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™

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

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

Another Test for Ho: p = po versus Ha: p ≠ po 1. The hypotheses are Ho: p1 = .04 p2 = .96 Ha: p1 ≠ .04 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

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

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

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

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

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

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

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

Multinomial Goodness-of-Fit Test Figure 13.2

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

Expected Frequencies Classification 1 Classification 2 1 2 3 4 c r C1 Cc R1 R2 R3 Rr E = ^ R2C3 n Figure 13.4

Chi-Square Test of Independence Estimating the Expected Frequencies Example Age Gender <30 30-45 >45 Total Male 60 (60) 20 (30) 40 (30) 120 Female 40 (40) 30 (20) 10 (20) 80 Total 100 50 50 200 Estimate for Male and Over 45 E = 200 • • = = 30 ^ 120 200 50 (120)(50)

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

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)

Chi-Square Test Example Figure 13.5

Chi-Square Test Example Figure 13.5

Test of Independence with Fixed Marginal Totals Figure 13.5