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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 T F
X2
Goodness of
Fit Test
OBS.
DATA
EXP.
DATA
45 15 12 47 51
42.5
25.5
8.5
34
59.5
H0: The distribution of absences is unchanged (same)
Ha: The distribution is different
X2  
Obs  Exp 
2
X2 = 12.10
P-Value = 0.0166
Exp
Since the p-val < .05, we Reject H0. There is sufficient evidence that the
distribution of absences has changed.
CONDITIONS
1. SRS – stated √
2. All Expected Counts are 5 or greater.
9/16 = 0.5625
3/16 = 0.1875
1/16 = 0.0625
x 100
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.
H0: The two variables are Independent (NO association)
Ha: The two variables are NOT Independent (They have a
relationship)
X 
2
 Obs  Exp 
Exp
2
P-Value = X2cdf (X2, E99, df)
RowTotal  ColumnTotal
Expected Counts 
TableTotal
Do Parental smoking habits affect student behavior?
5375 RANDOM students were surveyed.
Student
Smokes
Student does not
smoke
Both Parents smoke 400 | 332.49 1380 | 1447.51
One Parent smokes
Neither smokes
Expect Counts 
1780
2239
416 | 418.22 1823 | 1820.78
188 | 253.29 1168 | 1102.71 1356
1004
4371
5375 = TT
Row Total  Cols.Total
TableTotal
Expected
H0: There is NO relationship between Parental smoking
X2 Test of
habits and student smoking habits.
Independence
Ha: There is a relationship between Parental smoking
habits and student smoking habits.
Obs  Exp 

2
X2 = 37.566
X 
2
Exp
P-Value =
X2cdf (37.566, E99, 2)
=0
Student
Smokes
Student does not
smoke
Both Parents smoke 400 | 332.49 1380 | 1447.51
One Parent smokes
Neither smokes
416 | 418.22 1823 | 1820.78
188 | 253.29 1168 | 1102.71
Obs  Exp 

2
X2 = 37.566
X 
2
Exp
P-Value =
X2cdf (37.566, E99, 2)
=0
Since the P-Value is less than α = 0.05 REJECT H0 . There is a
relationship between the smoking habits of Parents and that of
students.
CONDITIONS
1. SRS – stated √
2. All Expected Counts are 5 or greater.
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
Violent Crime
Not Full
2
17
27
3
21
19
Domestic abuse 11
14
6
Property
Drugs/Alcohol
Other offenses
9
Full Moon
Not Full
Violent Crime
2
2.558 3
2.442
Property
17
19.442 21
18.558
Drugs/Alcohol
27
23.535 19
22.465
Domestic Abuse
11
12.791 14
12.209
X2 Test of
Independence
Other offenses
9
7.674 6
7.326
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.
Obs  Exp 

X2 = 2.904

2
X
2
Exp
P-Value =
X2cdf (2.904, E99, 4)
= 0.5740
Since the P-Value is greater than α = 0.05 Fail to REJECT H0 . There
is not enough evidence that the full moon has an influence over the
types of crimes committed.
CONDITIONS
1. SRS - Stated √
2. All Expected Counts are 5 or greater. X
Homework: Page 629; 12 e-i, 13a-e, 17, 18
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
1
Relapse No Relapse Proportion
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.
X 
2
 Obs  Exp 
2
Exp
To carry out the significance test with a two table you must
calculate the Expected Values.
Expected Counts 
Group Treatment
1
2
3
Row Total  Cols.Total
TableTotal
Relapse No Relapse Proportion
Desipramine 10 | 16 14
| 8
0.583
24
18 | 16 6
20 | 16 4
48
24
| 8
| 8
0.250
0.167
24
24
Lithium
Placebo
Expected
72 = TT
X2 Test of
Independence
10
14
18
20
6
4
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.
X2  
Obs  Exp 
Exp
2
X2 = 10.5
P-Value =
X2cdf (10.5, E99, 2)
= 0.0052
Since the P-Value is less than α = 0.05 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
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: According to the Statistical Abstract of the US, 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
Never
Married Widowed Divorced
distribution of marital status
Married
among males of that
Freq.
260
220
0
20
Age differs from the
EXP.
116.3
301.55
35
47.15
DATA
US adult population?
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.
Obs  Exp 

2
X2 = 250.24
X 
2
Exp
P-Value =
X2cdf (250.24, E99, 3) = 0
Conclusion???
Since the P-Value is less than α = 0.05 the data IS significant . There
is strong evidence to REJECT H0 . The Distribution of Marital Status of
US Males age 25-29 is Not equal to the that of all US adults.
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.
EXP.
DATA
116.3
301.55
35
47.15
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% - 99.7% Rule. Determine
whether the test scores are Appr. Normally Distributed.
TEST Scores
46
67
72
72
73
84
85
86
87
90
93
94
95
96
97
98
100
100
100
100
100
100
100
100
100
100
x  89.8 s  13.7
68%
n  26
.34
.34
.135
.025
-∞
62.4
OBS.
1
EXP.
.65
76.1
4
89.8
4
95%
.135
.025
117.2
103.5
17
99.7%
0
3.51 8.84 8.84 3.51
0
.65
∞
2.5%
13.5%
34%
34%
13.5%
2.5%
OBS.
1
4
4
17
0
0
EXP.
.65
X2
Goodness of
Fit Test
3.51 8.84 8.84 3.51
.65
H0: The data follows an approximately Normal
Distribution
Ha: The data does NOT follows an approximately
Normal Distribution
Obs  Exp 

2
X2 = 14.599
X 
2
Exp
P-Value =
X2cdf (14.599, E99, 5)
= 0.0122
Since the P-Value is less than α = 0.05 the data IS significant . There
is STRONG evidence to REJECT H0 . The data is NOT approximately
Normal.
CONDITIONS
1. SRS X
2. All Expected Counts are 1 or greater. X
3. No more than 20% of the Expected Counts are less than 5. X
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
X2
Goodness of
Fit Test
X2  
3
4
5
6
OBS.
DATA
45 33 18 33 36 27
EXP.
DATA
32
32
32
32
32
32
H0: The Die is Fair!
Ha: The Die is NOT Fair.
Obs  Exp 
Exp
2
2
X2 = 12.75
P-Value =
X2cdf (12.75, E99, 5)
= 0.0258
Since the P-Value is less than α = 0.05 REJECT H0 . There is strong
evidence that the die is weighted.
CONDITIONS
1. SRS – randomly tossed
2. All Expected Counts are 1 or greater.
3. No more than 20% of the Expected Counts are less than 5.
EXP.
DATA
32
32
32
32
32
32