John Loucks St . Edward’s 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 Independence



Comparing Multiple Proportions,Test of Independence and Goodness of Fit

In 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.



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

H0: p1 = p2 = . . . = pk

Ha: Not all population proportions are equal

Testing the Equality of Population Proportions

for Three or More Populations

The hypotheses for the equality of population proportions for k > 3 populations are as follows:



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.





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 customersatisfaction with the three home styles.



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 owners



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.





Home Owner Colonial Log A-Frame TotalLikely to Yes 100 81 83 264RepurchaseNo 35 20 41 96

Total 135 101 124 360

Observed Frequencies (sample results)





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.



(Row Total)(Column Total)Total Sample Sizeij

i je

Expected FrequenciesUnder the Assumption H0 is True





Home Owner Colonial Log A-Frame TotalLikely to Yes 99.00 74.07 90.93 264RepurchaseNo 36.00 26.93 33.07 96

Total 135 101 124 360

Expected Frequencies (computed)



22 ( )ij ij

i j ij

f e

e

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 jeij = expected frequency for the cell in row i and column j

under the assumption H0 is true

where:





Obs. Exp. Sqd. Sqd. Diff. /

Likely to Home Freq. Freq. Diff. Diff. Exp. Freq.

Repurch. Owner fij eij (fij - eij) (fij - eij)2 (fij - eij)2/eij

Yes Colonial 97 97.50 -0.50 0.2500 0.0026

Yes Log Cab. 83 72.94 10.06 101.1142 1.3862

Yes A-Frame 80 89.56 -9.56 91.3086 1.0196

No Colonial 38 37.50 0.50 0.2500 0.0067

No Log Cab. 18 28.06 -10.06 101.1142 3.6041

No A-Frame 44 34.44 9.56 91.3086 2.6509

Total 360 360 c2 = 8.6700


for Three or More Populations Computation of the Chi-Square Test Statistic.



where is the significance level and

there are k - 1 degrees of freedom

p-value approach:

Critical value approach:

Reject H0 if p-value < a

Rejection Rule

2 2 Reject H0 if





Rejection Rule (using a = .05)

22

5.991 5.991

Do Not Reject H0Do Not Reject H0 Reject H0Reject H0

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

Reject H0 if p-value < .05 or c2 > 5.991





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 .0131






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.



Multiple Comparisons Procedure

We begin by computing the three sample proportions.

We will use a multiple comparison procedure known as the Marascuillo procedure.

1100Colonial: .741135p

281Log Cabin: .802101p

383A-Frame: .669124p




Marascuillo Procedure

We compute the absolute value of the pairwise difference between sample proportions.

1 2 .741 .802 .061p p

1 3 .741 .669 .072p p

2 3 .802 .669 .133p p

Colonial and Log Cabin:

Colonial and A-Frame:

Log Cabin and A-Frame:




Critical Values for the Marascuillo Pairwise ComparisonFor each pairwise comparison compute a critical value as follows:

2; 1

(1 )(1 ) j ji iij k

i j

p pp pCV

n n

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




i jp p CVij

Significant if

Pairwise Comparison i jp p > CVij

Colonial vs. Log Cabin

Colonial vs. A-Frame

Log Cabin vs. A-Frame

.061

.072

.133

.0923

.0971

.1034

Not Significant

Not Significant

Significant

Pairwise Comparison Tests




ei j

ij (Row Total)(Column Total)

Sample Size

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 variable




22

( )f e

eij ij

ijji

5. Determine the rejection rule.

Reject H0 if p -value < a or .

2 2

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.



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)



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.

> $99,000 12 14 16 3< $99,000 18 6 19 12


Example: Finger Lakes Homes (B)



Hypotheses


H0: Price of the home is independent of the

style of the home that is purchasedHa: Price of the home is not independent of the

style of the home that is purchased



Expected Frequencies


Price Colonial Log Split-Level A-Frame Total

< $99K

> $99K

Total 30 20 35 15 100

12 14 16 3 45

18 6 19 12 55



Rejection Rule


2.05 7.815 With = .05 and (2 - 1)(4 - 1) = 3 d.f.,

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

22 2 218 16 5

16 56 11

113 6 75

6 75 ( . )

.( )

. .( . )

. .

