Can someone please help me, I have been working on this for 2 hours now and I keep getting a negative answer.
You are given the following data.
Number of Absences Final
Grade
0 99
1 94
2 79
3 73
4 68
5 53
- 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 data:
6 data pairs (x,y):
( 0.00 , 99.0 ); ( 1.00 , 94.0 ); ( 2.00 , 79.0 ); ( 3.00 , 73.0 ); ( 4.00 , 68.0 ); ( 5.00 , 53.0 );
y = a + bx where:
a= 100
b= -8.97
r = -0.986
In the regression equation, substitute 3.5 for x and solve for y to predict the final grade.
I hope this helps.
To find the correlation coefficient for the given data, we need to calculate the covariance and standard deviations for the two variables: Number of Absences and Final Grade. Here are the steps to do that:
1. Calculate the mean (average) for both Number of Absences and Final Grade. This can be done by adding up all the values for each variable and dividing by the total number of data points. In this case, we have:
Mean(Number of Absences) = (0 + 1 + 2 + 3 + 4 + 5) / 6 = 15 / 6 = 2.5
Mean(Final Grade) = (99 + 94 + 79 + 73 + 68 + 53) / 6 = 366 / 6 = 61
2. Calculate the deviation of each data point from their respective means. Subtract the mean of each variable from each data point. This will give us two new columns - Deviation(Number of Absences) and Deviation(Final Grade). For example:
Data Point: (Number of Absences, Final Grade)
(0, 99)
Deviation(Number of Absences) = 0 - 2.5 = -2.5
Deviation(Final Grade) = 99 - 61 = 38
Repeat this for all the data points to get the deviation values for both variables.
3. Calculate the product of the deviations for each data point. Multiply the deviation of Number of Absences with the deviation of Final Grade for each data point. This will give us a new column - Product of Deviations. For example:
Data Point: (Number of Absences, Final Grade)
(0, 99)
Product of Deviations = (-2.5) * 38 = -95
Repeat this for all the data points to get the product values.
4. Calculate the covariance. Sum up all the product values from the previous step and divide by the total number of data points minus 1. Using the formula:
Covariance = Σ(Product of Deviations) / (n - 1)
where Σ denotes the sum of all values and n is the number of data points.
In this case, we have:
Covariance = (-95 + (-44) + 7 + 20 + 25 + 4) / (6 - 1) = (-83) / 5 = -16.6
5. Calculate the standard deviation for each variable. This can be done by taking the square root of the sum of the squares of the deviations for each variable, divided by the total number of data points minus 1. Using the formula:
Standard Deviation = √(Σ(Deviation^2) / (n - 1))
In this case, we have:
Standard Deviation(Number of Absences) = √((6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25) / (6 - 1)) = √(17.25 / 5) = √3.45 ≈ 1.86
Standard Deviation(Final Grade) = √((1444 + 729 + 361 + 169 + 169 + 121) / (6 - 1)) = √(2973 / 5) = √594.6 ≈ 24.39
6. Calculate the correlation coefficient. Divide the covariance by the product of the standard deviations for each variable. Using the formula:
Correlation Coefficient = Covariance / (Standard Deviation(Number of Absences) * Standard Deviation(Final Grade))
In this case, we have:
Correlation Coefficient = -16.6 / (1.86 * 24.39) ≈ -0.378
Therefore, the correlation coefficient for the given data is approximately -0.378.
Now, let's move on to finding the equation for the regression line and predict the final grade for a student who misses 3.5 days.