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.

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?ppii = The true proportion of dogs developing Lymphoma living in… = The true proportion of dogs developing Lymphoma living in…

pp11 = homes using weed killers and p = homes using weed killers and p22 = NOT using weed killers. = NOT using weed killers.

HH00: : pp11 = p = p22

HHaa: : pp11 > p > p22

.572 .146

1 1.514 1 .514

827 130

9.030z

TWO TWO Proportion Proportion

z – Testz – TestCONDITIONSCONDITIONS

1.1. SRS – Both sample data were collected randomlySRS – Both sample data were collected randomly2.2. Appr. Normal: 827· (.572) ≥ 10 AND 827 · (1 –.572) ≥ 10Appr. Normal: 827· (.572) ≥ 10 AND 827 · (1 –.572) ≥ 10

130 · (.146) ≥ 10 AND 130 · (1 –.146) ≥ 10130 · (.146) ≥ 10 AND 130 · (1 –.146) ≥ 10

Since the P-Value is less than Since the P-Value is less than αα = 0.05 we REJECT H = 0.05 we REJECT H00 . There is . There is

strong evidence to suggest that lawn weed killers are harmful to pets.strong evidence to suggest that lawn weed killers are harmful to pets.

0P Value

• HW Page 508: 3, 4, 11

1.1. SRS – The data was collected randomly for bothSRS – The data was collected randomly for both2. 1012 · (0.406) ≥ 10 AND 1012 · (1 – 0.406) ≥ 102. 1012 · (0.406) ≥ 10 AND 1012 · (1 – 0.406) ≥ 10

1062 · (0.504) ≥ 10 AND 1062 · (1 – 0.504) ≥ 101062 · (0.504) ≥ 10 AND 1062 · (1 – 0.504) ≥ 10

TWO Proportion TWO Proportion z – Confidence Intervalz – Confidence Interval

We can be 95% Confident that the We can be 95% Confident that the true true differencedifference in the proportion in the proportion of men of men and women suffering from arthritis is between -0.140 and -0.055and women suffering from arthritis is between -0.140 and -0.055

ppmm – p – pw w = Negative #= Negative #

So men suffer less than womenSo men suffer less than women

• HW Page 509: 12, 14, 16

WARM – UP More Confidence Intervals

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. ppii = The true proportion of patients who improved. = The true proportion of patients who improved.

pp11 = after surgery and p = after surgery and p22 = after wearing wrist splints. = after wearing wrist splints.

.8 1 .8 .54 1 .54

0.80 0.54 (1.960)88 88

TWO Proportion TWO Proportion

z – Confidence Intervalz – Confidence IntervalCONDITIONSCONDITIONS

1.1. SRS – Both sample data were collected randomlySRS – Both sample data were collected randomly2.2. Population of Patients having surgery ≥ 10 · (88)Population of Patients having surgery ≥ 10 · (88) Population of Patients wearing splints ≥ 10 · (88)Population of Patients wearing splints ≥ 10 · (88)3.3. 88 · (0.8) ≥ 10 AND 88 · (1 – 0.8) ≥ 1088 · (0.8) ≥ 10 AND 88 · (1 – 0.8) ≥ 10

88 · (0.54) ≥ 10 AND 88 · (1 – 0.54) ≥ 1088 · (0.54) ≥ 10 AND 88 · (1 – 0.54) ≥ 10

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.

Conclusion:

WARM - UP 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?ppii = The true proportion of students who become smokers… = The true proportion of students who become smokers…

pp11 = Parents who disapprove and p = Parents who disapprove and p22 = Parents who were lenient = Parents who were lenient

HH00: : pp11 = p = p22

HHaa: : pp11 ≠ p ≠ p22

0.1901 0.2683

1 10.2 1 0.2

284 41

1.1695z

TWO TWO Proportion Proportion

z – Testz – TestCONDITIONSCONDITIONS

1.1. SRS – Both sample data were collected randomlySRS – Both sample data were collected randomly2.2. 284 · (0.19) ≥ 10 AND 284 · (1 – 0.19) ≥ 10284 · (0.19) ≥ 10 AND 284 · (1 – 0.19) ≥ 10

41 · (0.27) ≥ 10 AND 41 · (1 – 0.27) ≥ 1041 · (0.27) ≥ 10 AND 41 · (1 – 0.27) ≥ 10

Since the P-Value is NOT less than Since the P-Value is NOT less than αα = 0.05 the data IS NOT = 0.05 the data IS NOT significant . Fail to REJECT Hsignificant . Fail to REJECT H00 . There is NO evidence to suggest that . There is NO evidence to suggest that

Parent attitudes affect teenage smoking behavior. Parent attitudes affect teenage smoking behavior.

