/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Exercises 23 through 26 involve ... [FREE SOLUTION] | 91Ó°ÊÓ

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Exercises 23 through 26 involve the following setting. Some women would like to have children but cannot do so for medical reasons. One option for these women is a procedure called in vitro fertilization (IVF), which involves injecting a fertilized egg into the woman’s uterus. Acupuncture and pregnancy A study reported in the medical journal Fertility and Sterility sought to determine whether the ancient Chinese art of acupuncture could help infertile women become pregnant.\(^{18}\) One hundred sixty healthy women who planned to have IVF were recruited for the study. Half of the subjects (80) were randomly assigned to receive acupuncture 25 minutes before embryo transfer and again 25 minutes after the transfer. The remaining 80 women were assigned to a control group and instructed to lie still for 25 minutes after the embryo transfer. Results are shown in the table below. $$\begin{array}{ll}&{\text { Acupuncture group }} & {\text { Control group }} \\\ \text { Pregnant } & \quad\quad\quad\quad {34} & \quad\quad\quad {21} \\\ \text { Not Pregnant } & \quad\quad\quad\quad {46} & \quad\quad\quad {59} \\\ \text { Total } & \quad\quad\quad\quad {80} & \quad\quad\quad {80}\end{array}$$ Is the pregnancy rate significantly higher for women who received acupuncture? To find out, researchers perform a test of \(H_{0} : p_{1}=p_{2}\) versus \(H_{a} : p_{1}>p_{2},\) where \(p_{1}\) and \(p_{2}\) are the actual pregnancy rates for women like those in the study who do and don't receive acupuncture, respectively. (a) Name the appropriate test and check that the conditions for carrying out this test are met. (b) The appropriate test from part (a) yields a P-value of 0.0152. Interpret this P-value in context. (c) What conclusion should researchers draw at the \(\alpha=0.05\) significance level? Explain. (d) What flaw in the design of the experiment prevents us from drawing a cause-and-effect conclusion? Explain.

Short Answer

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(a) Two-proportion z-test is appropriate and conditions are met. (b) 1.52% chance under null hypothesis. (c) Reject null hypothesis; acupuncture increases pregnancy rate. (d) Lack of blinding prevents cause-and-effect conclusion.

Step by step solution

01

Identify the Appropriate Test

The scenario involves comparing two proportions: the pregnancy rates of women who received acupuncture and those who did not. The appropriate statistical test for this comparison is the two-proportion z-test.
02

Check Conditions for the Test

To carry out the two-proportion z-test, we must check two conditions: \(n_1\hat{p}_1 \geq 10\), \(n_1(1-\hat{p}_1) \geq 10\), \(n_2\hat{p}_2 \geq 10\), and \(n_2(1-\hat{p}_2) \geq 10\), where \(n_1 = n_2 = 80\), \(\hat{p}_1 = \frac{34}{80} = 0.425\), and \(\hat{p}_2 = \frac{21}{80} = 0.2625\). Each of these checks out to be greater than 10, so the test conditions are fulfilled.
03

Interpret the P-value

The P-value of 0.0152 indicates the probability of observing the data, or something more extreme, assuming the null hypothesis \(H_0 : p_1 = p_2\) is true. In the context of the problem, it means there is a 1.52% chance of seeing such a difference in pregnancy rates by random chance if there is no real effect of acupuncture.
04

Conclusion with Significance Level

Since the P-value (0.0152) is less than the significance level \(\alpha = 0.05\), we reject the null hypothesis \(H_0\). There is significant statistical evidence to suggest that the pregnancy rate is higher for women who received acupuncture compared to those who did not.
05

Discuss the Experimental Flaw

The flaw in this experimental design is the lack of blinding. Participants and possibly the practitioners knew who received acupuncture, introducing possible placebo effects or biases. Without blinding, a cause-and-effect relationship cannot be firmly established.

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Key Concepts

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

Two-Proportion Z-Test
In statistical hypothesis testing, comparing two groups often requires a careful choice of tests. The two-proportion Z-test is specifically designed for comparing the proportions of success between two independent groups. In this context, the success is defined as pregnancy after IVF treatment. The test begins with the null hypothesis \( H_0: p_1 = p_2 \), indicating no difference in the pregnancy rates between the acupuncture and control groups. On the other hand, the alternative hypothesis \( H_a: p_1 > p_2 \) suggests that the acupuncture group has a higher pregnancy rate.

