-
Notifications
You must be signed in to change notification settings - Fork 4
/
assignment2.R
1103 lines (920 loc) · 42.1 KB
/
assignment2.R
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882
883
884
885
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
#' ---
#' title: "Homework 2"
#' author: "Allan Kimaina"
#' date: "February 22, 2018"
#' header-includes:
#' - \usepackage{pdflscape}
#' - \newcommand{\blandscape}{\begin{landscape}}
#' - \newcommand{\elandscape}{\end{landscape}}
#' output:
#' pdf_document: default
#' html_document: default
#' ---
#'
#'
#knitr::opts_chunk$set(echo = F)
# load package
library(dplyr)
library(car)
library(sjPlot)
library(sjmisc)
library(sjlabelled)
library(ggpubr)
library(ggpmisc)
library("gridExtra")
library(stargazer)
library(e1071)
library(jtools)
library(effects)
library(multcompView)
library(ggplot2)
library(ggrepel)
library(MASS)
library(broom)
library(ggcorrplot)
library(leaps)
library(relaimpo)
library(olsrr)
# load all data
wine = read.csv("data/Wine.csv")
bush = read.csv("data/Bush.csv")
bloodBrain = read.csv("data/BloodBrain.csv")
# clean bush
colnames(bush)[1] <- "county" # name the 1st column using standard
col2cvt <- 2:ncol(bush) # get all numeric columns
bush[,col2cvt] <- lapply(bush[,col2cvt],function(x){as.numeric(gsub(",", "", x))}) # convert to numeric
# clean wine
colnames(wine)[1] <- "COUNTRY"
# clean wine
colnames(bloodBrain)[1] <- "BRAIN"
#'
#' \onecolumn
#'
#' # Question 1
#'
#' ## a. Using the wine data from the previous homework, create two new variables by logarithmically transforming both wine and mortality
#'
#'
wine$logMORTALITY <- log(wine$MORTALITY)
wine$logWINE <- log(wine$WINE)
# print
stargazer(wine,
header=F,
type = "latex",
no.space = T,
summary = F,
single.row = T
)
#'
#'
#' ## b. Plot mortality vs. wine; log mortality vs. wine, mortality vs. log wine and log mortality vs. log wine. Which one looks more linear?
#'
#'
wine.model.plot = ggplot(wine, aes(y=MORTALITY, x=WINE)) +
geom_point(alpha = .5) +
geom_point(color = "blue") +
labs(x='WINE', y='MORTALITY',
title = "Mortality vs Wine")+
# scale_x_log10() +
# scale_y_log10()+
theme_classic()+ guides(fill=FALSE)+
theme(
axis.title = element_text( size=8),
axis.title.x = element_text( size=8),
axis.title.y = element_text( size=8)
)
wine.model.log.x.plot=ggplot(wine, aes(y=MORTALITY, x=logWINE)) +
geom_point(alpha = .5) +
geom_point(color = "blue") +
labs(x='log(WINE)', y='MORTALITY',
title = "Mortality vs log(Wine)")+
#scale_x_log10() +
#scale_y_log10()+
theme_classic()+ guides(fill=FALSE)+
theme(
axis.title = element_text( size=8),
axis.title.x = element_text( size=8),
axis.title.y = element_text( size=8)
)
wine.model.log.y.plot=ggplot(wine, aes(y=logMORTALITY, x=WINE)) +
geom_point(alpha = .5) +
geom_point(color = "blue") +
labs(x='WINE', y='log(MORTALITY)',
title = "log(Mortality) vs Wine")+
# scale_x_log10() +
# scale_y_log10()+
theme_classic()+ guides(fill=FALSE)+
theme(
axis.title = element_text( size=8),
axis.title.x = element_text( size=8),
axis.title.y = element_text( size=8)
)
wine.model.log.xy.plot =ggplot(wine, aes(y=logMORTALITY, x=logWINE)) +
geom_point(alpha = .5) +
geom_point(color = "blue") +
labs(x='log(Wine)', y='log(Mortality)',
title = "log(Mortality) vs log(Wine)")+
#scale_x_log10() +
#scale_y_log10()+
theme_classic()+ guides(fill=FALSE)+
theme(
axis.title = element_text( size=8),
axis.title.x = element_text( size=8),
axis.title.y = element_text( size=8)
)
grid.arrange(wine.model.plot,
wine.model.log.x.plot,
wine.model.log.y.plot,
wine.model.log.xy.plot,
ncol =2.,
nrow = 2.)
#'
#'
#' - From the above scatter plot, the logarithmic transformation of both dependent variable (log of Mortality) and independent variable (log of Wine) results in the most linear relationship compared to the other plots.
#' - Transforming both predictor and response variable we get the best linear relationship
#'
#' ## c. Fit four linear regression models corresponding to the four plots and report the regression equation and R-squared. Which model do you think is best?
