admin管理员组

文章数量:1345134

Kia ora.

I have a data set where I am trying to generate 1000 samples of size n=5, where i am wanting to gather data only on measurements over 15m, to be produced to a sampling distribution for the sample mean.

this has been my code so far
d <- read.csv("name_of_file")

sample1 <- sample(d$hgt, size=5, replace=F)

proportion_taller_than_15m <- mean(d$hgt >15)

 

my.proportion_taller_tahn_15m <- numeric(1000)
for(i in 1:1000){sample1 <- sample(d$hgt, size=5), replace F}
my.proportion_taller_than_15m[i] <- mean(sample1)

However i only get the value zero. Anyone have any ideas?

本文标签： distributiongenerate samples of proportionsStack Overflow

版权声明：本文标题：distribution - generate samples of proportions - Stack Overflow 内容由网友自发贡献，该文观点仅代表作者本人， 转载请联系作者并注明出处：http://www.betaflare.com/web/1743803634a2541775.html， 本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌抄袭侵权/违法违规的内容，一经查实，本站将立刻删除。