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\( \quad(\mathrm{M} 1, \mathrm{M} 5, \mathrm{M} 6, \mathrm{P} 3)\) "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

Step by step solution

01

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.

02

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 \((p1-p2)/SE\).

03

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).

04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis

In hypothesis testing, the null hypothesis is a statement that suggests there is no effect or no difference. It acts as the default or neutral position. For example, in the context of a medical study, it would assume that a new treatment doesn’t lead to better outcomes compared to existing methods.

For our exercise, the null hypothesis, denoted as \(H_0\), proposes that the success rates from the experimental and standard treatments are equal. In mathematical terms, this is expressed as \(p_1 - p_2 = 0\), where \(p_1\) is the proportion of improvement with the experimental treatment and \(p_2\) for the standard treatment. By establishing this hypothesis, researchers can analyze data to determine whether observed differences are due to chance or if they suggest real effects.

Alternative Hypothesis

The alternative hypothesis offers a contrast to the null hypothesis. It proposes that there is an effect or a difference, suggesting that one treatment outperforms another or that a specific variable has an impact.

In our exercise, the alternative hypothesis, symbolized as \(H_1\), suggests that the proportion of improvements in patients is higher for the experimental group compared to the standard treatment group. So, mathematically, this is depicted as \(p_1 - p_2 > 0\). This hypothesis is what researchers aim to support or validate in their findings.

P-Value

A p-value helps determine the significance of your results in hypothesis testing. It represents the probability of observing the results given that the null hypothesis is true.

A lower p-value indicates that the observed data is less likely under the null hypothesis, providing stronger evidence against \(H_0\). In our study, once the test statistic is calculated, the p-value is derived from the Z distribution. If the p-value is less than a predetermined significance level (usually 0.05), it suggests that the data supports the alternative hypothesis. Always remember, the smaller the p-value, the more it challenges the notion that there is no difference.

Significance Level

The significance level, often denoted by alpha (\(\alpha\)), is a threshold set by the researcher which determines whether to reject the null hypothesis. Commonly, this is set at 0.05, marking a 5% risk of concluding that a difference exists when there is none (Type I error).

For our exercise, an alpha of 0.05 means that we accept a 5% chance of mistakenly claiming the experimental treatment improves outcomes over the standard treatment if, in reality, it does not. If the p-value calculated from the test is below this significance level, \(H_0\) is rejected, leading to the conclusion that there is a significant difference in treatment outcomes.