Tests of Goodness of Fit and Independencec

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Statistics

Tests of

Goodness of Fit and Independence

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Contents

STATISTICS in PRACTICE

Goodness of Fit Test: A MultinomialPopulation

Test of Independence

Goodness of Fit Test:

 Poisson and Normal Distributions

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STATISTICS in PRACTICE

United Way of Greater

Rochester is a nonprofit

organization dedicated

to improving the quality

of life for people. Because of enormousvolunteer involvement, United Way of GreaterRochester is able to hold its operating costs atjust eight cents of every dollar raised.

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STATISTICS in PRACTICE

One of statistical tests was to determine whetherperceptions of administrative expenses wereindependent of occupation.

In this chapter, you will learn how a statisticaltest of independence.

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Hypothesis (Goodness of Fit)Test for Proportions of aMultinomial Population

Consider the case that each element of apopulation is assigned to one and only oneof several classes or categories.

Such a population is a multinomialpopulation(多項母體).

We want to test if the population follows amultinominal distribution with specifiedprobabilities for each of the k categories.

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Hypothesis Test for Proportionsof a Multinomial Population--Procedures

1.Set up the null and alternative hypotheses.

2.Select a random sample and record theobserved frequency, fi , for each of the kcategories.

3.Assuming H0 is true, compute the expectedfrequency, ei , in each category bymultiplying the category probability by thesample size.

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4. Compute the value of the test statistic.

fi = observed frequency for category i

ei = expected frequency for category i

k = number of categories

where:

Hypothesis Test for Proportionsof a Multinomial Population--Procedures

Note: The test statistic has a chi-square distribution with (k – 1) dfprovided that the expected frequencies are 5 or more for allcategories.Note: The test statistic has a chi-square distribution with (k – 1) dfprovided that the expected frequencies are 5 or more for allcategories.

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where  is the significance level and

there are k - 1 degrees of freedom

p-value approach:

Critical value approach:

Reject H0 if p-value < Reject H0 if p-value < 

5. Rejection rule:

Reject H0 ifReject H0 if

Hypothesis Test for Proportionsof a Multinomial Population--Procedures

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Hypothesis (Goodness of Fit) Testfor Proportions of a MultinomialPopulation

Example:

The market share study being conducted by ScottMarketing Research. Over the past year market sharesstabilized at 30% for company A, 50% for company B,and 20% for company C.

Recently company C developed a “new and improved”product to replace its current entry in the market.

Company C retained Scott Marketing Research todetermine whether the new product will alter marketshares.

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Hypothesis (Goodness of Fit) Testfor Proportions of a MultinomialPopulation

Scott Marketing Research conduct a sample surveyand compute the proportion preferring eachcompany’s product.

A hypothesis test will then be conducted to seewhether the new product caused a change in marketshares.

The null and alternative hypotheses are

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Hypothesis (Goodness of Fit) Testfor Proportions of a MultinomialPopulation

Perform a goodness of fit test that willdetermine whether the sample of 200 customerpurchase preferences is consistent with the nullhypothesis.

Data

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Hypothesis (Goodness of Fit) Testfor Proportions of a MultinomialPopulation

Expected: the expected frequency for eachcategory is found by multiplying the samplesize of 200 by the hypothesized proportion forthe category.

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Hypothesis (Goodness of Fit) Testfor Proportions of a MultinomialPopulation

Computation of the Chi-square Test Statisticfor The Scott Marketing Research MarketShare Study

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Hypothesis (Goodness of Fit) Testfor Proportions of a MultinomialPopulation

p-Value Approach

The test statistic χ2 = 7.34 and 5.991 < 7.34 <7.378. Thus, the corresponding upper tail areaor p-value must be between .05 and .025. withp-value < .05, we reject H0.

Minitab or Excel can be used to show χ 2 = 7.34provides a p-value = .0255.

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Hypothesis (Goodness of Fit) Testfor Proportions of a MultinomialPopulation

Critical Value Approach

With α = .05 and 2 degrees of freedom, thecritical value for the test statistic is =5.991. The upper tail rejection rule becomes

Reject H0 if χ 2 > 5.991

With 7.34 > 5.991, we reject H0.

Comparisons of the observed and expectedfrequencies for the other two companiesindicate that company C’s gain in market sharewill hurt company A more than company B.

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Hypothesis (Goodness of Fit) Testfor Proportions of a MultinomialPopulation

Note:

The chi-square test is an approximate testand the test result may not be valid whenthe expected value for a category is lessthan 5.

