Question

\( \quad(\mathrm{M} 1, \mathrm{M} 5, \mathrm{M} 6, \mathrm{P} 3)\) "Doctors Praise Device That Aids Ailing Hearts" (Associated Press, November 9,2004 ) is the headline of an article that describes 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, while \(27 \%\) of the patients receiving the standard treatment had improved. Do these data provide evidence that the proportion of patients who improve is significantly higher for the experimental treatment than for the standard treatment? Test the relevant hypotheses using a significance level of 0.05

Short answer

The conclusion, i.e. whether the proportion of r Improved is significantly higher for the experimental treatment than for the standard treatment, depends on whether the calculated p-value is less than the significance level (0.05). If it is, we reject the null hypothesis. If not, we fail to reject it.

Step-by-step solution

Formulation of Hypotheses

We start by setting up the null and the alternative hypotheses. The null hypothesis assumes that there is no difference between the two proportions (i.e., the proportions of successes in both groups are equal) and the alternative hypothesis is that the proportion in the experimental group is higher than in the standard one. Specifically, the null hypothesis \(H_0: p_1 - p_2 = 0\) and alternative hypothesis \(H_1: p_1 - p_2 > 0\), where \(p_1\) and \(p_2\) are the proportions of improved patients in the experimental and standard groups respectively.

Calculate the Test Statistic

Next, we compute the test statistic (z), assuming the null hypothesis is true (i.e. \(p_1 - p_2 = 0\)). We first calculate the pooled proportion \(p\), which is the total successful outcomes divided by total outcomes. Then, the standard error (SE) is given by \( \sqrt{p(1-p)[(1/n1)+(1/n2)]}\) where \(n_1\) and \(n_2\) are the sizes of two groups. Finally, Z score is found by \((p1-p2)/SE\).

Compute the p-value

The p-value can be computed using a Z table or Z distribution calculator. The computed p-value is then compared with the significance level (0.05).

Conclusion

If the computed p-value is less than the significance level (0.05), then the null hypothesis would be rejected providing evidence that the proportion of patients who improve is significantly higher for the experimental treatment. Otherwise, we do not reject the null hypothesis, meaning there is not enough evidence to say that the experimental treatment leads to more improvements.