Quiz 3

2. Consider the mtcars data set.

Fit a model with mpg as the outcome that includes number of cylinders as a factor variable and weight as a possible confounding variable.

Compare the effect of 8 versus 4 cylinders on mpg for the adjusted and unadjusted by weight models.

Here, adjusted means including the weight variable as a term in the regression model and unadjusted means the model without weight included.

What can be said about the effect comparing 8 and 4 cylinders after looking at models with and without weight included?

The adjusted estimate for expected change in mpg comparing 8 cylinders to 4 is -6.0708597. This is the answer from the first question.

data("mtcars")
fit2 <- lm(mpg ~ factor(cyl), data = mtcars)
coefficients(fit2)

##  (Intercept) factor(cyl)6 factor(cyl)8 
##    26.663636    -6.920779   -11.563636

The adjusted estimate for expected change in mpg comparing 8 cylinders to 4 is -6.0708597. This is the answer from the first question. Summarizing the fit 33.990794, -4.2555824, -6.0708597, -3.2056133

The unadjusted estimate for expected change in mpg comparing 8 cylinders to 4 is -11.5636364. Summarizing the fit 26.6636364, -6.9207792, -11.5636364

From the above, we can say that adjusted model has less of an impacted on mpg than unadjusted.

3. Consider the mtcars data set. Fit a model with mpg as the outcome that considers number of cylinders as a factor variable and weight as confounder. Now fit a second model with mpg as the outcome model that considers the interaction between number of cylinders (as a factor variable) and weight. Give the P-value for the likelihood ratio test comparing the two models and suggest a model using 0.05 as a type I error rate significance benchmark.

data("mtcars")
fit3 <- lm(mpg ~ factor(cyl)*wt, data = mtcars)
coefficients(fit3)

##     (Intercept)    factor(cyl)6    factor(cyl)8              wt 
##       39.571196      -11.162351      -15.703167       -5.647025 
## factor(cyl)6:wt factor(cyl)8:wt 
##        2.866919        3.454587

summary(fit1)

## 
## Call:
## lm(formula = mpg ~ factor(cyl) + wt, data = mtcars)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -4.5890 -1.2357 -0.5159  1.3845  5.7915 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   33.9908     1.8878  18.006  < 2e-16 ***
## factor(cyl)6  -4.2556     1.3861  -3.070 0.004718 ** 
## factor(cyl)8  -6.0709     1.6523  -3.674 0.000999 ***
## wt            -3.2056     0.7539  -4.252 0.000213 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.557 on 28 degrees of freedom
## Multiple R-squared:  0.8374, Adjusted R-squared:   0.82 
## F-statistic: 48.08 on 3 and 28 DF,  p-value: 3.594e-11

summary(fit3)

## 
## Call:
## lm(formula = mpg ~ factor(cyl) * wt, data = mtcars)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -4.1513 -1.3798 -0.6389  1.4938  5.2523 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)       39.571      3.194  12.389 2.06e-12 ***
## factor(cyl)6     -11.162      9.355  -1.193 0.243584    
## factor(cyl)8     -15.703      4.839  -3.245 0.003223 ** 
## wt                -5.647      1.359  -4.154 0.000313 ***
## factor(cyl)6:wt    2.867      3.117   0.920 0.366199    
## factor(cyl)8:wt    3.455      1.627   2.123 0.043440 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.449 on 26 degrees of freedom
## Multiple R-squared:  0.8616, Adjusted R-squared:  0.8349 
## F-statistic: 32.36 on 5 and 26 DF,  p-value: 2.258e-10

anova(fit1, fit3)

## Analysis of Variance Table
## 
## Model 1: mpg ~ factor(cyl) + wt
## Model 2: mpg ~ factor(cyl) * wt
##   Res.Df    RSS Df Sum of Sq      F Pr(>F)
## 1     28 183.06                           
## 2     26 155.89  2     27.17 2.2658 0.1239

From the above anova test for the two models, we see that the P value of 0.1239 is higher than .05. So we will fail to reject.

