2. You are given the following data.
Number of Absences Final Grade
0 93
1 90
2 79
3 66
4 60
5 56
- Find the correlation coefficient for the data. - Find the equation for the regression line for the data, and predict the final grade of a student who misses 3.5 days. Show all work.
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To find the correlation coefficient and the equation for the regression line, you will need to follow these steps:
Step 1: Calculate the mean (average) of the number of absences and the mean of the final grades.
- Sum all the values of the number of absences and divide by the total number of data points to find the mean of absences.
- Sum all the values of the final grades and divide by the total number of data points to find the mean of final grades.
Let's calculate the means using the given data:
Number of Absences (x): 0, 1, 2, 3, 4, 5
Final Grades (y): 93, 90, 79, 66, 60, 56
Mean of absences (x̄) = (0 + 1 + 2 + 3 + 4 + 5) / 6 = 15 / 6 = 2.5
Mean of final grades (ȳ) = (93 + 90 + 79 + 66 + 60 + 56) / 6 = 444 / 6 = 74
Step 2: Calculate the sum of the products of the differences from the mean for both the number of absences and the final grades.
- Calculate (x - x̄) for each number of absences and (y - ȳ) for each final grade.
- Multiply these differences for each data point, and then sum them up.
Let's calculate the sum of the products of the differences from the mean:
(x - x̄) = (-2.5, -1.5, -0.5, 0.5, 1.5, 2.5)
(y - ȳ) = (19, 16, 5, -8, -14, -18)
(x - x̄) * (y - ȳ) = (-2.5 * 19, -1.5 * 16, -0.5 * 5, 0.5 * -8, 1.5 * -14, 2.5 * -18)
= (-47.5, -24, -2.5, -4, -21, -45)
Sum of (x - x̄) * (y - ȳ) = (-47.5) + (-24) + (-2.5) + (-4) + (-21) + (-45) = -144
Step 3: Calculate the sum of the squares of the differences from the mean for both the number of absences and the final grades.
- Calculate (x - x̄)² for each number of absences and (y - ȳ)² for each final grade.
- Square these differences for each data point, and then sum them up.
Let's calculate the sum of the squares of the differences from the mean:
(x - x̄)² = (6.25, 2.25, 0.25, 0.25, 2.25, 6.25)
(y - ȳ)² = (361, 256, 25, 64, 196, 324)
Sum of (x - x̄)² = 6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25 = 17.5
Sum of (y - ȳ)² = 361 + 256 + 25 + 64 + 196 + 324 = 1226
Step 4: Calculate the correlation coefficient (r) using the formula:
r = (Sum of (x - x̄) * (y - ȳ)) / √((Sum of (x - x̄)²) * (Sum of (y - ȳ)²))
Let's calculate the correlation coefficient:
r = (-144) / √((17.5) * (1226))
= -144 / √(21475)
≈ -144 / 146.514
≈ -0.982
The correlation coefficient is approximately -0.982.
Step 5: Calculate the equation for the regression line.
The equation for the regression line is y = mx + b, where m is the slope and b is the y-intercept.
The slope (m) is calculated using the formula:
m = r * (standard deviation of y) / (standard deviation of x)
The y-intercept (b) can be calculated using the formula:
b = ȳ - (m * x̄)
Since we do not have the standard deviation of x or y in the given data, we cannot calculate the exact equation for the regression line without additional information.
To predict the final grade of a student who misses 3.5 days using the regression line, you would need the slope (m) and y-intercept (b) values from Step 5.
However, without knowing those values, we cannot make an accurate prediction using the regression line.