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Please pick up a M&M activity sheet, form a group of 2-3 and choose a bag of M&Ms. Chapter 12. The Analysis of Categorical Data and Goodness of Fit Tests. - PowerPoint PPT Presentation
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Please pick up a M&M activity sheet, form a group of 2-3 and choose a bag of M&Ms
Chapter 12The 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 25A 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 ProcedureNull 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 observedX
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 of X2 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• c2 curves are skewed right• As df increases, the c2 curve shifts
toward the right and becomes more like a normal curve
Facts About c2 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 BirthsNew Moon 24 7680Waxing Crescent 152 48,442First Quarter 24 7579Waxing Gibbous 149 47,814Full Moon 24 7711Waning Gibbous 150 47,595Last Quarter 24 7733Waning Crescent 152 48,230
There are eight phases so k = 8.
Lunar Phases Continued . . .
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.
Lunar Phase Number of Days
Number of Births
Proportion of Days
Expected Number of Births
New Moon 24 7680 =24/699=.0343 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
Lunar Phases Continued . . .
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.
Lunar Phase Number of Days
Number of Births
Proportion of Days
Expected Number of Births
New Moon 24 7680 =24/699=.0343 Waxing Crescent 152 48,442 .217
First Quarter 24 7579 .0343 Waxing Gibbous 149 47,814 .213
Full Moon 24 7711 .0343 Waning Gibbous 150 47,595 .215
Last Quarter 24 7733 .0343 Waning Crescent 152 48,230 .217
Lunar Phases Continued . . .
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.
Lunar Phase Number of
Days
Number of Births
Proportion of Days
Expected Number of Births
New Moon 24 7680 0343 =.0343*222784=7641.49 Waxing Crescent 152 48,442 .217
First Quarter 24 7579 .0343 Waxing Gibbous 149 47,814 .213
Full Moon 24 7711 .0343 Waning Gibbous 150 47,595 .215
Last Quarter 24 7733 .0343 Waning Crescent 152 48,230 .217
Lunar Phases Continued . . .
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.
Lunar Phase Number of
Days
Number of Births
Proportion of Days
Expected Number of Births
New Moon 24 7680 0343 =.0343*222784=7641.49 Waxing Crescent 152 48,442 .217 48433.24
First Quarter 24 7579 .0343 7641.49 Waxing Gibbous 149 47,814 .213 47452.27
Full Moon 24 7711 .0343 7641.49 Waning Gibbous 150 47,595 .215 47809.44
Last Quarter 24 7733 .0343 7641.49 Waning Crescent 152 48,230 .217 48433.24
Lunar Phases Continued . . .
H0: p1 = .0343, p2 = .2175, p3 = .0343, p4 = .2132, p5 = .0343, p6 = .2146, p7 = .0343, p8 = .2175Ha: H0 is not trueTest Statistic:
P-value > .10 df = 7 a = .05Since the P-value > a, 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.
• Get your quizzes and homework from your folder
• Have your practice test out
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 2 3 or more Total
Soccer Players 45 25 11 10 91Non-Soccer Players 68 15 8 5 96
Non-Athletes 45 5 3 0 53Total 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 HomogeneityNull 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 observedX
The c2 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 of X2 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 2 3 or more Total
Soccer Players 45 25 11 10 91Non-Soccer Players 68 15 8 5 96
Non-Athletes 45 5 3 0 53Total 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 2 3 or more Total
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 Concussions0 1 2 or
more TotalSoccer Players 45 (59.9) 25 (17.1) 21
(14.0) 91Non-Soccer Players 68 (63.2) 15 (18.0) 13
(14.8) 96Non-Athletes 45 (34.9) 5 (10.0) 3 (8.2) 53Total 158 45 22 240
Soccer Players Continued . . .
Expected counts are shown in the parentheses
next to the observed counts.
df = 4Test 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( 222
X
P-value < .001 a = .05
Number of Concussions0 1 2 or
more TotalSoccer Players 45 (59.9) 25 (17.1) 21
(14.0) 91Non-Soccer Players 68 (63.2) 15 (18.0) 13
(14.8) 96Non-Athletes 45 (34.9) 5 (10.0) 3 (8.2) 53Total 158 45 22 240
Soccer Players Continued . . .
Since the P-value < a, 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 IndependenceNull 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 observedX
The c2 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 of X2 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 a = .01.
Body Piercing
OnlyTattoos
Only
Both Body Piercing
and Tattoos
No Body Art
Freshman 61 7 14 86Sophomore 43 11 10 64Junior 20 9 7 43Senior 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 86Sophomore 43 11 10 64Junior 20 9 7 43Senior 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 a = .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( 222
X
Body Art Continued . . .
Since the P-value < a, 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.