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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
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## 统计代写|生物统计分析代写Biological statistic analysis代考|Defining the Minimal Effect Size

The number of treatments $k$ is usually predetermined, and we then exploit the relation between noncentrality parameter, effect size, and sample size for the power analysis. A frequent challenge in practice concerns defining a reasonable minimal effect size $f_0^2$ or $b_0^2$ that we want to reliably detect. Using a minimal raw effect size also requires an estimate of the residual variance $\sigma_e^2$ from previous data or a dedicated preliminary experiment.

A simple method to provide a minimal effect size uses the fact that $f^2 \geq d^2 / 2 k$ for the standardized effect size $d$ between any pair of group means. The standardized difference $d_0=\delta_0 / \sigma=\left(\mu_{\max }-\mu_{\min }\right) / \sigma$ between the largest and smallest group means therefore provides a conservative minimal effect size $f_0^2=d_0^2 / 2 k$ (Kastenbaum et al. 1970 ).

We can improve on the inequality for specific cases, and Cohen proposed three patterns with minimal, medium, and maximal variability of treatment group differences $\alpha_i$, and provided their relation to the minimal standardized difference $d_0$ (Cohen 1988, p. 276ff).

• If only two groups show a deviation from the common mean, we have $\alpha_{\max }=$ $+\delta_0 / 2$ and $\alpha_{\min }=-\delta_0 / 2$ for these two groups, respectively, while $\alpha_i=0$ for the $k-2$ remaining groups. Then, $f_0^2=d_0^2 / 2 k$ and our conservative effect size is in fact exact.
• If the group means $\mu_i$ are equally spaced with distances $\delta_0 /(k-1)$, then the omnibus effect size is $f_0^2=d_0^2 / 4 \cdot(k+1) / 3(k-1)$. For $k=4$ and $d=3$, an example is $\mu_1=1, \mu_2=2, \mu_3=3$, and $\mu_4=4$.
• If half of the groups is at one extreme $\alpha_{\max }=+\delta_0 / 2$ while the other half is at the other extreme $\alpha_{\min }=-\delta_0 / 2$, then $f_0^2=d_0^2 / 4$ if $k$ is even and $f_0^2=d_0^2 / 4 \cdot(1-$ $1 / k^2$ ) if $k$ is odd. Again for $k=4$ and $d=3$, an example is $\alpha_1=\alpha_3=-1.5$, $\alpha_2=\alpha_4=+1.5$. For $\mu=10$ and $\sigma^2=1$, this corresponds to $\mu_1=\mu_3=8.5$ and $\mu_2=\mu_4=11.5$.

## 统计代写|生物统计分析代写Biological statistic analysis代考|Calculating Power

A simple power analysis function for $\mathrm{R}$ is given in Sect. $4.7$ for illustration, while the more flexible built-in procedure power. anova. test () directly provides the necessary calculations. This procedure accepts the number of groups $k$ ( $g$ roups=), the per-group sample size $n(\mathrm{n}=)$, the residual variance $\sigma_e^2$ (within. var=), the power $1-\beta$ (power =), the significance level $\alpha$ (sig . level=), and a modified version of the raw effect size $\nu^2=\sum_i \alpha_i^2 /(k-1)$ (between . var=) as its arguments. Given any four of these parameters, it will calculate the remaining one.
We look at a range of examples to illustrate the use of power analysis in $\mathrm{R}$. We assume that our previous analysis was completed and we intend to explore new experiments of the same type. This allows us to use the variance estimate $\hat{\sigma}_e^2$ as our assumed within-group variance for the power analyses, where we round this estimate to $\hat{\sigma}_e^2=1.5$ for the following calculations. In all examples, we set our false positive probability to the customary $\alpha=5 \%$.

From a minimal raw effect size $b_0^2$, we find the corresponding between. var argument as
$$\nu_0^2=\frac{k}{k-1} \cdot b_0^2 .$$
For example, we might consider an effect with $\alpha_1=\alpha_2=+1$ and $\alpha_3=\alpha_2=-1$, such that $D 1$ and $D 2$ yield identically higher average enzyme levels, while $D 3$ and $D 4$ yield correspondingly lower enzyme levels. Then our raw effect size is $b_0^2=\left((+1)^2+(+1)^2+(-1)^2+(-1)^2\right) / 4=1$ and we use between. var $=1.33$ and within.var $=1.5$ for our calculation. For a required power of $80 \%$, this yields $n=5$ mice per group. Using only $n=2$ mice per group, we achieve a power of $1-\beta=21 \%$.

# 生物统计代考

## 统计代写|生物统计分析代写Biological statistic analysis代考|定义最小效应大小

• 如果只有两组显示出与共同平均值的偏差，我们有 $\alpha_{\text {max }}=+\delta_0 / 2$ 和 $\alpha_{\min }=-\delta_0 / 2$ 分别为这两组，而 $\alpha_i=0$ 为了 $k-2$ 剩余的组。然后， $f_0^2=d_0^2 / 2 k$ 我们的 保守效应大小实际上是准确的。
• 如果组意味看 $\mu_i$ 等距的距离 $\delta_0 /(k-1)$ ，则综合效应大小为 $f_0^2=d_0^2 / 4 \cdot(k+1) / 3(k-1)$. 为了 $k=4$ 和 $d=3$, 一个例子是 $\mu_1=1, \mu_2=2, \mu_3=3$ ，和 $\mu_4=4$.
• 如果一半的组处于一个极端 $\alpha_{\max }=+\delta_0 / 2$ 而另一半则处于另一个极端 $\alpha_{\min }=-\delta_0 / 2$ ，然后 $f_0^2=d_0^2 / 4$ 如果 $k$ 是均匀的并且 $f_0^2=d_0^2 / 4 \cdot\left(1-1 / k^2\right)$ 如果 $k$ 很奇怪。再次为 $k=4$ 和 $d=3$ ，个例子是 $\alpha_1=\alpha_3=-1.5 ， \alpha_2=\alpha_4=+1.5$. 为了 $\mu=10$ 和 $\sigma^2=1$ ，这对应于 $\mu_1=\mu_3=8.5$ 和 $\mu_2=\mu_4=11.5$.

## 统计代写|生物统计分析代写Biological statistic analysis代考|计算功率

$$\nu_0^2=\frac{k}{k-1} \cdot b_0^2 .$$

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## MATLAB代写

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