This example will illustrate how to test for differences between survival functions estimated by the Kaplan-Meier product limit estimator. The tests all follow the methods described by Harrington and Fleming (1982) Link.
The first example will use as its outcome variable, the event of a child dying before age 1. The data for this example come from the model.data Demographic and Health Survey for 2012 children’s recode file. This file contains information for all births in the last 5 years prior to the survey.
The second example, we will examine how to calculate the survival function for a longitudinally collected data set. Here I use data from the ECLS-K. Specifically, we will examine the transition into poverty between kindergarten and fifth grade.
#load libraries
library(haven)
library(survival)
library(car)## Loading required package: carDatalibrary(muhaz)
model.dat<-read_dta("~/Google Drive/classes/dem7223/dem7223_19/data/zzkr62dt/ZZKR62FL.DTA")##Event - Infant Mortality In the DHS, they record if a child is dead or alive and the age at death if the child is dead. This can be understood using a series of variables about each child.
If the child is alive at the time of interview, then the variable B5==1, and the age at death is censored.
If the age at death is censored, then the age at the date of interview (censored age at death) is the date of the interview - date of birth (in months).
If the child is dead at the time of interview,then the variable B5!=1, then the age at death in months is the variable B7. Here we code this:
model.dat$death.age<-ifelse(model.dat$b5==1,
((((model.dat$v008))+1900)-(((model.dat$b3))+1900))
,model.dat$b7)
#censoring indicator for death by age 1, in months (12 months)
model.dat$d.event<-ifelse(is.na(model.dat$b7)==T|model.dat$b7>12,0,1)
model.dat$d.eventfac<-factor(model.dat$d.event); levels(model.dat$d.eventfac)<-c("Alive at 1", "Dead by 1")
table(model.dat$d.eventfac)##
## Alive at 1 Dead by 1
## 5434 534###Compairing Two Groups We will now test for differences in survival by characteristics of the household. First we will examine whether the survival chances are the same for children in relatively high ses (in material terms) households, compared to those in relatively low-ses households.
This is the equvalent of doing a t-test, or Mann-Whitney U test for differences between two groups.
library(survminer)## Loading required package: ggplot2## Registered S3 methods overwritten by 'ggplot2':
## method from
## [.quosures rlang
## c.quosures rlang
## print.quosures rlang## Loading required package: ggpubr## Loading required package: magrittrmodel.dat$highses<-Recode(model.dat$v190, recodes ="1:3 = 0; 4:5=1; else=NA")
fit1<-survfit(Surv(death.age, d.event)~highses, data=model.dat)
fit1## Call: survfit(formula = Surv(death.age, d.event) ~ highses, data = model.dat)
##
## n events median 0.95LCL 0.95UCL
## highses=0 4179 362 NA NA NA
## highses=1 1789 172 NA NA NAggsurvplot(fit1, xlim=c(0,12), ylim=c(.9, 1), conf.int=T, title="Survival Function for Infant Mortality - Low vs. High SES Households")## Warning: Removed 48 rows containing missing values (geom_path).## Warning: Removed 48 rows containing missing values (geom_point).## Warning: Removed 48 rows containing missing values (geom_path).## Warning: Removed 48 rows containing missing values (geom_point).
summary(fit1)## Call: survfit(formula = Surv(death.age, d.event) ~ highses, data = model.dat)
##
## highses=0
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 0 4179 134 0.968 0.00273 0.963 0.973
## 1 3992 18 0.964 0.00290 0.958 0.969
## 2 3914 28 0.957 0.00316 0.951 0.963
## 3 3808 24 0.951 0.00337 0.944 0.957
## 4 3709 10 0.948 0.00346 0.941 0.955
## 5 3625 15 0.944 0.00359 0.937 0.951
## 6 3520 20 0.939 0.00376 0.931 0.946
## 7 3414 14 0.935 0.00389 0.927 0.943
## 8 3325 21 0.929 0.00407 0.921 0.937
## 9 3238 17 0.924 0.00422 0.916 0.932
## 10 3159 3 0.923 0.00424 0.915 0.932
## 11 3090 10 0.920 0.00433 0.912 0.929
## 12 3015 48 0.906 0.00475 0.896 0.915
##
## highses=1
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 0 1789 75 0.958 0.00474 0.949 0.967
