Involves testing a hypothesis. There is no single parameter to estimate. Considers all categories to give an overall

idea of whether the observed distribution differs from the hypothesized one.

“All creatures have their determined time for giving birth and carrying fetus, only a man is born all year long, not in determined time, one in the seventh month, the other in the eighth, and so on till the beginning of the eleventh month.”

~Aristotle

Counted Data Condition◦ Check that the data are counts for the categories

of a categorical variable. Independence Assumption

◦ Check that the individuals counted in the cells are sampled independently from some population.

◦ If not, check the randomization condition – the individuals who have been counted should be a random sample from some population.

Sample Size Assumption◦ Expected cell frequency condition – expect to

observe at least 5 individuals in each cell.

Compare the observed counts in each cell with the expected counts.

Look at the differences between the observed and expected counts.

The test is always one-sided.

There is no direction to the rejection of the null model – we know it just doesn’t fit.

Chi- Square statistic refers to a family of sampling distribution models.

Number of degrees of freedom is n – 1, where n is the number of categories.

2

2

all cells

Obs Exp

Exp

Hypothesis◦ Ho: Births are

uniformly distributed over zodiac signs. (pAries=pTaurus=…)

◦ HA: Births are not uniformly distributed over zodiac signs.

Check Conditions:◦ COUNTED DATA CONDITION:

there are counts of the number of executives in categories.

◦ RANDOMIZATION CONDITION: this is a convenience sample, but no expectation of bias.

◦ EXPECTED CELL FREQUENCY CONDITION: the null hypothesis expects that of the 256 should occur in each sign.

1

12

1256 21.33

1221.33 5

•The sampling distribution of the test statistic is χ2 with 12 – 1 = 11 degrees of freedom.•Use a Chi-Square goodness-of-fit test.

The chi-square procedure:◦ Find the expected values.

Values come from the null hypothesis. Multiply the total number of observations by the

hypothesized proportion.◦ Compute the residuals, Observed – Expected.◦ Square the residuals.◦ Compute the component for each cell,◦ Find the sum of the components.◦ Find the degrees of freedom, the number of cells

minus 1.◦ Test the hypothesis: find the P-value.

2Observed Expected

Expected

◦ Enter counts in L1 and expected percentages in L2.

◦ Convert expected percentages to expected counts.

◦ Calculate chi-square in L3.

◦ Find the sum of L3.

◦ Find the P-value The probability of

finding a χ2 value at least as high as the one calculated from the data.

DISTR menu, χ2 cdf

P-value◦ Test is one-sided, only consider the right tail.◦ Large χ2 values correspond to small P-values,

leading to rejection of the null hypothesis.◦ The P-value is the area in the upper tail of the χ2

model for 11 degrees of freedom above the computed χ2 value.

Conclusion◦ The P-value of 0.926 means that an observed chi-

square value of 5.08 or higher would occur about 93% of the time.

◦ There is virtually no evidence that the distribution of zodiac signs among executives is not uniform.

Chi-square test for homogeneity Assumptions and Conditions

◦ Counted data condition◦ Check that the data are counts for the categories of

a categorical variable.◦ Independence Assumption: Randomization condition

◦ When we test for homogeneity, we often are not interested in some larger population so we don’t need to check the randomization condition.

◦ Sample Size Assumption◦ Expected cell frequency condition – expected count

in each cell must be at least 5 individuals.

Who: High school graduates

What: Post-graduation activities

When: 1980, 1990, 2000

Why: Regular survey for general information

1980 1990 2000 Total

College/Post-HS education

320 245 288 853

Employment 98 24 17 139

Military 18 19 5 42

Travel 17 2 5 24

Total 453 290 315 1058

Hypothesis◦ Have the choices made by high

school graduates in what they do after graduation changed?

◦ Ho: The post-high school choices made by the classes of 1980, 1990, and 2000 have the same distribution (homogeneous).

◦ HA: The post-high school choices made by the classes of 1980, 1990, and 2000 do not have the same distribution.

Check the conditions√ Counted data condition: there

are counts of the number of students in categories.

√ Randomization condition: No inference will be drawn to other high schools or other classes, so no need to check for a random sample.

√ Expected cell frequency condition: The expected values are all at least 5 (see table, later).

Under these conditions, the sampling distribution of the test statistic is χ2 with (4 – 1) X (3 – 1) = 6 degrees of freedom.Perform a chi-square test of homogeneity.

TI-84+ Steps:◦ Enter data in a

matrix.

