### 统计代写|生物统计学作业代写Biostatistics代考| Statistical Relationship

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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

## 统计代写|生物统计学作业代写Biostatistics代考|Statistical Relationship

The data from the cancer screening test of Example $1.4$ are reproduced here as Table 3.1. In this design, each member of the population is characterized by two variables: the test result $X$ and the true disease status $Y$. Following our definition above, the probability of a positive test result, denoted $\operatorname{Pr}(X=+)$, is
\begin{aligned} \operatorname{Pr}(X=+) &=\frac{516}{24,103} \ &=0.021 \end{aligned}
and the probability of a negative test result, denoted $\operatorname{Pr}(X=-)$, is
\begin{aligned} \operatorname{Pr}(X=-) &=\frac{23,587}{24,103} \ &=0.979 \end{aligned}
and similarly, the probabilities of having $(Y=+)$ and not having $(Y=-)$ the disease are given by

\begin{aligned} \operatorname{Pr}(Y=+) &=\frac{379}{24,103} \ &=0.015 \end{aligned}
and
\begin{aligned} \operatorname{Pr}(Y=-) &=\frac{23,724}{24,103} \ &=0.985 \end{aligned}
Note that the sum of the probabilities for each variable is unity:
\begin{aligned} &\operatorname{Pr}(X=+)+\operatorname{Pr}(X=-)=1.0 \ &\operatorname{Pr}(Y=+)+\operatorname{Pr}(Y=-)=1.0 \end{aligned}
This is an example of the addition rule of probabilities for mutually exclusive events: One of the two events $(X=+)$ or $(X=-)$ is certain to be true for a person selected randomly from the population.

Further, we can calculate the joint probabilities. These are the probabilities for two events – such as having the disease and having a positive test resultoccurring simultaneously. With two variables, $X$ and $Y$, there are four conditions of outcomes and the associated joint probabilities are
\begin{aligned} \operatorname{Pr}(X=+, Y=+) &=\frac{154}{24,103} \ &=0.006 \ \operatorname{Pr}(X=+, Y=-) &=\frac{362}{24,103} \ &=0.015 \ \operatorname{Pr}(X=-, Y=+) &=\frac{225}{24,103} \ &=0.009 \end{aligned}
and
\begin{aligned} \operatorname{Pr}(X=-, Y=-) &=\frac{23,362}{24,103} \ &=0.970 \end{aligned}
The second of the four joint probabilities, $0.015$, represents the probability of a person drawn randomly from the target population having a positive test result but being healthy (i.e., a false positive). These joint probabilities and the

marginal probabilities above, calculated separately for $X$ and $Y$, are summarized and displayed in Table 3.2. Observe that the four cell probabilities add to unity [i.e., one of the four events $(X=+, Y=+)$ or $(X=+, Y=-)$ or $(X=-, Y=+)$ or $(X=-, Y=-)$ is certain to be true for a randomly selected individual from the population]. Also note that the joint probabilities in each row (or column) add up to the marginal or univariate probability at the margin of that row (or column). For example,
\begin{aligned} \operatorname{Pr}(X=+, Y=+)+\operatorname{Pr}(X=-, Y=+) &=\operatorname{Pr}(Y=+) \ &=0.015 \end{aligned}

## 统计代写|生物统计学作业代写Biostatistics代考|Using Screening Tests

We have introduced the concept of conditional probability. However, it is important to distinguish the two conditional probabilities, $\operatorname{Pr}(X=+\mid Y=+)$ and $\operatorname{Pr}(Y=+\mid X=+)$. In Example 1.4, reintroduced in Section 3.1.3, we have
\begin{aligned} \operatorname{Pr}(X=+\mid Y=+) &=\frac{154}{379} \ &=0.406 \end{aligned}
whereas
\begin{aligned} \operatorname{Pr}(Y=+\mid X=+) &=\frac{154}{516} \ &=0.298 \end{aligned}
Within the context of screening test evaluation:

1. $\operatorname{Pr}(X=+\mid Y=+)$ and $\operatorname{Pr}(X=-\mid Y=-)$ are the sensitivity and specificity, respectively.
2. $\operatorname{Pr}(Y=+\mid X=+)$ and $\operatorname{Pr}(Y=-\mid X=-)$ are called the positive predictivity and negative predictivity.

