1. 15 Chi-Square Tests Chi-Square Test for Independence
Chapter
Chi-Square Test for Independence
Chi-Square Tests for Goodness-of-Fit
Uniform Goodness-of-Fit Test
Poisson Goodness-of-Fit Test
Normal Chi-Square Goodness-of-Fit Test
ECDF Tests (Optional)
McGraw-Hill/Irwin Copyright © 2009 by The McGraw-Hill Companies, Inc. All rights reserved.

2. Chi-Square Test for Independence
Contingency Tables
A contingency table is a cross-tabulation of n paired observations into categories.
Each cell shows the count of observations that fall into the category defined by its row (r) and column (c) heading. B A

3. Chi-Square Test for Independence
Contingency Tables
For example:
Table 15.1

4. Chi-Square Test for Independence
In a test of independence for an r x c contingency table, the hypotheses are H0: Variable A is independent of variable B H1: Variable A is not independent of variable B
Use the chi-square test for independence to test these hypotheses.
This non-parametric test is based on frequencies.
The n data pairs are classified into c columns and r rows and then the observed frequency fjk is compared with the expected frequency ejk.

5. Chi-Square Test for Independence
Chi-Square Distribution
The critical value comes from the chi-square probability distribution with n degrees of freedom.
n = degrees of freedom = (r – 1)(c – 1) where r = number of rows in the table c = number of columns in the table
Appendix E contains critical values for right-tail areas of the chi-square distribution.
The mean of a chi-square distribution is n with variance 2n.

6. Chi-Square Test for Independence
Chi-Square Distribution
Consider the shape of the chi-square distribution:
Figure 15.1

7. Chi-Square Test for Independence
Expected Frequencies
Assuming that H0 is true, the expected frequency of row j and column k is:
ejk = RjCk/n
where Rj = total for row j (j = 1, 2, …, r) Ck = total for column k (k = 1, 2, …, c) n = sample size

8. Chi-Square Test for Independence
Steps in Testing the Hypotheses
Step 1: State the Hypotheses
H0: Variable A is independent of variable B
H1: Variable A is not independent of variable B
Step 2: Specify the Decision Rule
Calculate n = (r – 1)(c – 1)
For a given a, look up the right-tail critical value (c2R) from Appendix E or by using Excel.
Reject H0 if c2R > test statistic.

9. Chi-Square Test for Independence
Steps in Testing the Hypotheses
For example, for n = 6 and a = .05, c2.05 =
Figure 15.2

10. Chi-Square Test for Independence
Steps in Testing the Hypotheses
Here is the rejection region.
Figure 15.3

11. Chi-Square Test for Independence
Steps in Testing the Hypotheses
Step 3: Calculate the Expected Frequencies
ejk = RjCk/n
For example,

12. Chi-Square Test for Independence
Steps in Testing the Hypotheses
Step 4: Calculate the Test Statistic
The chi-square test statistic is
Step 5: Make the Decision
Reject H0 if c2R > test statistic or if the p-value < a.
calc

13. Chi-Square Test for Independence
Small Expected Frequencies
The chi-square test is unreliable if the expected frequencies are too small.
Rules of thumb:
Cochran’s Rule requires that ejk > 5 for all cells.
Up to 20% of the cells may have ejk < 5
Most agree that a chi-square test is infeasible if ejk < 1 in any cell.
If this happens, try combining adjacent rows or columns to enlarge the expected frequencies.

14. Chi-Square Test for Independence
Cross-Tabulating Raw Data
Chi-square tests for independence can also be used to analyze quantitative variables by coding them into categories.
For example, the variables Infant Deaths per 1,000 and Doctors per 100,000 can each be coded into various categories:
Figure 15.6

15. Chi-Square Test for Goodness-of-Fit
Why Do a Chi-Square Test on Numerical
Data?
The researcher may believe there’s a relationship between X and Y, but doesn’t want to use regression.
There are outliers or anomalies that prevent us from assuming that the data came from a normal population.
The researcher has numerical data for one variable but not the other.

16. Chi-Square Test for Independence
3-Way Tables and Higher
More than two variables can be compared using contingency tables.
However, it is difficult to visualize a higher order table.
For example, you could visualize a cube as a stack of tiled 2-way contingency tables.
Major computer packages permit 3-way tables.

17. Chi-Square Test for Goodness-of-Fit
Purpose of the Test
The goodness-of-fit (GOF) test helps you decide whether your sample resembles a particular kind of population.
The chi-square test will be used because it is versatile and easy to understand.

