chapter 10 multinomial experiments

7/26/2019 Chapter 10 Multinomial Experiments

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1Chapter 10. Section 10.1 and 10.2 Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman

MARIO F. TRIOLAEIGHTH

EDITION

ELEMENTARY STATISTICSChapter 10 Multinomial Experiments and

Contingency Tables


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Chapter 10

Multinomial Experiments and

Contingency Tables

10-1 Overview10-2 Multinomial Experiments:

Goodness-of-fit

10-3 Contingency Tables:Independence and Homogeneity


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10-1 Overview

Focus on analysis of categorical (qualitative orattribute) data that can be separated into

different categories (often called cells)

Use the X2(chi-square) test statistic (Table A-4)

One-way frequency table (single row or column)

Two-way frequency table or contingency table

(two or more rows and columns)


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MARIO

F.

TRIOLA EIGHTH

EDITIO

ELEMENTARY STATISTICSSection 10-2 Goodness of Fit


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10-2 Multinomial Experiments:

Goodness-of-Fit

Assumptionswhen testing hypothesis that the population

proportion for each of the categories is as claimed:

1. The data have been randomly selected.

2. The sample data consist of frequency counts

for each of the different categories.

3. The expected frequency is at least 5. (There is

no requirement that the observed frequency

for each category must be at least 5.)


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Multinomial ExperimentAn experiment that meets the following conditions:

1. The number of trials is fixed.

2. The trials are independent.

3. All outcomes of each trial must beclassified into exactly one of several different

categories.

4. The probabilities for the differentcategories remain constant for each trial.

Definition


7/537Chapter 10. Section 10.1 and 10.2 Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman

Definition

Goodness-of-fit test

used to test the hypothesis that an

observed frequency distribution fits(or conforms to) some claimed

distribution



0 represents the observed frequencyof an outcome

E represents the expected frequency of an outcome

k represents the number of different categoriesor

outcomes

n represents the total number of trials

Goodness-of-Fit Test

Notation



Expected Frequencies

If all expected frequencies are equal:

the sum of all observed frequencies divided

by the number of categories

nE =k



Expected Frequencies

If all expected frequencies are not all equal:

each expected frequency is found by multiplying

the sum of all observed frequencies by the

probability for the category

E = n p



Goodness-of-fit Test in Multinomial Experiments

Test Statistic

Critical Values

1. Found in Table A-4 using k-1 degrees of

freedom

where k=number of categories

2. Goodness-of-fit hypothesis tests are always

right-tailed.

X2=

(O- E)2

E



A large disagreementbetween observed

and expected values will lead to a largevalue of X2and a small P-value.

A significantly largevalue of

2

will causea rejectionof the null hypothesis of no

difference between the observed and the

expected.

A close agreement between observed

and expected values will lead to a small

value of X2 and a large P-value.



Figure 10-3

Relationships Among

Components in

Goodness-of-Fit

Hypothesis Test



Categories with Equal

Frequencies

H0: p1= p2= p3= . . . = pk

H1: at least one of the probabilities is

different from the others

(Probabilities)


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H0: p1, p2, p3, . . . , pkare as claimed

H1: at least one of the above proportions

is different from the claimed value

Categories with Unequal

Frequencies(Probabilities)

E l


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Example: Mars, Inc. claims its M&M candies are distributed withthe color percentages of 30% brown, 20% yellow, 20% red, 10% orange,10% green, and 10% blue. At the 0.05 significance level, test the claimthat the color distribution is as claimed by Mars, Inc.

E l


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Claim: p1= 0.30, p2= 0.20, p3= 0.20, p4= 0.10,p5= 0.10, p6= 0.10

H0 : p1= 0.30, p2= 0.20, p3= 0.20, p4= 0.10,p5= 0.10, p6= 0.10

H1: At least one of the proportions is

different from the claimed value.

E l


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Brown Yellow Red Orange Green Blue

Observed frequency 33 26 21 8 7 5

Frequencies of M&Ms

n= 100

E l


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Frequencies of M&Ms

Brown E=np= (100)(0.30) = 30Yellow E=np= (100)(0.20) = 20

Red E=np= (100)(0.20) = 20Orange E=np= (100)(0.10) = 10Green E=np= (100)(0.10) = 10

Blue E=np= (100)(0.10) = 10

n= 100


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Frequencies of M&Ms

Expected frequency 30 20 20 10 10 10

(O -E)2/E 0.3 1.8 0.05 0.4 0.9 2.5


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Frequencies of M&Ms


(O -E)2/E 0.3 1.8 0.05 0.4 0.9 2.5

X2= = 5.95

(O- E)2

E

Test Statistic


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Frequencies of M&Ms


(O -E)2/E 0.3 1.8 0.05 0.4 0.9 2.5

X2= = 5.95

(O- E)2

E

Test StatisticCritical Value X

2=11.071

(with k-1 = 5 and = 0.05)


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Test Statistic does not fall within critical region;Fail to reject H0: percentages are as claimed

There is not sufficient evidence to warrant rejection of theclaim that the colors are distributed with the givenpercentages.

0

Sample data: X2= 5.95

= 0.05

X2= 11.071

Fail to Reject Reject


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Comparison of Claimed and Observed Proportions

0.30

0.20

0.10

0

Green

Yellow

Red

Orange

Brown

Blue

Claimed proportions

Observed proportions

Proportions


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MARIO F. TRIOLA EIGHTHEDITIO

ELEMENTARY STATISTICSSection 10-3 Contingency Tables: Independence

and Homogeneity


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Definition

Contingency Table (or two-way frequency table)

a table in which frequencies

correspond to two variables.

