# 统计代写|假设检验代写hypothesis testing代考|STAT101

## 统计代写|假设检验代写hypothesis testing代考|R Functions mean, tmean, and lloc

$\mathrm{R}$ has a built-in function that evaluates the trimmed mean. If observations are stored in the vector $x$, the $\mathrm{R}$ command
$$\operatorname{mean}(x, \text { trim }=0)$$
computes the $\gamma$-trimmed mean where the argument trim determines the amount of trimming. By default, the amount of trimming is 0 . For example, mean $(x, 0.2)$ returns the $20 \%$ trimmed mean. The value 283 is returned for the data in Table 3.2, assuming the data are stored in the $\mathrm{R}$ variable $\mathrm{x}$. Because it is common to use $20 \%$ trimming, for convenience the $\mathrm{R}$ function
$$\operatorname{tmean}(\mathrm{x}, \mathrm{tr}=0.2)$$
has been supplied, which computes a $20 \%$ trimmed mean by default using the data stored in the $\mathrm{R}$ variable $x$. The amount of trimming can be altered using the argument tr. So tmean(blob) will compute a $20 \%$ trimmed mean for the data stored in blob, and tmean(blob,tr=0.3) will use $30 \%$ trimming instead. For convenience, the function
$$\operatorname{lloc}(\mathrm{x}, \mathrm{est}=\operatorname{tmean}, \ldots)$$
is supplied for computing a trimmed mean when data are stored in list mode, in a data frame, or a matrix. If $\mathrm{x}$ is a matrix or data frame, lloc computes the trimmed mean for each column. Other measures of location can be used via the argument est. (For example, est=median will compute the median.) The argument … means that an optional argument associated with est can be used.

## 统计代写|假设检验代写hypothesis testing代考|Estimating the Standard Error of the Trimmed Mean

To have practical value when making inferences about $\mu_t$, properties of the sampling distribution of $\bar{X}_t$ need to be determined. This subsection takes up the problem of estimating $\sqrt{\operatorname{VAR}\left(\bar{X}_t\right)}$, the standard error of the sample trimmed mean.

At first glance the problem might appear to be trivial. The standard error of the sample mean is $\sigma / \sqrt{n}$, which is estimated with $s / \sqrt{n}$, where
$$s^2=\frac{1}{n-1} \sum\left(X_i-\bar{X}\right)^2$$
is the usual sample variance. A common mistake in applied work is to estimate the standard error of the trimmed mean by simply computing the sample standard deviation of the untrimmed observations, and then dividing by $\sqrt{n-2 g}$, the square root of the number of observations left after trimming. That is, apply the usual estimate of the standard error using the untrimmed values. To see why this simple idea fails, let $X_1, \ldots, X_n$ be any random variables, possibly dependent with unequal variances, and let $a_1, \ldots, a_n$ be any $n$ constants. Then the variance of $\sum a_i X_i$ is
$$\operatorname{VAR}\left(\sum a_i X_i\right)=\sum_{i=1}^n \sum_{i=1}^n a_i a_j \operatorname{COV}\left(X_i, X_j\right)$$
where $\operatorname{COV}\left(X_i, X_j\right)$ is the covariance between $X_i$ and $X_j$. That is,
$$\operatorname{COV}\left(X_i, X_j\right)=E\left{\left(X_i-\mu_i\right)\left(X_j-\mu_j\right)\right},$$
where $\mu_i=E\left(X_i\right)$. When $i=j, \operatorname{COV}\left(X_i, X_j\right)=\sigma_i^2$, the variance of $X_i$. When the random variables are independent, Eq. (3.2) reduces to
$$\operatorname{VAR}\left(\sum a_i X_i\right)=\sum_{i=1}^n a_i^2 \sigma_i^2$$

# 假设检验代考

## 统计代写|假设检验代写hypothesis testing代考|R Functions mean, tmean, and lloc

$mathrm{R}$有一个内置函数可以评估修剪后的平均值。如果观测值存储在向量$x$中，$mathrm{R}$命令
$$\ooperatorname{mean}(x, \text { trim }=0)$$

$$\ooperatorname{tmean}(mathrm{x},mathrm{tr}=0.2)$$

$$\operatorname{lloc}(\mathrm{x}, \mathrm{est}=\operatorname{tmean}, \ldots)$$

## 统计代写|假设检验代写hypothesis testing代考|Estimating the Standard Error of the Trimmed Mean

$$s^2=\frac{1}{n-1} \sum\left(X_i-\bar{X}\right)^2$$

$$\operatorname{VAR}\left(\sum a_i X_i\right)=\sum_{i=1}^n \sum_{i=1}^n a_i a_j operatorname{COV}\left(X_i, X_j\right)$$

$$\operatorname{COV}\left(X_i, X_j\right)=E\left{left(X_i-mu_i\right)\left(X_j-mu_j\right)\right}。$$

$$\operatorname{VAR}\left(sum a_i X_i\right)=sum_{i=1}^n a_i^2 \sigma_i^2$$

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