1.Consider the following data with x as the predictor and y as as the outcome.Give a P-value for the two sided hypothesis test of whether \(\beta_1\) from a linear regression model is 0 or not.

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)

fit <- lm(y~x)
beta <- coef(fit)[[2]]
n <- length(x)
sigma_squared <- sum(fit$residuals^2)/(n-2)
beta1_sigma <- sigma_squared/sum((x-mean(x))^2)
tbeta1 <- beta/sqrt(beta1_sigma) #test whether beta1 is zero
2*pt(tbeta1, n-2,lower.tail = FALSE)
## [1] 0.05296439

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

sqrt(sigma_squared)
## [1] 0.2229981

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?

library(UsingR)
data(mtcars)
y <- mtcars$mpg
x <- mtcars$wt
fit <- lm(y~x)
n <- length(y)

##Confidence interval for the expected mpg
##does not have the additional 1
sigma <- sqrt(sum(fit$residuals^2)/(n-2))
beta0_sigma <- sqrt(1/n)*sigma
t0025 <- qt(.025, df=n-2)
coef(fit)[[1]] + coef(fit)[[2]]*mean(x) + t0025*beta0_sigma
## [1] 18.99098
newx <- data.frame(x=mean(x))
predict(fit, newdata = newx, interval = ("confidence"))
##        fit      lwr      upr
## 1 20.09062 18.99098 21.19027

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?

Prediction interval for mpg has an additional 1.

newx <- data.frame(x=3)
predict(fit, newdata = newx, interval = ("prediction"))
##        fit      lwr      upr
## 1 21.25171 14.92987 27.57355

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.

fit2 <-lm(y ~I(x/2))
sumCoef <- summary(fit2)$coef
sumCoef[2,1] + qt(0.025,df=n-2)*sumCoef[2,2]
## [1] -12.97262

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, \(\sum(Y_i-\hat{Y}_i)^2\) when comparing a model with just an intercept (denominator) to the model with the intercept and slope (numerator)?

d <- sum((y-mean(y))^2)
beta0 <- summary(fit)$coef[1]
beta1 <- summary(fit)$coef[2]
ssy <- sum((y-beta1*x-beta0)^2) 
ssy/d
## [1] 0.2471672