If one or more categories have expectedvalues less than 5, you can combine themwith adjacent categories to achieve theminimum required expected value.

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Multinomial DistributionGoodness of Fit Test

 Example: Finger Lakes Homes (A)

Finger Lakes Homes manufactures four models ofprefabricated homes, a two-story colonial, a logcabin, a split-level, and an A-frame. To help

in production planning, management would like todetermine if previous customer purchases indicatethat there is a preference in the style selected.

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Split- A- Split- A-

Model Colonial Log Level FrameModel Colonial Log Level Frame

# Sold 30 20 35 15# Sold 30 20 35 15

The number of homes sold of each

model for 100 sales over the past two

years is shown below.

Multinomial DistributionGoodness of Fit Test

 Example: Finger Lakes Homes (A)

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Hypotheses

Multinomial DistributionGoodness of Fit Test

where:

pC = population proportion that purchase a colonial

pL = population proportion that purchase a log cabin

pS = population proportion that purchase a split-level

pA = population proportion that purchase an A-frame

H0: pC = pL = pS = pA = .25H0: pC = pL = pS = pA = .25

Ha: The population proportions are notHa: The population proportions are not

pC = .25, pL = .25, pS = .25, and pA = .25 pC = .25, pL = .25, pS = .25, and pA = .25

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Rejection Rule

χ2

7.815

Do Not Reject H0

Reject H0

With  = .05 andWith  = .05 and

k - 1 = 4 - 1 = 3 k - 1 = 4 - 1 = 3

degrees of freedom degrees of freedom

if p-value < .05 or > 7.815. Reject H0 if p-value < .05 or χ2 > 7.815.

Multinomial DistributionGoodness of Fit Test

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Expected Frequencies

Test Statistic

Multinomial DistributionGoodness of Fit Test

e1 = .25(100) = 25 e2 = .25(100) = 25e1 = .25(100) = 25 e2 = .25(100) = 25

e3 = .25(100) = 25 e4 = .25(100) = 25 e3 = .25(100) = 25 e4 = .25(100) = 25

= 1 + 1 + 4 + 4= 1 + 1 + 4 + 4

= 10= 10

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Multinomial DistributionGoodness of Fit Test

Conclusion Using the p-Value Approach

The p-value <  . We can reject the null hypothesis.

Because χ2 = 10 is between 9.348 and 11.345, thearea in the upper tail of the distribution is

between .025 and .01.

Area in Upper Tail .10 .05 .025 .01 .005Area in Upper Tail .10 .05 .025 .01 .005

2 Value (df = 3) 6.251 7.815 9.348 11.345 12.8382 Value (df = 3) 6.251 7.815 9.348 11.345 12.838

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Conclusion Using the Critical ValueApproach

Multinomial DistributionGoodness of Fit Test

We reject, at the .05 level of significance, theassumption that there is no home style preference.We reject, at the .05 level of significance, theassumption that there is no home style preference.

χ2 = 10 > 7.815

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Test of Independence:Contingency Tables(列聯表）

Another important application of the chi-square distribution involves using sampledata to test for the independence of twovariables.

Contingency table is a table that lists allpossible combinations of the two variables.

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Test of Independence: ContingencyTables

Example:

 A test of independence addresses the question ofwhether the beer preference (light, regular, or dark)is independent of the gender of the beer drinker(male, female).

If the independence assumption is valid, we arguethat the fraction of drinking light beer, regular beerand dark beer must be applicable to both male andfemale beer drinker.

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Test of Independence:Contingency Tables

1.Set up the null and alternative hypotheses.

2.Select a random sample and record theobserved frequency, fij , for each cell of thecontingency table.

3. Compute the expected frequency, eij , foreach cell.

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5. Determine the rejection rule.

Reject H0 if p -value <  or .Reject H0 if p -value <  or .

4. Compute the test statistic.

where  is the significance level and,

with n rows and m columns, there are

(n - 1)(m - 1) degrees of freedom.

Test of Independence:Contingency Tables

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Test of Independence: ContingencyTables

Example:

 A test of independence addresses the question ofwhether the beer preference (light, regular, or dark)is independent of the gender of the beer drinker(male, female).

