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. A study investigated rates of fatalities among patients with serious traumatic injuries. Among 61,909 patients transported by helicopter, 7813 died. Among 161,566 patients transported by ground services, 17,775 died (based on data from "Association Between Helicopter vs Ground Emergency Medical Services and Survival for Adults With Major Trauma," by Galvagno et al., Journal of the American Medical Association, Vol. \(307,\) No. 15 ). Use a 0.01 significance level to test the claim that the rate of fatalities is higher for patients transported by helicopter. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Considering the test results and the actual sample rates, is one mode of transportation better than the other? Are there other important factors to consider?

Step by step solution

01

- Define Null Hypothesis and Alternative Hypothesis

Let the proportion of fatalities for helicopter transport be denoted as \( p_1 \) and for ground transport be \( p_2 \). The null hypothesis \( H_0 \) and alternative hypothesis \( H_1 \) are:\( H_0: p_1 \, \leq \, p_2 \)\( H_1: p_1 \, > \, p_2 \)

02

- Calculate Sample Proportions

Compute the sample proportions for each group.Helicopter: \( \, \hat{p}_1 = \, \frac{7813}{61909} \approx \, 0.126 \)Ground: \( \, \hat{p}_2 = \, \frac{17775}{161566} \approx \, 0.110 \)

03

- Compute the Test Statistic

Use the formula for the test statistic for comparing two proportions:\[ z = \, \frac{\hat{p}_1 \, - \, \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1}\,+\,\frac{1}{n_2}\right)}} \]Where \( \hat{p} \) is the combined sample proportion:\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{7813 + 17775}{61909 + 161566} \approx 0.114 \]Then the test statistic is:\[ z = \, \frac{0.126 - 0.110}{\sqrt{0.114 \times (1 - 0.114) \times \left( \frac{1}{61909} + \frac{1}{161566} \right) }} \approx 7.03 \]

04

- Determine Critical Value and P-value

Using a significance level of 0.01, the critical value for a one-tailed test is:\[ z_{critical} = 2.33 \]The P-value for the calculated test statistic is practically 0 (P < 0.01).

05

- State the Conclusion About the Null Hypothesis

Since the test statistic (7.03) is greater than the critical value (2.33), and the P-value is less than 0.01, reject the null hypothesis \( H_0 \).

06

- Provide Final Conclusion Addressing the Original Claim

At a 0.01 significance level, there is enough evidence to support the claim that the rate of fatalities is higher for patients transported by helicopter.

07

- Construct a Confidence Interval

To construct a confidence interval for the difference in proportions, use the formula:\[ (\hat{p}_1 - \hat{p}_2) \, \pm \, Z_{\alpha/2} \times \sqrt{ \frac{\hat{p}_1 (1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1 - \hat{p}_2)}{n_2} } \]At a 99% confidence level, \( Z_{\alpha/2} = 2.575 \).The confidence interval is:\[ (0.126 - 0.110) \, \pm \, 2.575 \times \sqrt{ \frac{0.126 (0.874)}{61909} + \frac{0.110 (0.890)}{161566} } \approx \, 0.016 \, \pm \, 0.007 \]This results in the interval: [0.009, 0.023]

08

- Interpret Confidence Interval

Since the confidence interval [0.009, 0.023] does not include 0, it supports the conclusion that the fatality rate is higher for helicopter-transported patients.

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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, symbolized as \(H_0\), states that there is no significant effect or difference in the context of the study. It's a default position that assumes there is no relationship between the variables.
For this exercise, we are comparing the fatality rates of patients transported by helicopter and ground services. The null hypothesis is: \(H_0: p_1 \leq p_2\). This means that we assume there is no higher fatality rate for patients transported by helicopter compared to those transported by ground.
The null hypothesis is always tested with the aim of providing evidence to accept or reject it, which helps in making informed conclusions about the data.

alternative hypothesis

Contrary to the null hypothesis, the alternative hypothesis, noted as \(H_1\), suggests that there is a significant effect or difference. It is what you aim to prove.
In our example, the alternative hypothesis is: \(H_1: p_1 > p_2\). This posits that the fatality rate for patients transported by helicopter is higher than that for those transported by ground services.
To establish the alternative hypothesis, we need to provide sufficient evidence to show that the observed effect is true and not due to random chance. This is typically done through statistical tests that measure the probability of our results assuming the null hypothesis is true.

P-value

The P-value is a measure that helps you determine the significance of your results in a hypothesis test. It's the probability of obtaining test results at least as extreme as the observed results, assuming that the null hypothesis is true.
In this problem, the P-value is calculated from the test statistic (Z-score). The smaller the P-value, the stronger the evidence against the null hypothesis. In our case, the P-value is practically zero (\(P < 0.01\)), which is less than the significance level of 0.01. This leads us to reject the null hypothesis.
Essentially, a low P-value indicates that the observed data is highly unlikely under the null hypothesis, suggesting that the alternative hypothesis is likely true.

test statistic

A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It helps to determine whether to reject the null hypothesis.
For comparing two proportions like in this exercise, we use the Z-test. The formula for the test statistic for comparing two proportions is:
\(z = \frac{\text{prop}_1 - \text{prop}_2}{\text{SE}}\)
Where SE is the standard error calculated based on both sample proportions and sizes.
In this problem, the test statistic came out to be approximately 7.03. This high value, being far exceeding the critical value of 2.33 at a 0.01 significance level, indicates that the difference in fatality rates between helicopter and ground transportation is statistically significant.

confidence interval

A confidence interval (CI) gives an estimated range of values which is likely to include the parameter of interest. It is calculated from the sample data and provides a measure of the uncertainty around the sample estimate.
To construct a CI for the difference in proportions, we use the formula:
\[\text{diff} \, \pm \, Z_{\alpha/2} \, \times \, \text{SE} \]
Where diff is the difference between sample proportions, \(Z_{\alpha/2} \) is the critical value for the desired confidence level, and SE is the standard error.
For a 99% CI in this case, the result is [0.009, 0.023], which does not include zero. This supports our conclusion that there is a statistically significant difference in fatality rates, with helicopter transport having a higher fatality rate.

fatalities rate comparison

Fatality rate comparison looks at the mortality rates between two groups to understand if there is a significant difference. In this exercise, we compare the fatality rates of patients transported by helicopter versus those transported by ground services.
The sample proportions showed that 12.6% of patients transported by helicopter died, whereas only 11% of those transported by ground services died. The hypothesis testing and confidence interval calculated both supported the conclusion that the fatality rate is indeed higher for helicopter transport.
While the statistical analysis shows a higher fatality rate for helicopter transport, it's essential to consider other influencing factors like the severity of injuries, distance to the hospital, and availability of medical facilities. Statistical conclusions should always be complemented with contextual understanding to derive meaningful insights.