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Chapter 12
The Analysis of Categorical Data and Goodness of Fit Tests
There are six colors – so k = 6.
Suppose we wanted to determine if the proportions for the different colors in a large bag of M&M candies matches the proportions that the company claims is in their candies.
We could record the color of each candy in the bag.
This would be univariate,
categorical data.
How many categories for color would there
be?
k is used to denote the number of categories
for a categorical variable
M&M Candies Continued . . .
We could count how many candies of each color are in the bag.
A one-way frequency table is used to display the observed counts for
the k categories.
Red Blue Green Yellow Orange
Brown
23 28 21 19 22 25
A goodness-of-fit test will allow us to
determine if these observed counts are
consistent with what we expect to have.
Goodness-of-Fit Test Procedure
Null Hypothesis: H0: p1 = hypothesized proportion for Category 1
pk = hypothesized proportion for Category kHa: H0 is not true
Test Statistic:
. . .
cells all
22
count cell expectedcount cell expected - count cell observed
X
The goodness-of-fit statistic, denoted by X2, is a quantitative measure to the extent to which the observed counts differ from those expected when H0 is true.
The X2 value can never be negative.
Read “chi-squared”
The goodness-of-fit test is used to analysze univariate
categorical data from a single sample.
Goodness-of-Fit Test Procedure Continued . . .P-values: When H0 is true and all expected counts
are at least 5, X2 has approximately a chi-square distribution with df = k – 1. Therefore, the P-value associated with the computed test statistic value is the area to the right ofX under the df = k – 1 chi-square curve.
Assumptions:1) Observed cell counts are based on a random
sample2) The sample size is large enough as long as every
expected cell count is at least 5
• Different df have different curves• curves are skewed right• As df increases, the 2 curve shifts
toward the right and becomes more like a normal curve
Facts About2 distributions
df=3
df=5
df=10
A common urban legend is that more babies than expected are born during certain phases of the lunar cycle, especially near the full moon.
The table below shows the number of days in the eight lunar phases with the number of births in each phase for 24 lunar cycles.
Lunar Phase Number of Days Number of Births
New Moon 24 7680
Waxing Crescent 152 48,442
First Quarter 24 7579
Waxing Gibbous 149 47,814
Full Moon 24 7711
Waning Gibbous 150 47,595
Last Quarter 24 7733
Waning Crescent 152 48,230
There are eight phases so k = 8.
Lunar Phases Continued . . .
Let:
p1 = proportion of births that occur during the new moon
p2 = proportion of births that occur during the waxing crescent moon
p3 = proportion of births that occur during the first quarter moon
p4 = proportion of births that occur during the waxing gibbous moon
p5 = proportion of births that occur during the full moon
p6 = proportion of births that occur during the waning gibbous moon
p7 = proportion of births that occur during the last quarter moon
p8 = proportion of births that occur during the waning crescent moon
There is a total of 699 days in the 24 lunar cycles. If there is no relationship between the number of births and lunar phase, then the expected
proportions equal the number of days in each phase out of the
total number of days.
p1 = .0343 p2 = .2175 p3 = .0343 p4 = .2132
P5 = .0343 p6 = .2146 p7 = .0343 p8 = .2175
The hypothesis statements would be:
H0: p1 = .0343, p2 = .2175, p3 = .0343, p4 = .2132, p5 = .0343, p6 = .2146, p7 = .0343, p8 = .2175
Ha: H0 is not true
Lunar Phase Observed Number of Births
Expected Number of Births
New Moon 7680 7641.49
Waxing Crescent 48,442 48455.52
First Quarter 7579 7641.49
Waxing Gibbous 47,814 47,497.55
Full Moon 7711 7641.49
Waning Gibbous 47,595 47809.45
Last Quarter 7733 7641.49
Waning Crescent 48,230 48,455.52
Lunar Phases Continued . . .
There is a total of 222,784 births in the sample. If there is no relationship
between the number of births and lunar phase, then the expected
counts for each category would equal n(hypothesized proportion).
H0: p1 = .0343, p2 = .2175, p3 = .0343, p4 = .2132, p5 = .0343, p6 = .2146, p7 = .0343, p8 = .2175
Ha: H0 is not true
Lunar Phases Continued . . .