The hypotheses for this test of independence are:

H0: Beer preference is independent of the gender

of the beer drinker

Ha: Beer preference is not independent of the

gender of the beer drinker

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Sample Results

Expected Frequencies

Test of Independence:Contingency Tables

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80*1/3=26.67

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Test of Independence:Contingency Tables

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Computation of the Chi-square TestStatistic

p-value = .0468. At the .05 level ofsignificance, p-value < α= .05. Wereject the null hypothesis

Test of Independence:Contingency Tables

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Each home sold by Finger Lakes Homes can beclassified according to price and to style. FingerLakes’manager would like to determine if theprice of the home and the style of the home areindependent variables.

 Example: Finger Lakes Homes (B)

Test of Independence:Contingency Tables

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The number of homes sold for each modeland price for the past two years is shown below.For convenience, the price of the home is listedas either $99,000 or less or more than $99,000.

Price Colonial Log Split-Level A-Frame Price Colonial Log Split-Level A-Frame

> $99,000 12 14 16 3> $99,000 12 14 16 3

< $99,000 18 6 19 12< $99,000 18 6 19 12

 Example: Finger Lakes Homes (B)

Test of Independence:Contingency Tables

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Hypotheses

H0: Price of the home is independent of theH0: Price of the home is independent of the

style of the home that is purchased style of the home that is purchased

Ha: Price of the home is not independent of theHa: Price of the home is not independent of the

style of the home that is purchased style of the home that is purchased

Test of Independence:Contingency Tables

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Expected Frequencies

Price Colonial Log Split-Level A-Frame Total Price Colonial Log Split-Level A-Frame Total

< $99K< $99K

> $99K> $99K

Total Total

30 20 35 15 10030 20 35 15 100

12 14 16 3 4512 14 16 3 45

18 6 19 12 5518 6 19 12 55

Test of Independence:Contingency Tables

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Contingency Table(Independence) Test

Rejection Rule

Reject H0 if p-value < .05 or 2 > 7.815Reject H0 if p-value < .05 or 2 > 7.815

Test Statistic

With α = .05 and (2 - 1)(4 - 1) = 3 d.f., =7.815

= .1364 + 2.2727 + . . . + 2.0833 = 9.149

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Conclusion Using the p-ValueApproach

The p-value <  . We can reject the null

hypothesis.

Because χ2 = 9.145 is between 7.815 and 9.348,

the area in the upper tail of the distribution is

between .05 and .025.

Area in Upper Tail .10 .05 .025 .01 .005Area in Upper Tail .10 .05 .025 .01 .005

2 Value (df = 3) 6.251 7.815 9.348 11.345 12.8382 Value (df = 3) 6.251 7.815 9.348 11.345 12.838

Contingency Table(Independence) Test

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Conclusion Using the Critical ValueApproach

We reject, at the .05 level of significance, We reject, at the .05 level of significance,

the assumption that the price of the home isthe assumption that the price of the home is

independent of the style of home that isindependent of the style of home that is

purchased.purchased.

χ2 = 9.145 > 7.815

Contingency Table(Independence) Test

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Goodness of Fit Test: PoissonDistribution

 In general, the goodness of fit test can be usedwith any hypothesized probability distribution.

Here we will demonstrate the cases of Poissondistribution and Normal distribution.

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Goodness of Fit Test: PoissonDistribution

1. Set up the null and alternative hypotheses.

H0: Population has a Poisson probabilitydistribution

Ha: Population does not have a Poissondistribution

2. Select a random sample and

a. Record the observed frequency fi for eachvalue of the Poisson random variable.

b. Compute the mean number of occurrences .

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Goodness of Fit Test: PoissonDistribution

3. Compute the expected frequency of occurrences ei

for each value of the Poisson random variable.

4. Compute the value of the test statistic.

fi = observed frequency for category i

ei = expected frequency for category i

k = number of categories

where:

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where  is the significance level and

there are k - 2 degrees of freedom

p-value approach:

Critical value approach:

Reject H0 if p-value < Reject H0 if p-value < 

5. Rejection rule:

Reject H0 ifReject H0 if

Goodness of Fit Test: PoissonDistribution

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Goodness of Fit Test: PoissonDistribution—Example

Consider the arrival of customers atDubek’s Food Market in Tallahassee,Florida.

A statistical test conducted to seewhether an assumption of a Poissondistribution for arrivals is reasonable.

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Goodness of Fit Test: PoissonDistribution—Example

Hypotheses are

H0: The number of customers enteringthe store during 5-minute intervalshas a Poisson probability distribution

Ha: The number of customers entering thestore during 5-minute intervals doesnot have a Poisson distribution

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Goodness of Fit Test: PoissonDistribution—Example

The Poisson probability function,

where

--μ represents the expected number of customersarriving per 5-minute period,

--x is the random variable indicating the number ofcustomers arriving during a 5-minute period, and

--f (x) is the probability that x customers will arrive ina 5-minute interval.

