Transcript PPT Slides for Analysis of Count Data

Analysis of Count Data Chapter 26


Goodness of fit
Formulas and models for two-way tables
- tests for independence
- tests of homogeneity
Example 1: Car accidents and day of the week
A study of 667 drivers who were using a cell phone when they were involved
in a collision on a weekday examined the relationship between these
accidents and the day of the week.
Are the accidents equally likely to occur on any day of the working week?
Example 2: M & M Colors


Mars, Inc. periodically changes the M&M (milk chocolate) color
proportions. Last year the proportions were:
yellow 20%; red 20%, orange, blue, green 10% each; brown 30%
In a recent bag of 106 M&M’s I had the following numbers of each
color:
Yellow
29 (27.4%)

Red
Orange
23 (21.7%) 12 (11.3%)
Blue
Green
Brown
14 (13.2%)
8 (7.5%)
20 (18.9%)
Is this evidence that Mars, Inc. has changed the color distribution of
M&M’s?
Example 3: Are successful people more likely to be
born under some astrological signs than others?



256 executives of Fortune
400 companies have
birthday signs shown at
the right.
There is some variation in
the number of births per
sign, and there are more
Pisces.
Can we claim that
successful people are
more likely to be born
under some signs than
others?
Births
Sign
23
Aries
20
Taurus
18
Gemini
23
Cancer
20
Leo
19
Virgo
18
Libra
21
Scorpio
19
Sagittarius
22
Capricorn
24
Aquarius
29
Pisces
To answer these questions we use the chi-square
goodness of fit test
Data for n observations on a categorical variable with k possible
outcomes are summarized as observed counts, n1, n2, . . . , nk in k
cells.
2 hypotheses: null hypothesis H0 and alternative hypothesis HA
H0 specifies probabilities p1, p2, . . . , pk for the possible outcomes.
HA states that the probabilities are different from those in H0
The Chi-Square Test Statistic
The Chi-square test statistic is:
(Obs  Exp)
  
Exp
all cells
2
2
 where:
Obs = observed frequency in a particular cell
Exp= expected frequency in a particular cell if H0 is true
The expected frequency in cell i is npi
The Chi-Square Test Statistic (cont.)
The χ2 test statistic approximately follows a chi-squared
distribution with k-1 degrees of freedom, where k is the
number of categories.
 If the χ2 test statistic is large, this is evidence against
the null hypothesis.
2
(
Obs

Exp
)
2  
Exp
all cells

Decision
Rule:
2
2
χ

χ
.05 ,reject H0,
If
otherwise, do not reject
H0.
.05
0
Do not
reject H0
Reject H0
2.05
2
Car accidents and day of the week
H0 specifies that all days are equally likely for
car accidents  each pi = 1/5.
The expected count for each of the five days is npi = 667(1/5) = 133.4.
2
2
(count
133.4)
(observed
expected)
day
2  