4. Consider the mtcars data set. Fit a model with mpg as the outcome that includes number of cylinders as a factor variable and weight inlcuded in the model as: lm(mpg ~ I(wt * 0.5) + factor(cyl), data = mtcars)

How is the wt coefficient interpretted?

This is the estimated expected change in MPG per half ton increase in weight for a specific number of cylinders (4, 6, 8).

5.Consider the following data set

x <- c(0.586, 0.166, -0.042, -0.614, 11.72)

y <- c(0.549, -0.026, -0.127, -0.751, 1.344)

Give the hat diagonal for the most influential point

The most influential point in predictor x is 11.72. This value is furthest away from the rest of the x value group. we will identify the hat value of that influential point.

x <- c(0.586, 0.166, -0.042, -0.614, 11.72)
y <- c(0.549, -0.026, -0.127, -0.751, 1.344)

fit5 <- lm(y~x)
influence(fit5)$hat

##         1         2         3         4         5 
## 0.2286650 0.2438146 0.2525027 0.2804443 0.9945734

The Hat value is .995

6. Consider the following data set

x <- c(0.586, 0.166, -0.042, -0.614, 11.72)

y <- c(0.549, -0.026, -0.127, -0.751, 1.344)

Give the slope dfbeta for the point with the highest hat value.

The slope dfbeta is the second column. We need to identify that value for highest hat value.

x <- c(0.586, 0.166, -0.042, -0.614, 11.72)
y <- c(0.549, -0.026, -0.127, -0.751, 1.344)

fit6 <- lm(y~x)
influence.measures(fit5)

## Influence measures of
##   lm(formula = y ~ x) :
## 
##    dfb.1_     dfb.x     dffit cov.r   cook.d   hat inf
## 1  1.0621 -3.78e-01    1.0679 0.341 2.93e-01 0.229   *
## 2  0.0675 -2.86e-02    0.0675 2.934 3.39e-03 0.244    
## 3 -0.0174  7.92e-03   -0.0174 3.007 2.26e-04 0.253   *
## 4 -1.2496  6.73e-01   -1.2557 0.342 3.91e-01 0.280   *
## 5  0.2043 -1.34e+02 -149.7204 0.107 2.70e+02 0.995   *

The slope dfbeta is -1.34e+02 = -134

7. Consider a regression relationship between Y and X with and without adjustment for a third variable Z. Which of the following is true about comparing the regression coefficient between Y and X with and without adjustment for Z.

It is possible for the coefficient to reverse sign after adjustment. For example, it can be strongly significant and positive before adjustment and strongly significant and negative after adjustment.

Quiz 3

Ashley Machado

March 18, 2016

1. Consider the mtcars data set. Fit a model with mpg as the outcome that includes number of cylinders as a factor variable and weight as confounder.

Give the adjusted estimate for the expected change in mpg comparing 8 cylinders to 4.

2. Consider the mtcars data set.

Fit a model with mpg as the outcome that includes number of cylinders as a factor variable and weight as a possible confounding variable.

Compare the effect of 8 versus 4 cylinders on mpg for the adjusted and unadjusted by weight models.

Here, adjusted means including the weight variable as a term in the regression model and unadjusted means the model without weight included.

What can be said about the effect comparing 8 and 4 cylinders after looking at models with and without weight included?

4. Consider the mtcars data set. Fit a model with mpg as the outcome that includes number of cylinders as a factor variable and weight inlcuded in the model as: lm(mpg ~ I(wt * 0.5) + factor(cyl), data = mtcars)

How is the wt coefficient interpretted?

5.Consider the following data set

x <- c(0.586, 0.166, -0.042, -0.614, 11.72)

y <- c(0.549, -0.026, -0.127, -0.751, 1.344)

Give the hat diagonal for the most influential point

6. Consider the following data set

x <- c(0.586, 0.166, -0.042, -0.614, 11.72)

y <- c(0.549, -0.026, -0.127, -0.751, 1.344)

Give the slope dfbeta for the point with the highest hat value.

7. Consider a regression relationship between Y and X with and without adjustment for a third variable Z. Which of the following is true about comparing the regression coefficient between Y and X with and without adjustment for Z.