single_variable.Rmd

---
title: "single var"
author: "liuc"
date: '2022-07-12'
output: pdf_document
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```

## 单因素统计分析
单因素统计分析，既是只有一个变量分组的情况，类似于`y ~ x`，X的分组既可以是一组也可以是两组、多组。


### 单因素定性数据（名义变量）

##### *对于单组数据*
clopper pearson Binomial Test in r:
适用于n < 30的情况，其零假设为：true probability of success is not equal to 0.5，备择假设为：
```{r}
library(DescTools)

# 7 为success，21为trials
BinomCI(7, 21,
        conf.level = 0.95,
        method = "clopper-pearson")
binom.test(x=7, n=21) # same result


observed = c(7, 14)
total = sum(observed)
BinomCI(observed, total,
        conf.level = 0.95,
        method = "clopper-pearson")
```


单组数据n>30时可以考虑 *Z-test*

One Sample Z-Test in R
```{r}
library(BSDA)

#enter IQ levels for 20 patients
data = c(88, 92, 94, 94, 96, 97, 97, 97, 99, 99,
         105, 109, 109, 109, 110, 112, 112, 113, 114, 115)

#perform one sample z-test
z.test(data, mu=100, sigma.x=15)
```

Two Sample Z-Test in R
```{r}
library(BSDA)

#enter IQ levels for 20 individuals from each city
cityA = c(82, 84, 85, 89, 91, 91, 92, 94, 99, 99,
         105, 109, 109, 109, 110, 112, 112, 113, 114, 114)

cityB = c(90, 91, 91, 91, 95, 95, 99, 99, 108, 109,
         109, 114, 115, 116, 117, 117, 128, 129, 130, 133)

#perform two sample z-test
z.test(x=cityA, y=cityB, mu=0, sigma.x=15, sigma.y=15)
```

One Proportion Z-Test in R
```{r}
# If n ≤ 30: binom.test(x, n, p = 0.5, alternative = “two.sided”)
# If n> 30: prop.test(x, n, p = 0.5, alternative = “two.sided”, correct=TRUE)


prop.test(x=64, n=100, p=0.60, alternative="two.sided")

```



##### 单因素成组数据

既是X的分组大于或等于两组的情况。如上述的Two Sample Z-Test in R。

分组变量为有序多分类，响应变量为二分类时，除了可采用卡方检验外还可以采用Cochran-Armitage Test for Trend。
趋势性分析一般在实验组别中存在着计量梯度或暴露水平梯度时应用。
```{r}
library(DescTools)

dose <- matrix(c(10,9,10,7, 0,1,0,3), byrow=TRUE, nrow=2, dimnames=list(resp=0:1, dose=0:3))
Desc(dose)

CochranArmitageTest(dose)
CochranArmitageTest(dose, alternative="one.sided")


# 对于结果的阐释


```


_Mantel Haenszel_ 分层卡方
for 3-Dimensional Tables. The data are stratified so that each chi-square table is within one group or time.
One assumption of the test is that there are no three-way interactions in the data.  This is confirmed with a non-significant result from a test such as the Woolf test or Breslow–Day test.
Post-hoc analysis can include looking at the individual chi-square, Fisher exact, or G-test for association for each time or group.
```{r}
Input = ("
County       Sex     Result  Count
Bloom        Female  Pass     9
Bloom        Female  Fail     5
Bloom        Male    Pass     7
Bloom        Male    Fail    17
Cobblestone  Female  Pass    11
Cobblestone  Female  Fail    4
Cobblestone  Male    Pass    9
Cobblestone  Male    Fail    21
Dougal       Female  Pass     9
Dougal       Female  Fail     7
Dougal       Male    Pass    19
Dougal       Male    Fail     9
Heimlich     Female  Pass    15
Heimlich     Female  Fail     8
Heimlich     Male    Pass    14
Heimlich     Male    Fail    17
")

Data = read.table(textConnection(Input),header=TRUE)


### Order factors otherwise R will alphabetize them

Data$County = factor(Data$County,
                     levels=unique(Data$County))

Data$Sex    = factor(Data$Sex,
                     levels=unique(Data$Sex))

Data$Result = factor(Data$Result,
                     levels=unique(Data$Result))
Table = xtabs(Count ~ Sex + Result + County,
              data=Data)
ftable(Table)



stats::mantelhaen.test(Table,
                alternative = "two.sided",
                correct = TRUE, exact = FALSE, conf.level = 0.95)

vcd::woolf_test(Table)
### Woolf test for homogeneity of odds ratios across strata.
###   If significant, C-M-H test is not appropriate



rcompanion::groupwiseCMH(Table,
                         group   = 3,
                         fisher  = TRUE,
                         gtest   = FALSE,
                         chisq   = FALSE,
                         method  = "fdr",
                         correct = "none",
                         digits  = 3)
```