 8.49
expected
133.4
Following the chi-square distribution with 5 − 1 = 4 degrees of freedom.
p
df
0.25
0.2
0.15
0.1
0.05
0.025
0.02
0.01
0.005 0.0025
0.001 0.0005
1
1.32
1.64
2.07
2.71
3.84
5.02
5.41
6.63
7.88
9.14
10.83
12.12
2
2.77
3.22
3.79
4.61
5.99
7.38
7.82
9.21
10.60
11.98
13.82
15.20
3
4.11
4.64
5.32
6.25
7.81
9.35
9.84
11.34
12.84
14.32
16.27
17.73
4
5.39
5.99
6.74
7.78
9.49
11.14
11.67
13.28
14.86
16.42
18.47
20.00
5
6.63
7.29
8.12
9.24
11.07
12.83
13.39
15.09
16.75
18.39
20.51
22.11
6
7.84
8.56
9.45
10.64
12.59
14.45
15.03
16.81
18.55
20.25
22.46
24.10
Since
the
of the12.02
test statistic
is less
than18.48
the table
7
9.04value
9.808.49
10.75
14.07
16.01
16.62
20.28value
22.04of 9.49,
24.32 we
26.02
8
10.22
11.03
12.03
13.36
15.51
17.53
18.17
20.09
21.95
23.77
26.12
27.87
do9 not 11.39
reject H
0
12.24
13.29
14.68
16.92
19.02
19.68
21.67
23.59
25.46
27.88
29.67
10
12.55
13.44
14.53
15.99
18.31
20.48
21.16
23.21
25.19
27.11
29.59
31.42
11There
is
no
significant
evidence
of
different
car
accident
rates
for
different
13.70
14.63
15.77
17.28
19.68
21.92
22.62
24.72
26.76
28.73
31.26
33.14
12
14.85
15.81
16.99
18.55
21.03
23.34
24.05
26.22
28.30
30.32
32.91
34.82
weekdays
when
the
driver
was
using
a
cell
phone.
13
15.98
16.98
18.20
19.81
22.36
24.74
25.47
27.69
29.82
31.88
34.53
36.48
Using software
The chi-square function in Excel does not compute expected counts
automatically but instead lets you provide them. This makes it easy to test for
goodness of fit. You then get the test’s p-value—but no details of the X2
calculations.
=CHITEST(array of actual values, array of expected values)
with values arranged in two similar r * c tables
--> returns the p value of the Chi Square test
Note: Many software packages do not provide a direct way to compute the chisquare goodness of fit test. But you can find a way around:
Make a two-way table where the first column contains k cells with the
observed counts. Make a second column with counts that correspond
exactly to the probabilities specified by H0, using a very large number of
observations. Then analyze the two-way table with the chi-square function.
Example 2: M & M Colors

H0 : pyellow=.20, pred=.20, porange=.10, pblue=.10, pgreen=.10, pbrown=.30
Yellow
Red
Orange
Blue
Green
Brown
Total
Obs.
29
23
12
14
8
20
106
Exp.
21.2
21.2
10.6
10.6
10.6
31.8
106

Expected yellow = 106*.20 = 21.2, etc. for other expected counts.
(Obs  Exp ) 2 (29  21.2) 2 (23  21.2) 2
  



Exp
21.2
21.2
all cells
2
(12  10.6) 2 (14  10.6) 2 (8  10.6) 2 (20  31.8) 2



10.6
10.6
10.6
31.8
 2.87  0.153  0.185  1.091  0.638  4.379
 9.316
Example 2: M & M Colors (cont.)
 2  9.316;degrees of freedom  6 1  5
2
The test statistic is χ 2  9.316 ; χ0.05
with 5 d.f.  11.070
Decision Rule:
2
If χ 2  χ.05
,reject H0,
otherwise, do not reject
H0.
Here,
0.05
0
Do not
reject H0
Reject H0
20.05 = 11.070
χ
2
2
2
χ.05
= 9.316 <
= 11.070,
so we do not reject H0 and
conclude that there is not
sufficient evidence to conclude
that Mars has changed the color
proportions.
Models for two-way tables
The chi-square test is an overall technique for comparing any number
of population proportions, testing for evidence of a relationship between
two categorical variables. There are 2 types of tests:
1.
Test for independence: Take one SRS and classify the individuals in
the sample according to two categorical variables (attribute or condition)
 observational study, historical design.
2.
Compare several populations (tests for homogeneity): Randomly
select several SRSs each from a different population (or from a
population subjected to different treatments)  experimental study.
Both models use the X2 test to test of the hypothesis of no relationship.
Testing for independence
We have now a single sample from a single population. For each
individual in this SRS of size n we measure two categorical variables.
The results are then summarized in a two-way table.
The null hypothesis is that the row and column variables are
independent. The alternative hypothesis is that the row and column
variables are dependent.
Chi-square tests for independence
(Obs  Exp)
  
Exp
all cells
2
2

Expected cell frequencies:
row total  column total
Exp 
n
Where:
row total = sum of all frequencies in the row
column total = sum of all frequencies in the column
n = overall sample size
H0: The two categorical variables are independent
(i.e., there is no relationship between them)
H1: The two categorical variables are dependent
(i.e., there is a relationship between them)
Example 1: Parental smoking

Does parental smoking influence the incidence of smoking in
children when they reach high school? Randomly chosen high
school students were asked whether they smoked (columns) and
whether their parents smoked (rows).
Student
Smoke No smoke Total
Both smoke
400
1380
1780
416
1823
2239
Neither smokes
188
1168
1356
Total
1004
4371
5375
Parent One smokes



