1. X2 = 12.10 M T W F 45 15 12 47 51 P-Value = 0.0166 CONDITIONS
Warm-Up: A large school had a attendance problem in which 25% of the students were absent on Mondays, 15% on Tue., 5% on Wed, 20% on Thr. and 35% on Fridays. A new incentive plan was introduced to encourage attendance. Random weeks were selected and the absences are displayed. Did the plan change the distribution of absences? M T W F
OBS.
DATA 45 15 12 47 51
X2 Goodness of Fit Test
EXP.
DATA
42.5
25.5
8.5 34
59.5
H0: The distribution of absences is unchanged (same)
Ha: The distribution is different
X2 = 12.10
P-Value =
Since the p-val < .05, we Reject H0. There is sufficient evidence that the distribution of absences has changed.
CONDITIONS
SRS – stated √
All Expected Counts are 5 or greater.

3. 9/16 =
3/16 = x 100
1/16 =

4. df = (#Rows – 1) x (#Columns – 1)
Chapter 26 (continued)
The Chi-Square Test of
INDEPENDENCE
A test of whether TWO categorical variables are independent from each other. In other words, it examines the counts from a sample for evidence of an association between two categorical variables.
df = (#Rows – 1) x (#Columns – 1)
TWO-WAY TABLES
Two way tables are used to show a relationship among two Categorical Variables. Each Cell shows the counts of individuals that fit into both categories.

5. H0: The two variables are Independent (NO association)
Ha: The two variables are NOT Independent (They have a relationship)
P-Value = X2cdf (X2, E99, df)

6. Do Parental smoking habits affect student behavior?
5375 RANDOM students were surveyed.
Student
Smokes
Student does not smoke
Both Parents smoke
400
1380
One Parent smokes
416
1823
Neither smokes
188
1168 | |
1780
2239 | |
1356 | |
1004
4371
5375 = TT
Expected
H0: There is NO relationship between Parental smoking habits and student smoking habits.
Ha: There is a relationship between Parental smoking habits and student smoking habits.
X2 Test of Independence
P-Value =
X2cdf (37.566, E99, 2)
= 0
X2 =

7. Student
Smokes
Student does not smoke
Both Parents smoke
400
1380
One Parent smokes
416
1823
Neither smokes
188
1168 | | | | | |
P-Value =
X2cdf (37.566, E99, 2)
= 0
X2 =
Since the P-Value is less than α = REJECT H0 . There is a relationship between the smoking habits of Parents and that of students.
CONDITIONS
SRS – stated √
All Expected Counts are 5 or greater.

8. EXAMPLE: Some people believe that the full moon elicits unusual behavior in people. The following table shows the number of arrests in a town during weeks of 6 full moons and 6 other randomly selected weeks. Is there evidence that the phase of the moon is associated with degree of the crime?
Full Moon
Not Full
Violent Crime 2 3
Property 17 21
Drugs/Alcohol 27 19
Domestic abuse 11 14
Other offenses 9 6

9. X2 Test of Independence X2 = 2.904 P-Value = X2cdf (2.904, E99, 4)
Full Moon
Not Full
Violent Crime 2 3
Property 17 21
Drugs/Alcohol 27 19
Domestic Abuse 11 14
Other offenses 9 6
2.558
2.442
19.442
18.558
23.535
22.465
12.791
12.209
7.674
7.326
X2 Test of Independence
H0: A Full Moon is independent to the type of crime that is committed.
Ha: A Full Moon has an association to the type of crime that is committed.
P-Value =
X2cdf (2.904, E99, 4) =
X2 = 2.904
Since the P-Value is greater than α = Fail to REJECT H0 . There is not enough evidence that the full moon has an influence over the types of crimes committed.
CONDITIONS
SRS - Stated √
All Expected Counts are 5 or greater. X

10. Homework: Page 629; 12 e-i, 13a-e, 17, 18

14. Chapter 26 - The Chi-Square Test of INDEPENDENCE
EXAMPLE: The following example examines the relationship between the Drug treatment and Relapse status for an SRS of chronic cocaine users seeking help to stay off cocaine.
Group
Treatment
Relapse
No Relapse
Proportion 1
Desipramine 10 14
0.583 2
Lithium 18 6
0.250 3
Placebo 20 4
0.167
H0: Drug treatment is independent of a patient relapses or not.
Ha: There is an association of drug treatment to end
result. Not all treatments have the same result.

15. To carry out the significance test with a two table you must calculate the Expected Values.
Group
Treatment
Relapse
No Relapse
Proportion 1
Desipramine 10 14
0.583 2
Lithium 18 6
0.250 3
Placebo 20 4
0.167
| 16
| 8 24
| 16
| 8 24
| 16
| 8 24 48
Expected 24
72 = TT

16. X2 Test of Independence X2 = 10.5 10 14 18 6 20 4 P-Value =
H0: Drug treatment independent to whether a patient relapses or not.
Ha: There is an association of drug treatment to end result. Not all treatments have the same result.
P-Value =
X2cdf (10.5, E99, 2) =
X2 = 10.5
Since the P-Value is less than α = the data IS significant . There is strong evidence to REJECT H0 . There is strong evidence that suggests that the drug treatment is associated to relapse. (The treatments differ in terms of their effect.)
CONDITIONS
SRS - Stated √
All Expected Counts are 1 or greater. √
No more than 20% of the Expected Counts are less than 5. √

17. X2 = 250.24 Is there evidence the distribution of marital status
WARM-UP: According to the Statistical Abstract of the US, 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 of that
Age 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???

18. All Expected Counts are 1 or greater.
Since the P-Value is less than α = the data IS significant . There is strong evidence to 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 1 or greater.
No more than 20% of the Expected Counts are less than 5.
EXP.
DATA
116.3
301.55 35
47.15

19. EXAMPLE: To determine whether or not Data is Approximately Normal we have examined Bell shaped Curves in Histograms and Symmetry in Box Plots. You can also use a Chi-Squared GOF Test with the 68% - 95% % Rule. Determine whether the test scores are Appr. Normally Distributed.
TEST Scores 46 67 72 73 84 85 86 87 90 93 94 95 96 97 98
100
68%
95%
99.7%
.34
.34
.135
.135
.025
.025
- ∞ ∞
OBS. 1 4 17
EXP.
.65
3.51
8.84

20. 2.5%
13.5%
34%
OBS. 1 4 17
EXP.
.65
3.51
8.84
H0: The data follows an approximately Normal Distribution
Ha: The data does NOT follows an approximately Normal Distribution
X2 Goodness of Fit Test
P-Value =
X2cdf (14.599, E99, 5) =
X2 =
Since the P-Value is less than α = the data IS significant . There is STRONG evidence to REJECT H0 . The data is NOT approximately Normal.
CONDITIONS
SRS X
All Expected Counts are 1 or greater. X
No more than 20% of the Expected Counts are less than 5. X

21. Warm-Up: You suspect that a die at a casino craps table has been switched out with a weighted die by a cheating gambler.
Based on the following results, is the die in question fair? 1 2 3 4 5 6
OBS.
DATA 45 33 18 36 27
EXP.
DATA 32
X2 Goodness of Fit Test
H0: The Die is Fair!
Ha: The Die is NOT Fair.
P-Value =
X2cdf (12.75, E99, 5) =
X2 = 12.75

22. All Expected Counts are 1 or greater.
Since the P-Value is less than α = REJECT H0 . There is strong evidence that the die is weighted.
CONDITIONS
SRS – randomly tossed
All Expected Counts are 1 or greater.
No more than 20% of the Expected Counts are less than 5.
EXP.
DATA 32