AMS 5 Lecture 10
4/26/2017
(8:00-9:05)
Regression: Additional Comments
Interpreting the coefficients
• Example:
o x(j) = mathSAT score of jth student
o y(j) = Freshman GPA
_
o x = 550, SD(x) = 80
_
o y = 2.6, SD(y) = 0.6, r = 0.4
o Regression equation
▪ Slope coefficient: B1 = r*SD(y)/SD(x) = 0.003
_ _
▪ Intercept coefficient: B0 = y – B1*x
o Equation: ^y(j) = .95 + .003x(j) predicted average statistics of all students who
scored x(j) on math SAT.
▪ .95 and .003 = sample statistics
o B1 = .003 the average increase in GPA for every 1pt percentage in math SAT
score.
▪ Remember: Correlation does not mean causation
o B(0) = .95 Avg. GPA of all students who scored 0 on math SAT. that’s
meaningless, since you can’t score 0 on the SAT; lowest score is 200.
o Regression based on this window (ex: 200-800)…Doesn’t help with areas outside
the window aka, you can’t predict trends way outside your range with
regression
• Example:
o In a hypothetical BP on Cig/Day regression
▪ ^y(j) = 119.48 + .56x(j)
o y(j) = BP of jth man
o x(j) = # cigs/day smoked by jth man o 0.56 = avg. increase in BP for each cig./day smoked
o 119.48 = Avg BP of men who do not smoke
• Given a set of paired data, you can compute the correlation coefficient
o The correlation between x and y is the same as the correlation between y and x
over r.
▪ There are two regression equations!
• Example:
o SAT/GPA example
▪ ^y(j) = .95 + 0.003x(j)
▪ ^x(j) = U(0) + U(1)y(j)
▪ U(1) = r*SD(x)/SD(y)

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