1. Pick any two quantitative variables from a data set that interests you. If you are at a loss, look at the internal R datasets in base R and choose any.
data<-datasets::cars
head(data)
##   speed dist
## 1     4    2
## 2     4   10
## 3     7    4
## 4     7   22
## 5     8   16
## 6     9   10

You can continue to use the dataset from your last discussion, or pick up a new dataset.

  1. Tell us what are the dependent and independent variable.
#Dependent variable is "dist", measurements (in feet) of stopping distance.
#Independent variable is "speed", measuring the speed (in mph) of the car.

Type put your estimating equation.i.e. I am expecting to see subscripts i on your y, x and error term professionally done.

\[ y_i = \beta_0 + X_i\beta_1 + \epsilon_i \] \[y_i\] is the measurement (in feet) of the distance observation i. \[x_i\] is the speed of observation i. \[\beta_0\] is the y-intercept parameter \[\beta_1\] is the slope parameter \[\epsilon_i\] is the random error term for observation i.

Make sure to describe these two variables (y measures number of cars, x is income in 1000s of dollars).

  1. Estimate the linear regression in R using the lm() command.
model<-lm(dist~speed,data = data)
summary(model)
## 
## Call:
## lm(formula = dist ~ speed, data = data)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -29.069  -9.525  -2.272   9.215  43.201 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -17.5791     6.7584  -2.601   0.0123 *  
## speed         3.9324     0.4155   9.464 1.49e-12 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 15.38 on 48 degrees of freedom
## Multiple R-squared:  0.6511, Adjusted R-squared:  0.6438 
## F-statistic: 89.57 on 1 and 48 DF,  p-value: 1.49e-12
  1. Interpret the slope and intercept parameters.
model$coefficients
## (Intercept)       speed 
##  -17.579095    3.932409
#Intercept is -17.58. The interpretation is: if a car's speed is 0 mph, the predicted stopping distance is -17.58 feet. 
#Slope is 3.93. The interpretation is: for every 1 mph increase in a car's speed, the stopping distance increases by 3.93 feet.
  1. Replicate the slope and intercept parameter using the covariance/variance formulas

Slope: \[\hat{\beta_1} = \frac{Cov(X,Y)}{Var(X)}\] Intercept: \[\hat{\beta_0}=\bar{Y}-\hat{\beta_1}\bar{X}\]

 #Calculate means
x_bar<-mean(data$speed)
y_bar<-mean(data$dist)

#Calculate covariance and variance
cov_xy<-cov(data$speed,data$dist)
var_x<-var(data$speed)

#Calculate slope
beta_1<-cov_xy/var_x

#Calculate intercept
beta_0<-y_bar-beta_1*x_bar

beta_1
## [1] 3.932409
beta_0
## [1] -17.57909
#We get the same slope and intercept using the variance/covariance formulas as we do using the R lm() formula.