Goodness of Fit Tests Categorical Data and The Analysis...

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.

M&M Candies Continued . . .

We could count how many candies of each color are in the bag.

252219212823BrownOrangeYellowGreenBlueRed

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:

. . .

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• χ2 curves are skewed right• As df increases, the χ2 curve shifts

toward the right and becomes more like a normal curve

Facts About χ2 distributions

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

Lunar Phases Continued . . .Let:

p1 = proportion of births that occur during the new moonp2 = proportion of births that occur during the waxing crescent moonp3 = proportion of births that occur during the first quarter moonp4 = proportion of births that occur during the waxing gibbous moonp5 = 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 moonp8 = proportion of births that occur during the waning crescent moon

Lunar Phase Observed Number of Births

Expected Number of Births

New Moon 7680 7641.49Waxing Crescent 48,442 48455.52First Quarter 7579 7641.49Waxing Gibbous 47,814 47,497.55Full Moon 7711 7641.49Waning Gibbous 47,595 47809.45Last Quarter 7733 7641.49Waning Crescent 48,230 48,455.52

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

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 α = .05Since the P-value > α, we fail to reject H0. There is not sufficient evidence to conclude that lunar phases and number of births are related.

Type II

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.

240152245158Total5303545Non-Athletes96581568Non-Soccer Players9110112545Soccer Players

Total3 or more210

Number of Concussions

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:

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.

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 . . .

240152245158Total5303545Non-Athletes96581568Non-Soccer Players9110112545Soccer Players

Total3 or more210


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

240152245158Total530 (3.3)3 (4.9)5 (10.0)45 (34.9)Non-Athletes965 (6.0)8 (8.8)15 (18.0)68 (63.2)Non-Soccer Players9110 (5.7)11 (8.325 (17.1)45 (59.9)Soccer Players

Total3 or more210


2402245158Total533 (8.2)5 (10.0)45 (34.9)Non-Athletes9613 (14.8)15 (18.0)68 (63.2)Non-Soccer Players9121 (14.0)25 (17.1)45 (59.9)Soccer Players

Total2 or more10



df = 4Test Statistic:

2402245158Total533 (8.2)5 (10.0)45 (34.9)Non-Athletes9613 (14.8)15 (18.0)68 (63.2)Non-Soccer Players9121 (14.0)25 (17.1)45 (59.9)Soccer Players

Total2 or more10



Since the P-value < α, 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.

X2 Test for Independence

Null Hypothesis: H0: The two variables are independent

Alternative Hypothesis:Ha: The two variables are not independent

Test Statistic:

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.

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 α = .01.

Body Piercing Only Tattoos Only

Both Body Piercing and

Tattoos No Body ArtFreshman 61 7 14 86Sophomore 43 11 10 64Junior 20 9 7 43Senior 21 17 23 54

Body Art Continued . . .



Tattoos No Body ArtFreshman 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



Tattoos No Body ArtFreshman 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)


Test Statistic:

P-value < .001 α = .01





Since the P-value < α, we reject H0. There is sufficient evidence to suggest that class standing and the body art category are associated.




Goodness of Fit Tests Categorical Data and The Analysis...

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