please pick up a m&m activity sheet, form a group of 2-3 and choose a bag of m&ms

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

http://www.candywarehouse.com/mms.html

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.

http://www.candywarehouse.com/mms.html

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





days.

Lunar Phase Number of Days

Number of Births

Proportion of Days


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





days.

Lunar Phase Number of

Days

Number of Births

Proportion of Days


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





days.

Lunar Phase Number of

Days

Number of Births

Proportion of Days


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


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


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


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


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


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?


Test Statistic:

P-value < .001 a = .01

Body Piercing

OnlyTattoos

Only

Both Body Piercing

and Tattoos

No Body Art


48.290.58)0.5854(...7.49

)7.4961( 222

X


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


Which cell contributes the most to the X2 test

statistic?

Seniors having both body piercing and tattoos

contribute the most to the X2 statistic.

please pick up a m&m activity sheet, form a group of 2-3 and choose a bag of m&ms

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