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Chapter Fifteen

Frequency Distribution,Cross-Tabulation,

and Hypothesis Testing

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15-2

Chapter Outline

1) Overview

2) Frequency Distribution

3) Statistics Associated with Frequency Distribution

i. Measures of Locationii. Measures of Variability

iii. Measures of Shape

4) Introduction to Hypothesis Testing5) A General Procedure for Hypothesis Testing

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Chapter Outline

6) Cross-Tabulations

i. Two Variable Case

ii. Three Variable Case

iii. General Comments on Cross-Tabulations

7) Statistics Associated with Cross-Tabulation

i. Chi-Square

ii. Phi Correlation Coefficient

iii. Contingency Coefficient

iv. Cramers V

v. Lambda Coefficient

vi. Other Statistics

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Chapter Outline

8) Cross-Tabulation in Practice

9) Hypothesis Testing Related to Differences

10) Parametric Tests

i. One Sample

ii. Two Independent Samples

iii. Paired Samples

11) Non-parametric Tests

i. One Sampleii. Two Independent Samples

iii. Paired Samples

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Chapter Outline

12) Internet and Computer Applications

13) Focus on Burke

14) Summary

15) Key Terms and Concepts

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15-6

Respondent Sex Familiarity Internet Attitude Toward Usage of InternetNumber Usage Internet Technology Shopping Banking1 1.00 7.00 14.00 7.00 6.00 1.00 1.00

2 2.00 2.00 2.00 3.00 3.00 2.00 2.00

3 2.00 3.00 3.00 4.00 3.00 1.00 2.00

4 2.00 3.00 3.00 7.00 5.00 1.00 2.00

5 1.00 7.00 13.00 7.00 7.00 1.00 1.00

6 2.00 4.00 6.00 5.00 4.00 1.00 2.00

7 2.00 2.00 2.00 4.00 5.00 2.00 2.00

8 2.00 3.00 6.00 5.00 4.00 2.00 2.00

9 2.00 3.00 6.00 6.00 4.00 1.00 2.0010 1.00 9.00 15.00 7.00 6.00 1.00 2.00

11 2.00 4.00 3.00 4.00 3.00 2.00 2.00

12 2.00 5.00 4.00 6.00 4.00 2.00 2.00

13 1.00 6.00 9.00 6.00 5.00 2.00 1.00

14 1.00 6.00 8.00 3.00 2.00 2.00 2.00

15 1.00 6.00 5.00 5.00 4.00 1.00 2.00

16 2.00 4.00 3.00 4.00 3.00 2.00 2.00

17 1.00 6.00 9.00 5.00 3.00 1.00 1.00

18 1.00 4.00 4.00 5.00 4.00 1.00 2.00

19 1.00 7.00 14.00 6.00 6.00 1.00 1.0020 2.00 6.00 6.00 6.00 4.00 2.00 2.00

21 1.00 6.00 9.00 4.00 2.00 2.00 2.00

22 1.00 5.00 5.00 5.00 4.00 2.00 1.00

23 2.00 3.00 2.00 4.00 2.00 2.00 2.00

24 1.00 7.00 15.00 6.00 6.00 1.00 1.00

25 2.00 6.00 6.00 5.00 3.00 1.00 2.00

26 1.00 6.00 13.00 6.00 6.00 1.00 1.00

27 2.00 5.00 4.00 5.00 5.00 1.00 1.00

28 2.00 4.00 2.00 3.00 2.00 2.00 2.00

29 1.00 4.00 4.00 5.00 3.00 1.00 2.00

30 1.00 3.00 3.00 7.00 5.00 1.00 2.00

Internet Usage Data

Table 15.1

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15-7

Frequency Distribution

In a frequency distribution, one variable isconsidered at a time.

A frequency distribution for a variable produces atable of frequency counts, percentages, andcumulative percentages for all the values associated

with that variable.

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15-8Frequency Distribution of Familiaritywith the Internet

Table 15.2

Valid Cumulative

Value label Value Frequency (N) Percentage percentage percentage

Not so familiar 1 0 0.0 0.0 0.0

2 2 6.7 6.9 6.9

3 6 20.0 20.7 27.6

4 6 20.0 20.7 48.3

5 3 10.0 10.3 58.6

6 8 26.7 27.6 86.2

Very familiar 7 4 13.3 13.8 100.0

Missing 9 1 3.3

TOTAL 30 100.0 100.0

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Frequency HistogramFigure 15.1

2 3 4 5 6 70

7

4

3

2

1

6

5

Frequency

Familiarity

8

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The mean, or average value, is the most commonly usedmeasure of central tendency. The mean, ,is given by

Where,

Xi = Observed values of the variable X

n = Number of observations (sample size)

The mode is the value that occurs most frequently. It

represents the highest peak of the distribution. The modeis a good measure of location when the variable isinherently categorical or has otherwise been grouped intocategories.

Statistics Associated with Frequency DistributionMeasures of Location

X= Xi/nSi=1

nX

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15-11

The median of a sample is the middle value whenthe data are arranged in ascending or descendingorder. If the number of data points is even, themedian is usually estimated as the midpoint betweenthe two middle values by adding the two middle

values and dividing their sum by 2. The median isthe 50th percentile.

Statistics Associated with Frequency DistributionMeasures of Location

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The range measures the spread of the data. It issimply the difference between the largest andsmallest values in the sample. Range = XlargestXsmallest.

