# 统计代写|假设检验代写hypothesis testing代考|MATH214

## 统计代写|假设检验代写hypothesis testing代考|R Functions winmean, winvar, trimse, and winse

Included in the $\mathrm{R}$ functions written for this book is a function called winmean that computes the Winsorized mean. If the data are stored in the $\mathrm{R}$ variable $x$, it has the form
$$\text { winmean }(x, \operatorname{tr}=0.2) \text {. }$$
The optional argument tr is the amount of Winsorizing, which defaults to $0.2$ if unspecified. (The R function win also computes the Winsorized mean.) For example, the command winmean(dat) computes the $20 \%$ Winsorized mean for the data in the $\mathrm{R}$ vector dat. The command winmean $(x, 0.1)$ computes the $10 \%$ Winsorized mean. If there are any missing values (stored as NA in R), the function automatically removes them.
The function winvar computes the Winsorized sample variance, $s_w^2$. It has the form
$$\operatorname{winvar}(\mathrm{x}, \mathrm{tr}=0.2) \text {. }$$
Again, tr is the amount of Winsorization which defaults to $0.2$ if unspecified. The function
$$\text { trimse }(\mathrm{x}, \mathrm{tr}=0.2)$$
estimates the standard error of the trimmed mean and
$$\text { winse }(\mathrm{x}, \mathrm{tr}=.2)$$
estimates the standard error of the Winsorized mean. For example, the $\mathrm{R}$ command trimse $(x, 0.1)$ estimates the standard error of the $10 \%$ trimmed mean for the data stored in the vector $x$, and winvar $(\mathrm{x}, 0.1)$ computes the Winsorized sample variance using $10 \%$
Winsorization. The $\mathrm{R}$ command winvar(x) computes $s_w^2$ using $20 \%$ Winsorization.

## 统计代写|假设检验代写hypothesis testing代考|Estimating the Standard Error of the Sample Median, M

Trimmed means contain the usual sample median, $M$, as a special case where the maximum amount of trimming is used. When using $M$ and the goal is to estimate its standard error, alternatives to Eq. (3.9) should be used. Many methods have been proposed, comparisons of which were made by Price and Bonett (2001). In terms of hypothesis testing, an effective and fairly simple estimate appears to be one derived by McKean and Schrader (1984). To apply it, compute
$$k=\frac{n+1}{2}-z_{0.995} \sqrt{\frac{n}{4}},$$
where $k$ is rounded to the nearest integer and $z_{0.995}$ is the $0.995$ quantile of a standard normal distribution. Put the observed values in ascending order yielding $X_{(1)} \leq \cdots \leq X_{(n)}$. Then the McKean-Schrader estimate of the squared standard error of $M$ is
$$\left(\frac{X_{(n-k+1)}-X_{(k)}}{2 z_{0.995}}\right)^2 .$$
(Price \& Bonett, 2001 recommend a slightly more complicated estimator, but when computing a confidence interval for the median, currently it seems that their method offers little or no advantage.)

# 假设检验代考

## 统计代写|假设检验代写hypothesis testing代考|R Functions winmean, winvar, trimse, and winse

$$\纹理 { winmean }(x, \operatorname{tr}=0.2) 纹理 {. }$$

$$\left(\frac{X_{(n-k+1)}-X_{(k)}}{2 z_{0.995}}\right)^2 。$$
(Price\& Bonett, 2001推荐了一个稍微复杂的估计方法，但是当计算中位数的置信区间时，目前看来他们的方法几乎没有任何优势。)

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