## 1 1698 8 0.954 0.00498 0.944 0.963
## 2 1659 9 0.948 0.00524 0.938 0.959
## 3 1615 14 0.940 0.00564 0.929 0.951
## 4 1573 15 0.931 0.00604 0.919 0.943
## 5 1536 7 0.927 0.00622 0.915 0.939
## 6 1501 6 0.923 0.00638 0.911 0.936
## 7 1466 4 0.921 0.00648 0.908 0.934
## 8 1430 5 0.918 0.00662 0.905 0.931
## 9 1383 6 0.914 0.00679 0.900 0.927
## 10 1348 2 0.912 0.00684 0.899 0.926
## 11 1315 3 0.910 0.00693 0.897 0.924
## 12 1288 18 0.897 0.00746 0.883 0.912Gives us the basic survival plot.
Next we will use survtest() to test for differences between the two or more groups. The survdiff() function performs the log-rank test to compare the survival patterns of two or more groups.
#two group compairison
survdiff(Surv(death.age, d.event)~highses, data=model.dat)## Call:
## survdiff(formula = Surv(death.age, d.event) ~ highses, data = model.dat)
##
## N Observed Expected (O-E)^2/E (O-E)^2/V
## highses=0 4179 362 374 0.401 1.37
## highses=1 1789 172 160 0.940 1.37
##
## Chisq= 1.4 on 1 degrees of freedom, p= 0.2In this case, we see no difference in survival status based on household SES.
How about rural vs urban residence?
library(dplyr)##
## Attaching package: 'dplyr'## The following object is masked from 'package:car':
##
## recode## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionlibrary(car)
model.dat<-model.dat%>%
mutate(rural = Recode(v025, recodes ="2 = '0rural'; 1='1urban'; else=NA", as.factor = T))
fit2<-survfit(Surv(death.age, d.event)~rural, data=model.dat, conf.type = "log")
fit2## Call: survfit(formula = Surv(death.age, d.event) ~ rural, data = model.dat,
## conf.type = "log")
##
## n events median 0.95LCL 0.95UCL
## rural=0rural 4138 346 NA NA NA
## rural=1urban 1830 188 NA NA NAsummary(fit2)## Call: survfit(formula = Surv(death.age, d.event) ~ rural, data = model.dat,
## conf.type = "log")
##
## rural=0rural
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 0 4138 130 0.969 0.00271 0.963 0.974
## 1 3959 15 0.965 0.00286 0.959 0.971
## 2 3879 29 0.958 0.00314 0.952 0.964
## 3 3775 24 0.952 0.00336 0.945 0.958
## 4 3682 15 0.948 0.00349 0.941 0.955
## 5 3590 13 0.944 0.00360 0.937 0.951
## 6 3491 17 0.940 0.00375 0.932 0.947
## 7 3382 13 0.936 0.00387 0.929 0.944
## 8 3299 19 0.931 0.00404 0.923 0.939
## 9 3208 17 0.926 0.00419 0.918 0.934
## 10 3126 3 0.925 0.00422 0.917 0.933
## 11 3060 9 0.922 0.00430 0.914 0.931
## 12 2990 42 0.909 0.00469 0.900 0.918
##
## rural=1urban
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 0 1830 79 0.957 0.00475 0.948 0.966
## 1 1731 11 0.951 0.00506 0.941 0.961
## 2 1694 8 0.946 0.00528 0.936 0.957
## 3 1648 14 0.938 0.00566 0.927 0.949
## 4 1600 10 0.932 0.00592 0.921 0.944
## 5 1571 9 0.927 0.00615 0.915 0.939
## 6 1530 9 0.922 0.00637 0.909 0.934
## 7 1498 5 0.918 0.00650 0.906 0.931
## 8 1456 7 0.914 0.00668 0.901 0.927
## 9 1413 6 0.910 0.00683 0.897 0.924
## 10 1381 2 0.909 0.00689 0.895 0.922
## 11 1345 4 0.906 0.00700 0.893 0.920
## 12 1313 24 0.890 0.00764 0.875 0.905ggsurvplot(fit2, xlim=c(0,12), ylim=c(.9, 1), conf.int=T, title="Survival Function for Infant mortality - Rural vs Urban Residence")## Warning: Removed 48 rows containing missing values (geom_path).## Warning: Removed 48 rows containing missing values (geom_point).## Warning: Removed 48 rows containing missing values (geom_path).## Warning: Removed 48 rows containing missing values (geom_point).
