Section 12.3 Investigation 3.7: CPR vs. Chest Compressions
For many years, if a person experienced a heart attack and a bystander called 911, the dispatcher instructed the bystander in how to administer chest compression plus mouth-to-mouth ventilation (a combination known as CPR) until the emergency team arrived. Some researchers believe that giving instruction in chest compression alone (CC) would be a more effective approach.
In the 1990s, a randomized comparative experiment was conducted in Seattle involving 518 cases (Hallstrom, Cobb, Johnson, & Copass, New England Journal of Medicine, 2000): In 278 cases, the dispatcher gave instructions in standard CPR to the bystander, and in the remaining 240 cases the dispatcher gave instructions in CC alone. A total of 64 patients survived to discharge from the hospital: 29 in the CPR group and 35 in the CC group.
Checkpoint 12.3.1. Identify Study Components.
Checkpoint 12.3.2. Construct Two-Way Table.
Checkpoint 12.3.5. Calculate Difference.
Aside: Two-way Table Applet.
Checkpoint 12.3.6. Fisher’s Exact Test.
Use technology to carry out Fisher’s Exact Test (by calculating the corresponding hypergeometric probability) to assess the strength of evidence that the probability of survival is higher with CC alone as compared to standard CPR. Write out how to calculate this probability, report the p-value, and interpret what it is the probability of.
Let \(X\) represent
p-value = \(P(X\) ) =
Interpretation:
Solution.
Let \(X\) represent the number of survivors in the CPR group. Note, we are assessing the strength of evidence that the probability of survival is higher with CC alone.
Interpretation: If the two treatments were equally effective, there would be about a 9.7% chance of getting 29 or fewer survivors in the CPR group of 278 patients when randomly assigning all 518 patients to the two treatment groups.
Example 12.3.7. Applet output.
Example 12.3.8. R output.
Example 12.3.9. JMP output.
Because the sample sizes are large in this study, you should not be surprised that the probability distribution in checkpoint 6 is approximately normal. The large sample sizes allow us to approximate the hypergeometric distribution with a normal distribution. Thus, with large samples sizes (e.g., at least 5 successes and at least 5 failures in each group), an alternative to Fisher’s Exact Test is the two-sample z-test that you studied in Section 3.1.
Aside: Theory-Based Inference Applet.
Checkpoint 12.3.10. Two-Sample z-test.
Checkpoint 12.3.14. Improving Approximation.
Checkpoint 12.3.15. Optional: Continuity Correction.
Solution.
To make sure the probability mass at 29 is included, we could consider finding the probability for something just above 29 (e.g., 29.5 and 34.5) or applying an Adjusted Wald correction (adding one success and one failure to each sample), though that seems to work better for the confidence interval performance than matching the p-value to the exact. In this case, the adjusted Wald moves the statistic further from zero but with a larger se so the p-value doesn’t change much.
The p-value from the normal distribution is a little low but the continuity correction makes the approximation much more accurate.
Example 12.3.16. Applet.
Example 12.3.17. R.
Example 12.3.18. JMP.
Checkpoint 12.3.19. Significance at Different Levels.
Aside: Theory-Based Inference Applet.
Checkpoint 12.3.20. Confidence Interval.
An advantage to using the z-procedures is being able to easily produce a confidence interval for the parameter. Use technology to determine a 95% confidence interval for the parameter of interest, and then interpret this interval.
Solution.
Using the two-sample z-interval:
95% CI for \(\pi_{CPR} - \pi_{CC}\text{:}\) (-0.0988, 0.0158)
I’m 95% confident that the survival probability is up to 0.0988 higher for the CC group but could be up to 0.0158 higher for the CPR group.
Example 12.3.21. Applet (with Adjusted Wald).
Example 12.3.22. R.
Example 12.3.23. JMP (with Adjusted Wald).
Checkpoint 12.3.24. Reversing the Subtraction.
Solution.
(i) The observed statistic would now be positive 0.042
(ii) The standardized statistic would now be positive (but same magnitude)
(iii) The alternative hypothesis would now be \(\pi_{CC} - \pi_{CPR} > 0\)
(iv) The p-value, now the probability of a statistic above 0.042, would be the same
(v) The endpoints of the confidence interval will reverse sign
Note: Even with a 90% confidence level, zero is captured in the interval, even though we rejected the null hypothesis for a 10% level of significance. This can happen when you compare a one-sided p-value to a confidence interval which corresponds to a two-sided p-value. The other issue is even 0.0896 is not all that large of a value, so even if you consider the difference statistically significant, it’s not clear whether the CC treatment has a "large" benefit.
Subsection 12.3.1 Practice Problem 3.7A
Researchers in the CPR study also examined other response variables. For example, the 911 dispatcher’s instructions were completely delivered in 62% of episodes assigned to chest compression plus mouth-to-mouth compared to 81% of the episodes assigned to chest compression alone.
Checkpoint 12.3.25. Difference in Proportions.
Checkpoint 12.3.26. p-value Comparison.
Subsection 12.3.2 Practice Problem 3.7B
The above study was operationally identical to that of another study and the results of the two studies were combined. Of the 399 combined patients randomly assigned to standard CPR, 44 survived to discharge from the hospital. Of the 351 combined patients randomly assigned to chest compression alone, 47 survived to discharge.
Checkpoint 12.3.27. Combined Study Difference.
Checkpoint 12.3.28. Combined Study p-value.
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