Question

Patients with pleural infections (fluid buildup in the chest) were randomly assigned to placebo, the treatment "tPA" (tissue plasminogen activator), the treatment "DNase" (deoxyribonuclease), or a combination of tPA and DNase in a double-blind clinical trial. Some of the patients needed to be given penicillin. The following table shows a cross-classification of the data. \({ }^{40}\) Is there evidence of an association between treatment group and use of penicillin? $$ \begin{array}{|lrrrrr|} \hline & \text { Placebo } & \text { tPA } & \text { DNase } & \text { tPA+DNase } & \text { Total } \\ \hline \text { Penicillin } & 4 & 2 & 3 & 3 & 12 \\ \text { No penicillin } & 51 & 50 & 48 & 49 & 198 \\ \text { Total } & 55 & 52 & 51 & 52 & 210 \\ \hline \end{array} $$ Here is computer output for a chi-square test to compare the four groups. \(x\) -squared \(=0.59, d f=3, p\) -value \(=0.90\) (a) State the null and alternative hypotheses in context. (b) Compute the expected frequency of the Penicillin/ Placebo cell under the null hypothesis. (c) If \(\alpha=0.10\), what is your conclusion regarding \(H_{0}\) ?

Short answer

The null hypothesis is that there is no association between treatment group and penicillin usage, which cannot be rejected given the p-value of 0.90. The expected number of patients who received penicillin and a placebo is (12 * 55) / 210.

Step-by-step solution

State the Hypotheses

In context, the null hypothesis (\(H_0\)) states that there is no association between the treatment group and the use of penicillin. The alternative hypothesis (\(H_a\)) states that there is an association between the treatment group and the use of penicillin.

Compute Expected Frequency

To calculate the expected frequency for the Penicillin/Placebo cell, use the formula: Expected frequency = (Row total * Column total) / Grand total. So, here it is: Expected frequency for Penicillin/Placebo = (12 * 55) / 210.

Analyze the Chi-Square Test Output

With a chi-square test statistic of 0.59 and a p-value of 0.90, we compare the p-value to the significance level \(\alpha = 0.10\). Since the p-value (0.90) is greater than \(\alpha\), we fail to reject the null hypothesis, suggesting there is not enough evidence to conclude an association between treatment group and penicillin usage.

Key concepts

Null and Alternative Hypotheses

Understanding the null and alternative hypotheses is a fundamental part of any statistical test, including the chi-square test used in clinical trials. The null hypothesis (\(H_0\)) is the default assumption that there is no effect or no difference; it represents a statement of “no effect” or “no association.” In the context of the given exercise, the null hypothesis posits that there is no association between the type of treatment given to patients with pleural infections and whether they required penicillin.

The alternative hypothesis (\(H_a\text{ or }H_1\)), on the other hand, is what the researcher aims to prove. It suggests that there is an effect, a difference, or an association that is statistically significant. In our clinical trial example, the alternative hypothesis claims that the use of penicillin is indeed related to the type of treatment—placebo, tPA, DNase, or a combination of tPA and DNase.

It's crucial for students to understand these concepts because they frame the entire hypothesis testing process and determine the direction of statistical analysis. To examine the hypotheses accurately, researchers use data from the study and apply the chi-square test to determine whether the observed frequencies in the data significantly deviate from the expected frequencies under the null hypothesis.

Expected Frequency Calculation

The expected frequency is a cornerstone of the chi-square test and requires a clear understanding for accurate statistical analysis in clinical trials. It refers to the frequency of observations that we would expect to find in a particular category if the null hypothesis were true—meaning if there are no associations in our data set.

To calculate the expected frequency as shown in the exercise, we use the formula: \[ \text{Expected frequency} = \frac{(\text{Row total} \times \text{Column total})}{\text{Grand total}} \]
This formula helps determine what numbers we would expect if there were no association between the treatment groups and the usage of penicillin. For the Penicillin/Placebo cell in our example, this calculates to:\[ \text{Expected frequency} = \frac{(12 \times 55)}{210} \]
It's important for students to grasp that this calculation is essential for every cell in their contingency table to perform the chi-square test. These expected frequencies are then compared to the observed frequencies within the data, allowing us to investigate any significant differences that might indicate associations or effects that could be of clinical importance.

Associations in Clinical Data

Associations in clinical data are connections or relationships that may exist between different variables—such as treatment types and outcomes like medication usage. Discovering these associations can provide valuable insights into the effectiveness of treatments or possible side effects.

In the context of a chi-square test, we are particularly interested in the strength and significance of these associations. The chi-square test assesses whether the distribution of observed frequencies across different categories deviates from what would be expected under the null hypothesis of no association. In simpler terms, it lets us question if there’s a pattern or a relationship, or if our results could just as likely occur by random chance.

The exercise provided offers a clear example: researchers wanted to know if the type of treatment for pleural infections was linked to the need for penicillin. With a high p-value (0.90) in this case, suggesting a high probability that results as extreme as the observed data could occur under the null hypothesis, we do not find evidence of a significant association. When interpreted correctly, these statistical tests can guide critical decisions in healthcare and inform future research. It’s essential for students to understand how to evaluate such associations correctly to make informed decisions based on clinical data.