#'
#'
#' * **Mortality vs Wine:** \(HeartAttackMortality = 7.69- 0.0761*WineConsumption + \epsilon\)
#' * **Mortality vs log(Wine):** \(HeartAttackMortality = 10.280 - 1.771 *log(WineConsumption) + \epsilon\)
#' * **log(Mortality) vs Wine:** \(log(HeartAttackMortality) = 2.0453 - 0.0159 *WineConsumption + \epsilon\)
#' * **log(Mortality) vs log(Wine):** \(log(HeartAttackMortality) = 2.5556 - 0.3556*log(WineConsumption) + \epsilon\)
#'
#'
#'
wine.model.plot.1=ggplot(wine, aes(y=MORTALITY, x=WINE)) +
geom_point(alpha = .5) +
geom_point(color = "blue") +
stat_smooth(method = "lm", color="red")+
labs(x='WINE', y='MORTALITY',
title = "Mortality vs Wine")+
stat_poly_eq(aes(label = paste( ..rr.label.., ..adj.rr.label.., sep = "~~~")),
label.x.npc = "right", label.y.npc = 0.87, geom = "text",
formula = y ~x, parse = TRUE, size = 3)+
stat_poly_eq(aes(label = paste(..eq.label.., sep = "~~~")),
label.x.npc = "left", label.y.npc = 0.001, geom = "text",
formula = y ~x, parse = TRUE, size = 3)+
stat_fit_glance(method = "lm",
method.args = list(formula = y~x),
geom = "text", size = 3,
label.y.npc = 0.85, label.x.npc = "right",
aes(label = paste("P-value = ", signif(..p.value.., digits = 4), sep = "")))+
# scale_x_log10() +
# scale_y_log10()+
theme_classic()+ guides(fill=FALSE)+
theme(
axis.title = element_text( size=8),
axis.title.x = element_text( size=8),
axis.title.y = element_text( size=8)
)
wine.model.plot.2=ggplot(wine, aes(y=MORTALITY, x=logWINE)) +
geom_point(alpha = .5) +
geom_point(color = "blue") +
stat_smooth(method = "lm", color="red")+
labs(x='log(WINE)', y='MORTALITY',
title = "Mortality vs log(Wine)")+
stat_poly_eq(aes(label = paste( ..rr.label.., ..adj.rr.label.., sep = "~~~")),
label.x.npc = "right", label.y.npc = 0.87, geom = "text",
formula = y ~x, parse = TRUE, size = 3)+
stat_poly_eq(aes(label = paste(..eq.label.., sep = "~~~")),
label.x.npc = "left", label.y.npc = 0.03, geom = "text",
formula = y ~x, parse = TRUE, size = 3)+
stat_fit_glance(method = "lm",
method.args = list(formula = y~x),
geom = "text", size = 3,
label.y.npc = 0.85, label.x.npc = "right",
aes(label = paste("P-value = ", signif(..p.value.., digits = 4), sep = "")))+
#scale_x_log10() +
#scale_y_log10()+
theme_classic()+ guides(fill=FALSE)+
theme(
axis.title = element_text( size=8),
axis.title.x = element_text( size=8),
axis.title.y = element_text( size=8)
)
wine.model.plot.3=ggplot(wine, aes(y=logMORTALITY, x=WINE)) +
geom_point(alpha = .5) +
geom_point(color = "blue") +
stat_smooth(method = "lm", color="red")+
labs(x='WINE', y='log(MORTALITY)',
title = "log(Mortality) vs Wine")+
stat_poly_eq(aes(label = paste( ..rr.label.., ..adj.rr.label.., sep = "~~~")),
label.x.npc = "right", label.y.npc = 0.87, geom = "text",
formula = y ~x, parse = TRUE, size = 3)+
stat_poly_eq(aes(label = paste(..eq.label.., sep = "~~~")),
label.x.npc = "left", label.y.npc = 0.03, geom = "text",
formula = y ~x, parse = TRUE, size = 3)+
stat_fit_glance(method = "lm",
method.args = list(formula = y~x),
geom = "text", size = 3,
label.y.npc = 0.85, label.x.npc = "right",
aes(label = paste("P-value = ", signif(..p.value.., digits = 4), sep = "")))+
# scale_x_log10() +
#scale_y_log10()+
theme_classic()+ guides(fill=FALSE)+
theme(
axis.title = element_text( size=8),
axis.title.x = element_text( size=8),
axis.title.y = element_text( size=8)
)
wine.model.plot.4=ggplot(wine, aes(y=logMORTALITY, x=logWINE)) +
geom_point(alpha = .5) +
geom_point(color = "blue") +
stat_smooth(method = "lm", color="red")+
labs(x='log(Wine)', y='log(Mortality)',
title = "log(Mortality) vs log(Wine)")+
stat_poly_eq(aes(label = paste( ..rr.label.., ..adj.rr.label.., sep = "~~~")),
label.x.npc = "right", label.y.npc = 0.87, geom = "text",
formula = y ~x, parse = TRUE, size = 3)+
stat_poly_eq(aes(label = paste(..eq.label.., sep = "~~~")),
label.x.npc = "left", label.y.npc = 0.15, geom = "text",
formula = y ~x, parse = TRUE, size = 3)+
stat_fit_glance(method = "lm",
method.args = list(formula = y~x),
geom = "text", size = 3,
label.y.npc = 0.85, label.x.npc = "right",
aes(label = paste("P-value = ", signif(..p.value.., digits = 4), sep = "")))+
#scale_x_log10() +
#scale_y_log10()+
theme_classic()+ guides(fill=FALSE)+
theme(
axis.title = element_text( size=8),
axis.title.x = element_text( size=8),
axis.title.y = element_text( size=8)
)
grid.arrange(wine.model.plot.1,
wine.model.plot.2,
wine.model.plot.3,
wine.model.plot.4,
ncol =2.,
nrow = 2.)
