X2 = 10.6949 Based on the following results, is the die in Warm-Up: Quality control is a HUGE part of Casino operations. Gamblers can switch out Dice for loaded Dice to give them an edge in the odds. Casinos check and test their die on an hourly basis. A die is taken to the back room where it is rolled several times. Based on the following results, is the die in questions fair? 1 2 3 4 5 6 OBS. DATA 25 33 18 41 27 EXP. DATA 29.5 X2 Goodness of Fit Test H0: The Prop. Of 1’s =2’s = … = 6’s OR The Die is Fair! Ha: The Outcomes are not Uniformly distributed OR The Die is NOT Fair. P-Value = X2cdf (10.6949, E99, 5) = 0.0578 X2 = 10.6949

X2 = 10.6949 P-Value = X2cdf (10.6949, E99, 5) = 0.0578 CONDITIONS Since the P-Value is NOT less than α = 0.05 Fail to REJECT H0 . There is insufficient evidence that the die is weighted. CONDITIONS SRS – randomly tossed All Expected Counts are 5 or greater. EXP. DATA 29.5 P-Value = X2cdf (10.6949, E99, 5) = 0.0578 X2 = 10.6949

EXAMPLE: The NY Civil Liberties Union feel that the NYC Police Dept is not hiring an ‘ethnic composition’ representing the city. NYC is 29.2% White, 28.3% Black, 31.5% Latino, 9.1% Asian, and 2% other. If the NYC Police Dept. is composed of the following, does the Union have a case? White Black Latino Asian Other OBS. DATA 8560 7120 2762 1852 560 EXP. DATA 6089.4 5901.7 6569 1897.7 417.08 H0: The Police force represents the Population of NYC. Ha: The Police force Does NOT represents the Population of NYC. X2 Goodness of Fit Test P-Value = X2cdf (3510.3, E99, 4) = 0 X2 = 3510.3

P-Value = X2cdf (3510.3, E99, 4) = 0 X2 = 3510.3 Since the P-Value is less than α = 0.05 we REJECT H0 . The hiring practice of the NYC police dept. does NOT represent the ethnic composition of NYC. CONDITIONS SRS X All Expected Counts are 5 or greater. √ EXP. DATA 6089.4 5901.7 6569 1897.7 417.08

X2 = 250.24 Is there evidence the distribution of marital status EXAMPLE: According to the Statistical Abstract of the US in 1997, the marital status distribution of the US adult population was as follows: 23.26% Never Married, 60.31% Married, 7% Widowed, and 9.43% Divorced. An SRS of 500 US Males, aged 25-29, yielded the following frequency Distribution: Is there evidence the distribution of marital status among 25-29 males differs from the US adult population? Never Married Widowed Divorced Freq. 260 220 20 EXP. DATA 116.3 301.55 35 47.15 X2 Goodness of Fit Test H0: The Distribution of Marital Status of US Males age 25-29 is equal to that of all US adults. Ha: The Distribution of Marital Status of US Males age 25-29 is NOT equal to that of all US adults. P-Value = X2cdf (250.24, E99, 3) = 0 X2 = 250.24 Conclusion???

P-Value = X2cdf (250.24, E99, 3) = 0 CONDITIONS SRS – stated Since the P-Value is less than α = 0.05 we REJECT H0 . The Distribution of Marital Status of US Males age 25-29 is Not equal to the that of all US adults. CONDITIONS SRS – stated All Expected Counts are 5 or greater EXP. DATA 116.3 301.55 35 47.15

Lay out a large sheet of paper—you’ll be sorting M&Ms on this. Open up a bag of M&Ms and split them between the members of lab group. DO NOT EAT ANY OF THE M&M’S (for now) Separate the M&M’s into color categories and count the number of each color of M&M you have. Record your counts in Data Chart 1. Determine the Chi square value and p-value for your data. Data Chart Color Categories Brown Blue Orange Green Red Yellow Total Expected %   Observed Count (O) Expected Count (E) P-Value = X2 cdf(X2, E99, df):_________

Hypothesis Name of Test Obs/Exp. Counts X2 / p-Value Check Assumptions Conclusion % color Plain Peanut Crispy Minis Peanut Butter Almond Brown 13% 12% 17% 10% Blue 24% 23% 25% 20% Orange 16% Green 15% Red Yellow 14% H0: The Distribution of Colors in the packet of “Plain” M&M’s is equal to the Percentage indicated by the Mars Company. Ha: The Distribution of Colors in the packet of “Plain” M&M’s is NOT equal to the Percentage indicated by the Mars Company. Homework: Page 628: #5a, 8, 9

Homework: Page 628: #5a, 8, 9

Homework: Page 628: #5, 8, 9

Page 628: #9 Yel/Nor Yel/Sho Ebo/Nor Ebo/Sho OBS. DATA 59 20 11 10 X2 Goodness of Fit Test Expected 56.25 18.75 6.25 H0: The distribution of traits are consistent with the genetic theory. Ha: The distribution of traits are NOT consistent with the genetic theory. X2 = 5.671 P-Value = 0.1288 Fail to reject the null hypothesis. There is no evidence that the ratio of traits is different than the theoretical ratio predicted by the genetic model.

Homework: Page 628: #5, 8, 10