John Loucks St . Edward’s - PowerPoint Presentation

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John Loucks

St

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University

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Chapter 12 Comparing Multiple Proportions,Test of Independence and Goodness of Fit Testing the Equality of Population Proportions for Three or More Populations Goodness of Fit Test

Test of IndependenceSlide3

Comparing Multiple Proportions,Test of Independence and Goodness of FitIn this chapter we introduce three additional hypothesis-testing procedures.The test statistic and the distribution used are based on the chi-square (c2) distribution.

In all cases, the data are categorical.Slide4

Using the notation p1 = population proportion for population 1 p2 = population proportion for population 2 and pk = population proportion for population k

H

0

: p1 = p2 = . . . = pk Ha: Not all population proportions are equal

Testing the Equality of Population Proportions for Three or More PopulationsThe hypotheses for the equality of population proportions for k > 3 populations are as follows:Slide5

If H0 cannot be rejected, we cannot detect a difference among the k population proportions.If H0 can be rejected, we can conclude that not all k population proportions are equal.

Further analyses can be done to conclude which population proportions are significantly different from others.

Testing the Equality of Population

Proportions for Three or More PopulationsSlide6

Example: Finger Lakes Homes Finger Lakes Homes manufactures three models ofprefabricated homes, a two-story colonial, a log cabin,and an A-frame. To help in product-line planning, management

would like to compare the customers

atisfaction with the three home styles.

Testing the Equality of Population Proportions for Three or More Populations p1 = proportion likely to repurchase a Colonial for the population of Colonial owners p2 = proportion likely to repurchase a Log Cabin

for the population of Log Cabin owners p3 = proportion likely to repurchase an A-Frame for the population of A-Frame ownersSlide7

We begin by taking a sample of owners from each of the three populations.Each sample contains categorical data indicatingwhether the respondents are likely or not likely torepurchase the home.Testing the Equality of Population

Proportions for Three or More PopulationsSlide8

Home Owner Colonial Log A-Frame TotalLikely to Yes 100 81 83 264

Repurchase No

35

20 41 96 Total 135 101 124 360Observed Frequencies (sample results)Testing the Equality of Population Proportions for Three or More PopulationsSlide9

Next, we determine the expected frequencies under the assumption H0 is correct.If a significant difference exists between the observed and expected frequencies, H0 can be rejected.

Testing the Equality of Population Proportions

for Three or More Populations

Expected Frequencies

Under the Assumption H0 is TrueSlide10

Testing the Equality of Population Proportions for Three or More Populations Home Owner Colonial

Log A-Frame Total

Likely to

Yes 99.00 74.07 90.93 264Repurchase No 36.00 26.93 33.07 96 Total 135 101 124 360Expected Frequencies (computed)Slide11

Next, compute the value of the chi-square test statistic.

Note

: The test statistic has a chi-square

distribution with k – 1 degrees of freedom, provided the expected frequency is 5 or more for each cell.fij = observed frequency for the cell in row i and column j

eij = expected frequency for the cell in row i and column junder the assumption H0 is true where:Testing the Equality of Population Proportions for Three or More PopulationsSlide12

  Obs.Exp. Sqd.Sqd. Diff. /Likely to

HomeFreq.

Freq.

Diff.Diff.Exp. Freq.Repurch.Ownerfijeij

(fij - eij)(fij - eij)2(fij - eij)

2/eijYesColonial9797.50-0.500.25000.0026

YesLog Cab.8372.9410.06101.11421.3862YesA-Frame8089.56-9.5691.3086

1.0196NoColonial3837.500.500.25000.0067NoLog Cab.1828.06

-10.06101.11423.6041NoA-Frame4434.449.5691.30862.6509 Total

360360 c2 = 8.6700Testing the Equality of Population Proportions for Three or More Populations

Computation of the Chi-Square Test Statistic.Slide13

where  is the significance level andthere are k - 1 degrees of freedomp

-value approach:Critical value approach:

Reject

H0 if p-value < aRejection Rule

Reject H0 if Testing the Equality of Population Proportions for Three or More PopulationsSlide14

Rejection Rule (using a = .05)



2

5.991

Do Not Reject H0Reject H0

With  = .05 and k - 1 = 3 - 1 = 2 degrees of freedom Reject

H0 if p-value < .05 or c2 > 5.991Testing the Equality of Population Proportions for Three or More PopulationsSlide15

Conclusion Using the p-Value Approach The p-value < a

. We can reject the null hypothesis.