18. Chi-Square Test for Goodness-of-Fit
Multinomial GOF Test
A multinomial distribution is defined by any k probabilities p1, p2, …, pk that sum to unity.
For example, consider the following “official” proportions of M&M colors.
calc

19. Chi-Square Test for Goodness-of-Fit
Multinomial GOF Test
The hypotheses are
H0: p1 = .30, p2 = .20, p3 = .10, p4 = .10, p5 = .10, p6 = .20 H1: At least one of the pj differs from the hypothesized value
No parameters are estimated (m = 0) and there are c = 6 classes, so the degrees of freedom are n = c – m – 1 = 6 – 0 - 1

20. Chi-Square Test for Goodness-of-Fit
Hypotheses for GOF
The hypotheses are:
H0: The population follows a _____ distribution H1: The population does not follow a ______ distribution
The blank may contain the name of any theoretical distribution (e.g., uniform, Poisson, normal).

21. Chi-Square Test for Goodness-of-Fit
Test Statistic and Degrees of Freedom for GOF
Assuming n observations, the observations are grouped into c classes and then the chi-square test statistic is found using:
calc
where fj = the observed frequency of observations in class j
ej = the expected frequency in class j if H0 were true

22. Chi-Square Test for Goodness-of-Fit
Test Statistic and Degrees of Freedom for GOF
If the proposed distribution gives a good fit to the sample, the test statistic will be near zero.
The test statistic follows the chi-square distribution with degrees of freedom n = c – m – 1
where c is the no. of classes used in the test m is the no. of parameters estimated

23. Chi-Square Test for Goodness-of-Fit
Test Statistic and Degrees of Freedom for GOF

24. Chi-Square Test for Goodness-of-Fit
Data-Generating Situations
Instead of “fishing” for a good-fitting model, visualize a priori the characteristics of the underlying data-generating process.
Mixtures: A Problem
Mixtures occur when more than one data-generating process is superimposed on top of one another.

25. Chi-Square Test for Goodness-of-Fit
Eyeball Tests
A simple “eyeball” inspection of the histogram or dot plot may suffice to rule out a hypothesized population.
Small Expected Frequencies
Goodness-of-fit tests may lack power in small samples. As a guideline, a chi-square goodness-of-fit test should be avoided if n < 25.

26. Uniform Goodness-of-Fit Test
Uniform Distribution
The uniform goodness-of-fit test is a special case of the multinomial in which every value has the same chance of occurrence.
The chi-square test for a uniform distribution compares all c groups simultaneously.
The hypotheses are:
H0: p1 = p2 = …, pc = 1/c H1: Not all pj are equal

27. Uniform Goodness-of-Fit Test
Uniform GOF Test: Grouped Data
The test can be performed on data that are already tabulated into groups.
Calculate the expected frequency ej for each cell.
The degrees of freedom are n = c – 1 since there are no parameters for the uniform distribution.
Obtain the critical value c2a from Appendix E for the desired level of significance a.
The p-value can be obtained from Excel.
Reject H0 if p-value < a.

28. Uniform Goodness-of-Fit Test
Uniform GOF Test: Raw Data
First form c bins of equal width and create a frequency distribution.
Calculate the observed frequency fj for each bin.
Define ej = n/c.
Perform the chi-square calculations.
The degrees of freedom are n = c – 1 since there are no parameters for the uniform distribution.
Obtain the critical value from Appendix E for a given significance level a and make the decision.

29. Uniform Goodness-of-Fit Test
Uniform GOF Test: Raw Data
Maximize the test’s power by defining bin width as
As a result, the expected frequencies will be as large as possible.

30. Uniform Goodness-of-Fit Test
Uniform GOF Test: Raw Data
Calculate the mean and standard deviation of the uniform distribution as:
m = (a + b)/2
If the data are not skewed and the sample size is large (n > 30), then the mean is approximately normally distributed.
So, test the hypothesized uniform mean using
s = [(b – a + 1)2 – 1)/12

31. Poisson Goodness-of-Fit Test
Poisson Data-Generating Situations
In a Poisson distribution model, X represents the number of events per unit of time or space.
X is a discrete nonnegative integer (X = 0, 1, 2, …)
Event arrivals must be independent of each other.
Sometimes called a model of rare events because X typically has a small mean.

32. Poisson Goodness-of-Fit Test
The mean l is the only parameter.
Assuming that l is unknown and must be estimated from the sample, the steps are:
Step 1: Tally the observed frequency fj of each X-value.
Step 2: Estimate the mean l from the sample.
Step 3: Use the estimated l to find the Poisson probability P(X) for each value of X.

33. Poisson Goodness-of-Fit Test
Step 4: Multiply P(X) by the sample size n to get expected Poisson frequencies ej.
Step 5: Perform the chi-square calculations.
Step 6: Make the decision.
You may need to combine classes until expected frequencies become large enough for the test (at least until ej > 2).

