Optimization Week11

###### plot SS 
set.seed(1234)                        
x <- seq(0,5,length=30)
y <- 3 + 5*x + rnorm(length(x),0,1)
x.df <- data.frame(y,x)               # make a data.frame
res <- lm(y~x,data=x.df)              # do linear modeling
summary(res)                          # show results 

## 
## Call:
## lm(formula = y ~ x, data = x.df)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -2.0306 -0.5751 -0.2109  0.5522  2.7050 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.6800     0.3273   8.188 6.51e-09 ***
## x             5.0094     0.1124  44.562  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.9189 on 28 degrees of freedom
## Multiple R-squared:  0.9861, Adjusted R-squared:  0.9856 
## F-statistic:  1986 on 1 and 28 DF,  p-value: < 2.2e-16

set.seed used to Generate same set of random numbers. where x seq (generates a sequence of numbers). res is lm y~x. So p-value < alpha where (2.2e-16 < 0.05), then H0 will be rejected.

a1 <- cov(x,y)/var(x)              # analytical OLS solution 
a0 <- mean(y) - a1*mean(x)
c(a0=a0,a1=a1)                     # show OLS estimators a0,a1

##       a0       a1 
## 2.680009 5.009426

a1 is 2.68, we got it from the covariance of two variables x and y in data set divided by x variance

cor(x.df)                          # shows the Pearson  r.xy

##           y         x
## y 1.0000000 0.9930235
## x 0.9930235 1.0000000

correlation between x and y equal to 0.9930235 or 99,3%

sqrt(sum((y - predict(res))^2)/28) # residual standard error

## [1] 0.9188509

the residual standard error obtained is 0.9188509 or 91,8%

cor(x.df$y,predict(res))^2         # R2-value #1: cor(y,y.hat)^2; 

## [1] 0.9860958

The magnitude of the correlation between degrees of freedom x and y is 98.6%

ss <- anova(res)
names(ss)                          # what's in ss?

## [1] "Df"      "Sum Sq"  "Mean Sq" "F value" "Pr(>F)"

variables contained in ss namely sums of squares, mean squares, F-test, and p-value.

r2 <- ss[,2][1]/sum(ss[,2]) 
r2                              # R2-value #2: SS.model/ss.total

## [1] 0.9860958

R-aquared is ss divided by sum of ss, so we get 98,6%

F.test <- ss[,3][1]/ss[,3][2]   # Calculate F-value 
F.test

## [1] 1985.773

Calculate F-value is 1985.773

1-pf(F.test,ss[1,1],ss[1,2] )      # Prob(F>F.test)|H0

## [1] 0

The probability of the F value is greater than the F test of 0.