Consider the following dat with x as the predictor and y 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)Give a P-value for the two sided hypothesis test of whether \(\beta_{1}\) from a linear regression model is 0 or not.
Perform a linear regression model and then get the P-value from the summary
fit <- lm(y ~ x)
summary(fit)##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.27636 -0.18807 0.01364 0.16595 0.27143
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.1885 0.2061 0.914 0.391
## x 0.7224 0.3107 2.325 0.053 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.223 on 7 degrees of freedom
## Multiple R-squared: 0.4358, Adjusted R-squared: 0.3552
## F-statistic: 5.408 on 1 and 7 DF, p-value: 0.05296Consider the previous problem, give the estimate of the residual standard deviation
summary(fit)##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.27636 -0.18807 0.01364 0.16595 0.27143
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.1885 0.2061 0.914 0.391
## x 0.7224 0.3107 2.325 0.053 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.223 on 7 degrees of freedom
## Multiple R-squared: 0.4358, Adjusted R-squared: 0.3552
## F-statistic: 5.408 on 1 and 7 DF, p-value: 0.05296In 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")
y <- mtcars$mpg
x <- mtcars$wt
fit <- lm(y ~ x)
predict(fit, newdat = data.frame(x = mean(x)), interval = ("confidence"))## fit lwr upr
## 1 20.09062 18.99098 21.19027Refer to the previous question. Read the help file for mtcars. What is the weight coefficient interpreted as?
miles = Miles/(US) gallon wt = Weight (1000 lbs)
mpg per 1000 lbs, therefore for every 1k increase in weight will change the outcome of mpg
Consider again the mtcars data set and a linear regression with mpg as predicted by weight (1,000 lbs). A new car is coming, weighs 3000 pounds. Construct a 95% prediction interval for its mpg. What is the upper endpoint?
Using predict, get the predicted value.
newCarWeight <- 3000 / 1000
predict(fit, newdata = data.frame(x = newCarWeight), interval = ("prediction"))## fit lwr upr
## 1 21.25171 14.92987 27.57355or for calculating manually
sigma <- summary(fit)$sigma
yhat <- fit$coef[1] + fit$coef[2] * newCarWeight
predicted_tValues <- yhat + c(-1,1) * qt(.975, df=fit$df) * sigma * sqrt(1 + (1/length(y)) + ((newCarWeight - mean(x)) ^ 2 / sum((x - mean(x)) ^ 2)))
predicted_tValues## [1] 14.92987 27.57355Consider again the mtcars data set and a linear regression model with mpg as predicted by weight (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.
Calculate the new fit using short ton weight
shortTonWeight <- 2000/1000
fit_shortTon <- lm(y ~ I(x/shortTonWeight))Use the slope to calculate the change in mpg
sumCoef <- coef(summary(fit_shortTon))or
sumCoef <- summary(fit_shortTon)$coefficientsNow calculate:
sumCoef[2,1] + c(-1,1) * qt(.975, df=fit$df) * sumCoef[2,2]## [1] -12.97262 -8.40527If my X from a linear regression is measured in centimeters and I convert it to meters, what would happen to the slope coefficient?
The conversion from centimeters to meters is 100 cm = 1 m
x <- round(runif(n = 10, min = 0.1, max = 0.99),2)
x## [1] 0.75 0.70 0.59 0.62 0.89 0.30 0.61 0.65 0.36 0.95y <- round(runif(n = 10, min = 0.1, max = 0.99),2)
y## [1] 0.34 0.22 0.55 0.76 0.65 0.44 0.45 0.27 0.86 0.93fit <- lm(y ~ x)
fit$coefficients[2]## x
## 0.1251075Now convert to meters and check the results
fit <- lm(y ~ I(x/100))
fit$coefficients[2]## I(x/100)
## 12.51075I have an outcome, \(Y\), and a predictor, \(X\), and a fit a linear regression model with \(Y = \beta_{0} + \beta_{1}X + \epsilon\) to obtain \(\hat\beta_{0}\) and \(\hat\beta_{1}\). 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, \(c\)?
x <- round(runif(n = 10, min = 0.1, max = 0.99),2)
x## [1] 0.87 0.56 0.69 0.66 0.27 0.25 0.92 0.91 0.47 0.32y <- round(runif(n = 10, min = 0.1, max = 0.99),2)
y## [1] 0.71 0.16 0.64 0.91 0.20 0.53 0.68 0.45 0.62 0.47fit <- lm(y ~ x)
fit$coefficients## (Intercept) x
## 0.3103232 0.3829000Now add the constant to the equation, which basically has the new formula;
\(Y = \beta_{0} + \beta_{1}(X + c) + \epsilon\)
c = 100
fit_c <- lm(y ~ I(x + c))
fit_c$coefficients## (Intercept) I(x + c)
## -37.97968 0.38290fit$coefficients[1] - c * fit$coefficients[2]## (Intercept)
## -37.97968Refer back to the mtcars data set with mpg as an outcome and weight(wt) as the predictor. About what is the ratio of the sum of the squared errors, \(\sum^n_{i=1}(Y_{i} - \hat{Y})^2\) when comparing a model with just an intercept (denominator) to the model with the intercept and slope (numerator)?
This is simply one minus the \(R^2\) values
data(mtcars)
fit1 <- lm(mpg ~ wt, data = mtcars)
fit2 <- lm(mpg ~ 1, data = mtcars)
summary(fit1)$r.squared## [1] 0.7528328sse1 <- sum((predict(fit1) - mtcars$mpg)^2)
sse2 <- sum((predict(fit2) - mtcars$mpg)^2)
sse1/sse2## [1] 0.2471672Do the residuals always have to sum to 0 in linear regression?