Are parent smoking status and student smoking status related?
H0 : parent smoking status and student smoking status are
independent
HA : parent smoking status and student smoking status are not
independent
Example 1: Parental smoking (cont.)
Does parental smoking influence the incidence of smoking in children when
they reach high school? Randomly chosen high school students were asked
whether they smoked (columns) and whether their parents smoked (rows).
Examine the computer output for the chi-square test performed on these data.
What does it tell you?
Hypotheses?
Are data ok for 2 test? (All
expected counts 5 or more)
df = (rows-1)*(cols-1)=2*1=2
Interpretation? Since P-value is
less than .05, reject H0 and
conclude that parent smoking
status and student smoking status
are related.
Example 2: meal plan selection

The meal plan selected by 200 students is shown below:
Number of meals per week
Class
none
Standing 20/week 10/week
Total
Fresh.
24
32
14
70
Soph.
22
26
12
60
Junior
10
14
6
30
Senior
14
16
10
40
Total
70
88
42
200
Example 2: meal plan selection (cont.)

The hypotheses to be tested are:
H0: Meal plan and class standing are independent
(i.e., there is no relationship between them)
H1: Meal plan and class standing are dependent
(i.e., there is a relationship between them)
Example 2: meal plan selection (cont.)
Expected Cell Frequencies
Observed:
Class
Standing
Number of meals
per week
Expected cell frequencies if H0 is
true:
20/wk
10/wk
none
Total
Fresh.
24
32
14
70
Soph.
22
26
12
60
Junior
10
14
6
30
Senior
14
16
10
40
Class
Standing
Total
70
88
42
200
Example for one cell:
row total  column total
Exp 
n
30  70

 10.5
200
Number of meals
per week
20/wk
10/wk
none
Total
Fresh.
24.5
30.8
14.7
70
Soph.
21.0
26.4
12.6
60
Junior
10.5
13.2
6.3
30
Senior
14.0
17.6
8.4
40
70
88
42
200
Total
Example 2: meal plan selection (cont.)
The Test Statistic

The test statistic value is:
(Obs  Exp)2
  
Exp
all cells
2
(24  24.5)2 (32  30.8) 2



24.5
30.8
(10  8.4)2

 0.709
8.4
χ 0.2 05 = 12.592 from the chi-squared distribution
with (4 – 1)(3 – 1) = 6 degrees of freedom
Example 2: meal plan selection (cont.)
Decision and Interpretation
2
The test statistic is  2  0.709 ; 0.05
with 6 d.f.  12.592
Decision2 Rule:
If
> 12.592, reject H0, otherwise,
do not reject H0

Here,
0.05
0
Do not
reject H0
Reject H0
20.05=12.592
2
2
2
= 0.709 < χ 0.05 = 12.592,
so do not reject H0
Conclusion: there is not
sufficient evidence that meal
plan and class standing are
related.
Models for two-way tables
The chi-square test is an overall technique for comparing any number
of population proportions, testing for evidence of a relationship between
two categorical variables. There are 2 types of tests:
1.
Test for independence: Take one SRS and classify the individuals in the
sample according to two categorical variables (attribute or condition) 
observational study, historical design.
NEXT:
2. Compare several populations (tests for homogeneity): Randomly
select several SRSs each from a different population (or from a population
subjected to different treatments)  experimental study.
Both models use the X2 test to test of the hypothesis of no relationship.
Comparing several populations (tests for homogeneity)
Select independent SRSs from each of c populations, of sizes
n1, n2, . . . , nc. Classify each individual in a sample according to a
categorical response variable with r possible values. There are c
different probability distributions, one for each population.
The null hypothesis is that the distributions of the response variable are
the same in all c populations. The alternative hypothesis says that
these c distributions are not all the same.
Example: Cocaine addiction (test for homogeneity)
Cocaine produces short-term feelings of physical and mental well being. To
maintain the effect, the drug may have to be taken more frequently and at
higher doses. After stopping use, users will feel tired, sleepy, and depressed.
The pleasurable high followed by
unpleasant after-effects encourage
repeated compulsive use, which can
easily lead to dependency.
We compare treatment with an antidepressant (desipramine), a standard
treatment (lithium), and a placebo.
Population 1: Antidepressant treatment (desipramine)
Population 2: Standard treatment (lithium)
Population 3: Placebo (“sugar pill”)
Cocaine addiction
H0: The proportions of success (no relapse)
are the same in all three populations.
Observed
Expected
Expected relapse counts
No
Yes
Desipramine 25*26/74 ≈ 8.78
25*48/74 ≈ 16.22
Lithium
26*26/74 ≈ 9.14
26*48/74 ≈ 16.86
Placebo
23*26/74 ≈ 8.08
23*48/74 ≈ 14.92
Cocaine addiction
Table of counts:
“actual / expected,” with
three rows and two
columns:
No relapse
Relapse
Desipramine
15
8.78
10
16.22
Lithium
7
9.14
19
16.86
Placebo
4
8.08
19
14.92
df = (3−1)*(2−1) = 2
2
2