#Two- sample test
survdiff(Surv(death.age, d.event)~rural, data=model.dat)## Call:
## survdiff(formula = Surv(death.age, d.event) ~ rural, data = model.dat)
##
## N Observed Expected (O-E)^2/E (O-E)^2/V
## rural=0rural 4138 346 371 1.67 5.55
## rural=1urban 1830 188 163 3.79 5.55
##
## Chisq= 5.6 on 1 degrees of freedom, p= 0.02prop.table(table(model.dat$d.event, model.dat$rural), margin = 2)##
## 0rural 1urban
## 0 0.91638473 0.89726776
## 1 0.08361527 0.10273224chisq.test(table(model.dat$d.event, model.dat$rural))##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: table(model.dat$d.event, model.dat$rural)
## X-squared = 5.4595, df = 1, p-value = 0.01946Which shows a statistically significant difference in survival between rural and urban children, with rural children showing lower survivorship at all ages.
We can also compare the 95% survival point for rural and urban residents
quantile(fit2, probs=.05)## $quantile
## 5
## rural=0rural 4
## rural=1urban 2
##
## $lower
## 5
## rural=0rural 3
## rural=1urban 0
##
## $upper
## 5
## rural=0rural 6
## rural=1urban 3We can also calculate the hazard function for each group using the kphaz.fit function in the muhaz library.
haz2<-kphaz.fit(model.dat$death.age, model.dat$d.event, model.dat$rural)
haz2## $time
## [1] 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 10.5 11.5 0.5 1.5
## [15] 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 10.5 11.5
##
## $haz
## [1] 0.0038314014 0.0075716497 0.0064378810 0.0041197145 0.0036742403
## [6] 0.0049398949 0.0038930677 0.0058337248 0.0053636408 0.0009670736
## [11] 0.0029830591 0.0143373352 0.0064084477 0.0047764602 0.0086460188
## [16] 0.0063018400 0.0058154299 0.0059338979 0.0033648177 0.0048901211
## [21] 0.0042929999 0.0014588521 0.0030184080 0.0185979202
##
## $var
## [1] 9.786715e-07 1.976995e-06 1.727037e-06 1.131544e-06 1.038557e-06
## [6] 1.435586e-06 1.165912e-06 1.791334e-06 1.692389e-06 3.117631e-07
## [11] 9.887982e-07 4.894885e-06 3.733606e-06 2.852025e-06 5.339922e-06
## [16] 3.971474e-06 3.757836e-06 3.912526e-06 2.264454e-06 3.416455e-06
## [21] 3.071735e-06 1.064171e-06 2.277779e-06 1.441308e-05
##
## $strata
## [1] 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2plot(y=haz2$haz[1:12], x=haz2$time[1:12], col=1, lty=1, type="s")
lines(y=haz2$haz[13:24], x=haz2$time[13:24], col=2, lty=1, type="s")
This may be suggestive that children in urban areas may live in poorer environmental conditions.
###k- sample test Next we illustrate a k-sample test. This would be the equivalent of the ANOVA if we were doing ordinary linear models.
In this example, I use the v024 variable, which corresponds to the region of residence in this data. Effectively we are testing for differences in risk of infant mortality by region.
table(model.dat$v024, model.dat$d.eventfac)##
## Alive at 1 Dead by 1
## 1 2229 181
## 2 1435 141
## 3 631 94
## 4 1139 118fit3<-survfit(Surv(death.age, d.event)~v024, data=model.dat)
fit3## Call: survfit(formula = Surv(death.age, d.event) ~ v024, data = model.dat)
##
## n events median 0.95LCL 0.95UCL
## v024=1 2410 181 NA NA NA
## v024=2 1576 141 NA NA NA
## v024=3 725 94 NA NA NA
## v024=4 1257 118 NA NA NA#summary(fit3)
#quantile(fit3, probs=.05)
ggsurvplot(fit3,conf.int = T, risk.table = F, title = "Survivorship Function for Infant Mortality", xlab = "Time in Months", xlim = c(0,12), ylim=c(.8, 1))
survdiff(Surv(death.age, d.event)~v024, data=model.dat)## Call:
## survdiff(formula = Surv(death.age, d.event) ~ v024, data = model.dat)
##
## N Observed Expected (O-E)^2/E (O-E)^2/V
## v024=1 2410 181 215.5 5.52534 9.43344
## v024=2 1576 141 141.9 0.00537 0.00745
## v024=3 725 94 62.9 15.33021 17.70233
## v024=4 1257 118 113.7 0.16401 0.21218
##
## Chisq= 21.4 on 3 degrees of freedom, p= 9e-05Which shows significant variation in survival between regions. The biggest difference we see is between region 3 green) and region 1 (black line) groups.