The interquartile range is the difference between

the 75th and 25th percentile. For a set of datapoints arranged in order of magnitude, the pthpercentile is the value that has p% of the data pointsbelow it and (100 - p)% above it.

Statistics Associated with Frequency DistributionMeasures of Variability

15 13

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The variance is the mean squared deviation fromthe mean. The variance can never be negative.

The standard deviation is the square root of thevariance.

The coefficient of variation is the ratio of the

standard deviation to the mean expressed as apercentage, and is a unitless measure of relativevariability.

sx= (Xi-X)2

n-1Si=1

n

CV= sx/X

Statistics Associated with Frequency DistributionMeasures of Variability

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Skewness. The tendency of the deviations from the

mean to be larger in one direction than in the other.It can be thought of as the tendency for one tail ofthe distribution to be heavier than the other.

Kurtosis is a measure of the relative peakedness or

flatness of the curve defined by the frequencydistribution. The kurtosis of a normal distribution iszero. If the kurtosis is positive, then the distributionis more peaked than a normal distribution. Anegative value means that the distribution is flatter

than a normal distribution.

Statistics Associated with Frequency DistributionMeasures of Shape

15 15

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15-15

Skewness of a DistributionFigure 15.2

Skewed Distribution

Symmetric Distribution

Mean

MedianMode(a)

Mean Median Mode

(b)

15 16

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15-16

Steps Involved in Hypothesis TestingFig. 15.3

Draw Marketing Research Conclusion

Formulate H0 and H1

Select Appropriate Test

Choose Level of Significance

Determine ProbabilityAssociated with Test

Statistic

Determine CriticalValue of Test Statistic

TSCR

Determine if TSCRfalls into (Non)

Rejection Region

Compare with Levelof Significance,

Reject or Do not Reject H0

Collect Data and Calculate Test Statistic

15 17A G l P d f H h i T i

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15-17A General Procedure for Hypothesis TestingStep 1: Formulate the Hypothesis

Anull hypothesis is a statement of the status quo,one of no difference or no effect. If the nullhypothesis is not rejected, no changes will be made.

An alternative hypothesis is one in which somedifference or effect is expected. Accepting the

alternative hypothesis will lead to changes in opinionsor actions.

The null hypothesis refers to a specified value of thepopulation parameter (e.g., ), not a sample

statistic (e.g., ).

, ,

15 18A G l P d f H th i T ti

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A null hypothesis may be rejected, but it can neverbe accepted based on a single test. In classicalhypothesis testing, there is no way to determinewhether the null hypothesis is true.

In marketing research, the null hypothesis is

formulated in such a way that its rejection leads tothe acceptance of the desired conclusion. Thealternative hypothesis represents the conclusion forwhich evidence is sought.

0: 0.40

H1:>0.40

A General Procedure for Hypothesis TestingStep 1: Formulate the Hypothesis

15 19A G l P d f H th i T ti

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The test of the null hypothesis is a one-tailed test,because the alternative hypothesis is expresseddirectionally. If that is not the case, then a two-tailed test would be required, and the hypotheseswould be expressed as:

H0:=0.40

H

1:0.40

A General Procedure for Hypothesis TestingStep 1: Formulate the Hypothesis

15-20A G l P d f H th i T ti

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The test statistic measures how close the samplehas come to the null hypothesis.

The test statistic often follows a well-knowndistribution, such as the normal, t, or chi-squaredistribution.

In our example, the zstatistic, which follows thestandard normal distribution, would be appropriate.

A General Procedure for Hypothesis TestingStep 2: Select an Appropriate Test

=p -

pwhere

p

=

n


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15 21

Type I Error Type Ierror occurs when the sample results lead to

the rejection of the null hypothesis when it is in facttrue.

The probability of type I error ( ) is also called thelevel of significance.

Type II Error Type II error occurs when, based on the sample

results, the null hypothesis is not rejected when it isin fact false.

The probability of type II error is denoted by . Unlike , which is specified by the researcher, the

magnitude of depends on the actual value of thepopulation parameter (proportion).

A General Procedure for Hypothesis TestingStep 3: Choose a Level of Significance


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15 22

Power of a Test

The power of a test is the probability (1 - ) ofrejecting the null hypothesis when it is false andshould be rejected.

Although is unknown, it is related to . An

extremely low value of (e.g., = 0.001) will result inintolerably high errors.

So it is necessary to balance the two types of errors.

A General Procedure for Hypothesis TestingStep 3: Choose a Level of Significance

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15 23

Probabilities of Type I & Type II ErrorFigure 15.4

99% of

Total Area

Critical ValueofZ

= 0.40

= 0.45 = 0.01

= 1.645Z

= -2.33ZZ

Z

95% ofTotal Area = 0.05

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15 24

Probability of z with a One-Tailed Test

Unshaded Area

= 0.0301

Fig. 15.5

Shaded Area

= 0.9699

z = 1.880

15-25A General Procedure for Hypothesis Testing

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15 25

The required data are collected and the value of thetest statistic computed.

In our example, the value of the sample proportion is= 17/30 = 0.567.

