SOLUTION: You are given the following data. # of Absent/ Final grade 0----------- 93 1----------- 90 2----------- 79 3----------- 66 4----------- 60 5----------- 56

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Question 478490: You are given the following data.
# of
Absent/ 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.
Thank you

Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
You are given the following data.
# of
Absent/ Final grade
0------------93
1------------90
2------------79
3------------66
4------------60
5------------56
- Find the correlation coefficient for the data.
r = -0.9831
----------------------

- Find the equation for the regression line for the data, and predict the final grade of a student who misses 3.5 days.
y = -8.2286x + 94.5714
f(3.5) = 65.771
-----------------------------
Cheers,
Stan H.
=============

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
you should solve this by equation, but you can check your answer by using a calculator.
one such calculator can be found here:
http://www.alcula.com/calculators/statistics/linear-regression/
enter your data as:
0 1 2 3 4 5
93 90 79 66 60 56
click on SUBMIT DATA to see your linear regression graph
click on correlation to see what the correlation factor is
click on scatter plot to see what the data looks like without the regression line.
working it by using the formulas is a lot more labor intensive.
those formulas and how to use them are shown in the following references
http://easycalculation.com/statistics/learn-regression.php http://easycalculation.com/statistics/learn-correlation.php based on those formulas, i created a table in excel that shows you what the data would look like applied to your problem.
that table is shown below:
     x values   y values   x*y    x^2	y^2
     0	        93         0	  0	8649
     1	        90         90	  1	8100
     2	        79         158	  4	6241
     3	        66         198	  9	4356
     4	        60         240	  16	3600
     5	        56         280	  25	3136

based on that data the important parameters have been calculated to be:
sum of         x	y	x*y	x^2	y^2
equals         15	444	966	55	34082

n >>> 6
slope = -8.228571429
y-intercept = 94.57142857
correlation = -0.983101215
regression equation is y = mx + b
m = slope and b = y-intercept
-----
these numbers should agree with what you derive using the online calculator taking into account rounding to a comparable number of digits to the right of the decimal point.
the mean of x is equal to 15/6 = 2.5
the mean of y is equal to 444/6 = 74