1. Chapter 13 – Applications of the Chi-Square Statistic
Introduction to Business Statistics, 6e
Kvanli, Pavur, Keeling
Chapter 13 – Applications of the Chi-Square Statistic
Slides prepared by Jeff Heyl, Lincoln University
©2003 South-Western/Thomson Learning™

2. The Binomial Situation
The experiment consists of n repetitions called trials
The trials are independent
Each trial has two possible outcomes, success or failure
The probability of a success for each trial is p

3. Another Test for Ho: p = po versus Ha: p ≠ po
More than two possible outcomes: the multinomial situation
2 = ∑
(O - E)2 E

4. Another Test for Ho: p = po versus Ha: p ≠ po
1. The hypotheses are
Ho: p1 = p2 = .96
Ha: p1 ≠ p2 ≠ .96
2. The test statistic is
2 = ∑ =
(O - E)2 E
(O1 - E1)2 E1
(O2 - E2)2 E2
reject Ho if 2 > 2.05,1
3. Reject Ho if the chi-square test statistic lies in the right tail

5. Another Test for Ho: p = po versus Ha: p ≠ po
4. We have
2* = = 8.51
(13 - 6)2 6
( )2
144
O1 = 13 E1 = 6
O2 = 137 E2 = 144
5. The proportion of defectives (p1) is not .04

6. p-Value for Chi-Square Analysis2
2* = 8.51
Figure 13.1

7. Testing Ho: p = po versus Ha: p ≠ po
Using Z Test
Test statistic:
Z =
p - po
po(1 - po) n ^
Rejection region:
reject Ho if |Z| > Z/2

8. Testing Ho: p = po versus Ha: p ≠ po
Using 2 Test
Test statistic:
2 = ∑ =
(O - E)2 E
(O1 - E1)2 E1
(O2 - E2)2 E2
Rejection region:
reject Ho if 2 > ,1

9. The Multinomial Situation Assumptions
The experiment consists of n independent repetitions (trials)
Each trial outcome falls in exactly one of k categories
The probabilities of the k outcomes are denoted by p1, p2, ..., pk and remain the same on each trial. Further: p1 + p2, pk = 1

10. Hypothesis Testing for the Multinomial Situation
2 = ∑
(O - E)2 E
where:
1. The summation is across all categories
2. The O’s are the observed frequencies in each category using the sample
3. The E’s are the expected frequencies in each category if Ho is true
4. The df for the chi-square statistic are k-1, where k is the number of categories

11. p-Value for Hospital Example
10.6
38.4
Area = .005
Area = p-value 2
Figure 13.2

12. Multinomial Goodness-of-Fit Test
Figure 13.2

13. Chi-Square Test of Independence
Null and Alternative Hypothesis
Ho: the classifications are independent
Ha: the classifications are dependent
Estimating the Expected Frequencies
E = ^
(row total for this cell)•(column total for this cell) n

14. Expected Frequencies Classification 1 Classification 2 1 2 3 4 c r C1Cc R1 R2 R3 Rr
E = ^
R2C3 n
Figure 13.4

15. Chi-Square Test of Independence
Estimating the Expected Frequencies Example
Age
Gender < >45 Total
Male 60 (60) 20 (30) 40 (30) 120
Female 40 (40) 30 (20) 10 (20) 80
Total
Estimate for Male and Over 45
E = 200 • • = = 30 ^
120
200 50
(120)(50)

16. Chi-Square Test of Independence
The Testing Procedure
Ho: the row and column classifications are independent
Ha: the row and column classifications are dependent
reject Ho if 2 > 2.,df
where df = (r-1)(c-1)
2 = ∑
(O - E)2 E

17. Chi-Square Test of Independence
1. The summation is over all cells of the contingency table consisting of r rows and c columns
2. O is the observed frequency
3. E is the expected frequency ^
E =
(total of all cells)
total of row in
which the cell lies
total of column in •
4. The degrees of freedom are df = (r-1)(c-1)

18. Chi-Square Test Example
Figure 13.5

19. Chi-Square Test Example
Figure 13.5

20. Test of Independence with Fixed Marginal Totals
Figure 13.5