The value of can be determined as follows:

A General Procedure for Hypothesis TestingStep 4: Collect Data and Calculate Test Statistic

p

=(1 - )

n

=(0.40)(0.6)

30

= 0.089


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The test statistic zcan be calculated as follows:

p

pz

= 0.567-0.400.089

= 1.88

A General Procedure for Hypothesis TestingStep 4: Collect Data and Calculate Test Statistic

15-27

A General Procedure for Hypothesis Testing

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Using standard normal tables (Table 2 of the

Statistical Appendix), the probability of obtaining a zvalue of 1.88 can be calculated (see Figure 15.5).

The shaded area between - and 1.88 is 0.9699.Therefore, the area to the right ofz= 1.88 is 1.0000

- 0.9699 = 0.0301. Alternatively, the critical value ofz, which will give an

area to the right side of the critical value of 0.05, isbetween 1.64 and 1.65 and equals 1.645.

Note, in determining the critical value of the test

statistic, the area to the right of the critical value iseither or . It is for a one-tail test and

for a two-tail test.

A General Procedure for Hypothesis TestingStep 5: Determine the Probability (Critical Value)

//


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If the probability associated with the calculated or

observed value of the test statistic ( )is lessthan the level of significance ( ), the null hypothesisis rejected.

The probability associated with the calculated or

observed value of the test statistic is 0.0301. This isthe probability of getting a pvalue of 0.567 when= 0.40. This is less than the level of significance of0.05. Hence, the null hypothesis is rejected.

Alternatively, if the calculated value of the test

statistic is greater than the critical value of the teststatistic ( ), the null hypothesis is rejected.

A General Procedure for Hypothesis TestingSteps 6 & 7: Compare the Probability (Critical Value) and Makingthe Decision

TSCR

TSCA


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The calculated value of the test statistic z= 1.88 liesin the rejection region, beyond the value of 1.645.Again, the same conclusion to reject the nullhypothesis is reached.

Note that the two ways of testing the null hypothesis

are equivalent but mathematically opposite in thedirection of comparison.

If the probability of < significance level ( )then reject H0 but if > then reject H0.

A General Procedure for Hypothesis TestingSteps 6 & 7: Compare the Probability (Critical Value) and Makingthe Decision

TSCA

TSCA TSCR


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The conclusion reached by hypothesis testing mustbe expressed in terms of the marketing researchproblem.

In our example, we conclude that there is evidencethat the proportion of Internet users who shop via

the Internet is significantly greater than 0.40.Hence, the recommendation to the department storewould be to introduce the new Internet shoppingservice.

A General Procedure for Hypothesis TestingStep 8: Marketing Research Conclusion

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A Broad Classification of Hypothesis Tests

Median/Rankings

Distributions Means Proportions

Figure 15.6

Tests of

Association

Tests of

Differences

Hypothesis Tests

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Cross-Tabulation

While a frequency distribution describes one variableat a time, a cross-tabulation describes two or morevariables simultaneously.

Cross-tabulation results in tables that reflect the jointdistribution of two or more variables with a limited

number of categories or distinct values, e.g., Table15.3.

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Gender and Internet UsageTable 15.3

Gender

RowInternet Usage Male Female Total

Light (1) 5 10 15

Heavy (2) 10 5 15

Column Total 15 15

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Two Variables Cross-Tabulation

Since two variables have been cross classified,

percentages could be computed either columnwise,based on column totals (Table 15.4), or rowwise,based on row totals (Table 15.5).

The general rule is to compute the percentages in

the direction of the independent variable, across thedependent variable. The correct way of calculatingpercentages is as shown in Table 15.4.

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Internet Usage by GenderTable 15.4

Gender

Internet Usage Male Female

Light 33.3% 66.7%

Heavy 66.7% 33.3%

Column total 100% 100%

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Gender by Internet UsageTable 15.5

Internet Usage

Gender Light Heavy Total

Male 33.3% 66.7% 100.0%

Female 66.7% 33.3% 100.0%

15-37Introduction of a Third Variable in Cross-

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Refined Associationbetween the Two

Variables

No Associationbetween the Two

Variables

No Change inthe InitialPattern

Some Associationbetween the Two

Variables

Introduction of a Third Variable in CrossTabulation

Fig. 15.7

Some Associationbetween the Two

Variables

No Associationbetween the Two

Variables

Introduce a ThirdVariable

Introduce a ThirdVariable

Original Two Variables

15-38Three Variables Cross-Tabulation

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As shown in Figure 15.7, the introduction of a third

variable can result in four possibilities:

As can be seen from Table 15.6, 52% of unmarried respondentsfell in the high-purchase category, as opposed to 31% of themarried respondents. Before concluding that unmarriedrespondents purchase more fashion clothing than those who aremarried, a third variable, the buyer's sex, was introduced into the

analysis. As shown in Table 15.7, in the case of females, 60% of the

unmarried fall in the high-purchase category, as compared to 25%of those who are married. On the other hand, the percentages aremuch closer for males, with 40% of the unmarried and 35% of themarried falling in the high purchase category.

Hence, the introduction of sex (third variable) has refined therelationship between marital status and purchase of fashionclothing (original variables). Unmarried respondents are morelikely to fall in the high purchase category than married ones, andthis effect is much more pronounced for females than for males.