Because

c2 = 8.670 is between 9.210 and 7.378, the area in the upper tail of the distribution is between .01 and .025.Area in Upper Tail .10 .05 .025 .01 .005

c2 Value (df = 2) 4.605 5.991 7.378 9.210 10.597

Actual p-value is .0131Testing the Equality of Population Proportions for Three or More PopulationsSlide16

Testing the Equality of Population Proportions for Three or More PopulationsWe have concluded that the population proportions for the three populations of home owners are not equal.To identify where the differences between population proportions exist, we will rely on a multiple comparisons procedure.Slide17

Multiple Comparisons ProcedureWe begin by computing the three sample proportions.We will use a multiple comparison procedure known as the Marascuillo procedure.Slide18

Multiple Comparisons ProcedureMarascuillo ProcedureWe compute the absolute value of the pairwise difference between sample proportions.

Colonial and Log Cabin:

Colonial and A-Frame:

Log Cabin and A-Frame:Slide19

Multiple Comparisons ProcedureCritical Values for the Marascuillo Pairwise Comparison

For each pairwise comparison compute a critical value as follows:

For

a = .05 and k = 3: c2 = 5.991Slide20

Multiple Comparisons Procedure

CV

ij

Significant ifPairwise Comparison> CVijColonial vs. Log CabinColonial vs. A-FrameLog Cabin vs. A-Frame.061

.072.133.0923.0971.1034Not SignificantNot SignificantSignificantPairwise Comparison Tests Slide21

Test of Independence

1. Set up the null and alternative hypotheses.

2.

Select a random sample and record the observed frequency, fij , for each cell of the contingency table.3. Compute the expected frequency, eij , for each cell.

H0: The column variable is independent of the row variableHa: The column variable is not independent of the row variableSlide22

Test of Independence

5.

Determine the rejection rule.

Reject H0 if p -value < a 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.Slide23

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.Test of Independence

Example: Finger Lakes Homes (B)Slide24

Price Colonial Log Split-Level A-Frame The number of homes sold for each model andprice for the past two years is shown below. Forconvenience, the price of the home is listed as either

$99,000 or less or more than $99,000

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> $99,000 12 14 16 3< $99,000 18 6 19 12Test of Independence Example: Finger Lakes Homes (B)Slide25

HypothesesTest of IndependenceH0: Price of the home is

independent of the style of the home that is purchased

H

a: Price of the home is not independent of the style of the home that is purchasedSlide26

Expected Frequencies

Test of Independence

Price

Colonial Log Split-Level A-Frame Total< $99K> $99K Total30 20 35 15 10012 14 16 3 4518 6 19 12 55Slide27

Rejection RuleTest of Independence

With



= .05 and (2 - 1)(4 - 1) = 3 d.f.,Reject H0 if p-value < .05 or 2 > 7.815

= .1364 + 2.2727 + . . . + 2.0833 = 9.149Test StatisticSlide28

Conclusion Using the p-Value Approach The p-value < a

. We can reject the null hypothesis. Because

c

2 = 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 .005c2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838

Test of IndependenceActual p-value is .0274Slide29

Conclusion Using the Critical Value ApproachTest of Independence

We reject, at the .05 level of significance,

the assumption that the price of the home is

independent of the style of home that ispurchased.c2 = 9.145 > 7.815Slide30

Goodness of Fit Test:Multinomial Probability Distribution1. State the null and alternative hypotheses.

H

0

: The population follows a multinomial distribution with specified probabilities for each of the k categoriesHa: The population does not follow a multinomial distribution with specified probabilities for each of the k categoriesSlide31

Goodness of Fit Test:Multinomial Probability Distribution2. Select a random sample and record the observed

frequency, fi , for each of the

k

categories.3. Assuming H0 is true, compute the expected frequency, ei

, in each category by multiplying the category probability by the sample size.Slide32

Goodness of Fit Test:Multinomial Probability Distribution

4.

Compute the value of the test statistic.

