Assignment4

Example 1

For example 19 on Page 79 in the book, carry out the regression using R.

x <- -0.98 1.00 2.02 3.03 4.00 y <- 2.44 -1.51 -0.47 2.54 7.52

Hence, the regression equation is y = 0.9373x + 0.4038

x <- c(-0.98, 1.00, 2.02, 3.03, 4.00)
y <- c(2.44, -1.51, -0.47, 2.54, 7.52)

model1 <- lm(y~x)
model1

## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##      0.4038       0.9373

summary(model1)

## 
## Call:
## lm(formula = y ~ x)
## 
## Residuals:
##       1       2       3       4       5 
##  2.9547 -2.8511 -2.7671 -0.7037  3.3671 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept)   0.4038     2.2634   0.178    0.870
## x             0.9373     0.9058   1.035    0.377
## 
## Residual standard error: 3.481 on 3 degrees of freedom
## Multiple R-squared:  0.263,  Adjusted R-squared:  0.01739 
## F-statistic: 1.071 on 1 and 3 DF,  p-value: 0.3769

Example 2

Implement the nonlinear curve-fittting of Example 20 on Page 83 for the following data:

x <- 0.10 0.50 1.00 1.50 2.00 2.5 y <- 0.1 0.28 0.4 0.4 0.37 0.32

x <- c(0.10, 0.50, 1.00, 1.50, 2.00, 2.50)
y <- c(0.10, 0.28, 0.40, 0.40, 0.37, 0.32)

df1 <- data.frame(x, y)
eq1 <- function(x, a, b){x / (a + b * x^2)}
model2 <- nls(y ~ eq1(x, a, b), data = df1, start = list(a = 1, b = 1))
model2

## Nonlinear regression model
##   model: y ~ eq1(x, a, b)
##    data: df1
##     a     b 
## 1.485 1.002 
##  residual sum-of-squares: 0.00121
## 
## Number of iterations to convergence: 5 
## Achieved convergence tolerance: 3.899e-07

s <- seq(from = 0, to = 1, length = 50)
plot(x, y)
lines(s, predict(model2, list(x = s)))

Example 3

For the data with binary y values, try to fit the following data

x <- 0.1 0.5 1.0 1.5 2.0 2.5 y <- 0 0 1 1 1 0

to the linear function

y = 1/(1 + e^(a + bx)), starting with a = 1 and b = 1.

x <- c(0.1, 0.50, 1.0, 1.5, 2.0, 2.5)
y <- c(0, 1, 0, 1, 0, 1)

df2 <- data.frame(x, y)
eq2 <- function(x, a, b){1 / (1 + exp(-(a + b * x)))
}
model3 <- nls(y ~ eq2(x, a, b), data = df2, start = list(a = 1, b = 1))
model3

## Nonlinear regression model
##   model: y ~ eq2(x, a, b)
##    data: df2
##       a       b 
## -0.8294  0.6521 
##  residual sum-of-squares: 1.387
## 
## Number of iterations to convergence: 7 
## Achieved convergence tolerance: 1.879e-06

s <- seq(from = 0, to = 1, length = 50)
plot(x, y)
lines(s, predict(model3, list(x = s)))