Lastly, we examine comparing survival across multiple variables, in this case the education of the mother (secedu) and the rural/urban residence rural:
model.dat<-model.dat%>%
mutate(secedu=Recode(v106, recodes ="2:3 = 1; 0:1=0; else=NA"))
table(model.dat$secedu, model.dat$d.eventfac)##
## Alive at 1 Dead by 1
## 0 4470 430
## 1 964 104fit4<-survfit(Surv(death.age, d.event)~rural+secedu, data=model.dat)
#summary(fit4)
ggsurvplot(fit4,conf.int = T, risk.table = F, title = "Survivorship Function for Infant Mortality", xlab = "Time in Months", xlim = c(0,12), ylim=c(.8, 1))
#plot(fit4, ylim=c(.85,1), xlim=c(0,12), col=c(1,1,2,2),lty=c(1,2,1,2), conf.int=F)
#title(main="Survival Function for Infant Mortality", sub="Rural/Urban * Mother's Education")
#legend("topright", legend = c("Urban, Low Edu","Urban High Edu ", "Rural, Low Edu","Rural High Edu " ), col=c(1,1,2,2),lty=c(1,2,1,2))
# test
survdiff(Surv(death.age, d.event)~rural+secedu, data=model.dat)## Call:
## survdiff(formula = Surv(death.age, d.event) ~ rural + secedu,
## data = model.dat)
##
## N Observed Expected (O-E)^2/E (O-E)^2/V
## rural=0rural, secedu=0 3707 308 333.3 1.92685 5.22186
## rural=0rural, secedu=1 431 38 37.5 0.00632 0.00693
## rural=1urban, secedu=0 1193 122 107.1 2.07801 2.64671
## rural=1urban, secedu=1 637 66 56.1 1.76227 2.00529
##
## Chisq= 5.9 on 3 degrees of freedom, p= 0.1Which shows a marginally significant difference between at least two of the groups, in this case, I would say that it’s most likely finding differences between the Urban, low Education and the Rural low education, because there have the higher ratio of observed vs expected.
#Survival analysis using survey design
This example will cover the use of R functions for analyzing complex survey data. Most social and health surveys are not simple random samples of the population, but instead consist of respondents from a complex survey design. These designs often stratify the population based on one or more characteristics, including geography, race, age, etc. In addition the designs can be multi-stage, meaning that initial strata are created, then respondents are sampled from smaller units within those strata. An example would be if a school district was chosen as a sample strata, and then schools were then chosen as the primary sampling units (PSUs) within the district. From this 2 stage design, we could further sample classrooms within the school (3 stage design) or simply sample students (or whatever our unit of interest is).
A second feature of survey data we often want to account for is differential respondent weighting. This means that each respondent is given a weight to represent how common that particular respondent is within the population. This reflects the differenital probability of sampling based on respondent characteristics. As demographers, we are also often interested in making inference for the population, not just the sample, so our results must be generalizable to the population at large. Sample weights are used in the process as well.
When such data are analyzed, we must take into account this nesting structure (sample design) as well as the respondent sample weight in order to make valid estimates of ANY statistical parameter. If we do not account for design, the parameter standard errors will be incorrect, and if we do not account for weighting, the parameters themselves will be incorrect and biased.
In general there are typically three things we need to find in our survey data codebooks: The sample strata identifier, the sample primary sampling unit identifier (often called a cluster identifier) and the respondent survey weight. These will typically have one of these names and should be easily identifiable in the codebook.
