正态性假设 (Normality)：
方差齐性假设 (Homoscedasticity)：
独立性假设 (Independence)：
单样本t检验 (One-sample t-test): 用于比较单个样本的均值与已知的总体均值之间的差异。
独立样本t检验 (Independent samples t-test): 用于比较两个独立样本的均值差异。
配对样本t检验 (Paired samples t-test): 用于比较同一组受试者在两个不同条件下的均值差异。
library(tidyverse)
ggplot() + xlim(-10,10) +
geom_function(mapping = aes(color="normal Distribution"),
fun = dnorm, args = list(mean = 0, sd = 1),
)+
geom_function(mapping = aes(color="t Distribution"),
fun = dt, args = list(df = 1 ,ncp=0),
)+
scale_color_manual(values = c("normal Distribution" = "red",
"t Distribution" = "blue"))+
labs(color = "Distribution")
t 检验（Student‘s t test），主要用于小样本（n<30），标准差未知的正态分布总体。在进行t检验之前，可以先通过正态性检验 shapiro.test()
2017.临床研究中的统计分析和图形表达实例详解 第2版
在数据符合正态分布的前提下使用单样本t-test来比较一组样本的均值和已知(理论/总体)均值，所谓的已知均值能来自于之前的实验数据或者理论值。根据研究问题(原假设)的不同又分为双尾(不等)和单尾检验(大于或者小于)
\[t=\frac{\bar X-\mu_0}{S /\sqrt n} \sim t(\nu=n-1) \]
它是一种参数检验，用于检验样本均值是否可以合理地为总体均值或特定值。
Hb <- c(11.3,15.0,15.0,13.5,12.8,10.0,11.0,12.0,13.0,14.0)
shapiro.test(Hb)
#>
#> Shapiro-Wilk normality test
#>
#> data: Hb
#> W = 0.95907, p-value = 0.7752
t.test(Hb, mu=14.02)
#>
#> One Sample t-test
#>
#> data: Hb
#> t = -2.3623, df = 9, p-value = 0.04244
#> alternative hypothesis: true mean is not equal to 14.02
#> 95 percent confidence interval:
#> 11.55342 13.96658
#> sample estimates:
#> mean of x
#> 12.76t_test_two_sided <- function(data,mu,level=0.95,...){
bar_X <- mean(data,na.rm = T)
sd <- sd(data, na.rm = T)
n=length(data)
se=sd/sqrt(n)
t_statistic <- (bar_X-mu)/se
p_value <- 2*(1-pt(abs(t_statistic),df=n-1))
t_critical <- qt((1-(1-level)/2),df = n-1)
CI_lower <- bar_X - t_critical * se
CI_upper <- bar_X+ t_critical * se
CI <- paste0(level*100,"% confidence interval: ","[",CI_lower,",",CI_upper,"]",sep = "")
output <- data.frame(
均值=bar_X,
标准差=sd,
标准误=se,
t=t_statistic,
df=n-1,
p_value=p_value,
CI=CI
)
return(output)
}
t_test_two_sided(Hb, mu=14.02)
#> 均值 标准差 标准误 t df p_value
#> 1 12.76 1.68668 0.533375 -2.362315 9 0.04244005
#> CI
#> 1 95% confidence interval: [11.5534219269741,13.9665780730259]\[ H_0:\mu_{\bar d}=0 \] \[ t=\frac{\bar d- \mu_{\bar d}}{S_d /\sqrt n} \sim t(\nu) \] 其中\(d= X_2-X_1,\mu_{\bar d}=0\)。
df <- data.frame(
before = c(36,46,53,57,65,60,42,45,25,55,51,59),
after = c(45,64,66,57,70,55,70,45,50,80,60,60)
)
shapiro.test(df$before)
#>
#> Shapiro-Wilk normality test
#>
#> data: df$before
#> W = 0.95141, p-value = 0.6576
shapiro.test(df$after)
#>
#> Shapiro-Wilk normality test
#>
#> data: df$after
#> W = 0.96659, p-value = 0.872
t.test(df$after,df$before,paired = TRUE)
#>
#> Paired t-test
#>
#> data: df$after and df$before
#> t = 3.3053, df = 11, p-value = 0.007011
#> alternative hypothesis: true mean difference is not equal to 0
#> 95 percent confidence interval:
#> 3.563879 17.769455
#> sample estimates:
#> mean difference
#> 10.66667\[H_0:\frac{\sigma_1^2}{\sigma_2^2}=1\]
\[ F=\frac{\frac {(n_1-1)S_1^2}{\sigma_1^2}/(n_1-1)}{\frac {(n_2-1)S_2^2}{\sigma_2^2}/(n_2-1)}=(\frac{S_1^2}{S_2^2})(\frac{\sigma_2^2}{\sigma_1^2})=\frac{S_1^2}{S_2^2}\sim F(\nu_1,\nu_2),\nu_1=n_1-1,\nu_2=n_2-1 \]
test_group = c(10.2, 8.9, 10.1, 9.2, -0.8, 10.6, 6.5, 11.2, 9.3, 8.0, 10.7, 9.5, 12.7, 14.4, 11.9)
healthy_group = c(5.0, 6.7, -1.4, 4.0, 7.1, -0.6, 2.8, 4.3, 3.7, 5.8, 4.6, 6.0, 4.1, 5.1, 4.7)
# 检验两个样本的方差是否相等
var.test(test_group, healthy_group)
#>
#> F test to compare two variances
#>
#> data: test_group and healthy_group
#> F = 2.0812, num df = 14, denom df = 14, p-value = 0.1827
#> alternative hypothesis: true ratio of variances is not equal to 1
#> 95 percent confidence interval:
#> 0.6987342 6.1991582
#> sample estimates:
#> ratio of variances
#> 2.081241
# car::leveneTest()\[H_0:\mu_1-\mu_2=0\]
\[ t=\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{S_{\bar X_1-\bar X_2}}=\frac{\bar X_1-\bar X_2}{S_C\sqrt{(\frac{1}{n_1}+\frac{1}{n_2})}} \]
其中，\(S_c^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}\)
# 方差齐
t.test(test_group, healthy_group, conf.level = 0.95, var.equal = TRUE)
#>
#> Two Sample t-test
#>
#> data: test_group and healthy_group
#> t = 4.9875, df = 28, p-value = 2.869e-05
#> alternative hypothesis: true difference in means is not equal to 0
#> 95 percent confidence interval:
#> 3.162555 7.570779
#> sample estimates:
#> mean of x mean of y
#> 9.493333 4.126667\[H_0:\mu_1-\mu_2=0\]
\[ t'=\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}=\frac{\bar X_1-\bar X_2}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}\sim t(\nu ') \]
强调自由度
\[ v'=\frac{(\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2})^2}{\frac{(\frac{S_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{S_2^2}{n_2})^2}{n_2-1}}(舍入到最近整数) \]
# 假设方差不齐，使用Welch t检验
t.test(test_group, healthy_group, conf.level = 0.95)
#>
#> Welch Two Sample t-test
#>
#> data: test_group and healthy_group
#> t = 4.9875, df = 24.93, p-value = 3.878e-05
#> alternative hypothesis: true difference in means is not equal to 0
#> 95 percent confidence interval:
#> 3.150262 7.583071
#> sample estimates:
#> mean of x mean of y
#> 9.493333 4.126667