library(alr4)Loading required package: carLoading required package: carDataLoading required package: effectslattice theme set by effectsTheme()
See ?effectsTheme for details.library(smss)
library(stargazer)
Please cite as: Hlavac, Marek (2022). stargazer: Well-Formatted Regression and Summary Statistics Tables. R package version 5.2.3. https://CRAN.R-project.org/package=stargazer data(house.selling.price.2)
head(house.selling.price.2) P S Be Ba New
1 48.5 1.10 3 1 0
2 55.0 1.01 3 2 0
3 68.0 1.45 3 2 0
4 137.0 2.40 3 3 0
5 309.4 3.30 4 3 1
6 17.5 0.40 1 1 0The BEDS variable should be eliminated first because it has the highest p-value and is therefore the weakest predictor of price.
In forward selection, we add based on the strongest significance. The p-value for SIZE is 0 and the p-value for NEW is 0.000. I am inclined to believe that NEW appears this way because it has a value slightly higher than 0, so I would estimate that SIZE is the most statistically significant and should be added first.
I think the correlation between BEDS and PRICE is influenced by another variable, perhaps SIZE. On its own, BEDS may just be a weak indicator of PRICE,
m1 <- lm(P ~ S, house.selling.price.2)
m2 <- lm(P ~ Be, house.selling.price.2)
m3 <- lm(P ~ Ba, house.selling.price.2)
m4 <- lm(P ~ New, house.selling.price.2)
m5 <- lm(P~., house.selling.price.2)
m6 <- lm(P~.-Be, house.selling.price.2)
stargazer(m1, m2, m3, m4, m5, m6, type = 'text')
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Dependent variable:
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P
(1) (2) (3) (4) (5) (6)
----------------------------------------------------------------------------------------------------------------------------------------------------------------
S 75.607*** 64.761*** 62.263***
(3.865) (5.630) (4.335)
Be 42.969*** -2.766
(6.160) (3.960)
Ba 76.026*** 19.203*** 20.072***
(7.822) (5.650) (5.495)
New 34.158*** 18.984*** 18.371***
(9.383) (3.873) (3.761)
Constant -25.194*** -37.229* -49.248*** 89.249*** -41.795*** -47.992***
(6.688) (19.955) (15.644) (5.148) (12.104) (8.209)
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Observations 93 93 93 93 93 93
R2 0.808 0.348 0.509 0.127 0.869 0.868
Adjusted R2 0.806 0.341 0.504 0.118 0.863 0.864
Residual Std. Error 19.473 (df = 91) 35.861 (df = 91) 31.119 (df = 91) 41.506 (df = 91) 16.360 (df = 88) 16.313 (df = 89)
F Statistic 382.628*** (df = 1; 91) 48.660*** (df = 1; 91) 94.473*** (df = 1; 91) 13.254*** (df = 1; 91) 145.763*** (df = 4; 88) 195.313*** (df = 3; 89)
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Note: *p<0.1; **p<0.05; ***p<0.01For parts a and b above, I compare several models in one chart (I did not run all possible models; just a few that made sense based on previous information). Because a higher R² indicates a stronger fit, the model including all variables (m5) should be selected if using R² and the model including all except BEDS (m6) should be selected if using adjusted R².
#use PRESS function to test all models from previous questions
PRESS <- function(linear.model) {
pr <- residuals(linear.model)/(1-lm.influence(linear.model)$hat)
PRESS <- sum(pr^2)
return(PRESS)
}
cat(' ', 'PRESS', '\n',
'm1: ', PRESS(m1), '\n',
'm2: ', PRESS(m2), '\n',
'm3: ', PRESS(m3), '\n',
'm4: ', PRESS(m4), '\n',
'm5: ', PRESS(m5), '\n',
'm6: ', PRESS(m6), '\n') PRESS
m1: 38203.29
m2: 122984.3
m3: 95732.14
m4: 164039.3
m5: 28390.22
m6: 27860.05 If using PRESS as the primary criterion, model m6 is the best choice because it has the lowest PRESS score.
cat(' ', 'AIC', '\n',
'm1: ', AIC(m1), '\n',
'm2: ', AIC(m2), '\n',
'm3: ', AIC(m3), '\n',
'm4: ', AIC(m4), '\n',
'm5: ', AIC(m5), '\n',
'm6: ', AIC(m6), '\n') AIC
m1: 820.1439
m2: 933.7168
m3: 907.3327
m4: 960.908
m5: 790.6225
m6: 789.1366 m6 is the best choice here, but m5 is close.
cat(' ', 'BIC', '\n',
'm1: ', BIC(m1), '\n',
'm2: ', BIC(m2), '\n',
'm3: ', BIC(m3), '\n',
'm4: ', BIC(m4), '\n',
'm5: ', BIC(m5), '\n',
'm6: ', BIC(m6), '\n') BIC
m1: 827.7417
m2: 941.3146
m3: 914.9305
m4: 968.5058
m5: 805.8181
m6: 801.7996 Similar to AIC, m6 is again the best choice here, but m5 is close.
It seems to me that m6, which includes all variables but BEDS, is the best fit. It meets the criterion for best fit in most tests. Model m5, which includes all variables, would be a close 2nd choice.
Tree volume estimation is a big deal, especially in the lumber industry. Use the trees data to build a basic model of tree volume prediction. In particular,
tree <- lm(Volume ~ Girth + Height, data=trees)
tree
Call:
lm(formula = Volume ~ Girth + Height, data = trees)
Coefficients:
(Intercept) Girth Height
-57.9877 4.7082 0.3393 par(mfrow = c(2,3)); plot(tree, which = 1:6)
The Residuals vs Fitted plot violates the assumption of linearity. The line in the Scale-Location plot is not quite horizontal, but it is unclear which assumption this violates- perhaps homoskedasticity.
In the 2000 election for U.S. president, the counting of votes in Florida was controversial. In Palm Beach County in south Florida, for example, voters used a so-called butterfly ballot. Some believe that the layout of the ballot caused some voters to cast votes for Buchanan when their intended choice was Gore.
The data has variables for the number of votes for each candidate—Gore, Bush, and Buchanan.
vote <- lm(Buchanan ~ Bush, data=florida)
par(mfrow = c(2,3)); plot(vote, which = 1:6)
Palm Beach does appear to be an outlier as it falls way outside the lines on most plots. Dade is also an outlier on a few plots.
vote_log <- lm(log(Buchanan)~log(Bush),data=florida)
par(mfrow = c(2,3)); plot(vote_log, which = 1:6)
Palm Beach still appears to be a significant outlier, but now there are other counties which also stand out. Dade no longer appears to be an outlier on any plot.