Large Sample Confidence Interval for the Mean
Confidence Intervals for Proportions
library(proportion)
ciAllx(778,1497,0.05) method x LowerLimit UpperLimit LowerAbb UpperAbb ZWI
1 Wald 778 0.4943974 0.5450148 NO NO NO
2 ArcSine 778 0.4943829 0.5449787 NO NO NO
3 Likelihood 778 0.4943405 0.5449894 NO NO NO
4 Score 778 0.4943793 0.5449320 NO NO NO
5 Logit-Wald 778 0.4943685 0.5449427 NO NO NO
6 Wald-T 778 0.4943973 0.5450148 NO NO NOAlternative R package
library(binom)
binom.confint(778, 1497, conf.level=0.95, method="asymptotic") # Wald CI method x n mean lower upper
1 asymptotic 778 1497 0.5197061 0.4943974 0.5450148binom.coverage(0.5, 1500, conf.level = 0.95, method = "asymptotic") method p n coverage
1 asymptotic 0.5 1500 0.9472287binom.confint(778, 1497, conf.level=0.95, method="wilson") # Score CI method x n mean lower upper
1 wilson 778 1497 0.5197061 0.4943793 0.544932binom.coverage(0.5, 1500, conf.level = 0.95, method = "wilson") method p n coverage
1 wilson 0.5 1500 0.9472287Check Wilson’s method here: Binomial proportion confidence interval - Wikipedia
With a smaller value of \(n\)
binom.coverage(0.25, 20, conf.level=0.95, method="asymptotic") method p n coverage
1 asymptotic 0.25 20 0.8948752binom.coverage(0.25, 20, .level=0.95, method="wilson") method p n coverage
1 wilson 0.25 20 0.9347622binom.sim(M=1000, n=20, p=0.25, conf.level=0.95, methods="asymptotic") mean lower upper width
asymptotic 0.882 0.8620049 0.9019951 0.3642858binom.sim(M=1000, n=20, p=0.25, conf.level=0.95, methods="wilson") mean lower upper width
wilson 0.937 0.9202051 0.9504503 0.3493745Coverage
You can explore here: Web Apps (artofstat.com) - go to Explore Coverage app
Student’s t-distribution - Wikipedia
Example: Estimating Mean Weight Change for Anorexic Girls
Anorexia <- read.table("http://stat4ds.rwth-aachen.de/data/Anorexia.dat", header=TRUE)
head(Anorexia,3) # first3 lines in Anorexiadatafile subject therapy before after
1 1 cb 80.5 82.2
2 2 cb 84.9 85.6
3 3 cb 81.5 81.4change<-Anorexia$after - Anorexia$before
summary(change[Anorexia$therapy=="cb"]) # cb is cognitive behavioral therapy Min. 1st Qu. Median Mean 3rd Qu. Max.
-9.100 -0.700 1.400 3.007 3.900 20.900 sd(change[Anorexia$therapy=="cb"]) # standard deviation of weight changes[1] 7.308504hist(change[Anorexia$therapy=="cb"]) # histogram of weight changevalues
t.test(change[Anorexia$therapy=="cb"],conf.level=0.95)$conf.int[1] 0.2268902 5.7869029
attr(,"conf.level")
[1] 0.95t.test(change[Anorexia$therapy=="cb"],conf.level=0.99)$conf.int[1] -0.7432794 6.7570725
attr(,"conf.level")
[1] 0.99
The UN data set
zCI<- function(x, conf.level=0.95){
# x: vector of the sample values
mean(x)+c(-1,1)*qnorm((1+conf.level)/2)*sqrt(var(x)/length(x)) }
UN <- read.table("http://stat4ds.rwth-aachen.de/data/UN.dat", header=TRUE) # We revisit the UN data file
names(UN)[1] "Nation" "GDP" "HDI" "GII" "Fertility" "CO2"
[7] "Homicide" "Prison" "Internet" t.test(UN$GDP, conf.level=0.95)$conf.int[1] 22.05311 31.60879
attr(,"conf.level")
[1] 0.95zCI(UN$GDP)[1] 22.19406 31.46784library(DescTools)
MeanCI(UN$GDP, conf.level=0.95) # provides also the sampling mean mean lwr.ci upr.ci
26.83095 22.05311 31.60879 
Estimating \(\sigma\)
\(100 (1-\alpha) \%\) CI for \(\mu_1 - \mu_2\) is
Since the interval contains \(0\), it is plausible that the population means are identical.
Recall: Example for Anorexic Girls
Anor<-read.table("http://stat4ds.rwth-aachen.de/data/Anorexia.dat", header=TRUE)
head(Anor) subject therapy before after
1 1 cb 80.5 82.2
2 2 cb 84.9 85.6
3 3 cb 81.5 81.4
4 4 cb 82.6 81.9
5 5 cb 79.9 76.4
6 6 cb 88.7 103.6tail(Anor) subject therapy before after
67 67 c 85.5 88.3
68 68 c 84.4 84.7
69 69 c 79.6 81.4
70 70 c 77.5 81.2
71 71 c 72.3 88.2
72 72 c 89.0 78.8cogbehav<-Anor$after[Anor$therapy == "cb"]-Anor$before[Anor$therapy == "cb"]
control<-Anor$after[Anor$therapy == "c"]-Anor$before[Anor$therapy == "c"]
cbind(mean(cogbehav),sd(cogbehav),mean(control),sd(control)) [,1] [,2] [,3] [,4]
[1,] 3.006897 7.308504 -0.45 7.988705t.test(cogbehav,control,var.equal=TRUE,conf.level=0.95)
Two Sample t-test
data: cogbehav and control
t = 1.676, df = 53, p-value = 0.09963
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.680137 7.593930
sample estimates:
mean of x mean of y
3.006897 -0.450000 t.test(cogbehav,control) # not assuming equal population standard deviations
Welch Two Sample t-test
data: cogbehav and control
t = 1.6677, df = 50.971, p-value = 0.1015
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.7044632 7.6182563
sample estimates:
mean of x mean of y
3.006897 -0.450000 # Alternative approach
library(DescTools)
MeanDiffCI(cogbehav,control) meandiff lwr.ci upr.ci
3.4568966 -0.7044632 7.6182563 Confidence Interval Comparing Two Proportions
Example: Does Prayer Help Coronary Surgery Patients?
prop.test(c(315,304),c(604,597),conf.level=0.95, correct=FALSE)
2-sample test for equality of proportions without continuity correction
data: c(315, 304) out of c(604, 597)
X-squared = 0.18217, df = 1, p-value = 0.6695
alternative hypothesis: two.sided
95 percent confidence interval:
-0.04421536 0.06883625
sample estimates:
prop 1 prop 2
0.5215232 0.5092127 library(PropCIs) # uses binomial success count and n for each group
diffscoreci(315,604,304,597, conf.level=0.95) # 95% confidence interval for pi_1-pi_2
data:
95 percent confidence interval:
-0.04418699 0.06873177BinomDiffCI(315,604,304,597, method="wald") # (x1,n1,x2,n3) est lwr.ci upr.ci
[1,] 0.01231045 -0.04421536 0.06883625BinomDiffCI(315,604,304,597, method="score") est lwr.ci upr.ci
[1,] 0.01231045 -0.04408985 0.06860258