Note

: The test statistic has a chi-square distribution with k – 1 df provided that the expected frequencies are 5 or more for all categories.fi = observed frequency for category ie

i = expected frequency for category ik = number of categorieswhere:Slide33

Goodness of Fit Test:

Multinomial Probability Distribution

where

 is the significance level andthere are k - 1 degrees of freedomp-value approach:Critical value approach:

Reject H0 if p-value < a

5. Rejection rule:Reject H0 if Slide34

Multinomial Distribution Goodness of Fit Test Example: Finger Lakes Homes (A) Finger Lakes Homes manufactures four models of

prefabricated homes, a two-story colonial, a log cabin,a split-level, and an A-frame. To help in production

planning, management would like to determine if

previous customer purchases indicate that there is apreference in the style selected.Slide35

Split- A-Model Colonial Log Level Frame# 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 Distribution Goodness of Fit Test Example: Finger Lakes Homes (A)Slide36

HypothesesMultinomial Distribution Goodness of Fit Test

where:

p

C = 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-frameH

0: pC = pL = pS = pA = .25Ha: The population proportions are not pC = .25, pL

= .25, pS = .25, and pA = .25Slide37

Rejection Rule



2

7.815

Do Not Reject

H

0Reject H0

Multinomial Distribution Goodness of Fit Test

With  = .05 and k - 1 = 4 - 1 = 3 degrees of freedom Reject H0 if p-value <

.05 or c2 > 7.815.Slide38

Expected Frequencies

Test Statistic

Multinomial Distribution Goodness of Fit Test

e

1 = .25(100) = 25 e2 = .25(100) = 25 e3 = .25(100) = 25 e4 = .25(100) = 25

= 1 + 1 + 4 + 4 = 10Slide39

Multinomial Distribution Goodness of Fit TestConclusion Using the p-Value Approach

The

p

-value < a . We can reject the null hypothesis. Because c2 = 10 is between 9.348 and 11.345, the area in the upper tail of the distribution is between .025 and .01.

Area in Upper Tail .10 .05 .025 .01 .005c2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838Slide40

Conclusion Using the Critical Value ApproachMultinomial Distribution Goodness of Fit Test

We reject, at the .05 level of significance,

the assumption that there is no home style

preference.c2 = 10 > 7.815Slide41

Goodness of Fit Test: Normal Distribution1. State the null and alternative hypotheses.Compute the expected frequency, ei , for each interval. (Multiply the sample size by the probability of a

normal random variable being in the 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 frequenciesH0: The population has a normal distribution

Ha: The population does not have a normal distributionSlide42

4. Compute the value of the test statistic.Goodness of Fit Test: Normal Distribution

5. Reject

H0

if (where  is the significance level and there are k - 3 degrees of freedom).Slide43

Example: IQ Computers IQ Computers (one better than HP?) manufacturesand sells a general purpose microcomputer. As partof a study to evaluate sales personnel, managementwants to determine, at a .05 significance level, if theannual sales volume (number of units sold by asalesperson) follows a normal probabilitydistribution.Goodness of Fit Test: Normal DistributionSlide44

A simple random sample of 30 of the salespeoplewas taken and their numbers of units sold are listedbelow.Example: IQ Computers(mean = 71, standard deviation = 18.54)33 43 44 45 52 52 56 58 63 64

64 65 66 68 70 72 73 73 74 7583 84 85 86 91 92 94 98 102 105

Goodness of Fit Test: Normal DistributionSlide45

HypothesesHa: 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.Goodness of Fit Test: Normal DistributionSlide46

Interval Definition To satisfy the requirement of an expectedfrequency of at least 5 in each interval we willdivide the normal distribution into 30/5 = 6equal probability intervals.Goodness of Fit Test: Normal DistributionSlide47

Interval Definition

Areas

= 1.00/6

= .1667

71

53.02

71 - .43(18.54) = 63.0378.97

88.98 = 71 + .97(18.54)Goodness of Fit Test: Normal DistributionSlide48

Observed and Expected Frequencies 1-2 1 0-1 1

5

5

5 5 5 530 6 3 6 5 4 630Less than 53.02 53.02 to 63.03 63.03 to 71.00 71.00 to 78.97

78.97 to 88.98More than 88.98 i fi ei fi - eiTotalGoodness of Fit Test: Normal DistributionSlide49

Test Statistic

With 

= .05 and

k - p - 1 = 6 - 2 - 1 = 3 d.f.(where k = number of categories and p = numberof population parameters estimated), Reject H0

if p-value < .05 or 2 > 7.815.Rejection RuleGoodness of Fit Test: Normal DistributionSlide50

Conclusion Using the p-Value Approach The p-value > a . We cannot reject the null hypothesis. There is little evidence to support rejecting the assumption the population is normally distributed with

 = 71 and  = 18.54.

Because

c2 = 1.600 is between .584 and 6.251 in the Chi-Square Distribution Table, the area in the upper tailof the distribution is between .90 and .10. Area in Upper Tail .90 .10 .05 .025 .01 c2 Value (df = 3) .584 6.251 7.815 9.348 11.345

Goodness of Fit Test: Normal DistributionActual p-value is .6594Slide51

End of Chapter 12