Cross-Tabulations

Cross-TabsThe level of measurement used for cross-tabulations are mostly nominal. Even when continuous variables are used (such as age and income), they are converted to categorical variables.

When continuous variables are converted to categorical variables, important information (variation) is lost.

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Data Types

Data

Numerical(Quantitative)

Categorical(Qualitative)

Discrete Continuous

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Categorical Data

• Categorical random variables yield responses that classify– Example: Gender (female, male)

• Measurement reflects number in category• Nominal or ordinal scale

– Examples• Did you attend a community college? • Do you live on-campus or off-campus?

Why Concerned about Categorical Random Variables?

• Survey data tends to be categorical … hot/comfortable/cold, sunny/cloudy/fog/rain, yes/no…

• Know limitations– nature of relationship– causality

• Widely used in marketing for decision-making

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Cross-Tabs

The Chi-square, 2, statistic is used to test the null hypothesis.

[Unfortunately, Chi-square, like many other statistics that indicate statistical significance, tells us nothing about the

magnitude of the relation.]

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2 Test of Independence

• Shows whether a relationship exists between two categorical variables– One sample is drawn– Does not show nature of relationship– Does not show causality

• Used widely in marketing • Uses contingency table

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Upper Tail AreaDF .995 … .95 … .051 ... … 0.004 … 3.8412 0.010 … 0.103 … 5.991

Critical Value

20 5.991

Reject

What is the critical 2 value if table has 2 rows and 3 columns, =.05?

= .05df = (2 - 1)(3 - 1)

= 2

If fo = fe, 2 = 0.

Do not reject H0

2 Table (Portion)

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2 Test of Independence Hypotheses & Statistic

• Hypotheses– H0: Variables are not dependent

– H1: Variables are dependent (related)

• Test statistic

• Degrees of freedom: (r - 1)(c - 1)

cells all

22

e

eo

f

ff

Observed frequency

Expected frequency

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2 Test of Independence Expected Frequencies

• Statistical independence means joint probability equals product of marginal probabilities

– P(A and B) = P(A)·P(B)

• Compute marginal probabilities• Multiply for joint probability• Expected frequency is sample size times joint

probability

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Diet Pepsi Diet Coke No Yes Total

No 84 32 116 Yes 48 122 170 Total 132 154 286

You’re a marketing research analyst. You ask a random sample of 286 consumers if they purchase Diet Pepsi or Diet Coke. At the 0.05 level of significance, is there evidence of a relationship?

2 Test of Independence An Example

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Expected Frequencies

total Grand

totalRow total Column =frequency Expected

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Diet PepsiNo Yes

Diet Coke Obs. Exp. Obs. Exp. Total

No 84 53.5 32 62.5 116

Yes 48 78.5 122 91.5 170

Total 132 132 154 154 286

Expected Frequenciesfe 1 in all cells

132·170286

154·170286

132·116286

132·154286

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2 Test of Independence

Cell fo fe fo - fe (fo - fe)² (fo - fe)²/ fe

1,1 84 53.5 +30.5 930.25 17.3879

1,2 32 62.5 -30.5 930.25 14.8840

2,1 48 78.5 -30.5 930.25 11.8503

2,2 122 91.5 +30.5 930.25 10.1667

Total 286 286 54.2889

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2 Test of Independence

H0: Not Dependent

H1: Dependent

= .05

df = (2 - 1)(2 - 1) = 1

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Reject at = .05

There is evidence of a relationship

20 3.841

Reject = .05= .05

2889.54cells all

22

e

eo

f

ff

Cross-Tabs Please provide the requested information by checking (once)

in each category.   What is your:

 age ____ < 18 ___ 18 - 26 ____ > 26

 gender ____ male ____ female

 course load __ < 6 units __ 6 – 12 units __ > 12 units

 gpa __ < 2.0 __ 2.0 - 2.5 __ 2.6 - 3.0 __ 3.1 - 3.5 __ > 3.5 annual income __ < $15k __ $15k - $40k ___ > $40k  

Cross-Tabs

The information is coded and entered in the file student.sf by letting the first response be recorded as a 1, the second as a 2, etc.

Cross-Tabs

The hypothesis test generally referred to as

a test of dependence.

The researcher wishes to determine whether the variables are dependent, or, exhibit a relationship.

Cross-Tabs

Let’s investigate whether a relationship between a student’s gpa and units attempted exists.

H0: GPA and UNITS are not dependent

H1: GPA and UNITS are dependent.

Cross-Tabs

Chi-Square Test

------------------------------------------

Chi-Square Df P-Value

------------------------------------------

3.67 8 0.8853

------------------------------------------

Cross-Tabs

p-value = 0.8853, Retain H0

thus, GPA and UNITS are not dependent

[Based on our data, there is no evidence to support the concept that a relationship exists between gpa and units attempted.]

Cross-Tabs

Let’s investigate whether a relationship between a student’s age and units attempted exist.

H0: AGE and UNITS are not dependent

H1: AGE and UNITS are dependent.

Cross-Tabs

Chi-Square Test

------------------------------------------

Chi-Square Df P-Value

------------------------------------------

9.89 4 0.0423

------------------------------------------

Cross-Tabs

p-value = 0.0423, Reject H0

thus, AGE and UNITS are dependent

[Based on our data, there is sufficient evidence to support the concept that a relationship exists between age and units attempted.]

Cross-TabsFrequency Table for age by units

Units <6 6-12 >12 AGE Total --------------------------------------------------------

<18 | 10 | 19 | 17 | 46 | 17.24% | 20.88% | 33.33% | 23.00%

--------------------------------------------------------Age 18-26 | 24 | 22 | 16 | 62

| 41.38% | 24.18% | 31.37% | 31.00% --------------------------------------------------------

>26 | 24 | 50 | 18 | 92 | 41.38% | 54.95% | 35.29% | 46.00% --------------------------------------------------------

UNITS Total 58 91 51 200 29.00% 45.50% 25.50% 100.00%

Questions?

ANOVA