Three Variables Cross TabulationRefine an Initial Relationship

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Purchase of Fashion Clothing by Marital Status

Table 15.6

Purchase ofFashion

Current Marital Status

Clothing Married Unmarried

High 31% 52%

Low 69% 48%

Column 100% 100%

Number ofrespondents

700 300

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Purchase of Fashion Clothing by Marital Status

Table 15.7

Purchase ofFashion

SexMale Female

Clothing Married Not

Married

Married Not

Married

High 35% 40% 25% 60%

Low 65% 60% 75% 40%

Column

totals

100% 100% 100% 100%

Number ofcases

400 120 300 180


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Table 15.8 shows that 32% of those with college

degrees own an expensive automobile, as comparedto 21% of those without college degrees. Realizingthat income may also be a factor, the researcherdecided to reexamine the relationship betweeneducation and ownership of expensive automobiles inlight of income level.

In Table 15.9, the percentages of those with andwithout college degrees who own expensiveautomobiles are the same for each of the incomegroups. When the data for the high income and lowincome groups are examined separately, the

association between education and ownership ofexpensive automobiles disappears, indicating that theinitial relationship observed between these twovariables was spurious.

Three Variables Cross TabulationInitial Relationship was Spurious

15-42Ownership of Expensive Automobiles by

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Ownership of Expensive Automobiles byEducation Level

Table 15.8

Own ExpensiveAutomobile

Education

College Degree No College Degree

Yes 32% 21%

No 68% 79%

Column totals 100% 100%

Number of cases 250 750

15-43Ownership of Expensive Automobiles by

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Ownership of Expensive Automobiles byEducation Level and Income Levels

Table 15.9

Own

Expensive

Automobile

College

Degree

No

College

Degree

College

Degree

No College

Degree

Yes 20% 20% 40% 40%

No 80% 80% 60% 60%

Column totals 100% 100% 100% 100%

Number of

respondents

100 700 150 50

Low Income High Income

Income


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Table 15.10 shows no association between desire to travel

abroad and age. When sex was introduced as the third variable, Table

15.11 was obtained. Among men, 60% of those under 45indicated a desire to travel abroad, as compared to 40%of those 45 or older. The pattern was reversed forwomen, where 35% of those under 45 indicated a desire

to travel abroad as opposed to 65% of those 45 or older. Since the association between desire to travel abroad and

age runs in the opposite direction for males and females,the relationship between these two variables is maskedwhen the data are aggregated across sex as in Table15.10.

But when the effect of sex is controlled, as in Table 15.11,the suppressed association between desire to travelabroad and age is revealed for the separate categories ofmales and females.

Three Variables Cross TabulationReveal Suppressed Association

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Desire to Travel Abroad by AgeTable 15.10

Desire to Travel Abroad Age

Less than 45 45 or More

Yes 50% 50%

No 50% 50%


Number of respondents 500 500

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Desire to Travel Abroad by Age and Gender

Table 15.11

Desire toTravel

Abroad

SexMale

Age

Female

Age

< 45 >=45 =45

Yes 60% 40% 35% 65%

No 40% 60% 65% 35%

Column

totals

100% 100% 100% 100%

Number ofCases

300 300 200 200

15-47Three Variables Cross-Tabulations

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Consider the cross-tabulation of family size and the

tendency to eat out frequently in fast-foodrestaurants as shown in Table 15.12. No associationis observed.

When income was introduced as a third variable in

the analysis, Table 15.13 was obtained. Again, noassociation was observed.

No Change in Initial Relationship

15-48Eating Frequently in Fast-Food

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g q yRestaurants by Family Size

Table 15.12

Eat Frequently in Fast-Food Restaurants

Family Size

Small Large

Yes 65% 65%

No 35% 35%


Number of cases 500 500

15-49Eating Frequently in Fast Food-Restaurants

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Small Large Small Large

Yes 65% 65% 65% 65%

No 35% 35% 35% 35%

Column totals 100% 100% 100% 100%Number of respondents 250 250 250 250

Income

Eat Frequently in Fast-

Food Restaurants

Family size Family size

Low High

g q yby Family Size & Income

Table 15.13

15-50Statistics Associated with Cross-Tabulation

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To determine whether a systematic association

exists, the probability of obtaining a value of chi-square as large or larger than the one calculatedfrom the cross-tabulation is estimated.

An important characteristic of the chi-square statisticis the number of degrees of freedom (df) associatedwith it. That is, df = (r- 1) x (c-1).

The null hypothesis (H0) of no association betweenthe two variables will be rejected only when thecalculated value of the test statistic is greater than

the critical value of the chi-square distribution withthe appropriate degrees of freedom, as shown inFigure 15.8.

Chi-Square

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Chi-square DistributionFigure 15.8

Reject H0

Do Not RejectH0

CriticalValue

2


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Chi-Square

The chi-square statistic ( )is used to test the

statistical significance of the observed association ina cross-tabulation.

The expected frequency for each cell can becalculated by using a simple formula:

e =nrncn

where nr = total number in the row

nc = total number in the columnn = total sample size


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For the data in Table 15.3, the expected frequencies for

the cells going from left to right and from top to

bottom, are:

Then the value of is calculated as follows:

15 X 1530

=7.50 15 X 1530

=7.50

15 X 1530

=7.50 15 X 1530

=7.50

2 =(f

o- f

e)2

feSall

cells

Chi-Square


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For the data in Table 15.3, the value of is

calculated as:

= (5 -7.5)2 + (10 - 7.5)2 + (10 - 7.5)2 + (5 - 7.5)2

7.5 7.5 7.5 7.5

=0.833 + 0.833 + 0.833+ 0.833

= 3.333

Chi-Square


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The chi-square distribution is a skewed

distribution whose shape depends solely on thenumber of degrees of freedom. As the number ofdegrees of freedom increases, the chi-squaredistribution becomes more symmetrical.