Statistical software will have special routines for analyzing these types of data and you must be aware that the diversity of statistical routines that generally exists will be lower for analyzing complex survey data, and some forms of analysis may not be available!
In the DHS Recode manual, the sampling information for the data is found in variables v021 and v022, which are the primary sampling unit (PSU) and sample strata, respectively. The person weight is found in variable v005, and following DHS protocol, this has six implied decimal places, so we must divide it by 1000000, again, following the DHS manual.
library(survey)## Loading required package: grid## Loading required package: Matrix##
## Attaching package: 'survey'## The following object is masked from 'package:graphics':
##
## dotchartmodel.dat$wt<-model.dat$v005/1000000
#create the design: ids == PSU, strata==strata, weights==weights.
options(survey.lonely.psu = "adjust")
des<-svydesign(ids=~v021, strata = ~v022, weights=~wt, data=model.dat)
fit.s<-svykm(Surv(death.age, d.event)~rural, design=des, se=T)
#use svyby to find the %of infants that die before age 1, by rural/urban status
svyby(~d.event, ~rural, des, svymean)The plotting is a bit more of a challenge, as the survey version of the function isn’t as nice
plot(fit.s[[1]], ylim=c(.8,1), xlim=c(0,12),col=1, ci=F )
lines(fit.s[[2]], col=2)
title(main="Survival Function for Infant Mortality", sub="Rural vs Urban Residence")
legend("topright", legend = c("Urban","Rural" ), col=c(1,2), lty=1)
#test statistic
svylogrank(Surv(death.age, d.event)~rural, design=des)## Warning in regularize.values(x, y, ties, missing(ties)): collapsing to
## unique 'x' values## [[1]]
## score
## [1,] 28.80354 15.1323 1.903447 0.05698224
##
## [[2]]
## chisq p
## 3.62311066 0.05698224
##
## attr(,"class")
## [1] "svylogrank"And we see the p-value is larger than assuming random sampling.
#Using Longitudinal Data In this example, we will examine how to calculate the survival function for a longitudinally collected data set. Here I use data from the ECLS-K. Specifically, we will examine the transition into poverty between kindergarten and third grade.
First we load our data
load("~/Google Drive/classes/dem7223/dem7223_19/data/eclsk_k5.Rdata")
names(eclskk5)<-tolower(names(eclskk5))
#get out only the variables I'm going to use for this example
myvars<-c( "childid","x_chsex_r", "x_raceth_r", "x1kage_r","x4age", "x5age", "x6age", "x7age", "x2povty","x4povty_i", "x6povty_i", "x8povty_i","x12par1ed_i", "s2_id", "w6c6p_6psu", "w6c6p_6str", "w6c6p_20")
eclskk5<-eclskk5[,myvars]
eclskk5$age1<-ifelse(eclskk5$x1kage_r==-9, NA, eclskk5$x1kage_r/12)
eclskk5$age2<-ifelse(eclskk5$x4age==-9, NA, eclskk5$x4age/12)
#for the later waves, the NCES group the ages into ranges of months, so 1= <105 months, 2=105 to 108 months. So, I fix the age at the midpoint of the interval they give, and make it into years by dividing by 12
eclskk5$age3<-ifelse(eclskk5$x5age==-9, NA, eclskk5$x5age/12)
eclskk5$pov1<-ifelse(eclskk5$x2povty==1,1,0)
eclskk5$pov2<-ifelse(eclskk5$x4povty_i==1,1,0)
eclskk5$pov3<-ifelse(eclskk5$x6povty_i==1,1,0)
#Recode race with white, non Hispanic as reference using dummy vars
eclskk5$race_rec<-Recode (eclskk5$x_raceth_r, recodes="1 = 'nhwhite';2='nhblack';3:4='hispanic';5='nhasian'; 6:8='other';-9=NA", as.factor = T)
eclskk5$male<-Recode(eclskk5$x_chsex_r, recodes="1=1; 2=0; -9=NA")
eclskk5$mlths<-Recode(eclskk5$x12par1ed_i, recodes = "1:2=1; 3:9=0; else = NA")
eclskk5$mgths<-Recode(eclskk5$x12par1ed_i, recodes = "1:3=0; 4:9=1; else =NA") Now, I need to form the transition variable, this is my event variable, and in this case it will be 1 if a child enters poverty between the first wave of the data and the third grade wave, and 0 otherwise.