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Goodness of Fit Test: PoissonDistribution—Example

Sample Data

μ known, an estimateof μ is 640/128 = 5customers

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Goodness of Fit Test: PoissonDistribution—Example

Expected frequency

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Goodness of Fit Test: PoissonDistribution—Example

Computation of the Chi-square TestStatistic

p-value = .1403. With p-value > α= .05, we cannot reject H0.

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Example: Troy Parking Garage

In studying the need for an additionalentrance to a city parking garage, aconsultant has recommended an analysisapproach that is applicable only in situationswhere the number of cars entering during aspecified time period follows a Poissondistribution.

Goodness of Fit Test:Poisson Distribution

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A random sample of 100 one-minute timeintervals resulted in the customer arrivalslisted below. A statistical test must beconducted to see if the assumption of aPoisson distribution is reasonable.

Goodness of Fit Test:Poisson Distribution

# Arrivals 0 1 2 3 4 5 6 7 8 9 10 11 12# Arrivals 0 1 2 3 4 5 6 7 8 9 10 11 12

Frequency 0 1 4 10 14 20 12 12 9 8 6 3 1Frequency 0 1 4 10 14 20 12 12 9 8 6 3 1

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Hypotheses

Goodness of Fit Test:Poisson Distribution

Ha: Number of cars entering the garage during a

one-minute interval is not Poisson distributed

H0: Number of cars entering the garage during

a one-minute interval is Poisson distributed

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Estimate of Poisson ProbabilityFunction

Goodness of Fit Test:Poisson Distribution

otal Arrivals = 0(0) + 1(1) + 2(4) + . . . + 12(1) = 600

Hence,

Estimate of  = 600/100 = 6

Total Time Periods = 100

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Expected Frequencies

Goodness of Fit Test:Poisson Distribution

x f (x ) nf (x )x f (x ) nf (x )

13.77 13.77

10.33 10.33

6.88 6.88

4.13 4.13

2.25 2.25

2.01 2.01

100.00100.00

.1377 .1377

.1033 .1033

.0688 .0688

.0413 .0413

.0225 .0225

.0201 .0201

1.00001.0000

7 7

8 8

9 9

10 10

11 11

12+ 12+

TotalTotal

.0025.0025

.0149.0149

.0446.0446

.0892.0892

.1339.1339

.1606.1606

.25 .25

1.49 1.49

4.46 4.46

8.92 8.92

13.3913.39

16.0616.06

x f (x ) nf (x )x f (x ) nf (x )

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Observed and Expected Frequencies

Goodness of Fit Test:Poisson Distribution

i fi ei fi - ei i fi ei fi - ei

-1.20-1.20

1.08 1.08

0.61 0.61

3.94 3.94

-4.06-4.06

-1.77-1.77

-1.33-1.33

1.12 1.12

1.61 1.61

6.20 6.20

8.92 8.92

13.3913.39

16.0616.06

13.7713.77

10.3310.33

6.88 6.88

8.39 8.39

5 5

1010

1414

2020

1212

9 9

8 8

1010

0 or 1 or 20 or 1 or 2

3 3

4 4

5 5

6 6

7 7

8 8

9 9

10 or more10 or more

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Test Statistic

Goodness of Fit Test:Poisson Distribution

Reject H0 if p-value < .05 or 2 > 14.067.Reject H0 if p-value < .05 or 2 > 14.067.

Rejection Rule

With  = .05 and k - p - 1 = 9 - 1 - 1 = 7 d.f.

(where k = number of categories and p = number

of population parameters estimated),

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Conclusion Using the p-Value Approach

The p-value >  . We cannot reject the null hypothesis.There is no reason to doubt the assumption of a Poissondistribution.

Because χ2 = 3.268 is between 2.833 and 12.017 in theChi-Square Distribution Table, the area in the upper tailof the distribution is between .90 and .10.

Area in Upper Tail .90 .10 .05 .025 .01Area in Upper Tail .90 .10 .05 .025 .01

2 Value (df = 7) 2.833 12.017 14.067 16.013 18.4752 Value (df = 7) 2.833 12.017 14.067 16.013 18.475

Goodness of Fit Test:Poisson Distribution

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Goodness of Fit Test: NormalDistribution

1. Set up the null and alternative hypotheses.

3. Compute the expected frequency, ei , for each interval.

2. Select a random sample and

a. Compute the mean and standard deviation.

b. Define intervals of values so that the expected

frequency is at least 5 for each interval.

c. For each interval record the observed frequencies

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4. Compute the value of the test statistic.