Table 3 in the Statistical Appendix contains upper-tail

areas of the chi-square distribution for differentdegrees of freedom. For 1 degree of freedom theprobability of exceeding a chi-square value of 3.841is 0.05.

For the cross-tabulation given in Table 15.3, thereare (2-1) x (2-1) = 1 degree of freedom. Thecalculated chi-square statistic had a value of 3.333.Since this is less than the critical value of 3.841, thenull hypothesis of no association can not be rejectedindicating that the association is not statisticallysignificant at the 0.05 level.

Chi-Square


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The phi coefficient ( ) is used as a measure of the

strength of association in the special case of a tablewith two rows and two columns (a 2 x 2 table).

The phi coefficient is proportional to the square rootof the chi-square statistic

It takes the value of 0 when there is no association,which would be indicated by a chi-square value of 0as well. When the variables are perfectly associated,

phi assumes the value of 1 and all the observationsfall just on the main or minor diagonal.

Phi Coefficient

= 2

n


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While the phi coefficient is specific to a 2 x 2 table,

the contingency coefficient(C)can be used toassess the strength of association in a table of anysize.

The contingency coefficient varies between 0 and 1.

The maximum value of the contingency coefficientdepends on the size of the table (number of rowsand number of columns). For this reason, it shouldbe used only to compare tables of the same size.

Contingency Coefficient

C=

2

2 + n


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Cramer's Vis a modified version of the phi

correlation coefficient, , and is used in tableslarger than 2 x 2.

or

Cramers V

V=

2

min (r-1), (c-1)

V= 2

/nmin (r-1), (c-1)


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Asymmetric lambda measures the percentage

improvement in predicting the value of the dependentvariable, given the value of the independent variable. Lambda also varies between 0 and 1. A value of 0 means

no improvement in prediction. A value of 1 indicates thatthe prediction can be made without error. This happenswhen each independent variable category is associated

with a single category of the dependent variable. Asymmetric lambda is computed for each of the variables

(treating it as the dependent variable). Asymmetric lambda is also computed, which is a kind

of average of the two asymmetric values. The symmetriclambda does not make an assumption about whichvariable is dependent. It measures the overallimprovement when prediction is done in both directions.

Lambda Coefficient


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Other statistics like taub, tauc, and gamma are

available to measure association between twoordinal-level variables. Both tau band tau cadjustfor ties.

Taubis the most appropriate with square tables inwhich the number of rows and the number of

columns are equal. Its value varies between +1 and-1. For a rectangular table in which the number of rows

is different than the number of columns, taucshould be used.

Gamma does not make an adjustment for either tiesor table size. Gamma also varies between +1 and -1and generally has a higher numerical value than taubor tauc.

Other Statistics

15-61

C T b l ti i P ti

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Cross-Tabulation in Practice

While conducting cross-tabulation analysis in practice, it is useful to

proceed along the following steps.1. Test the null hypothesis that there is no association betweenthe variables using the chi-square statistic. If you fail to rejectthe null hypothesis, then there is no relationship.

2. IfH0 is rejected, then determine the strength of the associationusing an appropriate statistic (phi-coefficient, contingency

coefficient, Cramer's V, lambda coefficient, or other statistics),as discussed earlier.3. IfH0 is rejected, interpret the pattern of the relationship by

computing the percentages in the direction of the independentvariable, across the dependent variable.

4. If the variables are treated as ordinal rather than nominal, use

taub, tau c, or Gamma as the test statistic. IfH0 is rejected,then determine the strength of the association using themagnitude, and the direction of the relationship using the signof the test statistic.

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Hypothesis Testing Related to Differences

Parametric tests assume that the variables of

interest are measured on at least an interval scale. Nonparametric tests assume that the variables are

measured on a nominal or ordinal scale.

These tests can be further classified based on

whether one or two or more samples are involved. The samples are independent if they are drawn

randomly from different populations. For thepurpose of analysis, data pertaining to differentgroups of respondents, e.g., males and females, are

generally treated as independent samples. The samples are paired when the data for the two

samples relate to the same group of respondents.

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IndependentSamples

PairedSamples Independent

SamplesPaired

Samples* Two-Groupt test

* Z test

* Pairedt test * Chi-Square

* Mann-Whitney

* Median* K-S

* Sign* Wilcoxon

* McNemar* Chi-Square

Procedures for Examining Differences

Fig. 15.9 Hypothesis Tests

One Sample Two or MoreSamples

One Sample Two or MoreSamples

* t test* Z test

* Chi-Square* K-S* Runs* Binomial

Parametric Tests(Metric Tests)

Non-parametric Tests(Nonmetric Tests)

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Pa amet ic Tests

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Parametric Tests

The tstatistic assumes that the variable is normally

distributed and the mean is known (or assumed to beknown) and the population variance is estimatedfrom the sample.

