91Ó°ÊÓ

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. In a study of treatments for very painful "cluster" headaches, 150 patients were treated with oxygen and 148 other patients were given a placebo consisting of ordinary air. Among the 150 patients in the oxygen treatment group, 116 were free from headaches 15 minutes after treatment. Among the 148 patients given the placebo, 29 were free from headaches 15 minutes after treatment (based on data from "High-Flow Oxygen for Treatment of Cluster Headache," by Cohen, Burns, and Goads by, Journal of the American Medical Association, Vol. \(302,\) No. 22 ). We want to use a 0.01 significance level to test the claim that the oxygen treatment is effective. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, is the oxygen treatment effective?

Step by step solution

01

Identify the Hypotheses

The null hypothesis (H_0) is that the oxygen treatment is not more effective than the placebo (P_1 = P_2). The alternative hypothesis (H_a) is that the oxygen treatment is more effective than the placebo (P_1 > P_2).

02

Calculate the Test Statistic

Calculate the sample proportions: \[ \hat{p_1} = \frac{116}{150} = 0.7733 \] \[ \hat{p_2} = \frac{29}{148} = 0.1959 \]Next, calculate the combined sample proportion: \[ \hat{p} = \frac{116+29}{150+148} = 0.4865 \]Then, compute the test statistic using: \[ z = \frac{(\hat{p_1} - \hat{p_2})}{\sqrt{ \hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2}) }} = \frac{0.7733 - 0.1959}{\sqrt{0.4865(1-0.4865)(\frac{1}{150} + \frac{1}{148})}} = 9.21 \]

03

Determine Critical Value and P-Value

Using a 0.01 significance level, the critical value for a one-tailed test is z_c = 2.33. The P-value corresponding to a z value of 9.21 is much less than 0.01.

04

Decision on Null Hypothesis

Since the test statistic ( z = 9.21) is greater than the critical value ( z_c=2.33) and the P-value is less than 0.01, reject the null hypothesis.

05

Construct the Confidence Interval

Construct a 99% confidence interval for the difference in proportions: \[ (\hat{p_1} - \hat{p_2}) ± z_c \sqrt{ \frac{\hat{p_1}(1-\hat{p_1})}{n_1} + \frac{\hat{p_2}(1-\hat{p_2})}{n_2} } \] \[ 0.7733 - 0.1959 ± 2.33 \sqrt{ \frac{0.7733(1-0.7733)}{150} + \frac{0.1959(1-0.1959)}{148} } \] \[ 0.5774 ± 0.1377 \] So, the 99% confidence interval is (0.4397, 0.7151).

06

Conclusion

Since the confidence interval does not include 0, and we rejected the null hypothesis, conclude that the oxygen treatment is effective in reducing the number of headaches.

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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, often denoted as \( H_0 \), is a statement positing that there is no effect or no difference. It is the position of no change or status quo. For the given exercise, the null hypothesis is that the oxygen treatment is not more effective than the placebo. Mathematically, this can be written as \( P_1 = P_2 \), meaning that the proportion of patients relieved by the oxygen treatment is equal to the proportion relieved by the placebo.
The purpose of the null hypothesis is to provide a baseline or standard against which the evidence is measured. It is typically assumed true until evidence suggests otherwise. This hypothesis is tested against the alternative hypothesis, and the outcome of this test helps determine whether to reject or fail to reject the null hypothesis.

alternative hypothesis

The alternative hypothesis, designated as \( H_a \) or \( H_1 \), contrasts the null hypothesis. It asserts that there is an effect or a difference. In the exercise, the alternative hypothesis is that the oxygen treatment is more effective than the placebo. Formally, this can be written as \( P_1 > P_2 \), indicating that the proportion of patients obtaining relief from the oxygen treatment is higher than that from the placebo.
The alternative hypothesis is the statement we aim to support through our data. By establishing \( H_a \), we seek to demonstrate that the observed difference in treatment effectiveness is statistically significant and not due to random chance.

test statistic

A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It allows you to determine how far the observed data is from the null hypothesis. In the solution, the test statistic is calculated using the formula:
\[ z = \frac{(\hat{p_1} - \hat{p_2})}{\sqrt{ \hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2}) }} = 9.21 \]
Here, \( \hat{p_1} \) and \( \hat{p_2} \) are the sample proportions, initially computed as 0.7733 and 0.1959 respectively. \( \hat{p} \) represents the combined sample proportion, which is 0.4865. The test statistic tells us how far our sample statistic is from the null hypothesis.
A higher value of the test statistic indicates that the observed effect is less likely to be due to chance, leading us to consider rejecting the null hypothesis in favor of the alternative hypothesis.

P-value

The P-value is the probability of obtaining test results at least as extreme as the observed data, assuming that the null hypothesis is true. It quantifies the evidence against the null hypothesis. In this exercise, the P-value corresponding to a test statistic of 9.21 is very small, much less than the significance level of 0.01.
A smaller P-value signals stronger evidence against the null hypothesis. Typically, if the P-value is less than the chosen significance level (e.g., 0.01), we reject the null hypothesis. In this case, the very small P-value indicates a highly significant result, leading to the conclusion that the oxygen treatment is effective.

confidence interval

A confidence interval provides a range of values, derived from the sample data, that is likely to contain the population parameter. For the given exercise, a 99% confidence interval for the difference in proportions \( (P_1 - P_2) \) was constructed using:
\[ (\hat{p_1} - \hat{p_2}) ± z_c \sqrt{ \frac{\hat{p_1}(1-\hat{p_1})}{n_1} + \frac{\hat{p_2}(1-\hat{p_2})}{n_2} } \]Here, the confidence interval computed is (0.4397, 0.7151). This interval does not include 0, suggesting a significant difference between the treatment groups.
A confidence interval that does not contain the null value (in this context, 0) supports the rejection of the null hypothesis. It provides a range for the true difference in efficacy between the oxygen treatment and the placebo, further supporting the conclusion that the oxygen treatment is effective.