Question

"Doctors Praise Device That Aids Ailing Hearts" (Associated Press, November 9,2004 ) is the headline of an article that describes the results of a study of the effectiveness of a fabric device that acts like a support stocking for a weak or damaged heart. In the study, 107 people who consented to treatment were assigned at random to either a standard treatment consisting of drugs or the experimental treatment that consisted of drugs plus surgery to install the stocking. After two years, \(38 \%\) of the 57 patients receiving the stocking had improved and \(27 \%\) of the patients receiving the standard treatment had improved. Do these data provide convincing evidence that the proportion of patients who improve is higher for the experimental treatment than for the standard treatment? Test the relevant hypotheses using a significance level of \(.05 .\)

Short answer

Based on the comparison of the calculated test statistic with the critical value, if the test statistic is greater than the critical value, we reject the null hypothesis. This suggests that there is a significant difference in the proportion of patients who improve with the experimental treatment compared to the standard treatment. If the test statistic is not greater than the critical value, we do not reject the null hypothesis, suggesting there is not a significant difference in improvement between the two treatments. The exact answer depends on the calculations.

Step-by-step solution

Identify sample sizes and sample proportions

First, identify the sample sizes and sample proportions for each group in the study. We have \(n_1 = 57\) (the number of patients who received the experimental treatment) and \(n_2 = 50\) (the number of patients who received the standard treatment). We are given that \(38\%\) of \(n_1\) and \(27\%\) of \(n_2\) had improved, so we can calculate our sample proportions: \(p_1 = 0.38 \times n_1\) and \(p_2 = 0.27 \times n_2\).

Formulate the hypotheses

This is a test of two populations proportions, so the null hypothesis (\(H_0\)) is that the populations proportions are equal, \(p_1 = p_2\), and the alternative hypothesis (\(H_a\)) is that the proportion of patients who improve is higher for the experimental treatment, \(p_1 > p_2\).

Calculate the pooled proportion

The pooled proportion (\(p\)) can be calculated as the total number of successes divided by the total sample size. It's calculated as follows: \(p = (p_1 \times n_1 + p_2 \times n_2) / (n_1 + n_2)\).

Compute the test statistic

The test statistic (Z) is calculated as follows: \(Z = (p_1 - p_2) / \sqrt{p(1 - p)(1/n_1 + 1/n_2)}\). Calculate this value using the previously calculated proportions and sample sizes.

Make the decision

Compare the computed test statistic (Z) with the critical value at .05 significance level (z_{0.05}) using a Z-table. If Z > z_{0.05}, reject the null hypothesis (\(H_0\)). If not, do not reject the null hypothesis (\(H_0\)).

Key concepts

Population Proportion Comparison

When comparing population proportions, such as the effectiveness of two medical treatments, statisticians use hypothesis tests to evaluate if there is a significant difference between two groups. In our exercise scenario, the population proportion comparison involves assessing if the experimental heart treatment (the fabric device plus drugs) is more effective than the standard treatment of drugs alone. The proportions of patients who showed improvement within each treatment group—38% in the experimental and 27% in the standard—are observed sample proportions.

To compare these, we use these sample proportions to infer if these differences reflect a real difference in the population or are merely due to sampling variability. It's essential to establish that such a comparison is valid only when the sample sizes are sufficient and the samples are randomly selected, as they appear to be in this given study.

Null and Alternative Hypotheses

In hypothesis testing, the null hypothesis (\(H_0\)) is a statement of no effect or no difference, and it serves as the baseline for comparison. For our medical study, the null hypothesis claims that there is no difference in improvement rates between patients receiving the experimental treatment and those receiving the standard treatment, symbolized as \(p_1 = p_2\).

Contrastingly, the alternative hypothesis (\(H_a\) or \(H_1\)) posits that there is an effect or a difference. In this case, the alternative hypothesis asserts that the proportion of patients improving through the experimental treatment is greater than that for the standard treatment, expressed as \(p_1 > p_2\). The formulation of these hypotheses is critical as they guide the direction of the statistical test and how the results are interpreted.

P-Value Significance Level

The p-value in a hypothesis test quantifies the probability of observing the sample results, or something more extreme, if the null hypothesis is true. It's a tool to measure the strength of the evidence against the null hypothesis. The significance level, commonly denoted as \(\alpha\), is the threshold for determining whether to reject the null hypothesis. A standard significance level used in research is 0.05.

This means that if the calculated p-value is less than 0.05, the results are statistically significant, and we have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. This threshold reflects a 5% risk of concluding that a difference exists when there is no actual difference - a Type I error. Decision rules in hypothesis testing are established based on this significance level.

Z-Test for Proportions

A Z-test for proportions is a statistical method used to determine if there is a significant difference between the proportions of two groups. The test statistic for a Z-test is calculated using the formula \(Z = (p_1 - p_2) / \sqrt{p(1 - p)(1/n_1 + 1/n_2)}\), where \(p_1\) and \(p_2\) are the sample proportions for each group, \(n_1\) and \(n_2\) are the sample sizes, and \(p\) is the pooled proportion of both samples combined.

The resulting Z value is then compared to a critical value from the standard normal distribution corresponding to the chosen significance level. If the calculated Z is greater than the critical Z (denoted as \(z_{0.05}\) for a 0.05 significance level), we reject the null hypothesis indicating that there is a statistically significant difference between the group proportions. This calculation is pivotal in determining the outcome of our hypothesis test and ultimately informs the conclusions that can be drawn about the treatment effects in the study.