Assume that the random variable Xis normallydistributed, with mean and unknown population

variance , which is estimated by the samplevariance s2. Then, is tdistributed with n - 1

degrees of freedom. The tdistribution is similar to the normal

distribution in appearance. Both distributions arebell-shaped and symmetric. As the number ofdegrees of freedom increases, the tdistributionapproaches the normal distribution.

t = (X - )/s

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Hypothesis Testing Using the t Statistic

1. Formulate the null (H0) and the alternative (H1)

hypotheses.2. Select the appropriate formula for the tstatistic.

3. Select a significance level, , for testing H0.Typically, the 0.05 level is selected.

4. Take one or two samples and compute the meanand standard deviation for each sample.

5. Calculate the tstatistic assuming H0 is true.

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6. Calculate the degrees of freedom and estimate the

probability of getting a more extreme value of thestatistic from Table 4 (Alternatively, calculate thecritical value of the t statistic).

7. If the probability computed in step 5 is smaller thanthe significance level selected in step 2, reject H0.

If the probability is larger, do not reject H0.(Alternatively, if the value of the calculated tstatistic in step 4 is larger than the critical valuedetermined in step 5, reject H0. If the calculatedvalue is smaller than the critical value, do not rejectH0). Failure to reject H0 does not necessarily imply

that H0 is true. It only means that the true state isnot significantly different than that assumed by H0.

8. Express the conclusion reached by the ttest interms of the marketing research problem.

Hypothesis Testing Using the t Statistic

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For the data in Table 15.2, suppose we wanted to test

the hypothesis that the mean familiarity rating exceeds4.0, the neutral value on a 7 point scale. A significancelevel of = 0.05 is selected. The hypotheses may beformulated as:

t Test

H0: < 4.0

> 4.0

t= (X - )/s

sX= s/ n

s = 1.579/ 29= 1.579/5.385 = 0.293

t= (4.724-4.0)/0.293 = 0.724/0.293 = 2.471

H1:

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The degrees of freedom for the tstatistic to test the

hypothesis about one mean are n- 1. In this case,n- 1 = 29 - 1 or 28. From Table 4 in the StatisticalAppendix, the probability of getting a more extremevalue than 2.471 is less than 0.05 (Alternatively, the

critical t value for 28 degrees of freedom and asignificance level of 0.05 is 1.7011, which is less thanthe calculated value). Hence, the null hypothesis isrejected. The familiarity level does exceed 4.0.

t Test

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Note that if the population standard deviation was

assumed to be known as 1.5, rather than estimatedfrom the sample, a ztest would be appropriate. Inthis case, the value of the zstatistic would be:

where= = 1.5/5.385 = 0.279

and

z= (4.724 - 4.0)/0.279 = 0.724/0.279 = 2.595

z Test

z= (X - )/

1.5/ 29

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z Test

From Table 2 in the Statistical Appendix, the

probability of getting a more extreme value ofzthan2.595 is less than 0.05. (Alternatively, the critical zvalue for a one-tailed test and a significance level of0.05 is 1.645, which is less than the calculated

value.) Therefore, the null hypothesis is rejected,reaching the same conclusion arrived at earlier by thettest.

The procedure for testing a null hypothesis with

respect to a proportion was illustrated earlier in thischapter when we introduced hypothesis testing.

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Means

In the case of means for two independent samples, the

hypotheses take the following form.

The two populations are sampled and the means and variancescomputed based on samples of sizes n1 and n2. If bothpopulations are found to have the same variance, a pooledvariance estimate is computed from the two sample variancesas follows:

210

: H

211

: H

2

((

21

1 1

2

22

2

112

1 2

))

+

+

nn

XXXXs

n n

i iii or s

2=

(n1-1)s12

+(n2-1)s22

n1+n2-2

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The standard deviation of the test statistic can be

estimated as:

The appropriate value oftcan be calculated as:

The degrees of freedom in this case are (n1 + n2 -2).

Means

sX1 -X2 = s2 ( 1n1

+ 1n2)

t=(X1 -X2) - (1 - 2)

sX1 -X2

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An Ftest of sample variance may be performed if it is

not known whether the two populations have equal

variance. In this case, the hypotheses are:

H0: 12

= 22

H1: 12 2

2

F Test

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The Fstatistic is computed from the sample variances

as follows

where

n1 = size of sample 1n2 = size of sample 2

n1-1 = degrees of freedom for sample 1

n2-1 = degrees of freedom for sample 2

s12 = sample variance for sample 1

s22 = sample variance for sample 2

Using the data of Table 15.1, suppose we wanted to determinewhether Internet usage was different for males as compared tofemales. A two-independent-samples ttest was conducted. The

results are presented in Table 15.14.

F Statistic

(n1-1),(n2-1) =s1

2

s22

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Two Independent-Samples tTestsTable 15.14

Summary Statistics

Number Standard

of Cases Mean Deviation

Male 15 9.333 1.137Female 15 3.867 0.435

FTest for Equality of Variances

F 2-tailvalue probability

15.507 0.000

tTest

Equal Variances Assumed Equal Variances Not Assumed

t Degrees of 2-tail t Degrees of 2-tailvalue freedom probability value freedom probability

4.492 28 0.000 -4.492 18.014 0.000-

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The case involving proportions for two independent samples is also

illustrated using the data of Table 15.1, which gives the number ofmales and females who use the Internet for shopping. Is theproportion of respondents using the Internet for shopping thesame for males and females? The null and alternative hypothesesare:

AZtest is used as in testing the proportion for one sample.However, in this case the test statistic is given by:

Proportions

H0: 1 =2H1: 1 2

SPPpP

Z

21

21


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In the test statistic, the numerator is the difference between the

proportions in the two samples, P1 and P2. The denominator isthe standard error of the difference in the two proportions and isgiven by

where

+

nn

S PPpP21

21

11)1(

P =

n1P1 + n2P2

n1 + n2

Proportions


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A significance level of = 0.05 is selected. Given the data of

Table 15.1, the test statistic can be calculated as:

= (11/15) -(6/15)

= 0.733 - 0.400 = 0.333

P = (15 x 0.733+15 x 0.4)/(15 + 15) = 0.567

= = 0.181

Z= 0.333/0.181 = 1.84

PP 21

S pP 210.567x 0.433[ 1

15+ 1

15]

Proportions


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Given a two-tail test, the area to the right of the

critical value is 0.025. Hence, the critical value of thetest statistic is 1.96. Since the calculated value isless than the critical value, the null hypothesis cannot be rejected. Thus, the proportion of users (0.733

for males and 0.400 for females) is not significantlydifferent for the two samples. Note that while thedifference is substantial, it is not statisticallysignificant due to the small sample sizes (15 in eachgroup).

Proportions

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Paired Samples

The difference in these cases is examined by a paired samples t

test. To compute tfor paired samples, the paired differencevariable, denoted by D, is formed and its mean and variancecalculated. Then the tstatistic is computed. The degrees offreedom are n- 1, where nis the number of pairs. The relevantformulas are:

continued

H0: D = 0H1: D 0

tn-1 =D - D

sDn

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where,

In the Internet usage example (Table 15.1), a paired t test couldbe used to determine if the respondents differed in their attitudetoward the Internet and attitude toward technology. The resultingoutput is shown in Table 15.15.

D=

DiS

i=1

n

n

sD

=

(i- )

2

Si=1

n

n - 1

n

SS

D

D

Paired Samples

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Paired Samples tTest

Number Standard StandardVariable of Cases Mean Deviation Error

Internet Attitude 30 5.167 1.234 0.225Technology Attitude 30 4.100 1.398 0.255

Difference = Internet -Technology

Difference Standard Standard 2-tail t Degrees of 2-tailMean deviation error Correlation prob. value freedom probability

1.067 0.828 0.1511 0.809 0.000 7.059 29 0.000

Table 15.15

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Non-Parametric Tests

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Non Parametric Tests

Nonparametric tests are used when the independent

variables are nonmetric. Like parametric tests,nonparametric tests are available for testing variablesfrom one sample, two independent samples, or tworelated samples.

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Sometimes the researcher wants to test whether the

observations for a particular variable could reasonablyhave come from a particular distribution, such as thenormal, uniform, or Poisson distribution.

The Kolmogorov-Smirnov (K-S) one-sample test

is one such goodness-of-fit test. The K-S compares thecumulative distribution function for a variable with aspecified distribution.A

idenotes the cumulative

relative frequency for each category of the theoretical(assumed) distribution, and Oi the comparable value of

the sample frequency. The K-S test is based on themaximum value of the absolute difference betweenA

iand Oi. The test statistic is

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The decision to reject the null hypothesis is based on thevalue ofK. The larger the Kis, the more confidence wehave that H0 is false. For = 0.05, the critical value ofKfor large samples (over 35) is given by 1.36/Alternatively, Kcan be transformed into a normallydistributed zstatistic and its associated probabilitydetermined.

In the context of the Internet usage example, suppose wewanted to test whether the distribution of Internet usagewas normal. A K-S one-sample test is conducted, yieldingthe data shown in Table 15.16. Table 15.16 indicatesthat the probability of observing a Kvalue of 0.222, asdetermined by the normalized zstatistic, is 0.103. Since

this is more than the significance level of 0.05, the nullhypothesis can not be rejected, leading to the sameconclusion. Hence, the distribution of Internet usage doesnot deviate significantly from the normal distribution.

n

One Sample

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Normality of Internet Usage

Table 15.16

Test Distribution - Normal

Mean: 6.600Standard Deviation: 4.296

Cases: 30

Most Extreme DifferencesAbsolute Positive Negative K-S z 2-Tailed p0.222 0.222 -0.142 1.217 0.103


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The chi-square test can also be performed on a

single variable from one sample. In this context, thechi-square serves as a goodness-of-fit test.

The runs test is a test of randomness for thedichotomous variables. This test is conducted bydetermining whether the order or sequence in which

observations are obtained is random. The binomial test is also a goodness-of-fit test for

dichotomous variables. It tests the goodness of fit ofthe observed number of observations in each

category to the number expected under a specifiedbinomial distribution.

One Sample

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When the difference in the location of two populations is to becompared based on observations from two independent

samples, and the variable is measured on an ordinal scale, theMann-Whitney Utest can be used.

In the Mann-Whitney Utest, the two samples are combined andthe cases are ranked in order of increasing size.

The test statistic, U, is computed as the number of times ascore from sample or group 1 precedes a score from group 2.

If the samples are from the same population, the distribution ofscores from the two groups in the rank list should be random.

An extreme value ofUwould indicate a nonrandom pattern,pointing to the inequality of the two groups.

For samples of less than 30, the exact significance level for Uiscomputed. For larger samples, Uis transformed into a normally

distributed zstatistic. This zcan be corrected for ties withinranks.