H0: p1 = .0343, p2 = .2175, p3 = .0343, p4 = .2132, p5 = .0343, p6 = .2146, p7 = .0343, p8 = .2175
Ha: H0 is not true
Test Statistic:
P-value > .10 df = 7 = .05
Since the P-value > , we fail to reject H0. There is not sufficient evidence to conclude that lunar phases and number of births are related.
557.652.455,48
)52.455,48230,48(...
52.455,48)52.455,48442,48(
49.7641)49.76417680( 222
2
X
What type of error could we have potentially made with this decision? Type II
The X2 test statistic is smaller than the smallest entry in the df = 7 column
of Appendix Table 8.
A study was conducted to determine if collegiate soccer players had in increased risk of concussions over other athletes or students. The two-way frequency table below displays the number of previous concussions for students in independently selected random samples of 91 soccer players, 96 non-soccer athletes, and 53 non-athletes.
Number of Concussions
0 1 23 or
moreTotal
Soccer Players 45 25 11 10 91
Non-Soccer Players
68 15 8 5 96
Non-Athletes 45 5 3 0 53
Total 158 45 22 15 240
These values in green are the observed
counts.
Also called a contingency table.
These values in blue are the marginal
totals.
This value in red is the grand total.
This is univariate categorical data - number of concussions
- from 3 independent samples.
If there were no difference between these 3 populations in regards to the number of concussions, how many soccer players would you expect to
have no concussions?
We would expect (158/240)(91).
X2 Test for Homogeneity
Null Hypothesis: H0: the true category proportions are the same for all the populations or treatments
Alternative Hypothesis:Ha: the true category proportions are not all the same for all the populations or treatments
Test Statistic:
cells all
22
count cell expectedcount cell expected - count cell observed
X
The 2 Test for Homogeneity is used to analyze univariate
categorical data from 2 or more independent samples.
X2 Test for Homogeneity Continued . . .
Expected Counts: (assuming H0 is true)
P-value: When H0 is true and all expected counts are at least 5, X2 has approximately a chi-square distribution with df = (number of rows – 1)(number of columns – 1). The P-value associated with the computed test statistic value is the area to the right ofX under the appropriate chi-square curve.
total grandtotal) marginal umntotal)(col marginal (row
counts cell expected
X2 Test for Homogeneity Continued . . .
Assumptions:1) Data are from independently chosen
random samples or from subjects who were assigned at random to treatment groups.
2) The sample size is large: all expected cell counts are at least 5. If some expected counts are less than 5, rows or columns of the table may be combined to achieve a table with satisfactory expected counts.
Soccer Players Continued . . .
Number of Concussions
0 1 23 or
moreTotal
Soccer Players 45 25 11 10 91
Non-Soccer Players
68 15 8 5 96
Non-Athletes 45 5 3 0 53
Total 158 45 22 15 240
State the hypotheses.
H0: Proportions in each response category (number of concussions) are the same for all three groups
Ha: Category proportions are not all the same for all three groups
Df = (2)(3) = 6
To find df count the number of rows and columns – not including the
totals!df = (number of rows – 1)(number of columns
– 1)
Another way to find df – you can also cover one row and one column, then count the number of cells left
(not including totals)
Number of Concussions
0 1 23 or
moreTotal
Soccer Players 45 (59.9)
25 (17.1) 11 (8.3 10 (5.7) 91
Non-Soccer Players
68 (63.2)
15 (18.0) 8 (8.8) 5 (6.0) 96
Non-Athletes 45 (34.9)
5 (10.0) 3 (4.9) 0 (3.3) 53
Total 158 45 22 15 240
Number of Concussions
0 12 or
moreTotal
Soccer Players45 (59.9) 25 (17.1)
21 (14.0)
91
Non-Soccer Players68 (63.2) 15 (18.0)
13 (14.8)
96
Non-Athletes 45 (34.9) 5 (10.0) 3 (8.2) 53
Total 158 45 22 240
Soccer Players Continued . . .
Expected counts are shown in the parentheses
next to the observed counts.
df = 4
Test Statistic: Notice that NOT all the expected counts are at
least 5.
So combine the column for 2 concussions and the
column for 3 or more concussions.
This combined table has a df = (2)(2) = 4.
6.202.8
)2.83(...
5.59)9.5945( 22
2
X
P-value < .001 = .05
Number of Concussions
0 12 or
moreTotal
Soccer Players45 (59.9) 25 (17.1)
21 (14.0)
91
Non-Soccer Players68 (63.2) 15 (18.0)
13 (14.8)
96
Non-Athletes 45 (34.9) 5 (10.0) 3 (8.2) 53
Total 158 45 22 240
Soccer Players Continued . . .