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

Test Statistic





Because c2 = 9.145 is between 7.815 and 9.348, the area in the upper tail of the distribution is between .05 and .025.


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





Conclusion Using the Critical Value Approach


We reject, at the .05 level of significance,the assumption that the price of the home isindependent of the style of home that ispurchased.

c2 = 9.145 > 7.815



Goodness of Fit Test:Multinomial Probability Distribution

1. State the null and alternative hypotheses.

H0: 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 categories




2. 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.




22

1

( )f ee

i i

ii

k

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 iei = expected frequency for category i

k = number of categories

where:




where is the significance level and

there are k - 1 degrees of freedom

p-value approach:

Critical value approach:

Reject H0 if p-value < a

5. Rejection rule:

2 2 Reject H0 if



Multinomial Distribution Goodness of Fit Test

Example: Finger Lakes Homes (A)

Finger Lakes Homes manufactures four models ofprefabricated homes, a two-story colonial, a log cabin,a split-level, and an A-frame. To help in productionplanning, management would like to determine ifprevious customer purchases indicate that there is apreference in the style selected.



Split- A-Model Colonial Log Level Frame# Sold 30 20 35 15

The number of homes sold of each model for 100sales over the past two years is shown below.


Example: Finger Lakes Homes (A)



Hypotheses


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 = .25

Ha: The population proportions are not

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



Rejection Rule

22

7.815 7.815

Do Not Reject H0Do Not Reject H0 Reject H0Reject H0


With = .05 and k - 1 = 4 - 1 = 3 degrees of freedom

Reject H0 if p-value < .05 or c2 > 7.815.



Expected Frequencies

Test Statistic

22 2 2 230 25

2520 25

2535 25

2515 25

25 ( ) ( ) ( ) ( )


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

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

= 1 + 1 + 4 + 4 = 10






Because c2 = 10 is between 9.348 and 11.345, the area in the upper tail of the distribution is between .025 and .01.


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



Conclusion Using the Critical Value Approach


We reject, at the .05 level of significance,the assumption that there is no home stylepreference.

c2 = 10 > 7.815



Goodness of Fit Test: Normal Distribution

1. State the null and alternative hypotheses.

3. 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 anda. 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

H0: The population has a normal distribution

Ha: The population does not have a normal distribution



4. Compute the value of the test statistic.


22

1

( )f ee

i i

ii

k2

2

1

( )f ee

i i

ii

k

5. Reject H0 if (where is the significance level

and there are k - 3 degrees of freedom).

2 2



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.




A simple random sample of 30 of the salespeople

was taken and their numbers of units sold are listed

below.

Example: IQ Computers

(mean = 71, standard deviation = 18.54)

33 43 44 45 52 52 56 58 63 6464 65 66 68 70 72 73 73 74 7583 84 85 86 91 92 94 98 102 105




Hypotheses

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.




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.




Interval Definition

Areas = 1.00/6 = .1667

Areas = 1.00/6 = .1667

717153.0253.02

71 - .43(18.54) = 63.0371 - .43(18.54) = 63.0378.9778.9788.98 = 71 + .97(18.54)88.98 = 71 + .97(18.54)




Observed and Expected Frequencies

1-2 1 0-1 1

5 5 5 5 5 530

6 3 6 5 4 630

Less 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 - ei

Total




2 2 2 2 2 22 (1) ( 2) (1) (0) ( 1) (1)

1.6005 5 5 5 5 5

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), 2

.05 7.815

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

Rejection Rule





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.


c2 Value (df = 3) .584 6.251 7.815 9.348 11.345





End of Chapter 12

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John Loucks St . Edward’s University

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