Question

In an experiment to assess the effectiveness of a treatment to reduce blood pressure in heart patients, \(n\) independent pairs of heart patients are matched according to their sex, weight, smoking history, initial blood pressure, and so forth. Then one of each pair is selected at random and given the treatment. After a set time the blood pressures are again recorded, and it is desired to assess whether the treatment had any effect. A simple model for this is that the \(j\) th pair of final measurements, \(\left(Y_{j 1}, Y_{j 2}\right)\) is two independent normal variables with means \(\mu_{j}\) and \(\mu_{j}+\beta\), and variances \(\sigma^{2}\). It is desired to assess whether \(\beta=0\) or not. One approach is a \(t\) confidence interval based on \(Z_{j}=Y_{j 2}-Y_{j 1} .\) Explain this, and give the degrees of freedom for the \(t\) statistic. Show that the likelihood ratio statistic for \(\beta=0\) is equivalent to \(\bar{Z}^{2} / \sum\left(Z_{j}-\bar{Z}\right)^{2}\)

Short answer

The \( t \)-statistic uses \( n-1 \) degrees of freedom, and the likelihood ratio statistic is \( \frac{\bar{Z}^2}{\sum(Z_j - \bar{Z})^2} \).

Step-by-step solution

Understanding the Problem

The goal is to determine whether the treatment has an effect, i.e., if \( \beta eq 0 \), by analyzing the differences in post-treatment blood pressure measurements, \( Z_j = Y_{j2} - Y_{j1} \). The challenge is to ascertain if these differences are statistically significant or could just occur by random chance.

Formulating Hypothesis

We will perform a hypothesis test with the null hypothesis \( H_0: \beta = 0 \) against the alternative hypothesis \( H_a: \beta eq 0 \). \( \beta = 0 \) implies no difference between treated and untreated patients.

Constructing the Test Statistic

We define the sample mean \( \bar{Z} = \frac{1}{n}\sum Z_j \) and sample variance \( S^2 = \frac{1}{n-1} \sum (Z_j - \bar{Z})^2 \). The test statistic is \( t = \frac{\bar{Z} \sqrt{n}}{S} \), which follows a Student's t-distribution if \( \beta = 0 \).

Determining Degrees of Freedom

Since we have independent paired observations, each pair gives one degree of freedom. Hence, the degrees of freedom for the \( t \)-statistic is \( n-1 \), where \( n \) is the number of pairs.

Likelihood Ratio Statistic Equivalence

We use the likelihood ratio test statistic which is an alternative approach. The likelihood under \( \beta = 0 \) is compared with \( \beta eq 0 \). It can be shown that the likelihood ratio statistic is equivalent to \( \frac{\bar{Z}^2}{\sum(Z_j - \bar{Z})^2} \). This ratio measures the strength of \( \bar{Z} \) relative to the variance of \( Z_j \).

Key concepts

t-confidence interval

A t-confidence interval helps us understand the range in which we expect the true parameter of interest, like a difference in means, to fall. In this experiment, we're examining whether the treatment impacts blood pressure, represented by the parameter \( \beta \).
The confidence interval is constructed from the sample mean differences \( \bar{Z} \) and their standard deviation. It uses the formula:

• \( \bar{Z} \pm t \times \frac{S}{\sqrt{n}} \)
• Where \( t \) is the critical value from the t-distribution for \( n-1 \) degrees of freedom

This interval provides us with a range of plausible values for \( \beta \). If zero falls within this interval, it suggests that the treatment might not have an effect.

paired sample design

A paired sample design is used when each subject serves as their own control. This method allows for an effective comparison by eliminating variability due to inter-subject differences.
In this case, each heart patient has a matching counterpart. One from the pair receives treatment, while the other does not. The differences in their outcomes are then analyzed. Such a design reduces noise because:

• It controls for individual variability like sex, weight, and smoking history.
• This design focuses solely on the treatment effect \( \beta \).

This paired approach strengthens internal validity and increases the power of the statistical test.

likelihood ratio test

The likelihood ratio test is used to compare statistical models with and without a specific parameter, like \( \beta = 0 \) in this exercise. The goal is to determine if the parameter significantly improves the model.
To apply this test:

• The null model assumes \( \beta = 0 \).
• The alternative model assumes \( \beta eq 0 \).

The test statistic, \( \frac{\bar{Z}^2}{\sum(Z_j - \bar{Z})^2} \), measures the strength of the observed mean difference relative to variability. A higher value means \( \bar{Z} \) isn't likely due to chance, suggesting the treatment possibly has an effect.

degrees of freedom

Degrees of freedom (df) represent the number of values in the final calculation of a statistic that are free to vary. In hypothesis testing, they ensure that our statistical tests accurately reflect the variability in the data.
In paired samples, each patient pair contributes one piece of independent information. Thus, for \( n \) pairs:

• The formula is \( n-1 \) for degrees of freedom.
• This affects the critical value from the Student's t-distribution used in hypothesis testing.

Understanding df is crucial because it influences both the width of confidence intervals and the significance level of your test results.

Student's t-distribution

Student's t-distribution is essential for analyzing data with small sample sizes, especially when the population standard deviation is unknown. It is bell-shaped like the normal distribution but has thicker tails, which reflect more uncertainty.
Key features include:

• Dependence on degrees of freedom: As df increases, it approaches the normal distribution.
• Used to derive critical values for t-tests and confidence intervals.

In this experiment, the t-distribution helps us determine if the observed difference in \( \bar{Z} \) is significant, indicating a potential treatment effect on blood pressure. It's a crucial tool when working with smaller \( n \) due to increased variability.