1. WARM UP Pick the 4 papers from back of room and then check your answer to 12 (d-i)

2. WARM-UP: Examine whether the grade you got on the AP Statistics Test is independent of what class period you are in. ABCF 1 st Period5245 3 rd Period12551 4 th Period16843 X 2 Test of Independence P-Value = X 2 cdf (8.75, E99, 6) = 0.1881 H 0 : Test Grades and Class Period are independent. H a : The Grade you earn on the Test is associated to what class period you are in. X 2 = 8.751 7.53.43.02.1 10.84.94.33.0 14.66.65.84.0 Since the P-Value is NOT less than α = 0.05 you will Fail to reject H 0. There is no evidence to conclude that Test Grades and Class Period are related. 1.SRS - Stated X 2.All Expected Counts are 1 or greater. √ 3.No more than 20% of the Expected Counts are less than 5. X

3. The Chi-Square Test for Homogeneity A test comparing the distribution of counts for TWO or MORE Populations on the ONE categorical variable. -GOF tests only one Population on only ONE categorical variable. -Homogeneity represents Multiple GOF tests. df = (#Rows – 1) x (#Cols. – 1) H 0 : The distribution of the one variable is equivalent among the populations H a : The distribution of the one variable is NOT equivalent among the populations

4. P-Value = X 2 cdf (X 2, E99, df) NOTE: The Chi-Square Tests for Homogeneity and for Independence are performed exactly the same way!

5. WARM-UP: Examine whether the distribution of grades is equivalent for each period of AP Statistics. X 2 Test of HOMOGENEITY P-Value = X 2 cdf (8.75, E99, 6) =.1881 H 0 : The distribution of A,B,C, and F’s in all three periods is spread equivalently. H a : The distribution of A,B,C, and F’s in all three periods is NOT spread equivalently. X 2 = 8.75 Since the P-Value is NOT less than α = 0.05 we fail to reject H 0. There is no evidence to conclude that Test Grades are NOT distributed equally among the 3 classes. 1.SRS - Stated X 2.All Expected Counts are 1 or greater. √ 3.No more than 20% of the Expected Counts are less than 5. X ABCF 1 st Period5245 3 rd Period12551 4 th Period16843 7.53.43.02.1 10.84.94.33.0 14.66.65.84.0

6. An SRS of 120 voters from AR and an SRS of 115 voters from TX was taken to determine whether there was a significant difference in how people, as of that moment, would vote with regards to Obama. Definitely Would Mostly Likely Probably would Not Definitely Would Not Arkansas 35452812 Texas 30381730 X 2 = 11.277 X 2 Test ofHomogeneity P-Value = X 2 cdf (11.277, E99, 3) = 0.0103 H 0 : The Distribution of how people would vote today in the State of Arkansas is equal to that of Texas. H a : The Distribution of how people would vote today in the State of Arkansas is NOT equal to that of Texas. 33.1942.2822.9821.45 31.8140.6222.0220.55 Since the P-Value is less than α = 0.05 the data IS significant. REJECT H 0. Support is different between AR and TX. 1.SRS – stated 2.All Expected Counts are 1 or greater. 3.No more than 20% of the Expected Counts are less than 5.

7. #18 Medical researchers followed an SRS of 6272 Swedish men for 30 years to see if there was an association between the amount of fish in their diet and Prostate Cancer. Is there any evidence of such an association? Fish Consumption Total Subjects Prostate Cancer Never12414 Small part of diet2621201 Moderate part2978209 Large part54942 NO Prostate Cancer14110 2012420 2092769 42507 9.21114.79 194.742426.26 221.262756.74 40.79508.21 X 2 Test ofIndependence P-Value = X 2 cdf (3.677, E99, 3) = 0.2985 H 0 : There is NO relationship between fish consumption and the development of Prostate Cancer. H a : There is relationship between fish consumption and the development of Prostate Cancer. X 2 = 3.677

8. Fish Consumption Total Subjects Prostate Cancer Never12414 Small part of diet2621201 Moderate part2978209 Large part54942 NO Prostate Cancer14110 2012420 2092769 42507 9.21114.79 194.742426.26 221.262756.74 40.79508.21 X 2 Test ofIndependence P-Value = X 2 cdf (3.677, E99, 3) = 0.2985 H 0 : There is NO relationship between fish consumption and the development of Prostate Cancer. H a : There is relationship between fish consumption and the development of Prostate Cancer. X 2 = 3.677 Since the P-Value is NOT less than α = 0.05 there is NO evidence to reject H 0. There is NO relationship between fish consumption and Prostate Cancer. CONDITIONS 1.SRS - Stated √ 2.All Expected Counts are 1 or greater. √ 3.No more than 20% of the Expected Counts are less than 5. √ WARM – UP Medical researchers followed 6272 Swedish men for 30 years to see if there was an association between the amount of fish in their diet and Prostate Cancer. Is there any evidence of such an association?

9. WARM – UP Does ones regional location have an affect on their Political affiliation? To begin to investigate this situation data from 177 voters was analyzed. DemocratRepublican West3927 Northeast3515 Southeast1744 Political Affiliation Location a.) Find the Proportion of Democrats in each region. Democrats in each region. b.) Make a Bar Chart for the Prop. c.) Find the Expected Values for each cell. each cell. 0.591 0.700 0.279 % of Dem. 0 50 100 N NW SE Regional Location 32.07 24.29 29.64 33.93 25.71 31.36

10. Homework: Page 630: #15 omit h, 16, 17, 20, 21

13. EXAMPLE: Is the Distribution of colors in a package of PLAIN M&M’s statistically equivalent to the Distribution of colors in a package of PEANUT M&M’s? A random package of plain and peanut M&M’s are selected and analyzed. BrownBlueOrangeGreenRedYellow PLAIN432705 PEANUT132211 X 2 Test of Homogeneity P-Value = X 2 cdf (4.967, E99, 5) = 0.4200 H 0 : The Distribution of colors in the Plain Packet of M&M’s is equivalent to that of the Peanut M&M’s. H a : The Distribution of colors in the Plain Packet of M&M’s is NOT equivalent to that of the Peanut M&M’s. X 2 = 4.967 3.394.062.716.100.684.06 1.611.941.292.900.321.94

14. EXAMPLE: Is the Distribution of colors in a package of PLAIN M&M’s statistically equivalent to the Distribution of colors in a package of PEANUT M&M’s? A random package of plain and peanut M&M’s are selected and analyzed. BrownBlueOrangeGreenRedYellow PLAIN432705 PEANUT132211 X 2 Test of Homogeneity P-Value = X 2 cdf (4.967, E99, 3) = 0.4200 H 0 : The Distribution of colors in the Plain Packet of M&M’s is equivalent to that of the Peanut M&M’s. H a : The Distribution of colors in the Plain Packet of M&M’s is NOT equivalent to that of the Peanut M&M’s. X 2 = 4.967 Since the P-Value is NOT less than α = 0.05 there is NO evidence to reject H 0. No evidence that the Distributions are NOT equivalent. Although the results are uncertain. CONDITIONS 1.SRS - Stated √ 2.All Expected Counts are 1 or greater. X 3.No more than 20% of the Expected Counts are less than 5. X 3.394.062.716.100.684.06 1.611.941.292.900.321.94