#'
#'
#' Graphically we notice that the regression line for the log-transformed model (ln(y) ~ ln(x)) covers most data-points as well as having a more precise and uniform CI band compared with the other 3 models.
#'
#'
wine.model <- lm(MORTALITY~WINE, data=wine )
wine.model.log.x <- lm( MORTALITY~logWINE, data=wine )
wine.model.log.y <- lm( logMORTALITY~WINE, data=wine)
wine.model.log.xy <- lm( logMORTALITY~logWINE, data=wine )
stargazer(wine.model,wine.model.log.x,wine.model.log.y,wine.model.log.xy,
header=F,
type = "latex",
summary = F,
no.space = T,
ci = F,
#keep = c("\\bprecip\\b"),
title = "Comparison of the 4 Models",
column.labels = c( "y~x", "y~ln(x)","ln(y)~x", "ln(y) vs ln(x)"),
notes = "t-values have been hidden!",
dep.var.caption = "-" , # Bold
single.row = T
)
#'
#'
#' Looking at y~x and y~ln(x) models, we have really low explanatory power (R-squared). In y~x model, only 56% of the variability in heart attack mortality is explained by wine consumption. While in y~ln(x) model only 64% of the variability in the response is explained by the predictor.
#'
#' Comparing the explanatory power between the ln(y)~x model and ln(y)~ln(x) model, we get a higher explanatory power in ln(y)~ln(x) model. Over 73.6% of the variability in heart attack mortality is explained by wine consumption in ln(y)~ln(x) model while only 73.6% of the variability in heart attack mortality is explained by wine consumption in ln(y)~x model
#'
#' For these reasons, **the best fit model** is ln(y)~ln(x) model because it has the highest explanatory model compared to the other models. Furthermore, it has a p-value that is very significant (approximately 4.914e-06)
#'
#' ### Model Diagnostics
#'
#'
par( mfrow = c(2, 2)) # Split the plotting panel into a 2 x 2 grid
# model 1
plot(wine.model, main = "Mortality ~ Wine", which=c(1))
# model 2
plot(wine.model.log.x, main = "Mortality ~ ln(Wine)", which=c(1))
# model 3
plot(wine.model.log.y, main = "ln(Mortality) ~ Wine", which=c(1))
# model 4
plot(wine.model.log.xy, main = "ln(Mortality) ~ ln(Wine)", which=c(1))
#'
#'
#' Looking at the residual plots, ln(Mortality) ~ ln(Wine) model exhibits acceptable randomness and unpredictability. From the plot, the residuals bounce randomly around the 0 line and roughly forms a horizontal band around the 0 line without any noticeable pattern. This suggests that the variances of the error terms are approximately equal.
#'
#' The other models seem not to follow the rule of residual heteroskedasticity which is a crucial component of any regression model. Residuals in the other models deviate from linearity and equality of standard deviations assumptions. Hence this gives us more reason to select the model with logarithmic transformation of both predictor and response as the best fit model
#'
#' ### Conclusion
#'
#' In summary, the model with the natural log of heartAttackMortality as the response and the natural log of wineConsumption as the predictor exhibits a relationship with the highest explanatory power. This relationship appears to be linear and at the same time with error terms that appear independent and normally distributed with somewhat equal variances. Therefore we have sufficient evidence that this model [ln(y)~ln(x)] is the best.
