1. Chapter 22 Two-Sample Proportions Inference

2. Steps for doing a confidence interval: 1)State the parameter 2)Assumptions – 1) The samples are chosen randomly and were chosen independently of each other. OR Treatments were assigned randomly to individuals or objects (or vice-versa) 2) Both samples need to be large. n 1 p 1  10 and n 1 (1-p 1 )  10 and n 2 p 2  10 and n 2 (1-p 2 )  10 3) 10% rule – Both samples are less than 10% of the population. 3) Calculate the interval 4) Write a statement about the interval in the context of the problem.

3. Formula for confidence interval: Note: use p-hat when p is not known Standard error! Margin of error!

4. Example 1: At Community Hospital, the burn center is experimenting with a new plasma compress treatment. A random sample of 316 patients with minor burns received the plasma compress treatment. Of these patients, it was found that 259 had no visible scars after treatment. Another random sample of 419 patients with minor burns received no plasma compress treatment. For this group, it was found that 94 had no visible scars after treatment. What is a 95% confidence interval of the difference in proportion of people who had no visible scars between the plasma compress treatment & control group?

5. 3) The samples should be less than 10% of the population. The population should be at least 7350 burn patients which I will assume. p 1 = the true proportion of patients who had no visible scars with the plasma compress treatment State the parameters Justify the confidence interval needed (state assumptions) 1) The treatments were assigned randomly to individuals which is stated in the problem. 2) Both samples should be large. Since the conditions are satisfied a CI for proportions is appropriate. p 2 = the true proportion of patients who had no visible scars without the plasma compress treatment p 1 - p 2 = the true difference in proportion of patients who had no visible scars between the two treatments So both samples are large. Since the conditions are met a CI for the difference of proportions is appropriate.

6. Calculate the confidence interval. 95% CI Explain the interval in the context of the problem. We are 95% confident that the true difference in proportion of people who had no visible scars between the plasma compress treatment & control group is between 53.7% and 65.4%

7. Steps for a hypothesis test for the difference of two proportions: 1)Define the parameter 2)Hypothesis statements 3)Assumptions 4)Calculations (Find the p-value) 5)Conclusion, in context

8. Hypotheses Statements Null hypothesis H 0 : p 1 = p 2 or p 1 - p 2 =0 Alternative hypothesis H a : p 1 > p 2 H a : p 1 < p 2 H a : p 1 ≠ p 2

9. Assumptions 1)The samples are chosen randomly and were chosen independently of each other. OR Treatments were assigned randomly to individuals or objects (or vice- versa) 2) Both samples need to be large. n 1 p 1  10 and n 1 (1-p 1 )  10 and n 2 p 2  10 and n 2 (1-p 2 )  10 3) 10% rule – Both samples are less than 10% of the population.

10. Formula for Hypothesis test: where

11. A forest in Oregon has an infestation of spruce moths. In an effort to control the moth, one area has been regularly sprayed from airplanes. In this area, a random sample of 495 spruce trees showed that 81 had been killed by moths. A second nearby area receives no treatment. In this area, a random sample of 518 spruce trees showed that 92 had been killed by the moth. Do these data indicate that the proportion of spruce trees killed by the moth is different for these areas?

12. Parameters and Hypotheses p 1 = the true proportion of trees killed by moths in the treated area Assumptions (Conditions) Since the conditions are met, a z-test for the difference of proportions is appropriate. p 2 = the true proportion of trees killed by moths in the untreated area p 1 – p 2 = the true difference in proportion of trees killed by moths in the two areas. 3) The samples should be less than 10% of the population. The population should be at least 10130 spruce trees which I will assume. 1) Both samples must be random which is stated in the problem. 2) Both samples should be large. So both samples are large. H 0 : p 1 = p 2 H a : p 1 ≠ p 2

13. Calculations

14. Since the p-value > , I fail to reject the null hypothesis at the.05 level. Conclusion: Decision: There is not sufficient evidence to suggest that the proportion of spruce trees killed by the moth is different for these areas

15. Using the Calculator to find a 2 proportion Z-interval 1)Go to STAT, curser over to TESTS, scroll down to B:2-PropZInt 2)Enter the observed numbers (sometimes you may have to multiply np for each x value.) 3)Specify the desired confidence level, and calculate the result. 4)Example: X1 =.28 * 506 (p-hat = 28% and n = 506) [Round whole] N1 = 506 X2 =.14 * 520 (p-hat of second proportion and n of second) [Round] N2 = 520 C-level =.95 (95% confidence level) Calculate Answer: (.09101,.18948) P1 =.281 P2 =.14 N1 = 506Now explain what you see. We are 95% confident that the N2 = 520(relate back to problem)…is between 9.1% and 18.9%.

16. Using the Calculator to find a 2 Proportion Z-Test 1)Go to STAT, curser to TESTS, select 6: 2-PropZTest 2)Enter the observed numbers and sample sizes for both groups. (You may have to enter data by using np here also.) 3)Remember to indicate which inequality your alternative hypothesis test is using, then calculate. 4)Example: X1 = 318 N1 = 811 X2 = 48 N2 = 184 P1 = ≠ p2 Calculate Answer: p1 ≠ p2We interpret the result and state a conclusion. z = 3.333We see a z-score of 3.333 and the p-value of p =.0008594.00086. Such a small p-value indicates that the p1 =.3921observed difference is unlikely to be sampling p2 =.2609error.