15. Chi-Squared Test BA2606-05 (X2 Test) 1 Two Techniques... The first is a [ applied to data produced by a multinomial experiment, a generalization of a binomial experiment and is used to describe one population of data. The second uses data arranged in a [ to determine whether two classifications of a population of nominal data are statistically independent; this test can also be interpreted as a comparison of two or more populations. In both cases, we use the [ BA2606-05 (X2 Test) distribution. 2 Goodness-of-Fit test from multinomial experiment BA2606-05 (X2 Test) 3 Goodness-of-Fit test Hypothesis H0: p1 = a1, p2 = a2, ..., pk = ak where a1, a2, ..., ak are the values we want to test. H1: At least one pi is not equal to its specified value Test statistic: where fi = observed frequency, and ei = expected frequency = n * pi. Decision: If , then we reject H0. We have a strong evidence to infer that at least one pi is not equal to its specified value. BA2606-05 (X2 Test) 4 Ex 15.1 Two companies, A and B, have recently conducted aggressive advertising campaigns to increase their respective market shares (MS) for fabric softener. Before the campaigns, the MS was 45% for company A, 40% for company B, and 15% for other remaining competitors. To determine whether these MS changed after the campaigns, a marketing analyst collected a 200 random samples. The data showed that the preference was 102 for A's product, 82 for B's, and 16 for the other competitors. Can the analyst infer at the 5% significance level that customer preferences have changed from their levels before the advertising campaigns were launched? BA2606-05 (X2 Test) 5 Ex 15.1 (2) 1. Hypothesis H0: pA = 0.45, pB = 0.40, and pO = 0.15 H1: At least one pi is not equal to its specified value ( f i ei ) 2 ei i 1 k 2 2. Test statistic: with ei = n * pi. Observed Frequency Expected Frequency Delta Summation Component fi ei = n * pi (fi - ei) (fi - ei)2/ei A 102 90 12 1.60 B 82 80 2 0.05 30 -14 6.53 Company Others 16 Total 200 BA2606-05 (X2 Test) 8.18 = 2 6 Ex 15.1 (3) 3. Decision Criteria 2,k-1 = 2.05, 3-1 = (from Table 5, on p. B-11) 2 = 8.18> 2 Since our ,k-1, we have a sufficient evidence about the change in the MS. Instead we can get p-value (from Excel, not from Table). p-value = P(2 > 8.18) = CHIDIST(8.18, 2) = 0.0167 Another method to determine p-value in Excel A 1 2 3 4 5 B C Expected Observed Company Probability Frequency A 0.45 102 B 0.4 82 Others 0.15 16 Total 1 200 multiplied by n. =CHITEST(C2:C4,B2:B4*C5) Observed Frequency Expected Probability It will also give 0.0167. BA2606-05 (X2 Test) 7 Goodness-of-Fit test Hypothesis H0: p1 = a1, p2 = a2, ..., pk = ak H1: At least one pi is not equal to its specified value Test statistic: where fi = observed frequency, and ei = expected frequency = n * pi. Decision: If 2 > 2,k-1, then we reject H0. Required condition: Rule of Five = The sample size must be large enough so that the expected value for each cell is 5 or more. BA2606-05 (X2 Test) 8 Statistical Independence Test from Contingency Table BA2606-05 (X2 Test) 9 Statistical Independence The conditional probability of A given B, by P(A|B), is computed as follows: A B B P( A | B) Statistical Independence: P(A and B) P(B) B BC A An occurrence of an event does not affect the probability of another event. P(A|B) = Multiplication Rule for Independent Events BA2606-05 (X2 Test) AC 10 Statistical Independence Test The 2 test of a contingency table is used to determine whether there is enough evidence to infer that are related. Hypotheses H0: The two variables are independent. H1: The two variables are dependent. Test statistic: where fij = observed frequency, and eij = expected f. = nP(Ai & Bj) = d.f. = Requirement: BA2606-05 (X2 Test) 11 Ex 5.2 The MBA program was experiencing problems scheduling their courses. The demand for the program's optional courses and majors was quite variable from one year to the next. The Statistics professor believed that the problem relates to the academic background of the students, especially the undergraduate degree. As a start, he took a random sample of last year's MBA students and recorded the undergraduate degree and the major selected in the MBA program. The Xm 15-02 file was coded as [1-BA, 2BEng, 3-BBA, 4-Other] for the undergraduate degree, and [1ACCT, 2-FIN, 3-MKTG] for MBA majors. Can the statistician conclude that the undergraduate degree affects the choice of major? BA2606-05 (X2 Test) 12 Ex 5.2 (2) 1 Open and Execute Xm 15-02. 2 3 4 Steps 5 1. Change it to the original data, using 6 7 [1-BA, 2-BEng, 3-BBA, 4-Other] for degree, and [1-ACCT, 2-FIN, 3-MKTG] for MBA majors. 8 9 2. Pivot Table to the form of 10 11 Degree ACCT FIN MKTG Total 12 BA 13 14 BEng 15 BBA 16 Other 17 Total 18 19 A B Degree MBA Major 3 1 1 1 1 1 1 1 2 2 1 3 3 1 1 1 2 1 2 2 3 1 3 1 4 1 3 3 4 3 4 1 1 1 3 BA2606-05 (X2 Test) 3 13 Ex 5.2 (3) Steps 3. Using the Pivot Table, draw the bar graph You will get the similar graph as Picture 15.3 4. Copy the data. And then make two more tables, with the same format. 1) The first copy will be used for the expected frequency. (Use \"Anchored cell\" to compute eij = n(Ai)n(Bj)/n) 2) The second copy is used for computing (fi - ei)2/ei 5. Find the critical value for 2. And make a conclusion. (You may find the p-value from \"CHITEST\" function.) BA2606-05 (X2 Test) 14 Summary Two techniques multinomial experiment goodness-of-fit test H0: p1 = a1, p2 = a2, ..., pk = ak contingency table statistical independence test H0: The two variables are independent. 2 k Test statistic: 2 ( f i ei ) ei i 1 2 i where ei = n * pi. compared to 2,k-1 ( f ij eij ) 2 j eij 2, (r-1)(c-1) (p-value can be found by =CHIDIST() in Excel.) Requirement: Rule of Five BA2606-05 (X2 Test) 15 Answers #2: goodness-of-fit test #4: #6: ( f e )2 i i ei i 1 k chi-squared ( 2 ) contingency table 2 > 2,k-1 2 fi ei = n * p i (fi - ei) (fi - ei)2/ei A 102 90 12 1.60 B 82 80 2 0.05 Others 16 30 -14 6.53 Total 200 8.18 = 2 #7: 5.99 #8: {ei = npi 5 for all i} #10: P(A) P(A and B) = P(A) * P(B). #11: two nominal variables 2 i = nP(Ai)P(Bj)= n(Ai)n(Bj)/n BA2606-05 (X2 Test) j (r-1)(c-1) ( f ij e ij ) 2 eij Rule of Five 16 Scanned by CamScanner Method 1 2 3 4 1 2 2 2 1 2 2 4 3 3 1 2 4 1 1 1 1 1 2 2 2 4 1 2 3 1 2 4 4 2 4 1 1 1 4 2 2 2 1 1 3 1 2 4 3 2 2 1 1 3 2 4 1 2 2 2 2 2 2 1 2 1 1 1 1 4 2 1 4 1 2 3 2 4 2 2 1 1 2 2 4 4 4 2 1 2 4 Quit 1 1 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 2 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 1 1 1 2 1 2 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 2 2 1 1 1 2 4 1 3 2 4 2 2 2 2 2 2 2 1 4 2 3 4 4 2 1 2 1 2 4 4 1 2 3 4 4 1 1 2 2 1 2 4 1 2 4 2 4 1 1 4 1 3 3 2 3 4 2 3 2 2 1 1 1 4 2 2 1 2 1 1 1 1 1 1 2 2 1 2 1 3 1 2 1 2 2 2 1 1 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 2 1 1 2 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 2 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 3 4 1 2 1 3 3 1 1 2 2 2 3 1 2 2 3 3 4 1 2 3 2 2 2 2 1 1 3 4 1 2 1 2 1 2 3 2 2 2 2 4 2 1 1 1 1 1 2 2 1 4 1 3 2 1 4 1 3 2 4 1 2 3 2 4 4 4 2 1 1 2 1 4 2 1 2 1 1 1 2 4 4 2 1 2 2 2 3 2 3 2 1 1 2 1 1 1 1 1 1 2 1 2 1 1 2 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 2 1 1 1 1 1 1 2 1 1 1 2 4 1 2 2 4 1 2 1 1 2 2 1 1 1 3 4 2 2 4 1 3 2 4 1 1 2 3 2 1 1 2 2 3 4 1 1 3 2 2 2 2 2 1 4 2 2 1 4 2 1 1 1 1 2 2 1 4 4 2 1 2 4 3 1 3 2 2 2 2 2 4 2 1 2 1 1 4 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1