#'
#'
#' ## d. Interpret each regression model by describing the change in mortality for a given change in the predictor. Use an increase of 10 units of wine consumption for linear wine and a doubling of wine consumption for logarithmic wine
#'
#'
#' ### Mortality vs Wine
#'
#' - An increase of wine consumption by 10 Liters per person will decrease Ischemic heart attack mortality by 0.761 per 1000 person
#'
#' ### Mortality vs log(Wine)
#'
#' - Doubling wine consumption will decrease Ischemic heart attack mortality rate by 1.227 per 1000 person
#'
#' ### log(Mortality) vs Wine
#'
#' medianProportion = exp(-0.0159*10) = 0.8529964
#' %change = (exp(-0.0159*10)-1)*100 = -14.7%
#'
#' - An increase of wine consumption by 10 Liters per person will decrease Ischemic heart attack mortality by a median proportion of 0.8529964
#' - An increase of wine consumption by 10 Liters per person will decrease Ischemic heart attack mortality by 14.7%
#'
#' ### log(Mortality) vs log(Wine)
#'
#' %change = 1-exp(-0.3556*log(2)))*100 = -21.84555
#'
#' - Doubling wine consumption will decrease Ischemic heart attack mortality by 21.85%
#'
#' \onecolumn
#'
#' # Question 2: The Dramatic U.S. Presidential Election of 2000
#'
#' The U.S. presidential election of November 7, 2000 was one of the closest in history. As returns were counted on election night it became clear that the outcome in the state of Florida would determine the next president. At one point in the evening, television networks projected that the state was carried by the Democratic nominee, Al Gore, but a retraction of the projection followed a few hours later. Then, early in the morning of November 8, the networks projected that the Republican nominee, George W. Bush, had carried Florida and won the presidency. Gore called Bush to concede. While on route to his concession speech, though, the Florida count changed rapidly in his favor. The networks once again reversed their projection, and Gore called Bush to retract his concession. When the roughly 6 million Florida votes had been counted
#'
#'
#'
#'
#' ## a. The data in File Bush.xls contain the numbers of votes for Buchanan and Bush in all 67 counties in Florida. What evidence is there in the scatterplot of Display 8.25 that Buchanan received more votes than expected in Palm Beach County? Analyze the data without Palm Beach County results to obtain an equation for predicting Buchanan votes from Bush votes. Obtain a 95% prediction interval for the number of Buchanan votes in Palm Beach from this result-assuming the relationship is the same in this county as in the others. If it is assumed that Buchanan's actual count contains a number of votes intended for Gore, what can be said about the likely size of this number from the prediction interval? (Consider transformation.)
#'
#' 
#'
#' From the scatter plot (Display 8.25) we can vividly see that Palm Beach County is one of the outlier because it has been separated away from the trend of other observation. In fact, if we regress Buchanan on Bush using the full data set with this potential outlier, the studentized residual and cooks distance flags this datapoint because it influences the regression model to such an extent that the estimated regression equation is pulled towards this datapoint.
#'
#'
bush <- bush %>%
mutate(total96=clinton96+dole96+perot96+buchanan96p)
# transformation
bush.log <- bush
bush.log$reform.reg = bush.log$reform.reg+0.00000001
for(i in 2:ncol(bush.log)){
bush.log[,i] <- log(bush.log[,i])
}
bush.withPalmBeach.model <- lm(buchanan2000~bush2000, data=bush.log)
par( mfrow = c(1, 2))
# identify D values > 4/(n-k-1)
cutoff <- 4/((nrow(bush)-length(bush.withPalmBeach.model$coefficients)-2))
plot(bush.withPalmBeach.model, which=4, cook.levels=cutoff)
# Influence Plot
influencePlot(bush.withPalmBeach.model, id.method="identify", main="Influence Plot", sub="Circle size is proportial to Cook's Distance" )
#'
#'
#'
#'
#'
# What evidence is there in the scatterplot of Display 8.25 that Buchanan received more votes than expected in Palm Beach County?
# - Palm Beach County is an outlier
bushPalmBeach <- bush.log %>%
filter(county=='PALM BEACH')
bush.withoutPalmBeach <- bush.log %>%
filter(county!='PALM BEACH')
bush.withoutPalmBeach.model <- lm(buchanan2000~bush2000, data=bush.withoutPalmBeach)
stargazer(bush.withPalmBeach.model, bush.withoutPalmBeach.model,
header=F,
type = "latex",
summary = F,
no.space = T,
ci = TRUE,
keep = c("\\bprecip\\b"),
title = "Models With and without outlier",
column.labels = c( "Without Palm Beach", "With Palm Beach"),
notes = "Coefficients have been removed!",
dep.var.caption = "-" , # Bold
single.row = T
)
#'
#'
#' Furthermore, if we remove this potential outlier from the dataset and regress Buchanan on Bush, our regression model changes substantially in the sense that residual standard error is inflated from 413 to 371. This inflation highly increases the width of confidence and prediction intervals
#'
#'
bush.withoutPalmBeach.plot = ggplot(bush.withoutPalmBeach,
aes(y=buchanan2000, x=bush2000
)) +
geom_point(alpha = .5) +
geom_point(color = "black") +
stat_smooth(method = "lm", color="red")+
labs(x='log(bush2000)', y='log(buchanan2000)',
title = "Linear Model Without Palm Beach")+
# stat_poly_eq(aes(label = paste(..eq.label.., ..rr.label.., ..adj.rr.label.., sep = "~~~")),
# label.x.npc = "right", label.y.npc = 0.17, geom = "text",
# formula = y ~x, parse = TRUE, size = 3)+
# stat_fit_glance(method = "lm",
# method.args = list(formula = y~x),
# geom = "text", size = 3,
# label.y.npc = 0.15, label.x.npc = "right",
# aes(label = paste("P-value = ", signif(..p.value.., digits = 4), sep = "")))+
# scale_x_log10() +
# scale_y_log10()+
theme_classic()+ guides(fill=FALSE)+
theme(
axis.title = element_text( size=8),
axis.title.x = element_text( size=8),
axis.title.y = element_text( size=8)
)
bush.withPalmBeach.plot = ggplot(bush.log, aes(y=buchanan2000, x=bush2000, color=ifelse(((abs(buchanan2000)>8 )),"A", "B"))) +
geom_point(alpha = .5) +
geom_point() +
scale_color_manual(guide=FALSE, values=c("red", "black")) + #turn off the legend, define the colors
geom_text_repel(data = subset(bush.log, buchanan2000 > 8), aes(label = county))+
stat_smooth(method = "lm", color="red")+
labs(x='log(bush2000)', y='log(buchanan2000)',
title = "Linear Model with Palm Beach")+
# stat_poly_eq(aes(label = paste(..eq.label.., ..rr.label.., ..adj.rr.label.., sep = "~~~")),
# label.x.npc = "right", label.y.npc = 0.40, geom = "text",
# formula = y ~x, parse = TRUE, size = 3)+
# stat_fit_glance(method = "lm",
# method.args = list(formula = y~x),
# geom = "text", size = 3,
# label.y.npc = 0.25, label.x.npc = "right",
# aes(label = paste("P-value = ", signif(..p.value.., digits = 4), sep = "")))+
# scale_x_log10() +
# scale_y_log10()+
theme_classic()+ guides(fill=FALSE)+
theme(
axis.title = element_text( size=8),
axis.title.x = element_text( size=8),
axis.title.y = element_text( size=8)
)
# Obtain a 95% prediction interval for the number of Buchanan votes in Palm Beach from this result-assuming the relationship is the same in this county as in the others.
