date()
## [1] "Tue Oct 30 13:38:00 2012"
Due Date/Time: October 30, 2012, 1:45pm
The points per quesion are given in parentheses.
(1) The UScereal (MASS package) contains many variables regarding breakfast cereals. One variable is the amount of sugar per portion and another is shelf position (counting from the floor up). Create side-by-side box plots showing the distribution of sugar by shelf number. Perform a t test to determine if there is a significant difference in the amount of sugar in cereals on the first and second shelves. What do you conclude? (20)
(install.packages("MASS"))
## Error: trying to use CRAN without setting a mirror
require(MASS)
## Loading required package: MASS
attach(UScereal)
require(ggplot2)
## Loading required package: ggplot2
ggplot(UScereal, aes(x = factor(shelf), y = sugars)) + geom_boxplot()
t.test(sugars[shelf == 1], sugars[shelf == 2])
##
## Welch Two Sample t-test
##
## data: sugars[shelf == 1] and sugars[shelf == 2]
## t = -3.975, df = 30, p-value = 0.0004086
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -9.404 -3.021
## sample estimates:
## mean of x mean of y
## 6.295 12.508
detach(UScereal)
Since the p-value is 0.0004 and less and 0.01, there is moderate evidence to reject the null hypothesis. Therefore, there is a significant difference in the amount of sugar in cereals on the first and second shelves.
(2) The data set USmelanoma (HSAUR2 package) contains male mortality counts per one million inhabitants by state along with the latitude and longitude centroid of the state. (40)
a. Create a scatter plot of mortality versus latitude using latitude as the explanatory variable.
install.packages("HSAUR2")
## Error: trying to use CRAN without setting a mirror
require(HSAUR2)
## Loading required package: HSAUR2
## Loading required package: lattice
## Loading required package: scatterplot3d
require(ggplot2)
head(USmelanoma)
## mortality latitude longitude ocean
## Alabama 219 33.0 87.0 yes
## Arizona 160 34.5 112.0 no
## Arkansas 170 35.0 92.5 no
## California 182 37.5 119.5 yes
## Colorado 149 39.0 105.5 no
## Connecticut 159 41.8 72.8 yes
p = ggplot(USmelanoma, aes(x = latitude, y = mortality)) + geom_point()
p
b. Add the linear regression line to your scatter plot.
p = p + geom_smooth(method = "lm", se = FALSE)
p
c. Regress mortality on latitude and interpret the value of the slope coefficient.
lm(mortality ~ latitude, data = USmelanoma)
##
## Call:
## lm(formula = mortality ~ latitude, data = USmelanoma)
##
## Coefficients:
## (Intercept) latitude
## 389.19 -5.98
The slope coefficient is -5.98 meaning that as mortality increases 389.12 counts per one million inhabitants, latitude decreases 5.98 degrees.
d. Determine the sum of squared errors.
model = lm(mortality ~ latitude, data = USmelanoma)
deviance(model)
## [1] 17173
e. Use density and box plots to examine the model assumptions. What do you conclude?
boxplot(mortality ~ cut(latitude, breaks = quantile(latitude)), data = USmelanoma)
require(sm)
## Loading required package: sm
## Package `sm', version 2.2-4.1 Copyright (C) 1997, 2000, 2005, 2007, 2008,
## A.W.Bowman & A.Azzalini Type help(sm) for summary information
model = lm(mortality ~ latitude, data = USmelanoma)
res = residuals(model)
sm.density(res, xlab = "Model residuals", model = "Normal")
The assumptions of linearity and constant variance are suspect, however, there appears to be no evidence against normality.
(3) Davies and Goldsmith (1972) investigated the relationship between abrasion loss (abrasion) of samples of rubber (grams per hour) as a function of hardness (higher values indicate harder rubber) and tensile strength (kg/cm2 ). The data are in AbrasionLoss.txt. Input the data using AL = read.table(“http://myweb.fsu.edu/jelsner/AbrasionLoss.txt”, header=TRUE) (40)
a. Create a scatter plot matrix of the three variables. Based on the scatter of points in the plot of abrasion versus strength does it appear that tensile strength would be helpful in explaining abrasion loss?
AL = read.table("http://myweb.fsu.edu/jelsner/AbrasionLoss.txt", header = TRUE)
head(AL)
## abrasion hardness strength
## 1 372 45 162
## 2 206 55 233
## 3 175 61 232
## 4 154 66 231
## 5 136 71 231
## 6 112 71 237
pairs(AL, panel = panel.smooth)
No, it does not appear that tensile strength would be helpful in explaining abrasion loss. There appears to be little to no correlation (Row 1, Column 3).
b. Regress abrasion loss on hardness and strength. What is the adjusted R squared value? Is strength an important explanatory variable after accounting for hardness?
model1 = lm(abrasion ~ hardness + strength, data = AL)
summary(model1)
##
## Call:
## lm(formula = abrasion ~ hardness + strength, data = AL)
##
## Residuals:
## Min 1Q Median 3Q Max
## -79.38 -14.61 3.82 19.75 65.98
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 885.161 61.752 14.33 3.8e-14 ***
## hardness -6.571 0.583 -11.27 1.0e-11 ***
## strength -1.374 0.194 -7.07 1.3e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 36.5 on 27 degrees of freedom
## Multiple R-squared: 0.84, Adjusted R-squared: 0.828
## F-statistic: 71 on 2 and 27 DF, p-value: 1.77e-11
The adjusted R squared value is 82.84%. Strength is not an important explanatory variable after accounting for hardness due it's t-value of -7.073. Hardness has a t-value of -11.267 making it a more important explanatory variable.
c. On average how much additional abrasion is lost for every 1 kg/cm2 increase in tensile strength?
model1
##
## Call:
## lm(formula = abrasion ~ hardness + strength, data = AL)
##
## Coefficients:
## (Intercept) hardness strength
## 885.16 -6.57 -1.37
About 885 grams per hour are lost for every 1kg/cm2 increase in tensile strength.
d. Check the correlations between the explanatory variables. Could collinearity be a problem for interpreting the model?
cor(AL)
## abrasion hardness strength
## abrasion 1.0000 -0.7377 -0.2984
## hardness -0.7377 1.0000 -0.2992
## strength -0.2984 -0.2992 1.0000
Yes, collinearity could be a problem for interpreting the model regarding hardness. The value for hardness (-0.73) is almost double that of strength (-0.30).
e. Find the 95% prediction interval for the abrasion corresponding to a new rubber sample having a hardness of 60 units and a tensile strength of 200 kg/cm2.
predict(model1, data.frame(hardness = 60, strength = 200), level = 0.95, interval = "confidence")
## fit lwr upr
## 1 216 197.6 234.5