1. Business 205

2. Review Correlation MS5

3. Preview Chi-Square

4. Types of Tests Parametric Tests Assume normal and homogeneity of variance Require parameters Z-test, T-test, 2-Group T, ANOVA Non-parametric Tests Few (if any) assumptions about the population distribution Rarely state a hypotheses in terms of a specific parameter Chi-Square

5. Tests for “goodness of fit” Looks at sample data to test hypothesis about the shape or proportions of a population distribution How well does the sample proportions fit the population proportions?

6. Chi-Square Hypotheses Regular Hypothesis Specifies preferences Specifies differences in population Null Hypothesis Specifies no preferences Specifies no difference in population ABC ≠ 1/3 ABC 1/3

7. One-Sample Chi-Square χ2 Compares the goodness of fit of the data to that of the null hypothesis Compares observed frequencies against expected frequencies Looks at the difference between 1 IV with multiple levels

8. Observed Frequencies Number of individuals from the sample who are classified in a particular category How many times it occurs

9. Expected Scores/Frequencies The frequency value that is predicted from the null hypothesis and the sample size.

10. One-Sample Chi-Square

11. 1 Sample Chi-Square Example StudentObserved Scores 129 219 318 425 517 610 715 811

12. Fill out the chart Observed Scores (O) Expected Scores (E) O - E(O - E) 2 E 2918111216.72 191811.06 18 000 25187492.72 17181.06 1018-8643.56 1518-39.50 1118-7492.72

13. Chi-Square Chi-Square = (6.72+.06+0+2.72+.06+3.56+.50+2.72) = 16.34 Critical Chi-Square value at an alpha of.05 is 3.84. Accept that your sample is not from the population because 16.34 is larger than 3.84.

14. Two Sample Chi-Square Test Chi-Square Test for Independence 2 Factors Many levels Tests whether or not there is a relationship between 2 variables

15. Two Sample Chi-Square Test Chi-Square Test for Independence df = (number of rows - 1)(number of columns - 1)

16. Two Sample Chi-Square Test Chi-Square Test for Independence Three different drug treatments are used to control hypertension. At the end of treatment, the investigator classifies patients as having either a favorable or unfavorable response to the medication. Your hypothesis is that there is a different between treatments.

17. Two Sample Chi-Square Test Chi-Square Test for Independence Treatment 1Treatment 2Treatment 3Total in rows Favorable70160168 Unfavorable304032 Total in columnsN = 500 1.) Total up each column 2.) Total up each row 3.) Find the expected score for each cell

18. Two Sample Chi-Square Test Chi-Square Test for Independence Treatment 1Treatment 2Treatment 3Total in rows Favorable70 (E = 79.6) 160 (E = 159.2) 168 (E = 159.2) 398 Unfavorable30 (E = 20.4) 40 (E = 40.8) 32 (E = 40.8) 102 Total in columns100200 N = 500

19. Two Sample Chi-Square Test Chi-Square Test for Independence ObservedExpectedO-E(O- E) 2 (O- E) 2 /E 70(398)(100)/500 = 79.6-9.692.161.15 30(102)(100)/500 = 20.49.692.164.15 160(398)(200)/500 = 159.2.8.64.00 40(102)(200)/500 = 40.8-.8.64.01 168(398)(200)/500 = 159.28.877.4.48 32(102)(200)/500 = 40.8-8.877.41.89

20. Two Sample Chi-Square Test Chi-Square Test for Independence Chi-Square = (1.15+4.15+.00+.01+.48+1.89) = 8.02 Critical value for an alpha of.05 and at 2 df = 5.99 Ours falls above the critical region so we can accept the hypothesis and say that the treatments differ.

21. Reporting findings Chi-Square Test for Independence The chi-square value is in the critical region. Therefore, we can reject the null hypothesis. There is a relationship between drug treatments and responses towards medication (2, n =500) = 8.02, p <.05.

22. Chi-Square In Class Example Hotel 1Hotel 2Hotel 3Total in rows Favorable5010085 Unfavorable102016 Total in columns N = H1: There is a difference between hotels.

23. Excel: Chi-Square DO NOT USE: CHIDIST( )  Only returns a probability For Single Chi-Squares, you must program that in by hand CHITEST( )  Only returns a probability For a test of independence (2 sample) you must program that in by hand