grid.arrange(bush.withPalmBeach.plot,bush.withoutPalmBeach.plot , ncol = 1,
nrow = 2)
#'
#'
#'
#'
#' Using the regression model without palm beach to generate a 95% prediction interval for the number of Buchanan votes in the Palm Beach, we see that the expected votes for Buchanan given the corresponding votes for Bush (152,846) is 592.377 which highly deviates from the observed votes of 3407
#'
#'
bushPalmBeach.predict <- predict(bush.withoutPalmBeach.model,bushPalmBeach, interval = "prediction")
stargazer(exp(bushPalmBeach.predict), title="Prediction for Buchanana", type = "latex", header=FALSE)
#'
#'
#'
#' In fact, the observed votes for Buchanan (3407) is way far from our 95 % prediction interval of [250.8, 1399.164]. Therefore we have sufficient evidence that the Buchanan's actual count contains a number of votes intended for Gore because of the ballot paper candidate arrangement error. Using our prediction model, the extra votes for Buchanan (3407-592) was meant for Gore hence decreasing Buchanan votes by 592 and increasing Gores votes from 268945 to 268353
#'
#' ## b. Analyze the data in Ex1222 and write a statistical summary predicting the number of Buchanan, votes in Palm Beach Country that were not intended for him. It would be appropriate to describe any unverifiable assumptions used in applying the prediction equation for this purpose. (Suggestion: Find a model for predicting Buchanan's 2000 vote from other variables, excluding Palm Beach County, which is listed last in the data set. Consider a transformation of all counts.)
#'
#' We used Stepwise AIC for predictor variable selection and after evaluating the relative importance measure of each variable, we came up with the best model that is in line with the objective of this analysis.
#'
#' ### Variable Selection
#'
#' The full model containing all potential predictor variables were passed into the step function. The StepAIC function iteratively searched best predictors by dropping or adding one X predictor variable at a time. In each iteration, AIC of the models were calculated and the model that had the lowest AIC was retained for the next iteration.
#'
#' **Full Model: ** buchanan2000~bush2000+gore2000+nader2000+browne2000+total2000+clinton96+ dole96+perot96+buchanan96p+reform.reg+total.reg
#'
#' After the step iterations were completed the full model was reduced to a model with
#' 7 predictors:
#'
#' **StepWise AIC Model: **: buchanan2000 ~ nader2000 + browne2000 + total2000 + clinton96 + perot96 + buchanan96p + total.reg
#'
#' The rule of thumb for a good predictive accuracy has always been to look at no more variables than 10% of the total number of observation. This model is in check with this rule, therefore, there is no need to remove any more predictors. In fact looking at these variables individually with relation to 2000 and 1996 elections we have much more relevancy in context compared to the other predictors
#'
#'
bush.withoutPalmBeach.allSubset.model <- lm(buchanan2000~bush2000+gore2000+nader2000+browne2000+total2000+clinton96+
dole96+perot96+buchanan96p+reform.reg+total.reg, data=bush.withoutPalmBeach)
#plot(bush.withoutPalmBeach.stepAIC.model)
#summary(bush.withoutPalmBeach.allSubset.model)
#names(bush.withoutPalmBeach)
bush.withoutPalmBeach.stepAIC <- stepAIC(bush.withoutPalmBeach.allSubset.model, direction="both")
bush.withoutPalmBeach.stepAIC.anova <- bush.withoutPalmBeach.stepAIC$anova
# final model results
bush.WPB.stepAIC.best.model <-lm(buchanan2000 ~ nader2000 + browne2000 + total2000 + clinton96 +
perot96 + buchanan96p + total.reg, data=bush.withoutPalmBeach)
#'
#'
#'
stargazer(bush.WPB.stepAIC.best.model,
header=F,
type = "latex",
summary = F,
no.space = T,
ci = TRUE,
# keep = c("\\bprecip\\b"),
title = "Best Model",
# notes = "Coefficients have been removed!",
dep.var.caption = "-" , # Bold
single.row = T
)
#'
#'
#' This model has very high explanatory power, with R^2 of 0.898 and Adjusted R^2 of 0.886. Over 89.8% of the variability in Buchanan votes for 2000 is explained by selected predictors. Furthermore, most of the predictors have really low p-values except for browne2000
#'
#' ### Relative Importance Measure
#'
#' After building the model using stepwise variable selection, We went ahead and evaluated the relative importance of each predictor in the model.