NOTE I need to remove any children who are already in poverty age wave 1, because they are not at risk of experiencing this particular transition. Again, this is called forming the risk set
eclskk5<-subset(eclskk5, is.na(pov1)==F&is.na(pov2)==F&is.na(pov3)==F&is.na(age1)==F&is.na(age2)==F&is.na(age3)==F&pov1!=1)Now we do the entire data set. To analyze data longitudinally, we need to reshape the data from the current “wide” format (repeated measures in columns) to a “long” format (repeated observations in rows). The reshape() function allows us to do this easily. It allows us to specify our repeated measures, time varying covariates as well as time-constant covariates.
e.long<-reshape(data.frame(eclskk5), idvar="childid", varying=list(c("age1","age2"),
c("age2", "age3")),
v.names=c("age_enter", "age_exit"),
times=1:2, direction="long" )
e.long<-e.long[order(e.long$childid, e.long$time),]
e.long$povtran<-NA
e.long$povtran[e.long$pov1==0&e.long$pov2==1&e.long$time==1]<-1
e.long$povtran[e.long$pov2==0&e.long$pov3==1&e.long$time==2]<-1
e.long$povtran[e.long$pov1==0&e.long$pov2==0&e.long$time==1]<-0
e.long$povtran[e.long$pov2==0&e.long$pov3==0&e.long$time==2]<-0
#find which kids failed in earlier time periods and remove them from the second & third period risk set
failed1<-which(is.na(e.long$povtran)==T)
e.long<-e.long[-failed1,]
e.long$age1r<-round(e.long$age_enter, 0)
e.long$age2r<-round(e.long$age_exit, 0)
head(e.long, n=10)So, this shows us the repeated measures nature of the longitudinal data set.
#poverty transition based on mother's education at time 1.
fit<-survfit(Surv(time = time, event = povtran)~mlths, e.long)
summary(fit)## Call: survfit(formula = Surv(time = time, event = povtran) ~ mlths,
## data = e.long)
##
## 9 observations deleted due to missingness
## mlths=0
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 1 3774 114 0.97 0.00279 0.964 0.975
## 2 1830 56 0.94 0.00475 0.931 0.949
##
## mlths=1
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 1 212 34 0.84 0.0252 0.792 0.891
## 2 89 19 0.66 0.0415 0.584 0.747ggsurvplot(fit,conf.int = T, risk.table = F, title = "Survivorship Function for Poverty Transition", xlab = "Wave of survey")
survdiff(Surv(time = time, event = povtran)~mlths, e.long)## Call:
## survdiff(formula = Surv(time = time, event = povtran) ~ mlths,
## data = e.long)
##
## n=3986, 9 observations deleted due to missingness.
##
## N Observed Expected (O-E)^2/E (O-E)^2/V
## mlths=0 3774 170 211.7 8.2 167
## mlths=1 212 53 11.3 152.8 167
##
## Chisq= 167 on 1 degrees of freedom, p= <2e-16###poverty transition based on mother’s education at time 1 and child’s race/ethnicity
fit2<-survfit(Surv(time = time, event = povtran)~mlths+race_rec, e.long)
summary(fit2)## Call: survfit(formula = Surv(time = time, event = povtran) ~ mlths +
## race_rec, data = e.long)
##
## 9 observations deleted due to missingness
## mlths=0, race_rec=hispanic
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 1 698 48 0.931 0.00958 0.913 0.950
## 2 325 25 0.860 0.01636 0.828 0.892
##
## mlths=0, race_rec=nhasian
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 1 328 6 0.982 0.0074 0.967 0.996
## 2 161 4 0.957 0.0140 0.930 0.985
##
## mlths=0, race_rec=nhblack
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 1 250 16 0.936 0.0155 0.906 0.967
