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

Use the sample proportions, pooled sample proportion, and sample sizes to calculate the test statistic Z. Based on the comparison with the critical value \(z_{0.05}=1.645\), we decide whether to reject or fail to reject the null hypothesis \(H_0\). If the Z is higher, the conclusion would be that the data provides convincing evidence that the proportion of patients who improve is higher for the experimental treatment than for the standard treatment at a 5% level of significance.

Step-by-step solution

State the Hypotheses

The null hypothesis \(H_0\): \(p_1 = p_2\). The alternative hypothesis \(H_A\): \(p_1 > p_2\). Where \(p_1\) is the proportion of patients improving with the experimental treatment and \(p_2\) is the proportion of patients improving with standard treatment.

Calculate Sample Proportions

Given that 38% of the 57 patients in the experimental group improved, the sample proportion is \(p_1 = 0.38\). And that 27% of the 50 patients in the standard treatment group improved, the sample proportion is \(p_2 = 0.27\).

Statistical Test

We use the Z test for comparing two proportions. The test statistic is given by the formula: \(Z = \frac {(p_1 - p_2)}{\sqrt{p(1 - p)(\frac{1}{n_1} + \frac{1}{n_2})}}\) where \(p = \frac{p_1*n_1 + p_2*n_2}{n_1 + n_2}\) is the pooled sample proportion. So, substituting values we find: \(p = \frac{0.38*57 + 0.27*50}{57 + 50}\), and then calculate \(Z\).

Decision Rule and Conclusion

Compare the calculated test statistic with the critical value for a significance level of 0.05. If \(Z > z_{0.05}\), reject \(H_0\). If \(Z \leq z_{0.05}\), do not reject \(H_0\). \( z_{0.05} = 1.645\) for a one-tailed test. Use a Z table or suitable software to find the value of \(z_{0.05}\). If \(Z > z_{0.05}\), we can conclude that the data provides convincing evidence that the proportion of patients who improve is higher for the experimental treatment than for the standard treatment at a 5% level of significance.