t 检验的原假设是 。另一种选择是。

您将从具有 的正态分布总体中抽取大小为 20 的样本。您将使用 0.05 的显着性水平。

显示当总体的实际平均值从 350 变为 650（增量为 10）时，功效如何变化。

参数自抽样法：经验功效计算

统计问题：样本量如何影响功效？

我们如何回答：创建更多的功效曲线，因为实际均值在 350 到 650 之间变化，但使用大小为 n = 10、n = 20、n = 30、n = 40 和 n = 50 的样本生成它们。同一图上的所有 5 条功效曲线。

---

---

pvals <- replicate(m, pvalue())

power <- mean(pvals <= 0.05)





points(sequence,final2[2,],col="red",pch=1)



points(sequence,final2[3,],col="blue",pch=2)

参数自抽样法：经验置信水平

统计问题：在制作 95% CI 时，如果我们的样本很小并且不是来自正态分布，我们是否仍有 95% 的置信度？

我们如何回答它：根据样本为总体的平均值创建一堆置信区间 (95%)。

您的样本大小应为 16，取自具有 2 个自由度的卡方分布。

找出未能捕捉总体真实均值的置信区间的比例。（提醒：自由度为 \(k\) 的卡方分布的平均值为 \(k\)。）

for(i in 1:m){

  samp=rchisq(n,df=2)

  mean=mean(samp)

  sd=sd(samp)

  upper=mean+qt(0.975,df=15)*sd/4

非参数自抽样法置信区间

统计问题：基于一个样本，我们可以为总体相关性创建一个置信区间吗？

我们如何回答：为相关统计量创建一个 bootstrap t 置信区间估计。

boot.ti <-

  function(x, B = 500, R = 100, level = .95, stattic){

    

    x <- as.matrix(x)

 

library(boot)       #for boot and boot.ci



data(law, package = "bootstrap")



dat <- law



ci <- boot.t.ci(dat, statistic = stat, B=2000, R=200)

ci

自抽样法后的Jackknife

统计问题：R 的标准误差的 bootstrap 估计的标准误差是多少？

我们如何回答它： data(law) 像上一个问题一样使用。在 bootstrap 后执行 Jackknife 以获得标准误差估计的标准误差估计。（bootstrap 用于获得总体中 R 的 SE 的估计值。然后使用折刀法获得该 SE 估计值的 SE。）

indices <- matrix(0, nrow = B, ncol = n)



# 进行自举

for(b in 1:B){

    i <- sample(1:n, size = n, replace = TRUE)

    LSAT <- law$LSAT[i]

 

#  jackknife



for(i in 1:n){

    keepers <- function(k){

         !any(k == i)   

    }

自测题

Submit the rendered HTML file. Make sure all requested output (tables, graphs, etc.) appear in your document when you submit.

Parametric Bootstrap: Estimate MSE

Statistical question: What is the MSE of a level \(k\) trimmed mean?

How we can answer it: Estimate the MSE of the level \(k\) trimmed mean for random samples of size 20 generated from a standard Cauchy distribution (t-distribution w/df = 1). The target parameter \(\theta\) is the center or median. The mean does not exist for a Cauchy distribution. Summarize the estimates of MSE in a table for \(k = 1, 2, … 9\).

Parametric Bootstrap: Empirical Power Calculations

Statistical question: How does power change as the difference between the null hypothes and the reality changes?

How we can answer it: Plot an empirical power curve for a t-test.

The null hypothesis of the t-test is \(\mu = 500\). The alternative is \(\mu \ne 500\).

You will draw samples of size 20, from a normally distributed population with \(\sigma = 100\). You will use a significance level of 0.05.

Show how the power changes as the actual mean of the population changes from 350 to 650 (increments of 10).

On the y-axis will be the empirical power (estimated via bootstrap) and the x-axis will be the different values of \(\mu\) (350, 360, 370 … 650).

Parametric Bootstrap: Empirical Power Calculations

Statistical question: How does sample size affect power?

How we can answer it: Create more power curves as the actual mean varies from 350 to 650, but produce them for using samples of size n = 10, n = 20, n = 30, n = 40, and n = 50. Put all 5 power curves on the same plot.

Parametric Bootstrap: Empirical Confidence Level

Statistical question: When making a 95% CI, are we still 95% confident if our samples are small and do not come from a normal distribution?

How we can answer it: Create a bunch of Confidence Intervals (95%) for the mean of a population based on a sample.

\[\bar{x} \pm t^{*} \times \frac{s}{\sqrt{n}}\]

Your samples should be of size 16, drawn from a chi-squared distribution with 2 degrees of freedom.

Find the proportion of Confidence Intervals that fail to capture the true mean of the population. (Reminder: a chi-squared distribution with \(k\) degrees of freedom has a mean of \(k\).)

Non Parametric Bootstrap Confidence Interval

Statistical question: Based on one sample, can we create a confidence interval for the correlation of the population?

How we can answer it: Create a bootstrap t confidence interval estimate for the correlation statistic.

Jackknife after bootstrap

Statistical question: What is the standard error of the bootstrap estimate of the standard error of R?

How we can answer it: Use data(law) like the previous problem. Perform Jackknife after bootstrap to get a standard error estimate of the standard error estimate. (The bootstrap is used to get an estimate of the SE of R in the population. The jackknife is then used to get an SE of that SE estimate.)