#'
#'
bush.WPB.relimp<- calc.relimp(bush.WPB.stepAIC.best.model,type=c("lmg"), rela=TRUE)
# Bootstrap Measures of Relative Importance (1000 samples)
bush.WPB.stepAIC.best.boot <- boot.relimp(bush.WPB.stepAIC.best.model, b = 1000, type = c("lmg",
"last", "first", "pratt"), rank = TRUE,
diff = TRUE, rela = TRUE)
#booteval.relimp(bush.WPB.stepAIC.best.boot) # print result
plot(booteval.relimp(bush.WPB.stepAIC.best.boot,sort=TRUE)) # plot result
#'
#'
#' Using Lindeman, Merenda, and Gold method, all the predictors selected from the stepWise regression had significant contribution percentage into the model, therefore, we did not have any reason to remove any of the variables identified by stepwiseAIC.
#'
#'
#'
#' ### Prediction
#'
#'
#'
bush.WPB.stepAIC.predicted <-predict(bush.WPB.stepAIC.best.model,
bush.log[67,],
se.fit=T, interval="pred")
stargazer(exp(bush.WPB.stepAIC.predicted$fit), title="Best Prediction for Buchanana", type = "latex", header=FALSE)
#'
#'
#' ### Assumptions
#'
#' - Due to the fact we were modeling for predictive regression, we didn't assess multicollinearity
#' - Relative importance is not as effective when regressors are correlated. We did not evaluate collinearity
#' - We did not have a variable for total1996 and observations for other candidates in 1996. For us to standardize vote counts for the year 1996 and 2000, we needed these variables. Therefore we assumed that the data was standardized in the sense that the total number of votes for 1996 and 2000 were the same.
#'
#'
#' We used the above model to predict Buchanan's 2000 vote in Palm Beach County. Among the candidate predictors include in this model were 2000 votes for Browne and Nader, 1996 votes for Dole, Clinton, Perot, and Buchanan. With a 95% prediction interval of about [164.101, 1136.736], we got an expected number of votes for Buchanan in Palm Beach County as 431.903
#'
#' The Observed number of votes for Buchanan was 3407 which was way higher than expected number of votes of 432. Furthermore, this observation is way far from our 95 % prediction interval of about [164,1137]. This can only mean that Buchanan's actual count contains a number of votes intended for Gore because of the ballot paper candidate arrangement error. Using our prediction model, part of the extra votes for Buchanan (3407 - 432) was meant for Gore. From the physical evidence presented and the statistical inference made, we have sufficient evidence that the expected number of votes for Gore was supposed to be approximately 271917 (268945+2972) instead of 268945.
#'
#'
#'
#'
#'
#'
#' \onecolumn
#'
#' # Question 3
#'
#' ## a. Compute "Jittered" versions of-treatment, days after inoculation, and an indicator variable for females by adding small random numbers to each (uniform random numbers between -.15 and .15 work well). Or you could use the jitter function.
#'
#'
bloodBrain$brainLiver= bloodBrain$BRAIN/bloodBrain$LIVER
bloodBrain=bloodBrain%>%
mutate(
isTreatJitter=jitter(ifelse(TREAT=='BD',1,0),amount = 0.15),
daysJitter=jitter(DAYS,amount = 0.15),
femaleJitter=jitter(ifelse(SEX=='F',1,0),amount = 0.15)
)
bloodBrain$log_brainLiver= log(bloodBrain$brainLiver)
bloodBrain$log_Time= log(bloodBrain$TIME)
stargazer(bloodBrain[c(1,10:ncol(bloodBrain))],
header=F,
type = "latex",
no.space = T,
summary = F,
single.row = T
)
#'
#'
#' ## b. Obtain a matrix of scatter plots for the following variables: log sacrifice time, treatment (jittered), days after inoculation (jittered), sex (jittered), and the log of the brain tumor-to-liver antibody ratio. Use the function pairs in the graphics package or scatterplotMatrix in the car package.
#'
#'
#pairs(bloodBrain[11:15],pch=21)
scatterplotMatrix(~log_Time+isTreatJitter+daysJitter+femaleJitter+log_brainLiver, data=bloodBrain,
main="Variables Scatter Plot Matrix")
#'
#'
#' ## c. Obtain a matrix of the correlation coefficients among the same five variables (not jittered).