To ensure that this Z-test is valid, certain conditions must be met. Each group should have a minimum number of successes and failures, typically at least 10. In this study, with sizes \( n_1 = n_2 = 80 \), both groups fulfill this requirement. Successful pregnancies measure at least 10 for both cases \( n_1\hat{p}_1 = 34 \) and failures \( n_1(1-\hat{p}_1) = 46 \), ensuring the test's accuracy. This solid statistical foundation allows researchers to confidently draw inferences from the test results.
P-value Interpretation
Understanding the P-value is crucial for interpreting the results of a statistical test. In simple terms, the P-value represents the probability of obtaining test results at least as extreme as the observed ones, assuming the null hypothesis \( H_0 \) is true. In the context of this study, a P-value of 0.0152 suggests that there is a 1.52% chance that the observed difference in pregnancy rates could be due to random variation.

Smaller P-values indicate stronger evidence against the null hypothesis. When the P-value is less than the pre-set significance level (commonly \( \alpha = 0.05 \)), it implies that such extreme data is unlikely under \( H_0 \). Thus, researchers tend to reject \( H_0 \), signaling a potentially significant difference, as seen here where acupuncture seems to increase pregnancy rates. It's important to remember that a P-value alone does not prove a hypothesis but instead helps in deciding the likelihood of data under the null hypothesis.
Significance Level
The significance level, symbolized as \( \alpha \), is a threshold set by researchers to assess whether the results of their test are statistically significant. Commonly set at 0.05, this level indicates the probability of rejecting the null hypothesis when it is actually true—essentially, the risk of a false positive. In our test for acupuncture effectiveness, the significance level is 0.05.

Comparing the P-value (0.0152) to the significance level helps make a confident decision. Since 0.0152 is less than 0.05, we reject the null hypothesis, suggesting that acupuncture significantly impacts the pregnancy rate. Significance levels thus provide a clear boundary for decision-making in statistical tests, ensuring robust conclusions based on data evidence. However, the choice of \( \alpha \) is subjective and should reflect the context and consequences of potential errors.
Experimental Design Flaws
Even with sound statistics, experimental flaws can jeopardize the validity of conclusions. One major flaw in this study is the lack of blinding, which could introduce bias and confound the results. Blinding refers to the practice where participants and if possible, practitioners, are unaware of which group—treatment or control—participants belong to, thus preventing subconscious influence on the outcomes.

Without blinding, the placebo effect, or participant expectations, might distort the results. Participants knowing they received acupuncture could psychologically influence their body's response, displaying benefits not attributable to acupuncture itself. Moreover, unblinded practitioners might inadvertently bias the care they provide. These factors mean that while statistical evidence suggests a difference, we cannot conclusively attribute this to acupuncture without more rigorously controlled experimental conditions.

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Most popular questions from this chapter

How tall? 6.2 ) The heights of young men follow a Normal distribution with mean 69.3 inches and standard deviation 2.8 inches. The heights of young women follow a Normal distribution with mean 64.5 inches and standard deviation 2.5 inches. (a) Let \(M=\) the height of a randomly selected young man and \(W=\) the height of a randomly selected young woman. Describe the shape, center, and spread of the distribution of \(M-W\) (b) Find the probability that a randomly selected young man is at least 2 inches taller than a randomly selected young woman. Show your work.

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What’s wrong? A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state’s driver’s license exam. An incoming class of 100 students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30 of Instructor A’s students and 22 of Instructor B’s students pass the state exam. Do these results give convincing evidence that Instructor A is more effective? Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one. State: I want to perform a test of $$H_{0} : p_{1}-p_{2}=0$$ $$H_{a} : p_{1}-p_{2}>0$$ where \(p_{1}=\) the proportion of Instructor A's students that passed the state exam and \(p_{2}=\) the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use \(\sigma=0.05\) Plan: If conditions are met, I'll do a two-sample \(z\) test for comparing two proportions. \(\bullet\) Random The data came from two random samples of 50 students. \(\bullet\) Normal The counts of successes and failures in the two groups - \(30,20,22\) , and \(28-\) are all at least \(10 .\) \(\bullet\) Independent There are at least 1000 students who take this driving school's class. Do: From the data, \(\hat{p}_{1}=\frac{20}{50}=0.40\) and \(\hat{p}_{2}=\frac{30}{50}=0.60 .\) So the pooled proportion of successes is $$\hat{p}_{C}=\frac{22+30}{50+50}=0.52$$ \(\bullet\) Test statistic $$z=\frac{(0.40-0.60)-0}{\sqrt{\frac{0.52(0.48)}{100}+\frac{0.52(0.48)}{100}}}=-2.83$$ Conclude: The P-value, \(0.9977,\) is greater than \(\alpha=\) \(0.05,\) so we fail to reject the null hypothesis. There is not convincing evidence that Instructor A's pass rate is higher than Instructor B's.

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