# 统计代写|生物统计分析代写Biological statistic analysis代考|MPH701

## 统计代写|生物统计分析代写Biological statistic analysis代考|Unbalanced Data

When designing an experiment with a single treatment factor, it is rather natural to consider using the same number of experimental units for each treatment group. We saw in Sect. 3.2 that this also leads to the lowest possible standard error when estimating the difference between any two treatment means, provided that the ANOVA assumption of equal variance in each treatment group holds.

Sometimes, such a fully balanced design is not achieved, because the number of available experimental units is not a multiple of the number of treatment groups. For example, we might consider studying the four drugs again, but with only 30 mice at our disposal rather than 32 . Then, we either need to reduce each treatment group to seven mice, leaving two mice, or use eight mice in two treatment groups, and seven in the other two (alternatively eight mice in three groups and six in the remaining).
Another cause for missing full balance is that some experimental units fail to give usable response values during the experiment or their recordings go missing. This might happen because some mice die (or escape) during the experiment, a sample gets destroyed or goes missing, or some faulty readings are discovered after the experiment is finished.

In either case, the number of responses per treatment group is no longer the same number $n$, but each treatment group has its own number of experimental units $n_i$, with a total experiment size $N=n_1+\cdots+n_k$ for $k$ treatment groups.

## 统计代写|生物统计分析代写Biological statistic analysis代考|Estimating the Grand Mean

With unequal numbers $n_i$ of observations per cell, we now have two reasonable estimators for the grand mean $\mu$ : one estimator is the weighted mean $\bar{y} .$. , which is the direct equivalent to our previous estimator:
$$\bar{y}{. .}-\frac{y{\cdot}}{N}-\frac{1}{N} \sum_{i=1}^k \sum_{j=1}^{n_i} y_{i j}-\frac{n_1 \cdot \bar{y}{1 \cdot}+\cdots n_k \cdot \bar{y}_k \cdot}{n_1+\cdots+n_k}-\frac{n_1}{N} \cdot \bar{y}_1 \cdot \cdots+\frac{n_k}{N} \cdot \bar{y}_k .$$ This estimator weighs each estimated treatment group mean by the number of available response values and hence its value depends on the number of observations per group via the weights $n_i / N$. Its variance is $$\operatorname{Var}\left(\bar{y}{. .}\right)=\frac{\sigma^2}{N} .$$
The weighted mean is often an undesirable estimator, because larger groups then contribute more to the estimation of the grand mean. In contrast, the unweighted mean
$$\tilde{y}{. .}=\frac{1}{k} \sum{i=1}^k\left(\frac{1}{n_i} \sum_{j=1}^{n_i} y_{i j}\right)=\frac{\bar{y}{1 \cdot}+\cdots+\bar{y}_k .}{k}$$ first calculates the average of each treatment group based on the available observations, and then takes the mean of these group averages as the grand mean. This is precisely the estimated marginal mean, an estimator for the population marginal mean $\mu=\left(\mu_1+\cdots+\mu_k\right) / k$. In the direct extension to our discussion in Sect. 3.2, its variance is $$\operatorname{Var}\left(\tilde{y}{. .}\right)=\frac{\sigma^2}{k^2} \cdot\left(\frac{1}{n_1}+\cdots+\frac{1}{n_k}\right),$$ which is minimal if $n_1=\cdots=n_k$ and then reduces to the familiar $\sigma^2 / N$.
These two estimators yield the same result if the data are balanced and $n_1=\cdots=$ $n_k$, but their estimates differ for unbalanced data. For example, taking the first $4,2,1$, and 2 responses for $D 1, D 2, D 3$, and $D 4$ from Table $4.1$ yields a fairly unbalanced design in which $D 3$ (with comparatively low responses) is very underrepresented compared to $D 1$ (with high value), for example. The two estimators are $\bar{y} . .=12.5$ and $\tilde{y}_{. .}=11.84$. The average on the full data is $11.43$.

# 生物统计分析代考

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## 统计代写|生物统计分析代写生物统计分析代考|估计均数

$$\bar{y}{. .}-\frac{y{\cdot}}{N}-\frac{1}{N} \sum_{i=1}^k \sum_{j=1}^{n_i} y_{i j}-\frac{n_1 \cdot \bar{y}{1 \cdot}+\cdots n_k \cdot \bar{y}_k \cdot}{n_1+\cdots+n_k}-\frac{n_1}{N} \cdot \bar{y}_1 \cdot \cdots+\frac{n_k}{N} \cdot \bar{y}_k .$$ 该估计值通过可用响应值的数量对每个估计治疗组的平均值进行加权，因此其值取决于通过权重的每个组的观察数 $n_i / N$。方差是 $$\operatorname{Var}\left(\bar{y}{. .}\right)=\frac{\sigma^2}{N} .$$加权平均值通常是一个不受欢迎的估计量，因为较大的群体对总平均值的估计贡献更大。相比之下，未加权平均值
$$\tilde{y}{. .}=\frac{1}{k} \sum{i=1}^k\left(\frac{1}{n_i} \sum_{j=1}^{n_i} y_{i j}\right)=\frac{\bar{y}{1 \cdot}+\cdots+\bar{y}_k .}{k}$$ 首先根据现有的观察结果计算每个治疗组的平均值，然后取这些组平均值的平均值作为总平均值。这就是估计的边际均值，总体边际均值的估计量 $\mu=\left(\mu_1+\cdots+\mu_k\right) / k$。在我们3.2节讨论的直接延伸中，它的方差为 $$\operatorname{Var}\left(\tilde{y}{. .}\right)=\frac{\sigma^2}{k^2} \cdot\left(\frac{1}{n_1}+\cdots+\frac{1}{n_k}\right),$$ 如果是最小值 $n_1=\cdots=n_k$ 然后减少到熟悉的 $\sigma^2 / N$.

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