In class activity 7

Part 1: Linear Regression Recreation with challenger2

## Example: Space Shuttle Launch Data
launch <- read.csv("challenger2.csv")

# estimate beta manually
b <- cov(launch$temperature, launch$distress_ct) / var(launch$temperature)
b

[1] -0.03364796

# calculating the covariance between two variables

# estimate alpha manually
a <- mean(launch$distress_ct) - b * mean(launch$temperature)
a

[1] 2.814585

# calculates the difference between the mean of distress_ct and the product of b and the mean of temperature

# calculate the correlation of launch data
r <- cov(launch$temperature, launch$distress_ct) /
       (sd(launch$temperature) * sd(launch$distress_ct))
# the var (temp) we chose to explain these stress events in 33%, we need more variables
r

[1] -0.3359996

cor(launch$temperature, launch$distress_ct)

[1] -0.3359996

# 33% 

# computing the slope using correlation
r * (sd(launch$distress_ct) / sd(launch$temperature))

[1] -0.03364796

#  calculates the slope of the regression line that best fits the relationship between temperature and distress_ct

# confirming the regression line using the lm function (not in text)
model <- lm(distress_ct ~ temperature, data = launch)
model

Call:
lm(formula = distress_ct ~ temperature, data = launch)

Coefficients:
(Intercept)  temperature  
    2.81458     -0.03365  

# creates a linear regression model between the temperature and distress_ct variables in the launch data frame.

summary(model)

Call:
lm(formula = distress_ct ~ temperature, data = launch)

Residuals:
    Min      1Q  Median      3Q     Max 
-1.0649 -0.4929 -0.2573  0.3052  1.7090 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)  2.81458    1.24629   2.258   0.0322 *
temperature -0.03365    0.01815  -1.854   0.0747 .
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7076 on 27 degrees of freedom
Multiple R-squared:  0.1129,    Adjusted R-squared:  0.08004 
F-statistic: 3.436 on 1 and 27 DF,  p-value: 0.07474

# provides a summary with statistics

# creating a simple multiple regression function
reg <- function(y, x) {
  x <- as.matrix(x)
  x <- cbind(Intercept = 1, x)
  b <- solve(t(x) %*% x) %*% t(x) %*% y
  colnames(b) <- "estimate"
  print(b)
}

# This function can be used to fit a linear regression model between the x and y variables

# examine the launch data
str(launch)

'data.frame':   29 obs. of  4 variables:
 $ distress_ct         : int  0 1 0 0 0 0 0 0 1 1 ...
 $ temperature         : int  66 70 69 68 67 72 73 70 57 63 ...
 $ field_check_pressure: int  50 50 50 50 50 50 100 100 200 200 ...
 $ flight_num          : int  1 2 3 4 5 6 7 8 9 10 ...

# provides the structure of the launch data frame

# test regression model with simple linear regression
reg(y = launch$distress_ct, x = launch[2])

               estimate
Intercept    2.81458456
temperature -0.03364796

# use regression model with multiple regression
reg(y = launch$distress_ct, x = launch[2:4])

                          estimate
Intercept             2.239817e+00
temperature          -3.124185e-02
field_check_pressure -2.586765e-05
flight_num            2.762455e-02

# calculates the coefficients of the linear regression model

# confirming the multiple regression result using the lm function (not in text)
model <- lm(distress_ct ~ temperature + field_check_pressure + flight_num, data = launch)
model

Call:
lm(formula = distress_ct ~ temperature + field_check_pressure + 
    flight_num, data = launch)

Coefficients:
         (Intercept)           temperature  field_check_pressure            flight_num  
           2.240e+00            -3.124e-02            -2.587e-05             2.762e-02