#'
#'
#library(ggcorrplot)
#data(mtcars)
#corr <- cor(bloodBrain[c(10,4,5,6,3)])
# Second Correlogram Example
# Correlations with significance levels
#library(Hmisc)
#mtcars is a data frame
#rcorr(as.matrix(bloodBrain[c(10,4,5,6,3)]))
hist.panel = function (x, ...=NULL ) {
par(new = TRUE)
hist(x,
col = "light gray",
probability = TRUE,
axes = FALSE,
main = "",
breaks = "FD")
lines(density(x, na.rm=TRUE),
col = "red",
lwd = 1)
#lines(f, col="blue", lwd=1, lty=1) how to add gaussian normal overlay?
rug(x)
}
corr.panel <- function(x, y, digits=2, prefix="", cex.cor)
{
usr <- par("usr"); on.exit(par(usr))
par(usr = c(0, 1, 0, 1))
r <- abs(cor(x, y))
txt <- format(c(r, 0.123456789), digits=digits)[1]
txt <- paste(prefix, txt, sep="")
if(missing(cex.cor)) cex <- 0.8/strwidth(txt)
test <- cor.test(x,y)
# borrowed from printCoefmat
Signif <- symnum(test$p.value, corr = FALSE, na = FALSE,
cutpoints = c(0, 0.001, 0.01, 0.05, 0.1, 1),
symbols = c("***", "**", "*", ".", " "))
text(0.5, 0.5, txt, cex = 2)
text(.8, .8, Signif, cex=1.5, col=2)
}
pairs(bloodBrain[c(14,5,6,15,4)], lower.panel=corr.panel, upper.panel=corr.panel, diag.panel=hist.panel)
#'
#'
#'
#' ## d. On the basis of this, what can be said about the relationship between the covariates (sex and days after inoculation), the response, and the design variables (treatment and sacrifice time.
#'
#'
#' We have a very strong and very significant relationship between log response variable (ratio between antibodies in brain and liver) and the log of sacrifice time (r=.54). However, the relationship between the log of response and treatment is weak and insignificant. On the other hand, we have a moderately strong and significant relationship between log response variable and days after inoculation (r=.4). There is a strong and significant relationship between log response and sex (r=.56). We do not have a correlation between the covariates (sex and days after inoculation), their relationship is weak and insignificant. Also, the relationship between days after inoculation and log of sacrifice time is weak (r=.33) with a marginal p-value. We also have a weak insignificant signal between Days after inoculation and treatment. On the other hand, the relationship between sex and log of sacrifice time was moderately strong and very significant (r=.54). However, we do not have any signal between sex and treatment (r=0). Finally, there is no correlation between the design variable ( treatment and log of sacrifice time)
#'
#' In summary, we don't have collinearity within the design variables (treatment and log of sacrifice time) as well as within the covariates (sex and days after inoculation). However, we do have collinearity between sex and log of sacrifice time. The relationship between the log response variable and all the predictors are mostly strong and significant except for the treatment predictor variable.
#'
#' \onecolumn
#'
#' ## e. Fit the regression of the log response (brain tumor-to-liver antibody ratio) on an indicator variable for treatment and on sacrifice time treated as a factor with four levels (include three indicator variables, for sacrifice time == 3, 24, and 72 hours). Use the model to find the estimated mean of the log response at each of the eight treatment combinations (all combinations of the two infusions and the four sacrifice times).
#'
#'
bloodBrain$TIMEFactor<-as.factor(bloodBrain$TIME)
bloodBrain.model <- lm(log_brainLiver~TIMEFactor+TREAT,data=bloodBrain)
stargazer(bloodBrain.model,
header=F,
type = "latex",
summary = F,
no.space = T,
ci = TRUE,
# keep = c("\\bprecip\\b"),
title = "",
# notes = "Coefficients have been removed!",
dep.var.caption = "-" , # Bold
single.row = T
)
bloodBrain.treatment <- data.frame(
TIMEFactor=as.factor(rep(c(.5,3,24,72),2)),
TREAT=as.factor(c(rep("BD",4),rep("NS",4)))
)
bloodBrain.model.predicted <- cbind(bloodBrain.treatment,
data.frame(predict(bloodBrain.model,bloodBrain.treatment, interval = "prediction")))
stargazer(bloodBrain.model.predicted,
header=F,
type = "latex",
no.space = T,
summary = F,
single.row = T
)
#'
#' \onecolumn
#'
#' ## f. Let X represent log of sacrifice time. Fit the regression of the log response on an indicator variable for treatment, X, X2, and X3. Use the estimated model to find the estimated mean of the log response at each of the eight treatment combinations.
#'
#'
bloodBrain.model.2 <- lm(log_brainLiver~TREAT+log_Time+log_Time^2+log_Time^3,data=bloodBrain)
stargazer(bloodBrain.model.2,
header=F,
type = "latex",
summary = F,
no.space = T,
ci = TRUE,
# keep = c("\\bprecip\\b"),
title = "",
# notes = "Coefficients have been removed!",
dep.var.caption = "-" , # Bold
single.row = T
)
bloodBrain.treatment.2 <- data.frame(
log_Time=log(rep(c(.5,3,24,72),2)),
TREAT=as.factor(c(rep("BD",4),rep("NS",4)))
)
bloodBrain.model.2.predicted <- cbind(bloodBrain.treatment.2,
data.frame(predict(bloodBrain.model.2,bloodBrain.treatment.2, interval = "prediction")))
stargazer(bloodBrain.model.2.predicted,
header=F,
type = "latex",
no.space = T,
summary = F,
single.row = T
)
#'
#'
#' ## g. Why are the answers to parts (5) and (6) the same?
