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
最佳答案
请检查问题。 rnorm
的第一个参数应为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)
关于r - 计算R中的人口标准差,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44339070/