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"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 \(.05\).

Step by step solution

01

Identify the hypothesis and sample data

The null hypothesis \(H_0\) states that there's no difference in the proportions of patients who improve with experimental treatment and those who improve after the standard treatment. That is \(p_{1} - p_{2} = 0\). The alternate hypothesis \(H_1\) states that the proportion of patients who improve after the experimental treatment is higher than those improving after standard treatment i.e., \(p_{1} - p_{2} > 0\). Sample data extracted are: number of patients who improved with experimental treatment (n1) = 0.38 * 57 = 21.66, total number of patients who had experimental treatment (x1) = 57, number of patients who improved with standard treatment (n2) = 0.27 * 50 = 13.5, total patients who had standard treatment (x2) = 50.

02

Calculate Pooled Proportion and Standard Error

First compute the pooled sample proportion. That is \((n_1 + n_2) / (x_1 + x_2)\), which acts as an estimate of the population proportion. Standard error (SE) is computed as \(\sqrt{ p(1 - p) [ (1/x_1) + (1/x_2)] }\) where p is the pooled sample proportion.

03

Compute Test Statistic

After calculating SE, the next step is to compute the test statistic which is a z-score: it equals \((p_{1} - p_{2}) - 0 / SE\). In hypothesis testing, the denominator is always the standard error.

04

Find P-value

The p-value compares the z-score with a Normal distribution. If p-value is smaller than the significance level (alpha = 0.05), reject the null hypothesis.

05

Conclusion

Based on the calculated p-value and the significance level, state the conclusion. If the p-value is less than 0.05, it can be concluded that there is convincing evidence that the experimental treatment leads to a higher improvement in patients compared to the standard one. Otherwise, there's no significant evidence to suggest the experimental treatment is better.

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Key Concepts

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

Statistical Significance

Statistical significance is a determination made during hypothesis testing that provides us with a measure to decide whether to reject or not reject a null hypothesis. It involves a threshold called the significance level, often denoted by \( \alpha \), which is predefined by the researcher. The commonly used significance level is \( 0.05 \), meaning that there is only a 5% probability that the observed results would occur if the null hypothesis were true. Simplifying this, it's like saying there's a 95% confidence the results are not just by random chance.

P-value

The p-value is a vital concept in hypothesis testing. It quantifies the probability of obtaining results at least as extreme as the observed ones, given that the null hypothesis is true. A lower p-value indicates that the observed data is less likely to occur under the null hypothesis, while a higher p-value suggests the opposite. If the p-value is less than or equal to the significance level \( \alpha \), we reject the null hypothesis, which implies that our findings may be \'statistically significant\'.

Null Hypothesis

The null hypothesis, \( H_0 \), plays a core role in hypothesis testing as it represents a general statement or default position that there is no difference or effect. In the context of the provided exercise, the null hypothesis claims that there is no difference in the proportion of patients who improve between the experimental treatment and the standard treatment. It serves as the baseline for testing and is presumed true until evidence suggests otherwise.

Alternative Hypothesis

Contrasting the null hypothesis is the alternative hypothesis, labeled \( H_1 \). It represents what the researcher seeks to prove or is prepared to believe once there is sufficient evidence against the null hypothesis. In our example, the alternative hypothesis posits that the proportion of patients who improve is actually higher for the experimental treatment than for the standard treatment. It is a claim made when the null hypothesis fails to stand under scrutiny from our data.

Pooled Sample Proportion

The pooled sample proportion is a technique used when testing hypotheses about two population proportions. It creates a single, combined proportion by pooling the successes and total observations from both samples. The rationale behind this concept is to obtain a common estimate of the proportion that reflects both samples, which is then used to find the standard error and, subsequently, the test statistic. The formula that was used in the exercise is an example of how to calculate this combined measure.

Test Statistic

The test statistic is a value calculated from sample data during hypothesis testing which, when compared to a probability distribution, tells us how extreme the observed results are. The type of test statistic used depends on the distribution of the test. For proportions, a z-score is usually calculated, built upon the difference between sample proportion and null hypothesis value, divided by the standard error of the sampling distribution. If the test statistic lies in the extreme end of the distribution, it provides evidence against the null hypothesis.