```
# NOT RUN {
## EXAMPLE 1:
## From Scott et al. 2008, Table 1. A new diagnostic test was trialled
## on 1586 patients. Of 744 patients that were disease positive, 670 were
## test positive. Of 842 patients that were disease negative, 640 were
## test negative. What is the likeliood ratio of a positive test?
## What is the number needed to diagnose?
dat <- as.table(matrix(c(670,202,74,640), nrow = 2, byrow = TRUE))
colnames(dat) <- c("Dis+","Dis-")
rownames(dat) <- c("Test+","Test-")
rval <- epi.tests(dat, conf.level = 0.95)
print(rval); summary(rval)
## Test sensitivity is 0.90 (95% CI 0.88 -- 0.92). Test specificity is
## 0.76 (95% CI 0.73 -- 0.79). The likelihood ratio of a positive test
## is 3.75 (95% CI 3.32 to 4.24). The number needed to diagnose is
## 1.51 (95% CI 1.41 to 1.65). Around 15 persons need to be tested
## to return 10 positive tests.
## EXAMPLE 2:
## A biomarker assay has been developed to identify patients that are at
## high risk of experiencing myocardial infarction. The assay varies on
## a continuous scale, from 0 to 1. Researchers believe that a biomarker
## assay result of greater than or equal to 0.60 renders a patient test
## positive, that is, at elevated risk of experiencing a heart attack
## over the next 12 months.
## Generate data consistent with the information provided above. Assume the
## prevalence of high risk subjects in your population is 0.35:
set.seed(1234)
dat <- data.frame(out = rbinom(n = 200, size = 1, prob = 0.35),
bm = runif(n = 200, min = 0, max = 1))
## Classify study subjects as either test positive or test negative
## according to their biomarker test result:
dat$test <- ifelse(dat$bm >= 0.6, 1, 0)
## Generate a two-by-two table:
tab <- table(dat$test, dat$out)[2:1,2:1]
rval <- epi.tests(tab, conf.level = 0.95)
# What proportion of subjects are ruled out as being at high risk of
## myocardial infarction?
rval$elements$pro
# Answer: 0.61 (95% CI 0.54 to 0.68).
# What proportion of subjects are ruled in as being at high risk of
## myocardial infarction?
rval$elements$pri
# Answer: 0.38 (95% CI 0.32 to 0.45).
# What is the proportion of false positive results?
rval$elements$pfp
# Answer: 0.37 (95% CI 0.29 to 0.45).
# What is the proportion of false negative results?
rval$elements$pfn
# Answer: 0.58 (95% CI 0.44 to 0.70).
# }
```

Run the code above in your browser using DataCamp Workspace