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Chapter 11: Problem 46

"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

Expert verified
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

01

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.
02

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 \(p_2 = 0.27\).
03

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 \(Z\).
04

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.

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Key Concepts

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

Z test
A Z test is a type of statistical test that is used when you're dealing with large sample sizes and you want to compare averages or proportions. It's particularly handy when you believe your data follow a normal distribution, or your sample size is large enough that the Central Limit Theorem kicks in (usually 30 or more).

In this exercise, you're looking at the effectiveness of two different treatments on heart patients. The goal is to compare proportions: one for patients improving using an experimental treatment and the other for patients with standard treatment. Here, the Z test is just right because you're comparing two proportions to see if one is significantly higher than the other.

When you conduct a Z test for proportions, the hypothesis test results in a Z-score. This score tells you how many standard deviations away your observation is from the null hypothesis assumption. If this value is extreme enough (greater than the critical value), it suggests the observed difference in proportions isn't just due to random chance, and there's evidence supporting increased effectiveness for one treatment over another.
proportions comparison
Comparing proportions is a statistical method used when you have two groups, and you want to determine if a particular characteristic is more common in one group than in the other. This can often answer the question, "Is this treatment more effective than the other?"

In our exercise, we want to know whether the proportion of patients improving with an experimental treatment is higher than those improving with a standard treatment.

We started by calculating the sample proportions: one from the experimental group and one from the standard group. The calculation goes like this: divide the number of patients who improved by the total number of patients in each group. Here, it's 0.38 for the experimental and 0.27 for the standard.

Once we have these proportions, we use the Z test to run the comparison. The Z test considers the "pooled" proportion, which effectively blends both groups for a baseline comparison. With this approach, we can assess whether any observed difference in proportions between the two groups is statistically significant.
significance level
The significance level in hypothesis testing, often denoted by alpha (伪), represents the threshold for determining when to reject the null hypothesis. It's the probability of rejecting the null hypothesis when it is actually true (Type I error).

For this exercise, a significance level of 0.05 is used, which is quite common in hypothesis testing. It implies that you are willing to accept a 5% chance that your conclusion of a treatment being more effective is incorrect.

In simpler terms, with a significance level of 0.05, you鈥檙e looking for a less than 5% chance that your results are just due to random fluctuations in the sample, rather than a true difference in treatment effectiveness. This makes it a measure of the risk you鈥檙e willing to take in saying one treatment is better than the other.

It's also crucial to compare your Z-score against a critical value related to your significance level. For a one-tailed test at 0.05, this critical value is 1.645. If your calculated Z-value exceeds this, you reject the null hypothesis, suggesting there's convincing evidence that the experimental treatment has a higher improvement rate than the standard treatment.

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Most popular questions from this chapter

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