H0: p1 = p2 Ha: p1 > p2 WARM – UP In 1991, researchers at the National Cancer Institute released the results of a study that investigated the effects of a weed-killing herbicide on house pets. Of an SRS of 827 dogs in regular use herbicide settings, 473 of them developed lymphoma. Of an SRS of 130 dogs from homes with NO herbicide usage, 19 developed lymphoma. Is there evidence that treating lawns is harmful to pets? pi = The true proportion of dogs developing Lymphoma living in… p1 = homes using weed killers and p2 = NOT using weed killers. TWO Proportion z – Test H0: p1 = p2 Ha: p1 > p2 CONDITIONS 1. SRS – Both sample data were collected randomly 2. Appr. Normal: 827· (.572) ≥ 10 AND 827 · (1 –.572) ≥ 10 130 · (.146) ≥ 10 AND 130 · (1 –.146) ≥ 10 Since the P-Value is less than α = 0.05 we REJECT H0 . There is strong evidence to suggest that lawn weed killers are harmful to pets.

More Confidence Intervals z – Confidence Interval The painful wrist condition called Carpal Tunnel syndrome can be treated with surgery or wrist splints. In September 2002 a study of 176 random patients revealed that the half that had surgery, 80% showed improvement. Of the other half who use wrist splints, 54% showed improvement. Construct a 95% confidence interval for this difference. pi = The true proportion of patients who improved. p1 = after surgery and p2 = after wearing wrist splints. TWO Proportion z – Confidence Interval CONDITIONS 1. SRS – Both sample data were collected randomly 2. Population of Patients having surgery ≥ 10 · (88) Population of Patients wearing splints ≥ 10 · (88) 3. 88 · (0.8) ≥ 10 AND 88 · (1 – 0.8) ≥ 10 88 · (0.54) ≥ 10 AND 88 · (1 – 0.54) ≥ 10

Conclusion: We can be 95% Confident that the true difference in the proportion of patients who had surgery and showed improvement and the proportion of patients who wore wrist splints and showed improvement is between 0.116 and .384.

More with Tests A study done in 2001 ask random non-smoking teenagers about their parents attitudes about smoking. Three years later of the 284 students who said their parents disapproved of it, 54 of them were now smokers. Among the 41 students who said their parents were lenient about it, 11 became smokers. Is there evidence that parental attitudes influence teenagers’ behaviors? pi = The true proportion of students who become smokers… p1 = Parents who disapprove and p2 = Parents who were lenient TWO Proportion z – Test H0: p1 = p2 Ha: p1 ≠ p2 CONDITIONS 1. SRS – Both sample data were collected randomly 2. 284 · (0.19) ≥ 10 AND 284 · (1 – 0.19) ≥ 10 41 · (0.27) ≥ 10 AND 41 · (1 – 0.27) ≥ 10

H0: p1 = p2 Ha: p1 ≠ p2 Conclusion: Since the P-Value is NOT less than α = 0.05 the data IS NOT significant . Fail to REJECT H0 . There is NO evidence to suggest that Parent attitudes affect teenage smoking behavior.

z – Confidence Interval HW Page 508: 3, 4, 11 TWO Proportion z – Confidence Interval SRS – The data was collected randomly for both 2. 1012 · (0.406) ≥ 10 AND 1012 · (1 – 0.406) ≥ 10 1062 · (0.504) ≥ 10 AND 1062 · (1 – 0.504) ≥ 10 We can be 95% Confident that the true difference in the proportion of men and women suffering from arthritis is between -0.140 and -0.055 pm – pw = Negative # So men suffer less than women

HW Page 509: 12, 14, 16, 24

2005 Form A AP Statistic Exam Question #4 WARM – UP 2005 Form A AP Statistic Exam Question #4 Some boxes of a certain brand of breakfast cereal include a voucher for a free video rental inside the box. The company that makes the cereal claims that a voucher can be found in 20 percent of the boxes. However, based on their experiences eating this cereal at home, a group of students believe that the proportion of boxes with vouchers in less than 0.2. This group of students purchased 65 boxes of the cereal to investigate the company’s claim. The students found a total of 11 vouchers for free video rentals in the 65 boxes. Suppose it is reasonable to assume that the 65 boxes purchased by the students are a SRS of all boxes of this cereal. Based on this sample, is there support for the students’ belief that the proportion of boxes with vouchers is less than 0.2? Provide statistical evidence to support your answer.

WARM – UP 2005 Form A AP Statistic Exam Question #4

WARM – UP 2005 Form A AP Statistic Exam Question #4 - Page 12

z – Confidence Interval Chapter 22 (continued) WARM - UP In 1991, researchers at the National Cancer Institute released the results of a study that investigated the effects of a weed-killing herbicide on house pets. Of an SRS of 827 dogs in regular use herbicide settings, 473 of them developed lymphoma. Of an SRS of 130 dogs from homes with NO herbicide usage, 19 developed lymphoma. Construct a 95% Confidence Interval for the difference. TWO Proportion z – Confidence Interval CONDITIONS 1. SRS – Both sample data were collected randomly 2. Population of Dogs is ≥ 10 · (827) Population of Dogs is ≥ 10 · (130) 3. 827 · (0.5719) ≥ 10 AND 827 · (1 – 0.5719) ≥ 10 130 · (0.1462) ≥ 10 AND 130 · (1 – 0.1462) ≥ 10

Conclusion: We can be 95% Confident that the true difference in the proportion of Dogs developing lymphoma while living in homes that use the herbicide and the proportion of Dogs developing lymphoma while living in homes that use NO herbicides is between 0.356 and 0.495

EXAMPLE A random poll of conducted in 2002 asked if we should have or should not have gone to war in Iraq. 42 of 99 Independents said “Should Not have” while 75 of 98 Republicans said “Should have”. Is there evidence that there exists a significant difference in the proportion of Independent and Republicans that favored the war at that time? pi = The true proportion of the political party that supported the war. pI = Independents and pR = Republicans TWO Proportion z – Test H0: pI = pR Ha: pI ≠ pR

Population of Republicans is ≥ 10 · (98) pi = The true proportion of the political party that supported the war. pI = Independents and pR = Republicans TWO Proportion z – Test H0: pI = pR Ha: pI ≠ pR SRS – The data was collected randomly Population of Independents is ≥ 10 · (99) Population of Republicans is ≥ 10 · (98) 99 · (0.5758) ≥ 10 AND 99 · (1 – 0.5758) ≥ 10 98 · (0.7653) ≥ 10 AND 98 · (1 – 0.7653) ≥ 10 Since the P-Value is less than α = 0.05 the data IS significant . There is strong evidence to REJECT H0 . There is evidence to suggest that Independents and Republicans supported the war differently.

Population of Republicans is ≥ 10 · (98) Since the P-Value is less than α = 0.05 the data IS significant . There is strong evidence to REJECT H0 . There is evidence to suggest that Independents and Republicans supported the war differently. SRS – The data was collected randomly Population of Independents is ≥ 10 · (99) Population of Republicans is ≥ 10 · (98) 99 · (0.5758) ≥ 10 AND 99 · (1 – 0.5758) ≥ 10 98 · (0.7653) ≥ 10 AND 98 · (1 – 0.7653) ≥ 10

TWO PROPORTION Z – SIGNIFICANCE TEST H0: p1 = p2 Ha: p1 >, <, or ≠ p2 The Null Hypothesis states that there is NO Difference between the population proportions. If this is true then the observations really come from a single population, so instead of p1 and p2 separately we use pp = (p-pooled) in the test. TWO PROPORTION ASSUMPTIONS / CONDITIONS Same as Conf. Intervals