1. 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 ( , E99, 5) =
X2 =

2. X2 = 10.6949 P-Value = X2cdf (10.6949, E99, 5) = 0.0578 CONDITIONS
Since the P-Value is NOT less than α = 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 ( , E99, 5) =
X2 =

4. 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 =

5. P-Value =
X2cdf (3510.3, E99, 4)
= 0
X2 =
Since the P-Value is less than α = 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

6. 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: % 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 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 is equal to that of all US adults.
Ha: The Distribution of Marital Status of US Males age is NOT equal to that of all US adults.
P-Value =
X2cdf (250.24, E99, 3) = 0
X2 =
Conclusion???

7. P-Value = X2cdf (250.24, E99, 3) = 0 CONDITIONS SRS – stated
Since the P-Value is less than α = we REJECT H0 . The Distribution of Marital Status of US Males age 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

8. 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):_________

9. 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

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

11. Homework: Page 628: #5, 8, 9

12. 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 =
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.

13. Homework: Page 628: #5, 8, 10