#'
#'
augment(bloodBrain.model.2, bloodBrain) %>%
ggscatter(
x = ".fitted",
y = "log_brainLiver",
conf.int = T,
cor.coef = T,
cor.method = "pearson",
add = "reg.line"
)
augment(bloodBrain.model, bloodBrain) %>%
ggscatter(
x = ".fitted",
y = "log_brainLiver",
conf.int = T,
cor.coef = T,
cor.method = "pearson",
add = "reg.line"
)
#'
#'
#'
#'
bloodBrain.model.anova <- anova(bloodBrain.model, bloodBrain.model.2)
stargazer(bloodBrain.model.anova,
header=F,
type = "latex",
summary = F,
no.space = T,
single.row = T
)
# there is no significant difference between the 4 models beccause all the p-values are extremly high
#'
#'
#' Since the origin model is already linear, further log transformation wouldn't change any of the predictions by much. Transformation doesn't have much effect if the data points don't vary too much. In fact, plotting out a scatter plot for data of the 2 models indicates that the range of data points don't vary and spread a lot as we would have expected in a classical scenario where variable transformation was necessary.
#'
#' ## h. Fit the regression of the log response (brain tumor-to-liver antibody ratio) on all covariates, the treatment indicator, and sacrifice time, treated as a factor with four levels (include three indicator variables, for sacrifice time == 3, 24, and 72 hours).
#'
#'
bloodBrain.model.3 = lm(log_brainLiver~ TREAT+ TIMEFactor+ SEX+ WEIGHT+ DAYS+ LOSS + TUMOR, data=bloodBrain)
bloodBrain.model.3.df=augment(bloodBrain.model.3,bloodBrain)
#model4.df
stargazer(bloodBrain.model.3,
header=F,
type = "latex",
summary = F,
no.space = T,
ci = TRUE,
# keep = c("\\bprecip\\b"),
title = "",
# notes = "Coefficients have been removed!",
dep.var.caption = "-" , # Bold
single.row = T
)
#'
#'
#' ## i. Obtain a set of case influence statistics, including a measure of influence, the leverage, and the studentized residua1.
#'
#'
# influencial points
bloodBrain.model.3.influence <- influence.measures(bloodBrain.model.3)
bloodBrain.model.3.influence.df <- data.frame(bloodBrain.model.3.influence[[1]])
# studentized
bloodBrain.model.3.studres <- studres(bloodBrain.model.3)
bloodBrain.model.3.influence.df$studResidual<-bloodBrain.model.3.studres
as_tibble(bloodBrain.model.3.influence.df)%>%
mutate_if(is.numeric, funs(round(., 5))) %>%
dplyr::select(dffit,cov.r,cook.d,hat,studResidual)%>%
stargazer(
header=F,
type = "latex",
summary = F,
no.space = T,
single.row = T
)
#'
#'
#' \twocolumn
#'
#' ### Cook’s Distance
#'
#'
ols_cooksd_barplot(bloodBrain.model.3)
#'
#'
#'
#' ### DFFITS Plot - difference in fits
#'
#'
ols_dffits_plot(bloodBrain.model.3)
#'
#'
#'
#'
#' ### Studentized Residual Plot
#'
#'
ols_srsd_plot(bloodBrain.model.3)
#'
#'
#'
#' ### Leverage Plot
#'
#'
ols_rsdlev_plot(bloodBrain.model.3)
#'
#'
#'
#' ### Deleted Stud.Residual vs Fitted Values
#'
#'
ols_dsrvsp_plot(bloodBrain.model.3)
#'
#'
#'
#'
#' ### Hadi Plot
#'
#'
ols_hadi_plot(bloodBrain.model.3)
#'
#'
#'
#' \onecolumn
#'
#' ## * Discuss whether any influential observations or outliers occur with respect to this fit.
#'
#' - The cook's distance plot revealed that we have 3 regression outliers which strongly influences fitted values of our model. These outliers include observation 31, 33, and 34.
#'
#' - DFFIT diagnostics plot indicated that we have 6 influential data points. Observation 31, 33 and 34 had the strongest influence while observation 27, 22 and 8 had marginal (acceptable) influence on our model
#'
#' - From the Studentized residual plot, we found out that 1 observation which had studentized residual greater than 3 (absolute cutoff) indicating that observation 34 is an outlier that has a large effect on the overall residual
#'
#' - Leverage plot indicated that we did not have any leverage observations since we did not have any observation with an extreme value on a predictor variable
#'
#' - Deleted Studentized Residual vs Fitted Value plot indicated that we had 2 observations which pulled the estimated regression line towards themselves. This implies that if observation 31 and 34 are deleted from the model then the overall observed response would be much closer to the predicted response.
#'
#'