## 2 117 5 0.896 0.0229 0.852 0.942
##
## mlths=0, race_rec=nhwhite
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 1 2218 34 0.985 0.00261 0.980 0.990
## 2 1092 17 0.969 0.00449 0.961 0.978
##
## mlths=0, race_rec=other
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 1 280 10 0.964 0.0111 0.943 0.986
## 2 135 5 0.929 0.0190 0.892 0.966
##
## mlths=1, race_rec=hispanic
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 1 166 24 0.855 0.0273 0.804 0.911
## 2 71 18 0.639 0.0486 0.550 0.741
##
## mlths=1, race_rec=nhasian
## time n.risk n.event survival std.err
## 1.000 10.000 4.000 0.600 0.155
## lower 95% CI upper 95% CI
## 0.362 0.995
##
## mlths=1, race_rec=nhblack
## time n.risk n.event survival std.err
## 1.000 3.000 1.000 0.667 0.272
## lower 95% CI upper 95% CI
## 0.300 1.000
##
## mlths=1, race_rec=nhwhite
## time n.risk n.event survival std.err lower 95% CI upper 95% CI
## 1 31 5 0.839 0.0661 0.719 0.979
## 2 13 1 0.774 0.0870 0.621 0.965
##
## mlths=1, race_rec=other
## time n.risk n.event survival std.err lower 95% CI upper 95% CIcols<-RColorBrewer::brewer.pal(n=5, "Greys")
#ggsurvplot(fit2,conf.int = T, risk.table = F, title = "Survivorship Function for Infant Mortality", xlab = "Time in Months")
plot(fit2,
col=rep(cols,2),
lty=c(1,1,1,1,1,2,2,2,2,2),
ylim=c(.5,1), lwd=2 )
title(main="Survival function for poverty transition, K-5th Grade",
sub="By Race and Mother's Education",
xlab="Wave of survey", ylab="% Not Experiencing Transition")
legend("bottomleft",col=rep(cols,2),
lty=c(1,1,1,1,1,2,2,2,2,2),
lwd=2 ,
legend=c("Mom > HS & Hispanic",
"Mom > HS & Asian",
"Mom > HS & NH black",
"Mom > HS & NH white",
"Mom > HS & NH other",
"Mom < HS & Hispanic",
"Mom <HS & Asian",
"Mom < HS & NH black",
"Mom <HS & NH white",
"Mom < HS & NH other"), cex=.8)
survdiff(Surv(time = age2r, event = povtran)~mlths+race_rec, e.long)## Call:
## survdiff(formula = Surv(time = age2r, event = povtran) ~ mlths +
## race_rec, data = e.long)
##
## n=3986, 9 observations deleted due to missingness.
##
## N Observed Expected (O-E)^2/E (O-E)^2/V
## mlths=0, race_rec=hispanic 698 73 38.0317 32.1516 40.5023
## mlths=0, race_rec=nhasian 328 10 17.6537 3.3183 3.7611
## mlths=0, race_rec=nhblack 250 21 13.7490 3.8240 4.2518
## mlths=0, race_rec=nhwhite 2218 51 126.3543 44.9393 108.3126
## mlths=0, race_rec=other 280 15 15.9278 0.0540 0.0608
## mlths=1, race_rec=hispanic 166 42 8.7432 126.4996 137.3852
## mlths=1, race_rec=nhasian 10 4 0.4665 26.7623 27.9812
## mlths=1, race_rec=nhblack 3 1 0.1285 5.9098 6.1710
## mlths=1, race_rec=nhwhite 31 6 1.8595 9.2197 9.7703
## mlths=1, race_rec=other 2 0 0.0857 0.0857 0.0894
##
## Chisq= 264 on 9 degrees of freedom, p= <2e-16Which, again shows us that at least two of these groups are different from one another.
##With survey design
library(survey)
options(survey.lonely.psu = "adjust")
e.long<-e.long[complete.cases(e.long$w6c6p_6psu, e.long$race_rec, e.long$mlths),]
des2<-svydesign(ids = ~w6c6p_6psu, strata = ~w6c6p_6str, weights=~w6c6p_20, data=e.long, nest=T)
fit.s<-svykm(Surv(time, povtran)~race_rec, design=des2, se=T)
#use svyby to find the %of infants that die before age 1, by rural/urban status
svyby(~povtran, ~race_rec+time, des2, svymean)#test statistic
svylogrank(Surv(time, povtran)~mlths, design=des2)## [[1]]
## score
## [1,] 13277.47 3167.241 4.192126 2.763522e-05
##
## [[2]]
## chisq p
## 1.757392e+01 2.763522e-05
##
## attr(,"class")
## [1] "svylogrank"