Regression and Correlation

We are studying linear correlation coefficient, r, and the equation of the regression line. Please look at your field of study and apply the coefficient or the equation, a few examples are below:_x000D_
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•Business—using the regression line to anticipate the clients demands for the future, you help stock up on inventory_x000D_
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•Health Care-using the regression line to compare cholesterol or blood pressures for a patient over time_x000D_
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•IT—using the r coefficient to analyze traffic flow through the network or employee internet usage_x000D_
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•Programming-using the regression line to analyze scores on a game created, to determine the difficulty level_x000D_
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Are you looking for a coefficient r value close to 0 or 1, or for your analysis, should the value be in the middle? Do you want your analysis to be scattered, or fit closely to the regression line?_x000D_
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Regression and Correlation

Regression and Correlation

Question 1
Find the equation of the regression line for the given data. Predict the value of Y when X=-2? Predict the value of Y when X = 4?
x -5 -3 4 1 -1 -2 0 2 3 -4
y -10 -8 9 1 -2 -6 -1 3 6 -8
Y = a + bx (Retrieved from http://www.stat.yale.edu/Courses/1997-98/101/linreg.htm)
a = value of Y when X is 0 = -1
-10 = -1 + b(-5)
-10 + 1 = -5b
b = 1.8
The linear equation is therefore Y = -1 + 1.8X

When X = -2
Y = -1 + 1.8(-2) = – 4.6
When X = 4
Y = -1 + 1.8(4) = 6.2
Question 2
The data below are the final exam scores of 10 randomly selected statistics students and the number of hours they studied for the exam. Find the equation of the regression line for the given data. Predict the final exam score when a student studied for 4 hours. Predict the final exam score when a student studied for 6 hours.
Hours, x 3 5 2 8 2 4 5 4 6 3
Scores, y 65 80 60 88 66 78 85 90 90 71
b =(N?XY – (?X)(?Y)) / (N?X2 – (?X)2)
a = (?Y – b(?X)) / N
Retrieved from http://easycalculation.com/statistics/learn-regression.php
where

x y xy x2
3 65 195 9
5 80 400 25
2 60 120 4
8 88 704 64
2 66 132 4
4 78 312 16
5 85 425 25
4 90 360 16
6 90 540 36
3 71 213 9
42 773 3401 208

b = (34010 – 32466)/(2080 – 1764) = 1544/316 = 4.89
a = (773 – 205.2)/10 = 56.8
Y = 56.8 + 4.89X
When X = 4 Hours
Y = 56.8 + 19.6 = 76.4
When X = 6 Hours
Y = 56.8 + 29.3 = 86.1
Question 3
A manager wishes to determine the relationship between the number of miles (in hundreds of miles) the manager’s sales representatives travel per month and the amount of sales (in thousands of dollars) per month. Find the equation of the regression line for the given data. Predict the value of sales when the sales representative travel 8 miles. Predict the value of sales when the sales representative traveled 11 miles.

Miles traveled, x 2 3 10 7 8 15 3 1 11
Sales, y 31 33 78 62 65 61 48 55 120

x y xy x2
2 31 62 4
3 33 99 9
10 78 780 100
7 62 434 49
8 65 520 64
15 61 915 225
3 48 144 9
1 55 55 1
11 120 1320 121
60 553 4329 582

b =(N?XY – (?X)(?Y)) / (N?X2 – (?X)2)
a = (?Y – b(?X)) / N
b = (38961- 33180)/(5328 – 3600) = 5781/1728 = 3.35
a = (553 – 201)/ 9 = 39.1
Y = 39.1 + 3.35X
When X = 8 miles
Y = 39.1 + 26.8 = 65.9 Thousand dollars
When X = 11 miles
Y = 39.1 + 36.85 = 75.95 Thousand dollars
Question 4
Find the correlation coefficient between X and Y. Is there a weak or strong, positive or negative linear correlation between X and Y?

X -5 -3 4 1 -1 -2 0 2 3 -4
Y -10 -8 9 1 -2 -6 -1 3 6 -8
x y xy x2 y2
-5 -10 50 25 100
-3 -8 24 9 64
4 9 36 16 81
1 1 1 1 1
-1 -2 2 1 4
-2 -6 12 4 36
0 -1 0 0 1
2 3 6 4 9
3 6 18 9 36
-4 -8 32 16 64
-5 -16 181 85 396

r = Correlation coefficient

Formula retrieved from http://www.statisticshowto.com/articles/how-to-compute-pearsons-correlation-coefficients/

r = (1810 – 80)/ ?({850 – 25}{3960 – 256})
= 1730/?3055800 = 1730/1748 = 0.9897
There is therefore a strong positive correlation between X and Y
Question 5
The data below are the final exam scores of 10 randomly selected statistics students and the number of hours they studied for the exam. Find the correlation coefficient between hours studied and final exam scores. Is there a weak or strong, positive or negative correlation between hours studied and final exam scores?

Hours, x 3 5 2 8 2 4 4 5 6 3
Scores, y 65 80 60 88 66 78 85 90 90 71
x y xy x2 y2
3 65 195 9 4225
5 80 400 25 6400
2 60 120 4 3600
8 88 704 64 7744
2 66 132 4 4356
4 78 312 16 6084
4 85 340 16 7225
5 90 450 25 8100
6 90 540 36 8100
3 71 213 9 5041
42 773 3406 208 60875

r = (34060 – 32466)/?({2080 -1764}{608750 – 597529})
r = 1594/?3545836 = 1594/1833 = 0.8696
There is therefore a strong positive correlation between hours studied and the final exam scores.

References
n.a. Linear Regression. Retrieved from http://www.stat.yale.edu/Courses/1997-98/101/linreg.htm
n.a. Regression Tutorial. Retrieved from http://easycalculation.com/statistics/learn-regression.php
n.a. How to Compute Pearson’s Correlation Coefficients. Retrieved from http://www.statisticshowto.com/articles/how-to-compute-pearsons-correlation-coefficients/