15

8
.
78
10

16
.
22
2 

8.78
16.22
2
2


7  9.14
19  16.86


9.14
16.86
2
2


4  8.08
19  14.92


8.08
14.92
 10.74
2 components:
4 .4 1
0 .5 0
2 .0 6
2 .3 9
0 .2 7
1 .1 2
Cocaine addiction: Table χ
H0: The proportions of success (no relapse)
are the same in all three populations.
p
df
0.25
0.2
0.15
0.1
0.05
0.025
0.02
1
1.32
1.64
2.07
2.71
3.84
5.02
5.41
2
2.77
3.22
3.79
4.61
5.99
7.38
7.82
3
4.11
4.64
5.32
6.25
7.81
9.35
9.84
4
5.39
5.99
6.74
7.78
9.49
11.14
11.67
5
6.63
7.29
8.12
9.24
11.07
12.83
13.39
6
7.84
8.56
9.45
10.64
12.59
14.45
15.03
2
7
9.04
9.80
10.75
12.02
16.62
X =14.07
10.7116.01
> 5.99;
df
8
10.22
11.03
12.03
13.36
15.51
17.53
18.17
9
11.39
12.24
13.29
14.68
16.92
19.02the19.68
 reject
H0
10
12.55
13.44
14.53
15.99
18.31
20.48
21.16
11
13.70
14.63
15.77
17.28
19.68
21.92
22.62
12
14.85
15.81
16.99
18.55
21.03
23.34
24.05
13
15.98
16.98
18.20
19.81
22.36
24.74
25.47
The
proportions
of
success
are
not
the
same
in
14
17.12
18.15
19.41
21.06
23.68
26.12
26.87
18.25
19.31
20.60
22.31
25.00
27.49
28.26
all 15
three
populations
(Desipramine,
Lithium,
16
19.37
20.47
21.79
23.54
26.30
28.85
29.63
17
20.49
21.61
22.98
24.77
27.59
30.19
31.00
Placebo).
18
21.60
22.76
24.16
25.99
28.87
31.53
32.35
19
22.72
23.90
25.33
27.20
30.14
32.85
33.69
20
23.83
25.04
26.50
28.41
31.41
34.17
35.02
Desipramine
is
a
more
successful
treatment

21
24.93
26.17
27.66
29.62
32.67
35.48
36.34
0.01
6.63
9.21
11.34
13.28
15.09
16.81
= 18.48
2
20.09
21.67
23.21
24.72
26.22
27.69
29.14
30.58
32.00
33.41
34.81
36.19
37.57
38.93
0.005
7.88
10.60
12.84
14.86
16.75
18.55
20.28
21.95
23.59
25.19
26.76
28.30
29.82
31.32
32.80
34.27
35.72
37.16
38.58
40.00
41.40
0.0025
0.001 0.0005
9.14
10.83
12.12
11.98
13.82
15.20
14.32
16.27
17.73
16.42
18.47
20.00
18.39
20.51
22.11
20.25
22.46
24.10
22.04
24.32
26.02
23.77
26.12
27.87
25.46
27.88
29.67
27.11
29.59
31.42
28.73
31.26
33.14
30.32
32.91
34.82
31.88
34.53
36.48
Observed
33.43
36.12
38.11
34.95
37.70
39.72
36.46
39.25
41.31
37.95
40.79
42.88
39.42
42.31
44.43
40.88
43.82
45.97
42.34
45.31
47.50
43.78
46.80
49.01