Chapter 20 For Explaining Psychological Statistics, 4th ed. by B. Cohen 1 These tests can be used...

Chapter 20 For Explaining Psychological Statistics, 4th ed. by B. Cohen

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• These tests can be used when all of the data from a study has been measured on nominal scales — that is, the data are in the form of frequencies for different categories (e.g., How many consumers studied preferred each of the five most popular brands of toothpaste in the U.S.? How many professors are female and how many are male in 10 major academic fields in China?).

• You would need a multinomial distribution to determine the exact probabilities associated with random choices when dealing with more than two categories. It is much easier to use an approximation, instead. A chi-square statistic is calculated, which follows, approximately, the well-known chi-square distribution, when the null hypothesis is true.

Chapter 20: Chi-Square Tests


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The Chi-Square (χ2) Distribution– This distribution can be used to represent the

null hypothesis, and is therefore used to find the appropriate critical values for chi-square tests.

– Its shape depends on the number of degrees of freedom associated with the table of data.

– Because χ2 can’t be less than 0, all chi-square distributions are positively skewed, though the skewing decreases as the df get larger.

– With infinite df, the chi-square distribution becomes identical to the normal distribution.

• Similar to the way the F distribution is used for ANOVA, only one tail of the chi-square distribution is used to determine the statistical significance of a chi-square test.

– Differences larger than expected by chance lead to χ2 values in the positive tail; unusually small values are in the smaller, negative tail, near zero.

– Only large discrepancies from chance expectations are inconsistent with the null hypothesis.


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One-Way Chi-Square Tests

• “One way” means that all categories are considered levels of the same factor. The df equal one less than the number of categories (i.e., k – 1)

• Sometimes referred to as a “goodness of fit” test

Types of One-Way Χ2 Tests1. Expected frequencies are

hypothesized to be equal • Games of chance (with equal

frequencies of outcomes), or• Testing for equal preferences among

a set of categories• The expected frequency for each

category is the total number of obtained responses (N) divided by the number of categories (k)


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Types of One-Way Chi-

Square Tests (cont.)2. Population proportions are known• Sometimes we have good estimates of

population proportions.– # of voters registered in different political

parties.– % of population at different income levels.

3. The shape of a distribution is being tested• E.g., the distribution of annual income in a

sample. The expected frequencies are generated by the appropriate normal distribution. Rejection of the null implies, if the sample were truly random, that the underlying population is not normally distributed.

• If the population were known to be normal, rejection of the null would imply that either:

– The sample is not a random one, or– The sample has not been drawn from the

hypothesized population


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Types of One-Way Chi-

Square Tests (cont.)

4. A theoretical model is being tested

• Sometimes a well-formulated theory can make quantitative predictions in terms of frequencies expected in different categories.

• In such a case, you will want your observed frequencies to be as close as possible to the theoretically expected frequencies.

• For this type of chi-square test, you do not want to reject the null hypothesis, because the expected frequencies come from your research hypothesis.


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Try This Example…

Is one of the following three energy

drinks preferred in the U.S. more

often than chance? N = 90:

– df = k – 1; where k is the # of categories

– χ2crit increases as alpha decreases

– χ2crit increases as df increases

Red Bull Starbucks Espresso

5 Hour Energy

fo 38 28 24

fe 30 30 30

e

eo

f

ff 22


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Answer to the Example:

The df = k – 1 = 3 – 1 = 2.

The critical value is Х2.05 (2) = 5.99.

3.467 < 5.99, so we cannot reject the null hypothesis of equal prefer-ence in the population at the .05 level.

2 2 2

2 38 30 28 30 24 30

30 30 30

64 4 36 1043.467

30 30 30 30


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The Relationship Between the Binomial Test and the Chi-Square Test With Two Categories

– With two categories, the null hypothesis can be tested by using the binomial distribution, yielding a z score, or by performing a one-way χ2 Test. The χ2 statistic will equal the square of the z score.

– Squaring the normal distribution yields a chi-square distribution with one degree of freedom.


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Two-Way Chi-Square Tests

• The most interesting psychological questions involve the relationship between at least two variables rather than the distribution of just one variable.

• The two-way chi-square test is appropriate for quantifying the relationship between two categorical variables.

