计算R中的人口标准差

Question

寻找一种方法来计算R中的人口标准差-使用10个以上的样本。无法提取R中的源C代码以找到计算方法。

# Sample Standard Deviation 
# Note: All the below match with 10 or less samples
n <- 10 # 10 or greater it shifts calculation
set.seed(1)
x <- rnorm(n, 10)

# Sample Standard Deviation
sd(x)
# [1] 0.780586
sqrt(sum((x - mean(x))^2)/(n - 1))
# [1] 0.780586
sqrt(sum(x^2 - 2*mean(x)*x + mean(x)^2)/(n - 1)) # # Would like the Population Standard Deviation equivalent using this.
# [1] 0.780586
sqrt( (n/(n-1)) * ( ( (sum(x^2)/(n)) ) - (sum(x)/n) ^2 ) )
# [1] 0.780586

现在，“人口标准偏差”需要将sd（x）与100计数匹配。

# Population Standard Deviation 
n <- 100 
set.seed(1)
x <- rnorm(x, 10)

sd(x)
# [1] 0.780586

sqrt(sum((x - mean(x))^2)/(n))
# [1] 0.2341758

sqrt(sum(x^2 - 2*mean(x)*x + mean(x)^2)/(n)) 
# [1] 0.2341758

# Got this to work above using (eventual goal, to fix the below):
# https://en.wikipedia.org/wiki/Algebraic_formula_for_the_variance
sqrt( (n/(n-1)) * ( ( (sum(x^2)/(n)) ) - (sum(x)/n) ^2 ) )  # Would like the Population Standard Deviation equivalent using this.
# [1] 3.064027

Answer 1

请检查问题。的第一个参数rnorm应为n。

总体和样本标准差为：

sqrt((n-1)/n) * sd(x) # pop
## [1] 0.8936971

sd(x) # sample
## [1] 0.8981994

它们也可以这样计算：

library(sqldf)
library(RH2)

sqldf("select stddev_pop(x), stddev_samp(x) from X")
##   STDDEV_POP("x") STDDEV_SAMP("x")
## 1       0.8936971        0.8981994

注意：我们使用了以下测试数据：

set.seed(1)
n <- 100
x <- rnorm(n)
X <- data.frame(x)

Answer 2

我认为最简单的方法就是快速定义它sd：

sd.p=function(x){sd(x)*sqrt((length(x)-1)/length(x))}