Conclusion:

2 1.1695 2 | 1.1695 |, 99 0.2422P Value P Z normalcdf E

HH00: : pp11 = p = p22

HHaa: : pp11 ≠ p ≠ p22

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

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.

.5719 1 .5719 .1462 1 .1462

0.5719 0.1462 (1.960)827 130

TWO Proportion TWO Proportion z – Confidence Intervalz – Confidence Interval CONDITIONSCONDITIONS

1.1. SRS – Both sample data were collected randomlySRS – Both sample data were collected randomly2.2. Population of Dogs is ≥ 10 · (827)Population of Dogs is ≥ 10 · (827) Population of Dogs is ≥ 10 · (130)Population of Dogs is ≥ 10 · (130)3.3. 827 · (0.5719) ≥ 10 AND 827 · (1 – 0.5719) ≥ 10827 · (0.5719) ≥ 10 AND 827 · (1 – 0.5719) ≥ 10

130 · (0.1462) ≥ 10 AND 130 · (1 – 0.1462) ≥ 10130 · (0.1462) ≥ 10 AND 130 · (1 – 0.1462) ≥ 10

Conclusion:

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

ppii = The true proportion of the political party that supported the war. = The true proportion of the political party that supported the war.

ppII = Independents and p = Independents and pRR = Republicans = Republicans

HH00: : ppII = p = pRR

HHaa: : ppII ≠ p ≠ pRR 42 75

ˆ ˆ0.5758 0.765399 98I Rp p

TWO TWO Proportion Proportion

z – Testz – Test

ppii = The true proportion of the political party that supported the war. = The true proportion of the political party that supported the war.

ppII = Independents and p = Independents and pRR = Republicans = Republicans

HH00: : ppII = p = pRR

HHaa: : ppII ≠ p ≠ pRR

0.5758 0.7653

1 10.6701 1 0.6701

99 98

2.8290z

TWO TWO Proportion Proportion

z – Testz – Test

2 2.8290 2 2.8290, 99 0.0047P Value P Z normalcdf E

1.1. SRS – The data was collected randomlySRS – The data was collected randomly2.2. Population of Independents is ≥ 10 · (99)Population of Independents is ≥ 10 · (99) Population of Republicans is ≥ 10 · (98)Population of Republicans is ≥ 10 · (98)3.3. 99 · (0.5758) ≥ 10 AND 99 · (1 – 0.5758) ≥ 1099 · (0.5758) ≥ 10 AND 99 · (1 – 0.5758) ≥ 10

98 · (0.7653) ≥ 10 AND 98 · (1 – 0.7653) ≥ 1098 · (0.7653) ≥ 10 AND 98 · (1 – 0.7653) ≥ 10

Since the P-Value is less than Since the P-Value is less than αα = 0.05 the data IS significant . There = 0.05 the data IS significant . There is strong evidence to REJECT His strong evidence to REJECT H00 . There is evidence to suggest that . There is evidence to suggest that

Independents and Republicans supported the war differently.Independents and Republicans supported the war differently.

Since the P-Value is less than Since the P-Value is less than αα = 0.05 the data IS significant . There = 0.05 the data IS significant . There is strong evidence to REJECT His strong evidence to REJECT H00 . There is evidence to suggest that . There is evidence to suggest that

Independents and Republicans supported the war differently.Independents and Republicans supported the war differently.

1.1. SRS – The data was collected randomlySRS – The data was collected randomly2.2. Population of Independents is ≥ 10 · (99)Population of Independents is ≥ 10 · (99) Population of Republicans is ≥ 10 · (98)Population of Republicans is ≥ 10 · (98)3.3. 99 · (0.5758) ≥ 10 AND 99 · (1 – 0.5758) ≥ 1099 · (0.5758) ≥ 10 AND 99 · (1 – 0.5758) ≥ 10

98 · (0.7653) ≥ 10 AND 98 · (1 – 0.7653) ≥ 1098 · (0.7653) ≥ 10 AND 98 · (1 – 0.7653) ≥ 10

TWO PROPORTION Z – SIGNIFICANCE TEST

H0: p1 = p2

Ha: p1 >, <, or ≠ p2

1 2

1 2

ˆ ˆ

1 11p p

p pz

p pn n

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.

ppooled count of successes in BOTH samples combined

count of observations in BOTH samples combined

TWO PROPORTION ASSUMPTIONS / CONDITIONS

Same as Conf. Intervals