本次澳洲代写主要为实验设计与分析限时测试
问题1
（a）随机化用于最小化响应测量之间的依赖性，
符合独立性的假设[1分]。平衡的设计是其中
每种处理都经过相同次数的测试，而不平衡设计是一种
并非所有治疗方法都经过相同次数的测试[1分]。
（b）当治疗量大于同质药物时，需要封堵
实验单位[1分]。块被创建为以下级别的组合
每个区块内的实验单元均一的封闭因子[1
标记]。
（c）当建模效果彼此无法区分时，就会发生混淆[1
标记]。例如，当区块包含较少的实验数据时，就会发生混淆
单位比处理单位多，例如2个设计中的每个2个模块
只能包含2个实验单位[1分]。在这种情况下，
效果将与方块[1分]混淆。
（d）将治疗分配给实验单位[1分]，同时将响应变量
从测量单位[1标记]进行测量。例如，
评估每个地块上种植的西红柿的糖分含量的实验
接受不同的待遇。在这种情况下，地块是实验单位，但
西红柿本身就是度量单位[1分]。
问题2
（a）此实验的统计模型为
���=�+ 𝜏� +���
对于�∈{1，…，3}和�∈{1，…，4} [1 mark]其中
•“是方法“质量的第一个衡量标准。
•𝜏对第方法的质量有影响
•��是方法�的质量度量的第‧个误差。
[1分]
（b）的估计是
�=�̅..
= 100.66
而[𝜏1，，2，the3]的估计是
[𝜏1，𝜏2，𝜏3] = [�1。
，̅2。
，�̅3。
]-�̅..
= [93.10，102.68，106.21] − 100.66
= [-7.56，2.02，5.55]。
[2分]
Wèntí 1
Question 1
(a) Randomisation is used to minimise dependence between response measurement, in
line with the assumption of independence [1 mark]. A balanced design is one in which
each treatment is tested the same number of times while an unbalanced design is one
where not all treatments are tested the same number of times [1 mark].
(b) Blocking is required when there are more treatments than there are homogenous
experimental units [1 mark]. Blocks are created as combinations of the levels of
blocking factors where the experimental units within each block are homogenous [1
mark].
(c) Confounding occurs when modelling effects are indistinguishable from each other [1
mark]. For example, confounding occurs when blocks contain fewer experimental
units than treatments, such as in the case of 2� designs where each of the 2𝑚 blocks
contain only 2�−𝑚 experimental units [1 mark]. In this case some experimental
effects will be confounded with the blocks [1 mark].
(d) Treatments are assigned to experimental units [1 mark] while the response variable
is measured from the measurement units [1 mark]. An example of this could be an
experiment to assess the sugar content of tomatoes grown in plots where each plot
receives a different treatment. In this case the plots are the experimental units but the
tomatoes themselves are the measurement units [1 mark].
Question 2
(a) The statistical model for this experiment is
��� = � + 𝜏� + ���
for � ∈ {1,…,3} and � ∈ {1,…,4} [1 mark] where
• ��� is �-th measure of quality for method �
• 𝜏� is effect on quality of the �-th method
• ��� is �-th error in measure of quality for method �.
[1 mark]
(b) The estimate of � is
� = � ̅..
= 100.66
while the estimate for [𝜏1, 𝜏2, 𝜏3] is
[𝜏1, 𝜏2, 𝜏3] = [� ̅1.
, � ̅2.
, � ̅3.
] − � ̅..
= [93.10, 102.68, 106.21] − 100.66
= [−7.56, 2.02, 5.55].
[2 marks]