34. Poisson Goodness-of-Fit Test
Poisson GOF Test: Tabulated Data
Calculate the sample mean as:
Using this estimate mean, calculate the Poisson probabilities either by using the Poisson formula P(x) = (lxe-l)/x! or Excel. ^
l =
S xj fj c
j =1 n

35. Poisson Goodness-of-Fit Test
Poisson GOF Test: Tabulated Data
For c classes with m = 1 parameter estimated, the degrees of freedom are n = c – m – 1
Obtain the critical value for a given a from Appendix E.
Make the decision.

36. Normal Chi-Square Goodness-of-Fit Test
Normal Data Generating Situations
Two parameters, m and s, fully describe the normal distribution.
Unless m and s are know a priori, they must be estimated from a sample by using x and s.
Using these statistics, the chi-square goodness-of-fit test can be used.

37. Normal Chi-Square Goodness-of-Fit Test
Method 1: Standardizing the Data
Transform the sample observations x1, x2, …, xn into standardized values.
Count the sample observations fj within intervals of the form and compare them with the known frequencies ej based on the normal distribution.
x + ks

38. Normal Chi-Square Goodness-of-Fit Test
Method 1: Standardizing the Data
Advantage is a standardized scale.
Disadvantage is that data are no longer in the original units.
Figure 15.14

39. Normal Chi-Square Goodness-of-Fit Test
Method 2: Equal Bin Widths
To obtain equal-width bins, divide the exact data range into c groups of equal width.
Step 1: Count the sample observations in each bin to get observed frequencies fj.
Step 2: Convert the bin limits into standardized z-values by using the formula.

40. Normal Chi-Square Goodness-of-Fit Test
Method 2: Equal Bin Widths
Step 3: Find the normal area within each bin assuming a normal distribution.
Step 4: Find expected frequencies ej by multiplying each normal area by the sample size n.
Classes may need to be collapsed from the ends inward to enlarge expected frequencies.

41. Normal Chi-Square Goodness-of-Fit Test
Method 3: Equal Expected Frequencies
Define histogram bins in such a way that an equal number of observations would be expected within each bin under the null hypothesis.
Define bin limits so that ej = n/c
A normal area of 1/c in each of the c bins is desired.
The first and last classes must be open-ended for a normal distribution, so to define c bins, we need c – 1 cutpoints.

42. Normal Chi-Square Goodness-of-Fit Test
Method 3: Equal Expected Frequencies
The upper limit of bin j can be found directly by using Excel.
Alternatively, find zj for bin j using Excel and then calculate the upper limit for bin j as
Once the bins are defined, count the observations fj within each bin and compare them with the expected frequencies ej = n/c.
x + zjs

43. Normal Chi-Square Goodness-of-Fit Test
Method 3: Equal Expected Frequencies
Standard normal cutpoints for equal area bins.
Table 15.16

44. Normal Chi-Square Goodness-of-Fit Test
Histograms
The fitted normal histogram gives visual clues as to the likely outcome of the GOF test.
Histograms reveal any outliers or other non-normality issues.
Further tests are needed since histograms vary.
Figure 15.15

45. Normal Chi-Square Goodness-of-Fit Test
Critical Values for Normal GOF Test
Since two parameters, m and s, are estimated from the sample, the degrees of freedom are n = c – m – 1
At least 4 bins are needed to ensure 1 df.
Table 15.19

46. ECDF Tests Kolmogorov-Smirnov and Lilliefors Tests
There are many alternatives to the chi-square test based on the Empirical Cumulative Distribution Function (ECDF).
The Kolmogorov-Smirnov (K-S) test statistic D is the largest absolute difference between the actual and expected cumulative relative frequency of the n data values:
D = Max |Fa – Fe|
The K-S test is not recommended for grouped data.

47. ECDF Tests Kolmogorov-Smirnov and Lilliefors Tests
Fa is the actual cumulative frequency at observation i.
Fe is the expected cumulative frequency at observation i under the assumption that the data came from the hypothesized distribution.
The K-S test assumes that no parameters are estimated.
If parameters are estimated, use a Lilliefors test.
Both of these tests are done by computer.

48. ECDF Tests Kolmogorov-Smirnov and Lilliefors Tests
K-S test for uniformity.
Figure 15.20

49. ECDF Tests Kolmogorov-Smirnov and Lilliefors Tests
K-S test for normality.
Figure 15.21

50. ECDF Tests Anderson-Darling Tests
The Anderson-Darling (A-D) test is widely used for non-normality because of its power.
The A-D test is based on a probability plot.
When the data fit the hypothesized distribution closely, the probability plot will be close to a straight line.
The A-D test statistic measures the overall distance between the actual and the hypothesized distributions, using a weighted squared distance.

51. ECDF Tests
Anderson-Darling Tests with MINITAB
Figure 15.22

52. Applied Statistics in Business & Economics
End of Chapter 15
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