Question 1

Consider the following data with x as the predictor and y as as the outcome.

x <- c(0.61, 0.93, 0.83, 0.35, 0.54, 0.16, 0.91, 0.62, 0.62)
y <- c(0.67, 0.84, 0.6, 0.18, 0.85, 0.47, 1.1, 0.65, 0.36)

summary(lm(y ~ x))$coef
##              Estimate Std. Error   t value   Pr(>|t|)
## (Intercept) 0.1884572  0.2061290 0.9142681 0.39098029
## x           0.7224211  0.3106531 2.3254912 0.05296439

Answer

0.05296

Question 2

Consider the previous problem, give the estimate of the residual standard deviation.

summary(lm(y ~ x))$sigma
## [1] 0.2229981

Answer

0.223

Question 3

In the mtcars data set, fit a linear regression model of weight (predictor) on mpg (outcome). Get a 95% confidence interval for the expected mpg at the average weight. What is the lower endpoint?

data(mtcars)
fit <- lm(mpg ~ I(wt - mean(wt)), data = mtcars)
confint(fit)
##                      2.5 %    97.5 %
## (Intercept)      18.990982 21.190268
## I(wt - mean(wt)) -6.486308 -4.202635

Answer

18.991

Question 4

Refer to the previous question. Read the help file for mtcars. What is the weight coefficient interpreted as?

Answer

The estimated expected change in mpg per 1,000 lb increase in weight.

Question 5

Consider again the mtcars data set and a linear regression model with mpg as predicted by weight (1,000 lbs). A new car is coming weighing 3000 pounds. Construct a 95% prediction interval for its mpg. What is the upper endpoint?

fit <- lm(mpg ~ wt, data = mtcars)
predict(fit, newdata = data.frame(wt = 3), interval = "prediction")
##        fit      lwr      upr
## 1 21.25171 14.92987 27.57355

Answer

27.57

Question 6

Consider again the mtcars data set and a linear regression model with mpg as predicted by weight (in 1,000 lbs). A “short” ton is defined as 2,000 lbs. Construct a 95% confidence interval for the expected change in mpg per 1 short ton increase in weight. Give the lower endpoint.

fit <- lm(mpg ~ wt, data = mtcars)
confint(fit)[2, ] * 2
##     2.5 %    97.5 % 
## -12.97262  -8.40527
## Or equivalently change the units
fit <- lm(mpg ~ I(wt * 0.5), data = mtcars)
confint(fit)[2, ]
##     2.5 %    97.5 % 
## -12.97262  -8.40527

Answer

-12.973

Question 7

If my X from a linear regression is measured in centimeters and I convert it to meters what would happen to the slope coefficient?

Answer

It would get multiplied by 100.

Question 8

I have an outcome, Y, and a predictor, X and fit a linear regression model with Y=??0+??1X+?? to obtain ??^0 and ??^1. What would be the consequence to the subsequent slope and intercept if I were to refit the model with a new regressor, X+c for some constant, c?

Answer

The new intercept would be ??0???c??1

This is exactly covered in the notes. But note that if Y=??0+??1X+?? then Y=??0???c??1+??1(X+c)+?? so that the answer is that the intercept gets subtracted by c??1

Question 9

Refer back to the mtcars data set with mpg as an outcome and weight (wt) as the predictor. About what is the ratio of the the sum of the squared errors, ???ni=1(Yi???Y^i)2 when comparing a model with just an intercept (denominator) to the model with the intercept and slope (numerator)?

fit1 <- lm(mpg ~ wt, data = mtcars)
fit2 <- lm(mpg ~ 1, data = mtcars)
1 - summary(fit1)$r.squared
## [1] 0.2471672
sse1 <- sum((predict(fit1) - mtcars$mpg)^2)
sse2 <- sum((predict(fit2) - mtcars$mpg)^2)
sse1/sse2
## [1] 0.2471672

Answer

0.25

Question 10

Do the residuals always have to sum to 0 in linear regression?

Answer

If an intercept is included, then they will sum to 0.