1. Statistical Analysis Professor Lynne Stokes
Department of Statistical Science
Lecture #2
Chi-square Tests for Homogeneity, Chi-square Goodness of Fit Test,

2. Chi-square Tests Tests for independence in contingency tables
Tests for homogeneity

3. Binomial Samples (Product Binomial Sampling)
Genetic Theory: Ho: pW = vs. Ha: pW
Assumptions: 8 samples,
mutually independent counts
Hypothesis #1: Is pw = 0.5?
Binomial inference on p
Equivalently, overall goodness of fit (known p)
Hypothesis #2: Are all the pw equal?
Test for homogeneity (equal but unknown p)
Hypothesis #3: Is each pw = 0.5?
Goodness of fit (8 samples, known p)

4. Test of Homogeneity of k Binomial Samples, Specified p
Ho: p1 = p2 = … =p8 = vs. Ha: pj for some j
Does not assume
homogeneity
(see below)
X2 = , df = 8 , p = 0.003

5. Test of Homogeneity of k Binomial Samples: Unspecified p
Ho: p1 = p2 = … =p8 vs. Ha: pj pk for some (j,k)

6. Test of Homogeneity of k Binomial Samples: Unspecified p
Ho: p1 = p2 = … =p8 vs. Ha: pj pk for some (j,k)
X2 = , df = 7 , p = 0.005
Note: Only one of each pair of expected values is independently estimated
(k = 8, not 16)

7. Chi-square Tests Tests for independence in contingency tables
Tests for homogeneity
Goodness of fit tests

8. Chi-square Goodness of Fit Test: Specified Probabilities
Assumptions
n independent observations
k mutually exclusive possible outcomes
pj = Pr(outcome j) is the same on every trial
Sample size condition
All npj
At least 80% of the npj

9. Goodness of Fit Test: Specified Probabilities
Sample size: n
Observed count for outcome j : Oj
Expected count for outcome j : Ej = npj
Ho: Pr(outcome j) = pj for j = 1 , ... , k
Ha: Pr(outcome j) pj for at least one j
Reject Ho if X2 > Xa2
Xa2 = Chi-Square
df = k - 1

10. Sufficient Evidence of
Cognitive Learning
Path Chosen A B C D
Total
Number of rats
Expected number
Sufficient Evidence of
Cognitive Learning ?
p = 0.026
Using a significance level of a = 0.05, there is sufficient
evidence (p = 0.026) to reject the hypothesis that rats
choose the 4 doors with equal probability.

11. Mendelian Inheritance
Do the genotypes of a cross-breeding occur in the ratio
9:3:3:1 ?
Reject Ho if X2 > (a = 0.05)

12. Mendelian Inheritance
X2 =
= 2.66
There is insufficient evidence (p > 0.10) at a significance
level of 0.05 to conclude that the genotypes from this type
of cross-breeding occur in proportions that differ from
those predicted by Mendelian inheritance theory.

13. Chi-Square Goodness of Fit Test: Unknown Parameters
Estimate the parameters of the distribution
Divide range of data values into mutually exclusive and exhaustive classes
Discrete data: often use the values themselves
Continuous data: use k = n1/2 or k = log(n) classes
Estimate the probability of being in each class
Compare the observed (Oi) counts in each class with the estimated expected (Ei) counts

14. Chi-Square Goodness of Fit Test for the Poisson Distribution
Number of senders
(automated telephone
equipment) in use at
a given time
23 – 1 = 22 Categories
H0: number ~ Poisson
Ha: number not Poisson
Reject if X > C20.05(20) = 31.4
df: 22 – 1 (mutually exclusive &
exhaustive) – 1 (estimated
parameter) = 20

15. Chi-Square Goodness of Fit Test for the Normal Distribution
Divide the data into mutually exclusive and exhaustive (contiguous) classes
First and last classes are open-ended
( , U1), (L2,U2), (L3, U3) … (Lk, ) with Lj = Uj-1
Estimate the mean and standard deviation
Calculate z-scores for the limits of each class
Estimate the Probability Content for Each Class
pj = Pr(zLj < z < zUj)
Estimate the Expected Frequency for Each Class
Ej = npj

16. Chi-Square Goodness of Fit Test
Can be applied to any discrete or continuous probability distribution, only probabilities need be specified: Ei = npi
Asymptotic chi-square distribution
All Ei > 1 & at Least 80% of the Ei > 5
Does not have the highest power for specific distributions, against specific alternatives
Degrees of freedom (k classes)
If each class represents an independent sample (i.e, k replicate samples) and all parameters are known (i.e., known probabilities), df = k
If the classes represent mutually exclusive and exhaustive categories (i.e., expected frequencies must sum to n), data are independent and from a single sample
All parameters are known, df = k – 1
r parameters are estimated: df = k – r – 1
e.g., (n – 1)s2/s2 ~ C2(n – 1)

17. Goodness of Fit to the Binomial, Known p
Normal theory approximation
Chi-square tests

18. Binomial Sample, Specified p: Normal Theory Approximation
Genetic Theory: Ho: pW = vs. Ha: pW
Greater Power by Combining Samples
(Assuming Homogeneity)
p = 0.110

19. Alternative to the Binomial Test: Chi-square Goodness of Fit, Specified p
Genetic Theory: Ho: pW = vs. Ha: pW
p = 0.110

20. Overall Binomial Test vs. Test of Homogeneity, Specified p
Ho: p1 = p2 = … =p8 = vs. Ha: pj for some j
X2 = , df = 1 , p = 0.110
Greater Power if Homogeneous
X2 = , df = 8 , p = 0.003
Greater Power if Not Homogeneous

21. Homogeneity, unspecified p equivalent to independence
Binomial Samples
pw unspecified
Homogeneity, unspecified p equivalent to independence

22. Some Goodness of Fit Tests
Chi-square Goodness-of-fit test
Very general, can have little power
Kolmogorov-Smirnov goodness-of-fit test
Good general test, especially for continuous random variables
Wilk-Shapiro test for normality
Regarded as the best test for normality

23. Comparing Odds Ratios Across Categories

24. Race and Death Penalty Punishment
Are the results consistent across aggravation levels ?

25. Mantel-Haenszel Test Several 2 x 2 tables
Assuming a common odds ratio, test that the odds ratio = 1

26. Race and Death Penalty Punishment
Expected frequencies for chi-square test of independence
Note: None have sufficient sample sizes for tests of independence

27. Mantel-Haenszel Test Select one cell; e.g., upper-left
Calculate the excess for each table
Excess = Observed – Expected
e.g., Excess = O11 – E11
Calculate the variances of the excesses
Variance = R1R2C1C2/n2(n-1)

28. Race and Death Penalty Punishment
Conclusion:
Nearly 7 more white-victim murderers received the death penalty
than would be expected if the odds were the same for
white- and black-victim murderers

30. Estimating the Common Odds Ratio
Death Penalty and Race