-80.9) + (6-3.5)(70-80.9) + (1-3.5)(88-80.9) \][/tex]
[tex]\[ = -1.5 \times 6.1 + 3.5 \times -12.9 + -0.5 \times 9.1 + -2.5 \times 11.1 + -1.5 \times 7.1 + \][/tex]
[tex]\[ -0.5 \times -7.9 + 6.5 \times -20.9 + -3.5 \times 12.1 + 2.5 \times -10.9 + -2.5 \times 7.1 \][/tex]
[tex]\[ = -9.15 - 45.15 - 4.55 - 27.75 -10.65 + 3.95 - 135.85 - 42.35 - 27.25 - 17.75 = -316.5 \][/tex]
#### Squared Differences
[tex]\[ \text{Squared Difference for Absences} = \sum (a_i - \text{mean of absences})^2 \][/tex]
[tex]\[ = (2-3.5)^2 + (7-3.5)^2 + (3-3.5)^2 + (1-3.5)^2 + (2-3.5)^2 + (3-3.5)^2 + (10-3.5)^2 + \][/tex]
[tex]\[ (0-3.5)^2 + (6-3.5)^2 + (1-3.5)^2 = 90.5 \][/tex]
[tex]\[ \text{Squared Difference for Grades} = \sum (g_i - \text{mean of grades})^2 \][/tex]
[tex]\[ = (87-80.9)^2 + (68-80.9)^2 + (90-80.9)^2 + (92-80.9)^2 + (88-80.9)^2 + (73-80.9)^2 + \][/tex]
[tex]\[ (60-80.9)^2 + (93-80.9)^2 + (70-80.9)^2 + (88-80.9)^2 = 1274.9 \][/tex]
### Step 3: Calculate the Correlation Coefficient
The correlation coefficient is given by:
[tex]\[ r = \frac{\text{Numerator}}{\sqrt{\text{Squared Diff. for Absences} \times \text{Squared Diff. for Grades}}} \][/tex]
[tex]\[ = \frac{-316.5}{\sqrt{90.5 \times 1274.9}} \][/tex]
[tex]\[ = \frac{-316.5}{\sqrt{115309.45}} \][/tex]
[tex]\[ = \frac{-316.5}{339.57} \][/tex]
[tex]\[ = -0.931775665670006 \][/tex]
### Conclusion
With a correlation coefficient of approximately [tex]\( -0.9318 \)[/tex], there is a strong negative linear relationship between the number of absences and the students' grades. This suggests that as the number of absences increases, the students' grades tend to decrease significantly.
[tex]\[ \begin{array}{|l|c|c|} \hline \text{Student} & \text{Number of Absences} & \text{Algebra Grade (\%)} \\ \hline \text{Albert} & 2 & 87 \\ \hline \text{Brianna} & 7 & 68 \\ \hline \text{Carlos} & 3 & 90 \\ \hline \text{Diana} & 1 & 92 \\ \hline \text{Elaine} & 2 & 88 \\ \hline \text{Freddy} & 3 & 73 \\ \hline \text{Giana} & 10 & 60 \\ \hline \text{Helen} & 0 & 93 \\ \hline \text{Irene} & 6 & 70 \\ \hline \text{Jacob} & 1 & 88 \\ \hline \end{array} \][/tex]
### Step 1: Calculate the Mean of Absences and Grades
First, we need to calculate the mean (average) for the number of absences and the students' grades.
The data for absences are: [tex]\(2, 7, 3, 1, 2, 3, 10, 0, 6, 1\)[/tex]
The data for grades are: [tex]\(87, 68, 90, 92, 88, 73, 60, 93, 70, 88\)[/tex]
To find the mean:
[tex]\[ \text{Mean of Absences} = \frac{2+7+3+1+2+3+10+0+6+1}{10} = \frac{35}{10} = 3.5 \][/tex]
[tex]\[ \text{Mean of Grades} = \frac{87+68+90+92+88+73+60+93+70+88}{10} = \frac{809}{10} = 80.9 \][/tex]
### Step 2: Calculate the Terms Needed for the Correlation Coefficient
Next, we will compute the individual deviations from the mean for both absences and grades, then determine the following terms needed for the correlation coefficient: the numerator, and the sum of the squared deviations for absences and grades.
#### Numerator
[tex]\[ \text{Numerator} = \sum (a_i - \text{mean of absences}) \times (g_i - \text{mean of grades}) \][/tex]
[tex]\[ = (2-3.5)(87-80.9) + (7-3.5)(68-80.9) + (3-3.5)(90-80.9) + (1-3.5)(92-80.9) + (2-3.5)(88-80.9) + \][/tex]
[tex]\[ (3-3.5)(73-80.9) + (10-3.5)(60-80.9) + (0-3.5)(93-80.9) + (6-3.5)(70-80.9) + (1-3.5)(88-80.9) \][/tex]
[tex]\[ = -1.5 \times 6.1 + 3.5 \times -12.9 + -0.5 \times 9.1 + -2.5 \times 11.1 + -1.5 \times 7.1 + \][/tex]
[tex]\[ -0.5 \times -7.9 + 6.5 \times -20.9 + -3.5 \times 12.1 + 2.5 \times -10.9 + -2.5 \times 7.1 \][/tex]
[tex]\[ = -9.15 - 45.15 - 4.55 - 27.75 -10.65 + 3.95 - 135.85 - 42.35 - 27.25 - 17.75 = -316.5 \][/tex]
#### Squared Differences
[tex]\[ \text{Squared Difference for Absences} = \sum (a_i - \text{mean of absences})^2 \][/tex]
[tex]\[ = (2-3.5)^2 + (7-3.5)^2 + (3-3.5)^2 + (1-3.5)^2 + (2-3.5)^2 + (3-3.5)^2 + (10-3.5)^2 + \][/tex]
[tex]\[ (0-3.5)^2 + (6-3.5)^2 + (1-3.5)^2 = 90.5 \][/tex]
[tex]\[ \text{Squared Difference for Grades} = \sum (g_i - \text{mean of grades})^2 \][/tex]
[tex]\[ = (87-80.9)^2 + (68-80.9)^2 + (90-80.9)^2 + (92-80.9)^2 + (88-80.9)^2 + (73-80.9)^2 + \][/tex]
[tex]\[ (60-80.9)^2 + (93-80.9)^2 + (70-80.9)^2 + (88-80.9)^2 = 1274.9 \][/tex]
### Step 3: Calculate the Correlation Coefficient
The correlation coefficient is given by:
[tex]\[ r = \frac{\text{Numerator}}{\sqrt{\text{Squared Diff. for Absences} \times \text{Squared Diff. for Grades}}} \][/tex]
[tex]\[ = \frac{-316.5}{\sqrt{90.5 \times 1274.9}} \][/tex]
[tex]\[ = \frac{-316.5}{\sqrt{115309.45}} \][/tex]
[tex]\[ = \frac{-316.5}{339.57} \][/tex]
[tex]\[ = -0.931775665670006 \][/tex]
### Conclusion
With a correlation coefficient of approximately [tex]\( -0.9318 \)[/tex], there is a strong negative linear relationship between the number of absences and the students' grades. This suggests that as the number of absences increases, the students' grades tend to decrease significantly.