Two Independent Samples

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We examine again the difference in the Internet usage of malesand females. This time, though, the Mann-Whitney Utest is

used. The results are given in Table 15.17. One could also use the cross-tabulation procedure to conduct a

chi-square test. In this case, we will have a 2 x 2 table. Onevariable will be used to denote the sample, and will assume thevalue 1 for sample 1 and the value of 2 for sample 2. The othervariable will be the binary variable of interest.

The two-sample median test determines whether the twogroups are drawn from populations with the same median. It isnot as powerful as the Mann-Whitney Utest because it merelyuses the location of each observation relative to the median,and not the rank, of each observation.

The Kolmogorov-Smirnov two-sample test examines

whether the two distributions are the same. It takes intoaccount any differences between the two distributions, includingthe median, dispersion, and skewness.


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Internet Usage by Gender

Table 15.17

Sex Mean Rank Cases

Male 20.93 15Female 10.07 15

Total 30

Corrected for tiesU W z 2-tailed p

31.000 151.000 -3.406 0.001

NoteU= Mann-Whitney test statisticW= Wilcoxon W Statisticz = U transformed into a normally distributedzstatistic.

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The Wilcoxon matched-pairs signed-ranks test

analyzes the differences between the pairedobservations, taking into account the magnitude ofthe differences.

It computes the differences between the pairs ofvariables and ranks the absolute differences.

The next step is to sum the positive and negativeranks. The test statistic, z, is computed from thepositive and negative rank sums.

Under the null hypothesis of no difference, zis a

standard normal variate with mean 0 and variance 1for large samples.

Paired Samples

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The example considered for the paired ttest,

whether the respondents differed in terms of attitudetoward the Internet and attitude toward technology,is considered again. Suppose we assume that boththese variables are measured on ordinal rather thaninterval scales. Accordingly, we use the Wilcoxontest. The results are shown in Table 15.18.

The sign test is not as powerful as the Wilcoxonmatched-pairs signed-ranks test as it only comparesthe signs of the differences between pairs ofvariables without taking into account the ranks.

In the special case of a binary variable where the

researcher wishes to test differences in proportions,the McNemar test can be used. Alternatively, thechi-square test can also be used for binary variables.

Paired Samples

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Internet with Technology

(Technology - Internet) Cases Mean rank

-Ranks 23 12.72

+Ranks 1 7.50

Ties 6

Total 30

z = -4.207 2-tailed p= 0.0000

Table 15.18

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Related to Differences

Table 15.19

Contd.

Sample Application Level of Scaling Test/Comments

One Sample

One Sample Distributions Nonmetric

K-S and chi-square for

goodness of fit

Runs test for randomness

Binomial test for goodness of

fit for dichotomous variables

One Sample Means Metric t test, if variance is unknown

z test, if variance is known

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Related to Differences

Table 15.19 cont.


Two independent samples Distributions Nonmetric K-S two-sample test for examining theequivalence of two distributions

Two independent samples Means Metric Two-group ttestFtest for equality of variances

Two independent samples Proportions Metric ztestNonmetric Chi-square test

Two independent samples Rankings/Medians Nonmetric Mann-Whitney U test is morepowerful than the median test

Paired Samples

Paired samples Means Metric Paired ttest

Paired samples Proportions Nonmetric McNemar test for binary variablesChi-square test

Paired samples Rankings/Medians Nonmetric Wilcoxon matched-pairs ranked-signstest is more powerful than the sign test

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SPSS Windows

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The main program in SPSS is FREQUENCIES. It

produces a table of frequency counts, percentages,and cumulative percentages for the values of eachvariable. It gives all of the associated statistics.

If the data are interval scaled and only the summarystatistics are desired, the DESCRIPTIVES procedure

can be used. The EXPLORE procedure produces summary statistics

and graphical displays, either for all of the cases orseparately for groups of cases. Mean, median,variance, standard deviation, minimum, maximum,and range are some of the statistics that can becalculated.

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SPSS Windows

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To select these procedures click:

Analyze>Descriptive Statistics>FrequenciesAnalyze>Descriptive Statistics>DescriptivesAnalyze>Descriptive Statistics>Explore

The major cross-tabulation program is CROSSTABS.

This program will display the cross-classification tablesand provide cell counts, row and column percentages,the chi-square test for significance, and all themeasures of the strength of the association that havebeen discussed.

To select these procedures click:

Analyze>Descriptive Statistics>Crosstabs

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SPSS Windows

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The major program for conducting parametric

tests in SPSS is COMPARE MEANS. This program canbe used to conduct ttests on one sample orindependent or paired samples. To select theseprocedures using SPSS for Windows click:

Analyze>Compare Means>Means Analyze>Compare Means>One-Sample T Test Analyze>Compare Means>Independent-

Samples T Test Analyze>Compare Means>Paired-Samples T

Test

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The nonparametric tests discussed in this chapter can

be conducted using NONPARAMETRIC TESTS.

To select these procedures using SPSS for Windowsclick:

Analyze>Nonparametric Tests>Chi-Square Analyze>Nonparametric Tests>Binomial Analyze>Nonparametric Tests>Runs Analyze>Nonparametric Tests>1-Sample K-S

Analyze>Nonparametric Tests>2 IndependentSamples Analyze>Nonparametric Tests>2 Related

Samples

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