With positive predictivity (or positive predictive value), the question is: Given that the test $X$ suggests cancer, what is the probability that, in fact, cancer is present? Rationales for these predictive values are that a test passes through several stages. Initially, the original test idea occurs to a researcher. It must then go through a developmental stage. This may have many aspects (in biochemistry, microbiology, etc.) one of which is in biostatistics: trying the test out on a pilot population. From this developmental stage, the efficiency of the test is characterized by its sensitivity and specificity. An efficient test will then go through an applicational stage with an actual application of $X$ to a target population; and here we are concerned with its predictive values. The simple example given in Table $3.3$ shows that unlike sensitivity and specificity, the positive and negative predictive values depend not only on the efficiency of the test but also on the disease prevalence of the target population. In both cases, the test is $90 \%$ sensitive and $90 \%$ specific. However:

1. Population A has a prevalence of $50 \%$, leading to a positive predictive value of $90 \%$.
2. Population B has a prevalence of $10 \%$, leading to a positive predictive value of $50 \%$.

## 统计代写|生物统计学作业代写Biostatistics代考|Measuring Agreement

Many research studies rely on an observer’s judgment to determine whether a disease, a trait, or an attribute is present or absent. For example, results of ear examinations will certainly have effects on a comparison of competing treatments for ear infection. Of course, the basic concern is the issue of reliability. Sections $1.1 .2$ and 3.1.4 dealt with an important aspect of reliability, the validity of the assessment. However, to judge a method’s validity, an exact method for classification, or gold standard, must be available for the calculation of sensitivity and specificity. When an exact method is not available, reliability can only be judged indirectly in terms of reproducibility; the most common way for doing that is measuring the agreement between examiners.

For simplicity, assume that each of two observers independently assigns each of $n$ items or subjects to one of two categories. The sample may then be enumerated in a $2 \times 2$ table (Table $3.4$ ) or in terms of the cell probabilities (Table 3.5). Using these frequencies, we can define:

1. An overall proportion of concordance:
$$C=\frac{n_{11}+n_{22}}{n}$$
2. Category-specific proportions of concordance:
\begin{aligned} C_{1} &=\frac{2 n_{11}}{2 n_{11}+n_{12}+n_{21}} \ C_{2} &=\frac{2 n_{22}}{2 n_{22}+n_{12}+n_{21}} \end{aligned}

The distinction between concordance and association is that for two responses to be associated perfectly, we require only that we can predict the category on one response from the category of the other response, while for two responses to have a perfect concordance, they must fall into the identical category. However, the proportions of concordance, overall or categoryspecific, do not measure agreement. Among other reasons, they are affected by the marginal totals. One possibility is to compare the overall concordance,
$$\theta_{1}=\sum_{i} p_{i i}$$
where $p$ ‘s are the proportions in the second $2 \times 2$ table above, with the chance concordance,
$$\theta_{2}=\sum_{i} p_{i+} p_{+i}$$
which occurs if the row variable is independent of the column variable, because if two events are independent, the probability of their joint occurrence is the product of their individual or marginal probabilities (the multiplication rule). This leads to a measure of agreement,
$$\kappa=\frac{\theta_{1}-\theta_{2}}{1-\theta_{2}}$$
called the kappa statistic, $0 \leq \kappa \leq 1$, which can be expressed as
$$\kappa=\frac{2\left(n_{11} n_{22}-n_{12} n_{21}\right)}{n_{1+} n_{+2}+n_{+1} n_{2+}}$$
and the following are guidelines for the evaluation of kappa in clinical research:
$\kappa>0.75: \quad$ excellent reproducibility
$0.40 \leq \kappa \leq 0.75: \quad$ good reproducibility
$0 \leq \kappa<0.40: \quad$ marginal/poor reproducibility
In general, reproducibility that is not good indicates the need for multiple assessment.

## 统计代写|生物统计学作业代写Biostatistics代考|Using Screening Tests

1. 公关⁡(X=+∣是=+)和公关⁡(X=−∣是=−)分别是敏感性和特异性。
2. 公关⁡(是=+∣X=+)和公关⁡(是=−∣X=−)分别称为正预测和负预测。

1. 人群 A 的患病率为50%，导致阳性预测值为90%.
2. 人群 B 的流行率为10%，导致阳性预测值为50%.

## 统计代写|生物统计学作业代写Biostatistics代考|Measuring Agreement

1. 一致性的总体比例：
C=n11+n22n
2. 特定类别的一致性比例：
C1=2n112n11+n12+n21 C2=2n222n22+n12+n21

θ1=∑一世p一世一世

θ2=∑一世p一世+p+一世

ķ=θ1−θ21−θ2

ķ=2(n11n22−n12n21)n1+n+2+n+1n2+

ķ>0.75:出色的重现性
0.40≤ķ≤0.75:重现性好
0≤ķ<0.40:可重复性边缘/差

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