Goodness of Fit Test: NormalDistribution

5. Reject H0 if (where  is the significance

level and there are k - 3 degrees of freedom).

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Goodness of Fit Test: NormalDistribution--Example

Chemline EmployeeAptitude Scores

To test the nullhypothesis that thepopulation of testscores has a normaldistribution.

Sample Data

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Goodness of Fit Test: NormalDistribution—Example

Estimates of μ and are = 68.42, s = 10.41.

Hypotheses

H0: The population of test scores has anormal distribution with mean68.42 and standard deviation 10.41.

Ha: The population of test scores does nothave a normal distribution with mean68.42 and standard deviation 10.41



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Goodness of Fit Test: NormalDistribution—Example

Expected Frequency

Normal Distribution with 10 Equal-Probability Intervals (The ChemilineExample)

Note: 55.10 = 68.42 - 1.28(10.41)

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Goodness of Fit Test: NormalDistribution—Example

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Goodness of Fit Test: NormalDistribution—Example

Computation of the Chi-square TestStatistic

p-value = .4084 >α = .10. Cannot rejectH0.

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Normal DistributionGoodness of Fit Test

Example: IQ Computers

IQIQ

IQ Computers (one better than

HP?) manufactures and sells a

general purpose microcomputer. As part of

a study to evaluate sales personnel, management

wants to determine, at a .05 significance level, if theannual sales volume (number of units sold by asalesperson) follows a normal probabilitydistribution.

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A simple random sample of 30 of thesalespeople was taken and theirnumbers of units sold are below.

Normal DistributionGoodness of Fit Test

Example: IQ Computers

(mean = 71, standard deviation = 18.54)

33 43 44 45 52 52 56 58 63 6433 43 44 45 52 52 56 58 63 64

64 65 66 68 70 72 73 73 74 7564 65 66 68 70 72 73 73 74 75

83 84 85 86 91 92 94 98 102 10583 84 85 86 91 92 94 98 102 105

IQIQ

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Hypotheses

Normal DistributionGoodness of Fit Test

Ha: The population of number of units sold

does not have a normal distribution with

mean 71 and standard deviation 18.54.

H0: The population of number of units sold

has a normal distribution with mean 71

and standard deviation 18.54.

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Interval Definition

Normal DistributionGoodness of Fit Test

To satisfy the requirement of an expected

frequency of at least 5 in each interval we

will divide the normal distribution into

30/5 = 6 equal probability intervals.

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Interval Definition

Areas

= 1.00/6

= .1667

Areas

= 1.00/6

= .1667

53.02

71  .43(18.54) = 63.03

78.97

88.98 = 71 + .97(18.54)

Normal DistributionGoodness of Fit Test

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Observed and Expected Frequencies

Normal DistributionGoodness of Fit Test

1 1

-2-2

1 1

0 0

-1-1

1 1

5 5

3030

6 6

3 3

6 6

5 5

4 4

6 6

3030

Less than 53.02Less than 53.02

53.02 to 63.03 53.02 to 63.03

63.03 to 71.00 63.03 to 71.00

71.00 to 78.97 71.00 to 78.97

78.97 to 88.98 78.97 to 88.98

More than 88.98More than 88.98

i fi ei fi - ei i fi ei fi - ei

TotalTotal

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Test Statistic

Reject H0 if p-value < .05 or 2 > 7.815.Reject H0 if p-value < .05 or 2 > 7.815.

Rejection Rule

Normal DistributionGoodness of Fit Test

With  = .05 and k - p - 1 = 6 - 2 - 1 = 3 d.f.

(where k = number of categories and p = number

of population parameters estimated),

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Normal DistributionGoodness of Fit Test

Conclusion Using the p-Value Approach

 Because χ2 = 1.600 is between .584 and 6.251 in the Chi-Square Distribution Table, the area in the upper tail of thedistribution is between .90 and .10. The p-value >  . Wecannot reject the null hypothesis.

There is little evidence to support rejecting theassumption the population is normally distributed with  =71 and  = 18.54.

Area in Upper Tail .90 .10 .05 .025 .01Area in Upper Tail .90 .10 .05 .025 .01

2 Value (df = 3) .584 6.251 7.815 9.348 11.3452 Value (df = 3) .584 6.251 7.815 9.348 11.345