sampSEs #ditto

Question

我有一个家庭作业问题，要求我使用R中的模拟测试置信区间（作为上一个问题的一部分发现）的覆盖概率。

我的代码正尝试从我拥有的样本数据中生成1000个随机样本（有替换样本），有效地将原始样本视为新种群。随机样本的大小与我的总体相同。然后，我想找到每个随机样本的95％置信区间，看看有多少样本包含“真实平均值”（在问题陈述中给出）与“人口平均值”（原始样本的平均值）。

set.seed(1987)

iq <- rnorm(1000,91.08065,14.40393)

pop_mean <- mean(iq) #the mean of my sample is now considered the population mean
true_mean <- 100 #the true mean is 100, specified in question

sampSEs <- numeric() #create an empty vector to put the sample SEs in
sampMeans <- numeric() #create an empty vector to put the sample means in

get_conf_interval <- function(sample_measurements) {
  iqSE_samp <- 15/sqrt(length(iq)) #find the SE using an sd of 15
  iqMean_samp <- mean(sample_measurements) #take the mean of each sample
  upper <- iqMean_samp + 1.96*iqSE_samp #find the upper bound for a 95% CI
  lower <- iqMean_samp - 1.96*iqSE_samp #find the lower bound for a 95% CI
  list(lower=lower, upper=upper)
}

interval_contains_true_mean <- function(interval) { #check if the interval contains the true mean
  true_mean >= interval$lower && true_mean <= interval$upper
}
interval_contains_population_mean <- function(interval) { #check if the interval contains the population mean
  pop_mean >= interval$lower && pop_mean <= interval$upper
}

samples <- replicate(1000, sample(iq, size = 124, replace = T)) #take 1000 samples with replacement from my iq data

for(i in 1:1000) { #for each sample taken
  sampMeans[i] <- mean(samples[i]) #put the mean of it in the vector created previously
  sampSEs[i] <- 15/sqrt(length(iq)) #put the SE in a vector... these are all the same bc not finding the sample sd
}

intervals <- apply(samples, FUN=get_conf_interval, MARGIN=2) #call the function to find the confidence intervals

sampMeans #just check if worked
#sampSEs #ditto

percent_intervals_with_true_mean <- mean(sapply(intervals, FUN=interval_contains_true_mean)) * 100
cat("% Intervals Containing True Mean: ", percent_intervals_with_true_mean, "%\n")

percent_intervals_with_pop_mean <- mean(sapply(intervals, FUN=interval_contains_population_mean)) * 100
cat("% Intervals Containing Population Mean: ", percent_intervals_with_pop_mean, "%")

此代码报告我样本的0％置信区间包含真实均值。这是不正确的。我查看了样本均值，其中一些是真实均值。

Answer 1

1.-我有两个解决方案，第一个解决方案是在'mean（samples [，i]）'和

中放入逗号

'set.seed（1987）

sigma_M = 14.40393

mu_M = 91.08065

m = 10

iq <-rnorm（m，mu_M，sigma_M）

pop_mean <-mean（iq）＃我的样本均值现在视为总体均值

样本<-复制（m，样本（iq，大小= 4，替换= T））＃从我的iq数据中提取m个样本并进行替换

sampSEs <-numeric（）＃创建一个空向量以将示例SE放入

sampMeans <-numeric（）＃创建一个空向量，将样本均值放入

for（i in 1：m）{＃每个取样样本

sampMeans [i] <-mean（samples [，i]）＃将其平均值放入先前创建的向量中

sampSEs [i] <-15 / sqrt（length（iq））＃将SE放入向量中...这些都是相同的bc，没有找到样本sd }

get_conf_interval <-函数（sample_measurements）{

iqSE_samp <-15 / sqrt（length（iq））＃使用15的sd查找SE

iqMean_samp <-mean（sample_measurements）＃取每个样本的平均值

上部<-iqMean_samp + 1.96 * iqSE_samp＃找到95％CI的上限

lower <-iqMean_samp-1.96 * iqSE_samp＃找到95％CI的下限

列表（下=下，上=上） }

interval_contains_population_mean <-function（interval）{＃检查间隔是否包含总体平均值

pop_mean> = interval $ lower && pop_mean <= interval $ upper }

间隔<-apply（samples，FUN = get_conf_interval，MARGIN = 2）＃调用函数以查找置信区间

sampMeans＃只是检查是否有效

sampSEs #ditto

percent_intervals_with_pop_mean <-平均值（sapply（intervals，FUN = interval_contains_population_mean））* 100

cat（“％包含人口的区间平均值：”，percent_intervals_with_pop_mean，“％”）'

2.-第二个解决方案是更改代码，但是我只做了'pop_mean'人口平均（并计算标准偏差）

'set.seed（1987）

sigma_M = 14.40393

mu_M = 91.08065

m = 10

iq <-rnorm（m，mu_M，sigma_M）

pop_mean <-mean（iq）＃我的样本均值现在视为总体均值

样本<-复制（m，样本（iq，大小= 4，替换= T））＃从我的iq数据中提取m个样本并进行替换

sampMeans = apply（样本，2，均值）

iqSE_samp <-15 / sqrt（length（iq））＃使用15的sd查找SE

iqMean_samp <-sampMeans＃取每个样本的平均值

上部<-iqMean_samp + 1.96 * iqSE_samp＃找到95％CI的上限

lower <-iqMean_samp-1.96 * iqSE_samp＃找到95％CI的下限

intervals = cbind（下，上）

percent_intervals_with_pop_mean =平均值（apply（intervals，1，findInterval，x = pop_mean）== 1）* 100

cat（“％包含人口的区间平均值：”，percent_intervals_with_pop_mean，“％”）'

决赛对我来说是80％

在R中使用模拟测试置信区间的覆盖概率

1 个答案:

sampSEs #ditto