(One variable is used to categorize rows,and a second variable is used to

categorize columns.)


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Definition

Contingency Table (or two-way frequency table)

a table in which frequencies

correspond to two variables.

(One variable is used to categorize rows,and a second variable is used to

categorize columns.)

Contingency tables have at least tworows and at least two columns.


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Test of Independence

tests the null hypothesis that

the row variable and columnvariable in a contingency table arenot related. (The null hypothesis

is the statement thatthe row and column variables areindependent.)

Definition


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Assumptions

1. The sample data are randomly selected.

2. The null hypothesis H0is the statement that

the row and column variables

are independent; the alternative

hypothesis H1is the statement that the row

and column variables are dependent.

3. For every cell in the contingency table, the

expectedfrequency E is at least 5. (There is

no requirement that everyobserved

frequency must be at least 5.)


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Test of Independence

Test Statistic

Critical Values

1. Found in Table A-4 using

degrees of freedom = (r - 1)(c - 1)

r is the number of rows and c is the number of columns

2. Tests of Independence are always right-tailed.

X2=

(O- E)2

E


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(row total) (column total)

(grand total)E=

Total number of all observed frequencies

in the table


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Tests of Independence

H0: The row variable is independent of thecolumn variable

H1: The row variable is dependent (related to)

the column variable

This procedure cannot be used to establish adirect cause-and-effect link between variables inquestion.

Dependence means only there is a relationshipbetween the two variables.

E t d F f


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Expected Frequency for

Contingency Tables

E t d F f


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E= row total column total

grand total


Contingency Tables

grand totalgrand total

E t d F f


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n p


grand total


Contingency Tables


(probability of a cell)

E pected Freq enc for


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n p


grand total


Contingency Tables





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n p


grand total


Contingency Tables



E= (row total) (column total)(grand total)

I th t f i i d d t f h th th


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Is the type of crime independent of whether thecriminal is a stranger?

Stranger

Acquaintance

or Relative

12

39

379

106

727

642

Homicide Robbery Assault



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Row Total

Column Total

Stranger

Acquaintance

or Relative

1118

787

1905

12

39

51

379

106

485

727

642

1369





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Row Total

Column Total


Stranger

Acquaintance

or Relative



1118

787

1905

12

39

51

379

106

485

727

642

1369



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Row Total

(29.93)

Column Total


E= (1118)(5

1) 1905= 29.93

Stranger

Acquaintance

or Relative



1118

787

1905

12

39

51

379

106

485

727

642

1369



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Row Total

(29.93)

(21.07)

(284.64)

(200.36)

(803.43)

(565.57)

Column Total


E= (1118)(5

1) 1905= 29.93 E= (1118)(485)

1905= 284.64

etc.

Stranger

Acquaintance

or Relative



1118

787

1905

12

39

51

379

106

485

727

642

1369

Is the t pe of crime independent of hether the


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12

39

379

106

727

642

Homicide Robbery Forgery

(29.93)

(21.07)

(284.64)

(200.36)

(803.43)

(565.57

[10.741]Stranger

Acquaintance

or Relative

X2=

(O - E )2E

(O -E )2

EUpper left cell: = = 10.741

(12 -29.93)2

29.93

(E)

(O - E )2

E


Is the type of crime independent of whether the


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12

39

379

106

727

642


(29.93)

(21.07)

[15.258]

(284.64)

[31.281]

(200.36)

[44.439]

(803.43)

[7.271]

(565.57)

[10.329]

[10.741]Stranger

Acquaintance

or Relative

X2=

(O - E )2E

(O -E )2

EUpper left cell: = = 10.741

(12 -29.93)2

29.93

(E)

(O - E )2

E


Is the type of crime independent of whether the


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12

39

379

106

727

642


(29.93)

(21.07)

[15.258]

(284.64)

[31.281]

(200.36)

[44.439]

(803.43)

[7.271]

(565.57)

[10.329]

[10.741]Stranger

Acquaintance

or Relative

X2=

(O - E )2E

(E)

(O - E )2

E


Test Statistic X2= 10.741 + 31.281 + ... + 10.329 =

119.319

Test Statistic X2

= 119 319


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Test Statistic X=119.319

with = 0.05 and (r -1) (c-1) = (2 -1) (3 -1) = 2 degrees offreedomCritical Value X

2=5.991 (from Table A-4)

Test Statistic X2= 119 319


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with = 0.05 and (r -1) (c-1) = (2 -1) (3 -1) = 2 degrees offreedom

0

= 0.05

X2= 5.991

RejectIndependence

Critical Value X2=5.991 (from Table A-4)

Sample data: X2=119.319

Fail to RejectIndependence



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0

= 0.05

X2= 5.991

RejectIndependence


Reject independence





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0

= 0.05

X2= 5.991

RejectIndependence


Reject independence



Claim: The type of crime and knowledge of criminal are independentHo: The type of crime and knowledge of criminal are independentH1: The type of crime and knowledge of criminal are dependent



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It appears that the type of crime andknowledge of the criminal are related.

0

= 0.05

X2= 5.991

RejectIndependence


Reject independence



Relationships Among Components in X2 Test


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Figure 10-8

Relationships Among Components in XTest

of Independence


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Definition

Test of Homogeneity

test the claim that di f ferent popu lat ions

have the same proportions of somecharacteristics


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How to distinguish between a

test of homogeneity and a

test for independence:

Werepredetermined

sample sizesused for different populations (test of

homogeneity), or was one big sample

drawn so both row and column totalswere determined randomly (test of

independence)?

chapter 10 multinomial experiments

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