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)

Give a P-value for the two sided hypothesis test of whether \(\beta\)1 from a linear regression model is 0 or not.

Answer: from the linear model using x as predictor and y as regressor we can obtain the p-value displayed at the bottom line.

asw<-lm(y~x)
summary(asw)
## 
## 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: 0.05296

0.05296

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

Answer: as we can read at the third line from bottom, the residual error is 0.223

0.223

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?

Answer: Using the Predict method for Linear Model (see help predict.lm in R Studio) we can set interval to “confidence”. Obs: mean vaue must be a data.frame

data(mtcars)
head(mtcars)
##                    mpg cyl disp  hp drat    wt  qsec vs am gear carb
## Mazda RX4         21.0   6  160 110 3.90 2.620 16.46  0  1    4    4
## Mazda RX4 Wag     21.0   6  160 110 3.90 2.875 17.02  0  1    4    4
## Datsun 710        22.8   4  108  93 3.85 2.320 18.61  1  1    4    1
## Hornet 4 Drive    21.4   6  258 110 3.08 3.215 19.44  1  0    3    1
## Hornet Sportabout 18.7   8  360 175 3.15 3.440 17.02  0  0    3    2
## Valiant           18.1   6  225 105 2.76 3.460 20.22  1  0    3    1
carslm<-lm(mpg~wt, data=mtcars)
meanvalue = data.frame(wt=mean(mtcars$wt))
predict(carslm, meanvalue, interval="confidence")
##        fit      lwr      upr
## 1 20.09062 18.99098 21.19027

18.991

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

Answer: From the help, in R Studio, we know that wt represents the weight in thousand pounds. So, as the model uses wt as predictor for the miles per gallon, we can say that is

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

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?

Answer: a simple application of the prediction, giving wt as 3 once wt is the weight in thousand pounds.

newcarweight = data.frame(wt=3)
predict(carslm, newcarweight, interval="predict")
##        fit      lwr      upr
## 1 21.25171 14.92987 27.57355

27.57

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.

Answer: in this case we will scalate the previous predictor by 2.

newpredictor <-lm(mpg~I(wt/2), data=mtcars)
newsummary <-summary(newpredictor)$coefficients
newsummary
##              Estimate Std. Error   t value     Pr(>|t|)
## (Intercept)  37.28513   1.877627 19.857575 8.241799e-19
## I(wt/2)     -10.68894   1.118202 -9.559044 1.293959e-10
confint(newpredictor)
##                 2.5 %   97.5 %
## (Intercept)  33.45050 41.11975
## I(wt/2)     -12.97262 -8.40527

-12.973

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: when you multiply a regression variable by a factor, the slope will be divided by this factor. So, multiplying x by 1/100 wil divide \(\beta\)1 by 1/100 that means multiply by 100.

It would get multiplied by 100.

8. I have an outcome, Y, and a predictor, X and fit a linear regression model with Y=\(\beta\)0 +\(\beta\)1 1X+\(\epsilon\) to obtain \(\widehat{\beta}\)0 and \(\widehat{\beta}\)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?

Answer: If you only shift the predictor, the slope wil remain the same and the intercept will move in the x predictor line multiplying the \(\widehat{\beta}\)1 factor by the value of the constant with opposite signal

The new intercept would be \(\widehat{\beta}\)0 - c\(\widehat{\beta}\)1

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)?

Answer: 

InterSlope<-lm(mpg~wt, data=mtcars)
JustInter<-lm(mpg~1, data=mtcars)

num<-sum((predict(InterSlope)-mtcars$mpg)^2)
den<-sum((predict(JustInter)-mtcars$mpg)^2)        
num/den         
## [1] 0.2471672

0.25

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.