1. Statistical Inference
Chapter 17
Statistical Inference
For Frequency Data
I Three Applications of Pearson’s 2
 Testing goodness of fit
 Testing independence
 Testing equality of proportions

2. A. Testing Goodness of Fit 1. Statistical hypotheses
H0: OPop 1 = EPop 1, , OPop k = EPop k
H1: OPop j ≠ EPop j for some j and j
2. Randomization Plan
 One random sample of n elements
 Each element is classified in terms of
membership in one of k mutually exclusive
categories

3. B. Testing Independence 1. Statistical hypotheses
H0: p(A and B) = p(A)p(B)
H1: p(A and B) ≠ p(A)p(B)
2. Randomization Plan
 One random sample of n elements
 Each element is classified in terms of
two variables, denoted by A and B, where
each variable has two or more categories.

4. C. Testing Equality of Proportions 1. Statistical hypotheses
H0: p1 = p2 = = pc
H1: pj ≠ pj for some j and j
2. Randomization Plan
 c random samples, where c ≥ 2
 For each sample, elements are classified in
terms of membership in one of r = 2 mutually
exclusive categories

5. II Testing Goodness of Fit
A. Chi-Square Distribution

6. B. Pearson’s chi-square statistic
1. Oj and Ej denote, respectively, observed and
expected frequencies. k denotes the number of
categories.
2. Critical value of chi square is with  = k – 1
degrees of freedom.

7. 1. Is the distribution of grades for summer-school
C. Grade-Distribution Example
1. Is the distribution of grades for summer-school
students in a statistics class different from that for
the fall and spring semesters?
Fall and Spring Summer
Grade Proportion Obs. frequency
A B C D F

8. 2. The statistical hypotheses are
H0: OPop 1 = EPop 1, , OPop 5 = EPop 5
H1: OPop j ≠ EPop j for some j and j
3. Pearson’s chi-square statistic is
4. Critical value of chi square for  = .05, k = 5
categories, and  = 5 – 1 = 4 degrees of freedom is

9. Table 1. Computation of Pearson’s Chi-Square for n = 72
Table 1. Computation of Pearson’s Chi-Square for n = Summer-School Students
(1) (2) (3) (4) (5) (6)
Grade Oj pj npj = Ej Oj – Ej
A (.12) =
B (.23) =
C (.47) = 33.8 –
D (.13) = 9.4 –
F (.05) = 3.6 –
2 = *
*p < .025

10. 5. Degrees of freedom when e parameters of a
theoretical distribution must be estimated is
k – 1 – e.
D. Practical Significance
1. Cohen’s w
observed and and expected proportions in the jth
category.

11. 2. Simpler equivalent formula for Cohen’s
3. Cohen’s guidelines for interpreting w
 0.1 is a small effect
 0.3 is a medium effect
 0.5 is a large effect

12. 1. When  = 1, Yates’ correction can be applied to
E. Yates’ Correction
1. When  = 1, Yates’ correction can be applied to
make the sampling distribution of the test statistic
for Oj – Ej , which is discrete, better approximate
the chi-square distribution.

13. F. Assumptions of the Goodness-of-Fit Test
1. Every observation is assigned to one and only
one category.
2. The observations are independent
3. If  = 1, every expected frequency should be at
least 10. If  > 1, every expected frequency should
be at least 5.

14. A. Statistical Hypotheses
III Testing Independence
A. Statistical Hypotheses
H0: p(A and B) = p(A)p(B)
H1: p(A and B) ≠ p(A)p(B)
B. Chi-Square Statistic for an r  c Contingency
Table with i = 1, , r Rows and j = 1, , c Columns

15. C. Computational Example: Is Success on an
C. Computational Example: Is Success on an Employment-Test Item Independent of Gender?
Observed Expected
b1 b2 b1 b2
Fail Pass Fail Pass
a1 Man
a2 Women

16. p(ai and bj) = p(ai)p(bj)
D. Computation of expected frequencies
1. A and B are statistically independent if
p(ai and bj) = p(ai)p(bj)
2. Expected frequency, for the cell in
row i and column j

17. Observed Expected
b1 b2 b1 b2 a a    

18. E. Degrees of Freedom for an r  c Contingency Table
df = k – 1 – e
= rc – 1 – [(r – 1) + (c – 1)]
= rc – 1 – r + 1 – c + 1
= rc – r – c + 1
= (r – 1)(c – 1)
= (2 – 1)(2 – 1) = 1

19. F. Strength of Association and Practical Significance
where s is the smaller of the number of rows and
columns.

20. 3. For a contingency table, an alternative formula foris

21. 1. Motivation and education of conscientious
G. Three-By-Three Contingency Table
1. Motivation and education of conscientious
objectors during WWII
High Grade
College School School Total
Coward
Partly Coward
Not Coward
Total

22. 2. Strength of Association, Cramér’s
3. Practical significance

23. H. Assumptions of the Independence Test
1. Every observation is assigned to one and only
one cell of the contingency table.
2. The observations are independent
3. If  = 1, every expected frequency should be at
least 10. If  > 1, every expected frequency should
be at least 5.

24. IV Testing Equality of c ≥ 2 Proportions
A. Statistical Hypotheses
H0: p1 = p2 = = pc
H1: pj ≠ pj for some j and j
1. Computational example: three samples of n = 100
residents of nursing homes were surveyed.
Variable A was age heterogeneity in the home;
variable B was resident satisfaction.

25. Table 2. Nursing Home Data
Age Heterogeneity
Low b1 Medium b2 High b3
Satisfied a1 O = 56 O = 58 O = 38
E = E = E = 50.67
Not Satisfied a2 O = 44 O = 42 O = 52
E = E = E = 49.33

26. B. Assumptions of the Equality of Proportions Test
1. Every observation is assigned to one and only
one cell of the contingency table.

27. C. Test of Homogeneity of Proportions
2. The observations are independent
3. If  = 1, every expected frequency should be at
least 10. If  > 1, every expected frequency should
be at least 5.
C. Test of Homogeneity of Proportions
1. Extension of the test of equality of
proportions when variable A has r > 2 rows

28. 2. Statistical hypotheses
for columns j and j'