Question

A research physician compared the effectiveness of two blood pressure drugs \(A\) and \(B\) by administering the two drugs to each of four pairs of identical twins. Drug \(A\) was given to one member of a pair; drug \(B\) to the other. If, in fact, there is no difference in the effects of the drugs, what is the probability that the drop in the blood pressure reading for drug \(A\) exceeds the corresponding drop in the reading for drug \(B\) for all four pairs of twins? Suppose drug \(B\) created a greater drop in blood pressure than drug \(A\) for each of the four pairs of twins. Do you think this provides sufficient evidence to indicate that drug \(B\) is more effective in lowering blood pressure than drug \(A\) ?

Short answer

Answer: The probability is 0.0625.

Step-by-step solution

Enumerate all possible scenarios for twin pairs.

Each twin pair can have two possible outcomes, either drug \(A\) has a greater blood pressure drop than drug \(B\) or drug \(B\) has a greater blood pressure drop than drug \(A\). Since there are 4 twin pairs, we evaluate \(2^4 = 16\) possible scenarios of these outcomes.

Count the successful outcomes.

In order to find the required probability, we need to count all the scenarios in which drug \(A\) has a greater blood pressure drop than drug \(B\) for all 4 pairs of twins, which means having drug \(A\) outperforming drug \(B\) in all 4 twin pairs. There is only 1 such scenario.

Calculate the probability.

The probability of drug \(A\) having a greater blood pressure drop than drug \(B\) for all 4 pairs of twins can now be calculated by dividing the number of successful outcomes (1) by the total possible outcomes (16), i.e., Probability = \(\frac{1}{16}\) = 0.0625.

Discuss the implications of the observed results.

In the given exercise, it is mentioned that drug \(B\) created a greater drop in blood pressure than drug \(A\) for all four pairs of twins. This means that we have observed the worst possible outcome for drug \(A\) (the least likely situation). While this alone doesn't necessarily prove that drug \(B\) is more effective than drug \(A\), it does provide strong evidence in favor of drug \(B\) being more effective in lowering blood pressure than drug \(A\). Additionally, it would be recommended to perform further studies with larger samples to strengthen the conclusions derived from this experiment.

Key concepts

Hypothesis Testing

In the realm of medicine, hypothesis testing is a crucial statistical method. It helps researchers determine whether there is a significant effect or difference due to treatment or intervention. In the given exercise, the hypothesis tested is whether there is a difference in the effectiveness between drugs A and B for reducing blood pressure. The null hypothesis, which assumes no difference between the drugs, is implicitly tested.

Hypothesis testing involves several key steps:

• Establishing the null and alternative hypotheses
• Choosing a significance level, commonly 0.05
• Calculating the test statistic from observed data
• Determining the probability (p-value) of observing such data if the null hypothesis is true
• Making a decision to reject or fail to reject the null hypothesis based on the p-value

In this exercise, if the observed outcome (where drug B performed better in all cases) occurs under the assumption of the null hypothesis, it might indicate significant evidence against it.
The p-value is the probability calculated, 0.0625 in this case, suggesting that the observed result is somewhat unlikely yet possible under the null hypothesis.

Probability Calculation

Calculating probabilities effectively is a core component of analyzing results in medical research. The method utilized in solving the exercise is a basic example of probability calculation.

The probability in question was to determine how likely it is for drug A to outperform drug B in all twin pairs if there is no inherent effectiveness difference. With four twin pairs and two possible outcomes per pair (A is better or B is better), the total number of outcomes is calculated as \( 2^4 = 16 \).

• The only scenario favoring drug A outperforming in all pairs is just one out of the possible 16.
• This gives a probability of \( P = \frac{1}{16} \) or approximately 0.0625.

Understanding these probabilities is essential for determining how "surprising" or "expected" certain experimental results might be. In this specific example, the low probability calculated indicates that seeing drug A outperform drug B in all cases would be unlikely if the drugs are equally effective.

Experimental Design in Medicine

A well-thought-out experimental design is vital in medical studies to draw valid conclusions. The exercise demonstrates a specific type of experimental design using paired samples, in this case, identical twins.

This design ensures as much control over confounding variables as possible, as identical twins have inherently similar genetics.

• Each twin in a pair receives a different drug to minimize individual differences.
• By comparing within twin pairs, researchers can focus heavily on the drug effect.

Such designs are powerful in reducing variability and enhancing the reliability of results. The twin-pair model also prevents the skew from environmental factors, which could otherwise introduce bias.
By conducting experiments with controlled pairs like twins, results gain immense credibility, especially when variability is naturally minimized. Yet, medical research often requires larger sample sizes to confirm results. Other designs might include randomization or crossover studies to further ensure unbiased findings, which often accompany paired designs in comprehensive research endeavors.