◦ Do the chi-square test of homogeneity.

◦ Matrix Edit [B] Note that all expected

counts are at least 5.

Conclusion◦ The P-value is very small.

Observed pattern is very unlikely to occur by chance. Reject the null hypothesis. The choices made by high school graduates have

changed over the two decades examined.

Examine the Residuals◦ Standardized Residuals

Divide the cell’s residual by the square root of its expected value.

Values are the square root of the components calculated for each cell, with + or – to show whether we observed more or less cases than expected.

What trends do you see?1980

1990 2000

College/Post-HS education

-2.366 0.732 2.136

Employment 4.989 -2.284 -3.791

Military 0.004 2.207 -2.122

Travel 2.098 -1.785 -0.803

Obs Exp

Expc

Chi-Square Test for Independence◦ Data categorize

subjects from a single group on two categorical variables.

◦ Contingency Tables Categorize counts on

two or more variables. Decide whether the

distribution of counts on one variable is contingent on the other.

Assumptions and Conditions

◦ Counted data condition◦ Check that the data are counts

for the categories of a categorical variable.

◦ Independence Assumption: Randomization condition

◦ When we test for independence, we are interested in generalizing to some larger population.

◦ Sample Size Assumption◦ Expected cell frequency

condition – expected count in each cell must be at least 5 individuals.

Who: Patients being treated for non-blood-related disorders

What: Tattoo status and hepatitis C status

When: 1991, 1992 Where: Texas

Hepatitis C

No Hepatitis C

Total

Tattoo, Parlor

17 35 52

Tattoo, elsewhere

8 53 61

None 22 491 513

Total 47 579 626

Hypothesis◦ Are the categorical

variables “tattoo status” and “hepatitis C status” statistically independent?

◦ H0: Tattoo status and hepatitis C status are independent.

◦ HA: Tattoo status and hepatitis C status are not independent.

Check the conditions√ Counted data condition: there are

counts of individuals in categories of two categorical variables.

√ Randomization condition: Although not an SRS, the data were selected to avoid biases and should be representative of the general population.

√ Expected cell frequency condition: The expected values do not meet the condition that all are

greater than 5. Continue with caution – be sure to check the residuals.

Under these conditions, the sampling distribution of the test statistic is χ2 with (3 – 1) X (2 – 1) = 2 df.Perform a chi-square test for independence.

TI-84+ Steps:◦ Enter data in a matrix.

◦ Do the chi-square test of independence.

◦ Matrix Edit [B] Note that not all

expected counts are at least 5.

Conclusion:◦ The P-Value is very small, indicating that if these

variables were independent, the pattern seen would be very unlikely to occur by chance.

◦ The hepatitis C status is not independent of the tattoo status.

◦ HOWEVER, check the two cells with the small expected counts to determine if they did or did not influence the result too greatly.

Remember: A complete solution must include additional analysis, recalculation, and a final conclusion.

Analysis of Residuals◦ Too small an expected

frequency can arbitrarily inflate the residual, leading to an inflated chi-square statistic.

◦ In this case, the standardized residual for the hepatitis C and Tattoo, Parlor cell is large ⇒ Inflated chi-square statistic?

Standardized Residuals

Hepatitis C No Hepatitis C

Tattoo, Parlor

6.628 -1.888

Tattoo, elsewhere

1.598 -0.455

None -2.661 0.758

Options:◦ Based upon concerns, choose

not to report the results.◦ Include a warning when

reporting the results.◦ Combine the appropriate

categories to larger sample size and expected frequencies.

Recalculation:

Recalculation (continued):

Conclusion:◦ The tattoo status and hepatitis C

status are not independent. The data suggest that tattoo parlors may be a particular problem, but we do not have enough data to draw that conclusion.

Hepatitis C

No Hepatitis C

Ttl

None 22 491 513

Tattoo 25 88 113

Total 47 579 626

A failure of independence between two categorical variables does not show a cause-and-effect relationship between them.

There is no way to differentiate the direction of any possible causation from one variable to another.

Lurking variables could be responsible for the observed lack of independence.

Don’t use chi-square methods unless the data are counts.◦ Data reported as proportions or

percentages can be used if they are converted to counts.

◦ Just because data are reported in a two-way table does not mean they are suitable for chi-square procedures.

Beware large samples.◦ The degrees of freedom for the chi-

square tests do not grow with sample size.

◦ With a sufficiently large sample size, a chi-square test can always reject the null hypothesis.

◦ There are no confidence intervals to help in determining the effect size.