• It is often referred to as Pearson’s Chi-Square Test of Association (or Independence).

• The Null Hypothesis (H0) for the two-way test is: There is no association between the two variables; that is, the way one of the variables is distributed into categories does not change at different levels of the second variable.

• The Alternative Hypothesis (HA) is simply: The null hypothesis is not true; that is, the two variables are not independent.


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Contingency Tables – The data for a two-way chi-square test

are usually arranged in a two-way contingency table, also known as a cross-classification table. (Therefore, the data are often referred to as cross-classified categorical data.)

– The df for a two-way chi-square test are equal to: (R – 1)(C – 1), where R = the number of rows in the table, and C = the number of columns.

The following is an example of a two-way conting-ency table for a study investigating the relationship between musical ability (high or low), and confidence (high, medium, or low) in 10-year-old children.

High Medium Low

Musical High 17 32 11 60

Ability Low 13 43 34 90

30 75 45 150


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Fill in the expected frequencies in the empty parentheses in the table below for the musical ability / confidence example, using the formula shown below the table (where N is the total of all the frequencies in the table).

After all the fe’s are found, the two-way chi-square statistic is calculated using the same formula as in the one-way case, applied to each cell of the table.

Confidence

High Medium Low

High 17 ( ) 32 ( ) 11 ( ) 60

Low 13 ( ) 43 ( ) 34 ( ) 90

30 75 45 150

Try This Example…

N

sumcolumnsumrowfe


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The df = (3 – 1) (2 – 1) = 2

The critical value is Х2.05 (2) = 5.99.

8.23 > 5.99, so we can reject the null hypothesis that musical ability is unrelated to levels of confidence in 10-year-old children.

Confidence

High Medium Low

High 7 (12) 32 (30) 11 (18) 60

Low 13 (18) 43 (45) 34 (27) 90

30 75 45 150

Answer to the Example:

2 2 2

2 7 12 32 30 11 18

12 30 18

2 2 213 18 43 45 34 27

8.2318 45 27


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Simplified Formula for the

2 x 2 Case

where the frequencies of the 2 x 2 table are labeled as: A B

C D

Strength of Association– Can also be called effect size

• For a 2 x 2 contingency table, it is appropriate to calculate Pearson’s r for the two variables, but first you have to assign arbitrary values (e.g., 1 and 2) to the two levels of each variable.

• Pearson’s r for two dichotomous variables is called the phi coefficient, and it can be obtained directly from the chi-square statistic, as shown next.

dbcadcba

bcadN

2

2


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Phi (φ) Coefficient (can also be called the fourfold point correlation)

• Measure of Effect Size for the 2 x 2 case only:– Because the square root is positive, the value

of φ ranges between 0 and +1.0.– The square of φ is a measure of the

proportion of variance accounted for in one variable by the other.

Cramer’s V or phi (φC)– For contingency tables larger than 2 x 2– L = # of rows or columns, whichever is

smaller, or either if both are the same– Ranges from 0 to +1.0

N

2

2

1C N L


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Assumptions of the χ2 Test1. The categories are mutually exclusive and exhaustive (all cases fall into one and only one category).

2. Independence of observations:• This assumption is usually violated

when the same subject is categorized more than once.

• A violation of this assumption seriously undermines the validity of the test.

3. Minimal size of expected frequencies:• The use of the chi-square distribution as

an approximation becomes inaccurate if the fe’s are too low.

• Strict rule: fe for each cell should be at least 5 when df > 1; 10 when df = 1

• Less strict: no fe < 1, and no more than 20% of the fes < 5


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Some Uses for the χ2 Test for Independence

– It can measure the strength of the relationship between two categorical variables.

– It can be used for a study with a quantitative DV, originally designed to be analyzed with a t test or ANOVA.• If the distribution of the DV is very far

from normal, and N is not very large, it can be desirable to transform the DV to a few distinct categories (e.g., annual income < $30K; between $30K and $50K; $50K to $80K; > $80K).

• Some power is lost by throwing away most of the quantitative information, but the validity of the chi-square test does not depend on making an assumption about the distribution of the DV.

Chapter 20 For Explaining Psychological Statistics, 4th ed. by B. Cohen 1 These tests can be used...

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