Since the P-value < , we reject H0. There is strong evidence to suggest that the category
proportions for the number of concussions is not the same
for the 3 groups.Is that all I can say – that there is a difference in
proportions for the groups?
We can look at the chi-square contributions – which of the
cells above have the greatest contributions to the value of
the X2 statistic?
These cells had the largest contributions to the X2 test
statistic.
X2 Test for Independence
Null Hypothesis: H0: The two variables are independent
Alternative Hypothesis:Ha: The two variables are not independent
Test Statistic:
cells all
22
count cell expectedcount cell expected - count cell observed
X
The 2 Test for Independence is used to analyze bivariate
categorical data from a single sample.
X2 Test for Independence Continued . . .
Expected Counts: (assuming H0 is true)
P-value: When H0 is true and assumptions for X2 test are satisfied, X2 has approximately a chi-square distribution with df = (number of rows – 1)(number of columns – 1). The P-value associated with the computed test statistic value is the area to the right ofX under the appropriate chi-square curve.
total grandtotal) marginal umntotal)(col marginal (row
counts cell expected
X2 Test for Independence Continued . . .
Assumptions:1) The observed counts are based on data
from a random sample. 2) The sample size is large: all expected cell
counts are at least 5. If some expected counts are less than 5, rows or columns of the table may be combined to achieve a table with satisfactory expected counts.
The paper “Contemporary College Students and Body Piercing” (Journal of Adolescent Health, 2004) described a survey of 450
undergraduate students at a state university in the southwestern region of the United States. Each student in the sample was classified according to class standing (freshman, sophomore, junior, senior) and body art category (body piercing only, tattoos only, both tattoos and body piercing, no body art). Is there evidence that there is an association between class standing and response to the body art question? Use = .01.
Body Piercing
OnlyTattoos
Only
Both Body Piercing
and Tattoos
No Body Art
Freshman 61 7 14 86
Sophomore 43 11 10 64
Junior 20 9 7 43
Senior 21 17 23 54
State the hypotheses.
Body Art Continued . . .
Body Piercing
OnlyTattoos
Only
Both Body Piercing
and Tattoos
No Body Art
Freshman 61 7 14 86
Sophomore 43 11 10 64
Junior 20 9 7 43
Senior 21 17 23 54
H0: class standing and body art category are independent
Ha: class standing and body art category are not independent
df = 9
Assuming H0 is true, what are the expected counts?
Body Piercing
OnlyTattoos
Only
Both Body Piercing
and Tattoos
No Body Art
Freshman 61 (49.7) 7 (15.1) 14 (18.5) 86 (84.7)
Sophomore 43 (37.9) 11 (11.5) 10 (14.1) 64 (64.5)
Junior 20 (23.4) 9 (7.1) 7 (8.7) 43 (39.8)
Senior 21 (34.0) 17 (10.3) 23 (12.7) 54 (58.0)
How many degrees of freedom does this two-
way table have?
Body Art Continued . . .
Test Statistic:
P-value < .001 = .01
Body Piercing
OnlyTattoos
Only
Both Body Piercing
and Tattoos
No Body Art
Freshman 61 (49.7) 7 (15.1) 14 (18.5) 86 (84.7)
Sophomore 43 (37.9) 11 (11.5) 10 (14.1) 64 (64.5)
Junior 20 (23.4) 9 (7.1) 7 (8.7) 43 (39.8)
Senior 21 (34.0) 17 (10.3) 23 (12.7) 54 (58.0)
48.290.58
)0.5854(...
7.49)7.4961( 22
2
X
Body Art Continued . . .
Since the P-value < , we reject H0. There is sufficient evidence to suggest that class standing and the body art category are associated.
Body Piercing
OnlyTattoos
Only
Both Body Piercing
and Tattoos
No Body Art
Freshman 61 (49.7) 7 (15.1) 14 (18.5) 86 (84.7)
Sophomore 43 (37.9) 11 (11.5) 10 (14.1) 64 (64.5)
Junior 20 (23.4) 9 (7.1) 7 (8.7) 43 (39.8)
Senior 21 (34.0) 17 (10.3) 23 (12.7) 54 (58.0)
Which cell contributes the most to the X2 test
statistic?
Seniors having both body piercing and tattoos
contribute the most to the X2 statistic.