## data-generating process: linear vs. quadratic, homoscedastic vs. heteroscedastic type <- sample(c("constant", "linear"), size = 1, prob = c(0.35, 0.65)) d <- data.frame(x = runif(100, -1, 1)) a <- 0 b <- if(type == "constant") 0 else sample(c(-1, 1), 1) * runif(1, 0.6, 0.9) d$y <- a + b * d$x + rnorm(100, sd = 0.25) write.csv(d, "linreg.csv", row.names = FALSE, quote = FALSE) ## model and interpretation m <- lm(y ~ x, data = d) ahat <- coef(m)[1] bhat <- coef(m)[2] bpvl <- summary(m)$coefficients[2, 4] bsol <- c(bpvl >= 0.05, (bpvl < 0.05) & (bhat > 0), (bpvl < 0.05) & (bhat < 0)) bint <- c("`x` and `y` are not significantly correlated", "`y` increases significantly with `x`", "`y` decreases significantly with `x`") bint <- bint[bsol]
Theory: Consider a linear regression of y
on x
. It is usually estimated with
which estimation technique (three-letter abbreviation)?
This estimator yields the best linear unbiased estimator (BLUE) under the assumptions of the Gauss-Markov theorem. Which of the following properties are required for the errors of the linear regression model under these assumptions?
Application: Using the data provided in linreg.csv estimate a
linear regression of y
on x
. What are the estimated parameters?
Intercept: ##ANSWER3##
Slope: ##ANSWER4##
In terms of significance at 5% level:
x
and y
are not significantly correlatedy
increases significantly with x
y
decreases significantly with x
Theory: Linear regression models are typically estimated by ordinary least squares (OLS). The Gauss-Markov theorem establishes certain optimality properties: Namely, if the errors have expectation zero, constant variance (homoscedastic), no autocorrelation and the regressors are exogenous and not linearly dependent, the OLS estimator is the best linear unbiased estimator (BLUE).
Application: The estimated coefficients along with their significances are reported in the
summary of the fitted regression model, showing that r bint
(at 5% level).
summary(m)
Code: The analysis can be replicated in R using the following code.
## data d <- read.csv("linreg.csv") ## regression m <- lm(y ~ x, data = d) summary(m) ## visualization plot(y ~ x, data = d) abline(m)
exname: Linear regression
extype: cloze
exsolution: OLS|01001|r fmt(ahat, 3)
|r fmt(bhat, 3)
|r mchoice2string(bsol)
exclozetype: string|mchoice|num|num|schoice
extol: 0.01
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