You are given the following data.

Number of Absences Final
Grade
0 95
1 91
2 82
3 69
4 67
5 58


- 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.

Using an online calculator, I have the following:

6 data pairs (x,y):
( 0.00 , 95.0 ); ( 1.00 , 91.0 ); ( 2.00 , 82.0 ); ( 3.00 , 69.0 ); ( 4.00 , 67.0 ); ( 5.00 , 58.0 );

r = -0.987 -->correlation coefficient

predicted y = 96.3 - 7.71x -->regression equation

To find the final grade of a student who misses 3.5 days, substitute that value for x and solve for predicted y.

I hope this will help.

To find the correlation coefficient for the given data, we can use the formula:

r = (∑xy - n(mean(x) * mean(y))) / sqrt((∑x^2 - n * (mean(x))^2) * (∑y^2 - n * (mean(y))^2))

First, let's calculate the necessary values:

x: Number of Absences
y: Final Grade

n: Total number of data points (n = 6)

∑x: Sum of all x values = 0 + 1 + 2 + 3 + 4 + 5 = 15
∑y: Sum of all y values = 95 + 91 + 82 + 69 + 67 + 58 = 462

∑xy: Sum of the product of x and y values = (0*95) + (1*91) + (2*82) + (3*69) + (4*67) + (5*58) = 977

∑x^2: Sum of squared x values = (0^2) + (1^2) + (2^2) + (3^2) + (4^2) + (5^2) = 55
∑y^2: Sum of squared y values = (95^2) + (91^2) + (82^2) + (69^2) + (67^2) + (58^2) = 40083

mean(x): Mean of x values = ∑x / n = 15 / 6 = 2.5
mean(y): Mean of y values = ∑y / n = 462 / 6 = 77

Now, let's substitute these values into the correlation coefficient formula:

r = (977 - 6(2.5)(77)) / sqrt((55 - 6(2.5)^2) * (40083 - 6(77)^2))
r = (977 - 1155) / sqrt((55 - 37.5) * (40083 - 28194))
r = -178 / sqrt(17.5 * 11889)
r = -178 / sqrt(208057.5)
r ≈ -0.908

So, the correlation coefficient for the given data is approximately -0.908. This indicates a strong negative correlation between the number of absences and the final grade.

To find the equation for the regression line, we can use the formula:

y = a + bx

Where:
y: Dependent variable (Final Grade)
x: Independent variable (Number of Absences)
a: Y-intercept (constant)
b: Slope

Using the following formulas:

b = (n(∑xy) - (∑x)(∑y)) / (n(∑x^2) - (∑x)^2)
a = mean(y) - b(mean(x))

Substituting the values we previously calculated:

b = (6(977) - (15)(462)) / (6(55) - (15)^2)
b = (5862 - 6930) / (330 - 225)
b = -1068 / 105
b ≈ -10.17

a = 77 - (-10.17)(2.5)
a ≈ 77 + 25.42
a ≈ 102.42

So, the equation for the regression line is:

y = 102.42 - 10.17x

Now, to predict the final grade of a student who misses 3.5 days, we substitute x = 3.5 into the regression equation:

y = 102.42 - 10.17(3.5)
y ≈ 102.42 - 35.595
y ≈ 66.825

Therefore, the predicted final grade for a student who misses 3.5 days is approximately 66.825.