Homework #4 Problem 3

Use the purity-hydrocarbon data from Homework #2 and R to answer the following questions. The data are hosted on my github and the linke is provided on the WISE page.

The purity of oxygen produced by a fractionation process is thought to be related to the percentage of hydrocarbons in the main condenser of the processing unity. Data for twenty samples are contained in the dataset. The dataset contains two columns: Purity (%) and Hydrocarbon (%).

# Load Data

oxygen <- read.csv("https://raw.githubusercontent.com/kitadasmalley/sp21_MATH376LMT/main/data/oxygenPurity.csv",
                   header = TRUE)

# Explanatory: Percent Hydrocarbon

x<- oxygen$hydro

# Response: Purity of Oxygen

y <- oxygen$purity

Use R to report the least squares regression line equation. Please submit your R code for both parts.

a) First program it from scratch using the least squares formulas that we derived.

#First Attempt

beta_1 <- sum((x-(mean(x))*(y-mean(y))))/(sum(x-mean(x))^2)
beta_0 <- mean(y)-beta_1*mean(x)

Verifying:

beta_1

## [1] 5.315003e+30

beta_0

## [1] -6.284991e+30

xy <- lm(y~x)
xy

## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##       77.86        11.80

These two aren’t the same at all.

This became too complicated, so I decided to break it up into separate parts.

# Second Attempt

x2 <- sum((x-mean(x))*(y-mean(y)))
x3 <- sum((x-mean(x))^2)

beta_1 <- x2/x3
beta_0 <- mean(y) - beta_1*mean(x)

beta_0

## [1] 77.86328

beta_1

## [1] 11.80103

b) Verify your solutions using the lm() function built into R.

xy <- lm(y~x)
xy

## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##       77.86        11.80

That looks better!