12/5/2018
MLR is used when we want to build a model with more than one variable
This is singular LR and models a line. \[ y_i = \beta_0 + \beta_i x_i + \epsilon_i i = 1,...,n \]
This is a MLR model that descrbies a parabola \[ y_i = \beta_0 + \beta_i x_1 + \beta_2 x_2 + \epsilon_i i = 1,...,n \]
This is a single row of data
bwt ges par age ht wt sk
1 120 285 0 27 62 100 0
This is the model resulting from said data
est std er t p
intercept -80.41 14.35 -5.60 0.0000
gestation 0.44 0.03 15.26 0.0000
parity -3.33 1.13 -2.95 0.0033
age -0.01 0.09 -0.10 0.9170
height 1.15 0.21 5.63 0.0000
weight 0.05 0.03 1.99 0.0471
smoke -8.40 0.95 -8.81 0.0000
Our model is built from the coefficient column of that data
\[ y_i = -8041 + .44 x_1 -3.33x_2 - .01x_3 + 1.15 x_4 + .05 x_5 -8.40 x_6+ \epsilon_i \]
This slope represents the average increase in \( y \) over \( x \)
Gestation: the average birthweight increases by .44 ounces per day of pregnacy
AGE: The average birth weight decreease by about 3.33 ounces per year of the mother's age
The coefficient for parity in the previous exercise was -1.93.
These coefficients are different because our model includes more variables.
To calculate the residual, we can use R right here in the presentation! This is the value predicted by our model
[1] 120.58
To find the residual, we subtract the predicted y value from the actual y value.
[1] -0.58
Hey! That's pretty good.
To calculate the residual, we can use R right here in the presentation!
This is the value predicted by our model
var.e = 249.28
var.y = 332.57
pop = 1236
k = 6
r.square = 1-(var.e/var.y)
r.square.adjusted = 1 -(var.e/var.y) * (pop-1/pop-k-1)
r.square
[1] 0.2504435
r.square.adjusted
[1] -920.2043