Chapter 18 Chi-Square Tests

2 Distribution Let x1, x2, .. xn be a random sample from a normal distribution with  and 2, and let s2 be the sample variance, then the random variable (n-1)s2/2 has 2 distribution with n-1 degrees of freedom. Probability Density Function, with k degrees of freedom, Mean Variance Mode = k-2 (when k 3)

fr.academic.ru/pictures/frwiki/67/Chi-square_.. 2 Distribution fr.academic.ru/pictures/frwiki/67/Chi-square_..

2 Distribution Goodness-of-fit Tests Tests of Independence Tests of Homogeneity

Multinominal Experiments A Multinomial experiment is a statistical experiment that has the following properties: It consists of n repeated trials (repetitions). Each trial can result in one of k possible outcomes. The trials are independent. The probabilities of the various events remain constant for each trial.

Goodness-of-fit Test Observed Frequencies (Oi): Frequencies obtained from the actual performance of an experiment. Goodness-of-fit Test: Test of null hypothesis that the observed frequencies follow certain pattern or theoretical distributions, expressed by the Expected Frequencies (Ei).

Goodness-of-fit Test for Multinominal Experiments Degree of freedom = k -1, where k is the number of categories Chi-square goodness-of-fit test is always a right-tailed test Sample size should be large enough so that the expected frequency for each category is at least 5.

Goodness-of-fit Test for Multinominal Experiments Null Hypothesis: H0: the observed frequencies follow certain pattern Test statistic: Degree of Freedom = k -1 Alt. Hypothesis P-value Rejection Criterion H1 P(2>02) 02 > 2,k-1

Goodness-of-fit Test for Multinominal Experiments -- Example 18.1 Department stores in shopping mall H0: p1 = p2 = p3 = p4 = p5 = .20 H1: At least 2 of pi  .20  = .01, df = 5 -1 = 4 Test statistic: Critical value: .01,4 = 13.276 P-value = .000549 Reject H0 Oi Ei Oi-Ei (Oi-Ei)2 (Oi-Ei)2/Ei 1 214 200 14 196 0.98 2 231 31 961 4.81 3 182 -18 324 1.62 4 219 19 361 1.81 5 154 -46 2116 10.58 19.79

Goodness-of-fit Test for Multinominal Experiments -- Example 18.2 Market shares H0: p1=.144, p2=.181, p3=.248, p4=.141, p5=.149, p6 =.137 H1: At least 2 of pi are different  = .025, df = 6 -1 = 5 Test statistic: Critical value: .025,5 = 12.8325 P-value = .1252 Fail to reject H0 Oi Ei Oi-Ei (Oi-Ei)2 (Oi-Ei)2/Ei 1 270 288 -18 324 1.1250 2 382 362 20 400 1.1050 3 467 496 -29 841 1.6956 4 317 282 35 1225 4.3440 5 298 -10 100 0.3356 6 276 274 0.0146 2000 8.6197

Test of Independence for a Contingency Table Columns 1 2 … c Rows O11 O12 O1c u1 O21 O22 O2c u2 r Or1 Or2 Orc ur 1 2 c

Test of Independence for a Contingency Table Null Hypothesis: H0: The two attributes are independent Alt. Hypothesis: H1: The two attributes are dependent Test statistic: Degree of Freedom = (r-1)(c-1) Alt. Hypothesis P-value Rejection Criterion H1 P(2>02) 02 > 2,df

Test of Independence for a Contingency Table – Example 18.3 Oij Support Against No Opinion Total ui Female 87 32 6 125 .4167 Male 93 70 12 175 .5833 180 102 18 300 j .60 .34 .06 Eij Support Against No Opinion Female 75 42.5 7.5 Male 105 59.5 10.5

Test of Independence for a Contingency Table – Example 18.4 Null Hypothesis: H0: The two attributes are independent Alt. Hypothesis: H1: The two attributes are dependent Test statistic: Degree of Freedom = (r-1)(c-1) = (2-1)(3-1) = 2 Critical value: .025,2 = 7.3778 P-value = .0161 Reject H0

Test of Independence for a Contingency Table – Example 18.5 Oij Good Defective Total ui Mac 1 109 11 120 .6 Mac 2 66 14 80 .4 175 25 200 j .875 .125 Eij Good Defective Mac 1 105 15 Mac 2 70 10

Test of Independence for a Contingency Table – Example 18.5 Null Hypothesis: H0: The two attributes are independent Alt. Hypothesis: H1: The two attributes are dependent Test statistic: Degree of Freedom = (r-1)(c-1) = (2-1)(2-1) = 1 Critical value: .01,1 = 6.6349 P-value = .0809 Fail to reject H0

Test of Homogeneity Similar to the test of independence If row/column totals are fixed, perform a test of homogeneity Columns 1 2 … c Rows O11 O12 O1c u1 O21 O22 O2c u2 r Or1 Or2 Orc ur 1 2 c

Test of Homogeneity Null Hypothesis: H0: two sets of data are homogeneous Alt. Hypothesis: H1: sets of data are not homogeneous Test statistic: Degree of Freedom = (r-1)(c-1) Alt. Hypothesis P-value Rejection Criterion H1 P(2>02) 02 > 2,df

Test of Independence for a Contingency Table – Example 18.6 Oij Calif. NY Total ui Very Satis. 60 75 135 .15 Somewhat Sat. 100 125 225 .25 Somewhat Dissat. 184 140 324 .36 Very Dissatis. 156 216 .24 500 400 900 j .556 .444 Eij Calif. NY Very Satis. 75 60 Somewhat Sat. 125 100 Somewhat Dissat. 180 144 Very Dissatis. 120 96

Test of Independence for a Contingency Table – Example 18.6 Null Hypothesis: H0: The two states are homogeneous Alt. Hypothesis: H1: The two states are not homogeneous Test statistic: Degree of Freedom = (r-1)(c-1) = (4-1)(2-1) = 3 Critical value: .025,3 = 9.3484 P-value = 3.1424E-9 Reject H0