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## Please cite as:## Hlavac, Marek (2018). stargazer: Well-Formatted Regression and Summary Statistics Tables.## R package version 5.2.2. https://CRAN.R-project.org/package=stargazer## Loading required package: grid## Loading required package: Matrix## Loading required package: survival##
## Attaching package: 'survey'## The following object is masked from 'package:graphics':
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## Attaching package: 'dplyr'## The following object is masked from 'package:car':
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## intersect, setdiff, setequal, unionload(url("https://github.com/coreysparks/data/blob/master/brfss_2017.Rdata?raw=true"))nams<-names(brfss_17)
head(nams, n=10)## [1] "dispcode" "statere1" "safetime" "hhadult" "genhlth" "physhlth"
## [7] "menthlth" "poorhlth" "hlthpln1" "persdoc2"newnames<-tolower(gsub(pattern = "_",replacement = "",x = nams))
names(brfss_17)<-newnamesa.) Define a binary outcome variable of your choosing and define how you recode the original variable
brfss_17$chcocncr<-Recode(brfss_17$chcocncr, recodes ="7:9=NA; 1=1;2=0")
#I selected chcocncr, Have you ever been told you have other types of cancer? Where 1=1 is yes, 2=0 is no. b.) State a research question about what factors you believe will affect your outcome variable.
#Does gender correlate with cancer?c.) Define at least 2 predictor variables, based on your research question. For this assignment, it’s best if these are categorical variables.
#Income, zip code, race/ethnicity, age, and gender. For assignment purposes I will select gender and health insurance which are both categorical variables.
#gender
brfss_17$sex<-as.factor(ifelse(brfss_17$sex==1, "Male", "Female"))
#insurance
brfss_17$ins<-Recode(brfss_17$hlthpln1, recodes ="7:9=NA; 1=1;2=0")d.) Perform a descriptive analysis of the outcome variable by each of the variables you defined in part c. (e.g. 2 x 2 table, 2 x k table). Follow a similar approach to presenting your statistics as presented in Sparks 2009 (in the Google drive). This can be done easily using the tableone package!
#Texas
brfss_17$tx<-NA
brfss_17$tx[grep(pattern = "TX", brfss_17$mmsaname)]<-1
##Remove Non-Responses
brfss_17<-brfss_17%>%
filter(tx==1, is.na(mmsawt)==F, is.na(chcocncr)==F)
dc<-CreateTableOne(vars = c("sex", "hlthpln1"), strata = "chcocncr", test = T, data = brfss_17)
print(dc, format="p")## Stratified by chcocncr
## 0 1 p test
## n 7805 815
## sex = Male (%) 42.3 35.8 <0.001
## hlthpln1 (mean (SD)) 1.17 (0.57) 1.06 (0.30) <0.001print(dc)## Stratified by chcocncr
## 0 1 p test
## n 7805 815
## sex = Male (%) 3305 (42.3) 292 (35.8) <0.001
## hlthpln1 (mean (SD)) 1.17 (0.57) 1.06 (0.30) <0.001d-a.) Calculate descriptive statistics (mean or percentages) for each variable using no weights or survey design, as well as with full survey design and weights.
#with sex variable weights and no weights
#counts
#raw, unweighted table
table(brfss_17$chcocncr, brfss_17$sex)##
## Female Male
## 0 4500 3305
## 1 523 292csex<-wtd.table(brfss_17$chcocncr, brfss_17$sex, weights = brfss_17$mmsawt)
print(csex)## Female Male
## 0 7395078.7 7207020.8
## 1 498331.8 285301.3#proportions
prop.table(wtd.table(brfss_17$chcocncr, brfss_17$sex, weights = brfss_17$mmsawt), margin=2)## Female Male
## 0 0.93686737 0.96192085
## 1 0.06313263 0.03807915#compare that with the original, unweighted proportions
prop.table(table(brfss_17$chcocncr, brfss_17$sex), margin=2)##
## Female Male
## 0 0.89587896 0.91882124
## 1 0.10412104 0.08117876d-a #with insurance variable weights and no weights
#No weights
table(brfss_17$chcocncr, brfss_17$hlthpln1)##
## 1 2 7 9
## 0 6685 1087 16 17
## 1 775 39 1 0cins<-wtd.table(brfss_17$chcocncr, brfss_17$hlthpln1, weights = brfss_17$mmsawt)
print(cins)## 1 2 7 9
## 0 1.104828e+07 3.490948e+06 4.170268e+04 2.116835e+04
## 1 7.055679e+05 7.806041e+04 4.705983e+00 0.000000e+00#proportions
prop.table(wtd.table(brfss_17$chcocncr, brfss_17$hlthpln1, weights = brfss_17$mmsawt), margin=2)## 1 2 7 9
## 0 0.9399713253 0.9781282661 0.9998871667 1.0000000000
## 1 0.0600286747 0.0218717339 0.0001128333 0.0000000000#compare that with the original, unweighted proportions
prop.table(table(brfss_17$chcocncr, brfss_17$hlthpln1), margin=2)##
## 1 2 7 9
## 0 0.89611260 0.96536412 0.94117647 1.00000000
## 1 0.10388740 0.03463588 0.05882353 0.00000000d-b.) Calculate percentages, or means, for each of your independent variables for each level of your outcome variable and present this in a table, with appropriate survey-corrected test statistics. (table one package helps)
library(tableone)
#not using survey design
t1<-CreateTableOne(vars = c("sex", "hlthpln1"), strata = "chcocncr", test = T, data = brfss_17)
#t1<-print(t1, format="p")
print(t1,format="p")## Stratified by chcocncr
## 0 1 p test
## n 7805 815
## sex = Male (%) 42.3 35.8 <0.001
## hlthpln1 (mean (SD)) 1.17 (0.57) 1.06 (0.30) <0.001#using survey design
des<-svydesign(ids = ~1, strata = ~ststr, weights = ~mmsawt, data = brfss_17)
options(survey.lonely.psu = "adjust")
st1<-svyCreateTableOne(vars = c("sex", "hlthpln1"), strata = "chcocncr", test = T, data = des)
#st1<-print(st1, format="p")
print(st1, format="p")## Stratified by chcocncr
## 0 1 p test
## n 14602099.4 783633.0
## sex = Male (%) 49.4 36.4 0.005
## hlthpln1 (mean (SD)) 1.27 (0.60) 1.10 (0.30) <0.001d-c.) Are there substantive differences in the descriptive results between the analysis using survey design and that not using survey design?
#Yes, there are substantive differences. Numbers with no survey design were higher while numbers with a survey design were slightly lower.