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# Confidence Interval for the Difference of 2 Proportions

Confidence Interval for the Difference of 2 Proportions. Problem 12.22 In-line Skaters.

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## Confidence Interval for the Difference of 2 Proportions

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1. Confidence Interval for the Difference of 2 Proportions Problem 12.22 In-line Skaters

2. Problem: A study of the injuries to in-line skaters used data from the National Electronic Injury Surveillance System, which collects data from a random sample of hospital emergency rooms. In the six-month study period, 206 people came to the sample hospitals with injuries from in-line skating. We can think of these people as an SRS of all people injured while skating. Researchers were able to interview 161 of these people. Wrist injuries (mostly fractures) were the most common. The interviews found that 53 people were wearing wrist guards and 6 of these had wrist injuries. Of the 108 who did not wear wrist guards, 45 had wrist injuries. Find a 95% confidence interval for the difference between the two population proportions of wrist injuries.

3. Step 1: Step 2: Z interval for 2 proportions Assumptions: • The original sample was random, as stated in the problem. Dividing this sample into two groups based on whether or not the injured person had worn wrist guards should not prevent our samples from being independent. • The population of injured in-line skaters wearing wrist guards is at least 10 times the sample size. Likewise, for the population of injured in-line skaters not wearing wrist guards.

4. Let Pop1= in-line skaters with injuries who did not wear wrist guards, and Pop2= those who did wear wrist guards For n1: For n2: Recall that with two proportions, np and n(1-p) should be greater than 5, rather than 10. Note that there are 4 combinations to check.

5. Step 3: or

6. We are 95% confident that the true difference in the proportions of wrist injuries between Step 4: persons who did not and those who did wear wrist guards among in-line skaters with injuries seen in emergency rooms is 18% and 43%. That is, among in-line skaters seen in emergency rooms those not wearing wrist guards were between 18% and 43% more likely to have wrist injuries.

7. Step 1: H0: p1= p2 The difference, in the proportions of wrist injuries (among in-line skaters with injuries seen in an emergency room) between those wearing and not wearing wrist guards is zero. Problem: You are selling wrist guards and would like to make the claim that wrist guards prevent injuries. Perform a test and make the strongest statement that you can based on this data. Ha: p1 > p2 The difference, in the proportions of wrist injuries (among in-line skaters with injuries seen in an emergency room) between those not wearing and wearing wrist guards, is positive. (where pop1= in-line skaters with injuries who did not wear wrist guards, and pop2= those who did wear wrist guards)

8. Step 2: Assumptions: • We have independent random samples, as stated. • The populations are more than 10 times the sample sizes. • np and n(1-p) are greater than 5 for both populations (previously shown).

9. Step 4: The area to shade is so small that it cannot be seen on this scale.

10. Step 5: Step 6: Step 7: We have evidence that the rate of wrist injuries is lower when wearing wrist guards among persons who will be injured while in-line skating. This analysis was made of persons injured seriously enough to seek treatment at a hospital emergency room. P-value = P(z >3.889) = .0000503 Reject H0, a value this extreme will rarely occur by chance alone.

11. THE END

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