Part IVB

1. Part IVB Chi Square Tests Dr. Stephen H. Russell Weber State University

2. The Chi Square Technique A Non-Parametric way of testing hypotheses (Non-parametric means no assumption is required regarding the population’s distribution.) The technique uses “count data” in cells Chi Square tests are always right-tailed

3. Overview: We’ll test hypotheses relating to: • Goodness-of-Fit for equal population proportions • Goodness-of-Fit multiple population proportions • The alleged independence of two qualitative variables An aside: This technique uses a Greek Symbol for a statistic: 2 (Chi-Square)

4. A “Goodness-of-Fit” Example I set up a booth in the Mall to sell baseball cards of six Hall of Fame players: Tom Seaver, Nolan Ryan, Ty Cobb, George Brett, Hank Aaron, and Johnny Bench. I believe that card sales will reveal that these Hall of Famers are equally popular with the card-buying public. At the end of the day I sold 120 cards. If my supposition is true, I would have sold 20 cards on each player. What I actually sold by player:

5. A Contingency Table: O (E) Step 1: State Hypotheses: HO: Players are equally popular HA: Players are not equally popular

6. A Contingency Table: O (E) * The Null hypothesis really states that there is no difference between the observed frequencies and the expected frequencies—or that the differences noted can be attributed to sampling error (chance). * This is one row of cell data. Chi-square Goodness-of-Fit problems always have just one row of cell data.

7. The Baseball Card problem continued Step 2: Set alpha (We’ll make  = .05) Step 3: Collect data Step 4: Compute a test statistic

8. Computing a test statistic What is the expected value of 2 if the Null is true?

9. Making a Decision In Step 2 we set  = .05. Let’s now define this .05 tail of rejection. The Chi-Square Table (MINITAB invcdf function) Degrees of Freedom = number of categories – 1 = k - 1 = 6 – 1 in this problem In a Chi Square distribution for 5 df, what number places .05 of the distribution to the right? 11.07

10. A “Picture” of our decision: The Chi-Square Distribution is skewed (similar to the F distribution) α = .05 34.30 11.07 We conclude that the differences between O’s and E’s is not due to chance. These results are statistically significant. We reject the idea that these baseball players are equally popular.

11. One final note on the baseball player problem Our test statistic was 34.40 and we had 5 degrees of freedom in this problem. Use the cdf function in MINITAB to figure out the p value in this problem. It’s zero.

12. Let’s try a Multiple Proportions problem • A lawyer wants to know if the educational levels of those called for jury duty in Weber County reflect the educational makeup of the county population that is eligible for jury duty. The educational level of the last 100 jurors chosen was recorded. • We have the following data that we will test at the .10 level of significance:

13. Highest Educational Level County-wide: .50 have high school or less .35 have some college .15 has a college degree The last 100 jurors: 53 have high school or less 39 have some college 8 have a college degree What are the O’s and the E’s?

14. The Hypotheses: HO: high school or less = .50 some college = .35 college degree = .15 HA: Not all of the above proportions are true Contingency Table H.S. or Less Some College College Degree

15. Calculating the Chi Square Statistic H.S. or Less Some College College Degree What Chi Square value defines the tail of rejection?

16. Calculating the Chi Square Statistic H.S. or Less Some College College Degree 4.605 What is our decision?

17. Our Decision: H.S. or Less Some College College Degree We are not in the right tail. We do not have sufficient evidence to reject the proposition that our jury pool reflects the county.

18. A note about MINITAB • MINITAB does not include a routine to compute Chi Square Goodness of Fit tests. You’ll have to do them by hand. • MINITAB does include commands to do Chi Square Tests of Independence, which we will now consider.

19. Chi Square Tests of Independence • Goodness of Fit tests are one-dimensioned: We only have one row in the contingency table • Chi Square tests of independence necessarily are twodimensioned—that is they have at least two rows. This is true because we classify the data with a column characteristic and a row characteristic. • We ask the question: Are the column characteristic and the row characteristic independent or dependent (related)? • Consider this example:

20. Chi Square Test of Independence Example • The Federal Corrections Agency wants to investigate whether a male released from federal prison makes a different adjustment to civilian life if he returns to his hometown as compared to being relocated to a new city. To put it another way, is there a relationship between adjustment to civilian life and place of residence after release from prison? • In an experiment, 200 released prisoners are tracked for 15 months. • HO: There is no relationship between behavior after release and where the individual lives after release HA: Behavior after release and where the individual lives are dependent. Note: ALWAYS the null states independence The Alternative states dependence I

21. Prisoner problem: Adjustment to Free Society After Release These Table values are the O’s. Can you figure the joint probabilities required to calculate the 8 E’s we need for a Chi Square Test?

22. Prisoner problem: Adjustment to Free Society After Release Let’s solve this problem with MINITAB

23. Prisoner Problem • This is a Chi-Square Test of Independence • As in all Chi-Square Test of Independence problems, this problem is two-dimensioned. The column characteristic is behavior; the row characteristic is location upon release from prison. • Notice the Chi-Square statistic is 5.729 and has an associated P-Value of .126. What is our decision at an alpha of .10? • We don’t “accept” the Null but we cannot reject it. We say “At the 10% level, we have no statistically significant evidence that behavior and location are related.”

24. Interpreting the Chi-Square Statistic • Since deviations between the observed and expected values are squared, chi-square can never be negative. • The chi-square procedure magnifies large deviation relative to small deviations. • Large squared numbers indicate likely dependence between 2 qualitative variables. • Dividing each squared difference by the expected frequency converts absolute into relative squared deviations and thereby puts all cells on an equal footing.

25. Degrees of Freedom • The nature of the sampling distribution of chi-square depends on the number of degrees of freedom associated with the problem under investigation. • Number of degrees of freedom for a test of independence is (number of rows – 1) x (number of columns – 1)

26. Summary • The chi-square technique can help us test hypotheses when we cannot use the normal or t distributions. • We can construct a chi-square statistic by comparing observed frequencies of some phenomenon against theoretical expected frequencies (based on independent samples) on the other hand. • We introduced two types of Chi Square Problems: Goodness of Fitand Tests of Independence • Different distributions of the chi-square statistic appear based on different numbers of degrees of freedom.