Problema Solution
In 2000, 40% of college seniors graduated from college on time. In 2005, 43% graduated on time. In 2010, 50% graduated on time. In 2003, 48% graduated on time.
A) find the equation for the line of best for for the data.
B) based on your model, what percent of the seniors graduated on time in 2009?
C) predict what percent of the seniors will graduate on time in 2020.
Answer provided by our tutors
A) we will find Linear Regression Equation in the form of y = a + bx
the x-axis will have the years after 2000
the y -axis will have the % of graduated on time
x: 0, 3, 5, 10
sum of all x is Sx = 0 + 3 + 5 + 10 = 18
x^2: 0, 9, 25, 100
sum of all x^2 is Sx^2 = 134
y: 0.40, 0.48, 0.43, 0.50
sum of all y: Sy = 1.8
y^2: 0.40^2. 0.48^2, 0.43^2, 0.50^2
sum of all y^2 is Sy^2 = 0.8253
xy:0*40, 3*0.40, 5*0.48, 10*0.50
sum of all xy: Sxy = 0*40 + 3*0.40 + 5*0.48 + 10*0.50 = 8.6
a = ((Sy)(Sx^2) - (Sx)(Sxy))/(n(Sx^2) - (Sx)^2)
n is the sample size in our case n = 4
a = (1.8*134 - 18*8.6)/(4*134 - 18^2)
a = 0.4075
b = (n(Sxy) - (Sx)(Sy))/(n(Sx^2) - (Sx)^2))
b = (4*8.6 - 18*1.8)/(4*134 - 18^2)
b = 0.009434
the equation for the line of best for the data is y = 0.475 + 0.009434x
B) in 2009 we have x = 9 and for y we have
y = 0.475 + 0.009434*9
y = 0.5599 or 55.99% graduated in 2009
C) in 200 we have x = 20 and for y we have
y = 0.475 + 0.009434*20
y = 0.66368 or 66.37% will graduate on time