Prestige <-
  read.table('http://math.uttyler.edu/nathan/classes/statistics/data/prestige.data',header=TRUE)

attach(Prestige)

plot(education,prestige)

# want m and b so that we can take a number of years of eductaion,
# evaluate m*education + b, and obtain a guess at the prestige level

# prestige = m*education + b (+ err)

### The easy way:

mod.1 <- lm(prestige ~ education)  #### lm(y~x)
summary(mod.1)

#Call:
#lm(formula = prestige ~ education)
#
#Residuals:
#     Min       1Q   Median       3Q      Max 
#-26.0397  -6.5228   0.6611   6.7430  18.1636 
#
#Coefficients:
#            Estimate Std. Error t value Pr(>|t|)    
#(Intercept)  -10.732      3.677  -2.919  0.00434 ** 
#education      5.361      0.332  16.148  < 2e-16 ***
#---
#Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 
#
#Residual standard error: 9.103 on 100 degrees of freedom
#Multiple R-squared: 0.7228,	Adjusted R-squared:  0.72 
#F-statistic: 260.8 on 1 and 100 DF,  p-value: < 2.2e-16 

### prestige = 5.361*education - 10.732 ( + err) 

### making a predication:
5.361*16-10.732

abline(mod.1,col='red')

#### slope = 5.361 -- every extra year of education amounts to 5.361
####                  extra prestige points
#### intercept = -10.732 -- zero education = -10 prestige (kind of meaningless)
####
#### std error = 9.103 -- average residual is 9.103
####
#### r^2 = .7228 -- 72.28% of the variability in prestige is explained
####                by prestige's linear relationship with education

x <- rnorm(30)
y <- x*x
plot(x,y)

### question1 == what percent of the vaiablility in y is explained by x?

summary(lm(y~x))