1. Chapter 12 – Chi Square Tests We have done tests on population means and population proportions. However, it is possible to test for population variances. Also, it is necessary to test if an observed frequency is different from an expected frequency and if a variable is independent of another variable. These three tests are covered in Chapter 12. When performing these tests, we use a different type of distribution. This is known as chi-square (pronounced as kai-square) distribution.

2. A chi-square distribution A chi-square distribution is shown in figure 12-1, on page 372. The shape of the curve depends on sample size. Appendix F (pages 482-483) presents probabilities of chi-square distribution.

3. Tests for population variance The test statistic for these tests is: 2 2 (n – 1) s X = ------------- 2 σ The two hypotheses are: 2 2 H : σ = H : σ > < ≠

4. Tests for pop. var. (Cont) The decision rule is the same as before example problem 12-1 page 310. 2 X = 11.25 Critical x 2 from appendix F for 9 Degrees of freedom & alpha of 0.01 is 21.666. Decision: Do not reject H 0.

5. Interpretation: The variance in breaking strength of the cable has remained the same (at most 40,000 pounds). Exercises from the book Problem 10, page 312 Given: n=6 s 2 =91.6 σ 2 =100 Alpha =0.1 H0: σ 2 =100 H1: σ 2 <100

6. Decision Rule 2 Reject H 0 if x < critical x 2 Critical x 2 from appendix F is 1.61031 Be careful of how to find critical x 2 from appendix F

7. Decision: Do not reject H 0 Interpretation: The weight gain for the specific breed of sheep is equally uniform (That is the variance in weight gain is equal to 100). Problems for you to do: Problem #6 p.312, #8 p.312, #11 p.313

8. Testing for goodness of fit In some experiments, we are interested in knowing if observed frequencies are same as expected frequencies. For example, we toss a coin 100 times. We expect that we will get 50 heads and 50 tails. Now, suppose we observed 45 heads and 55 tails. Can we still say that the coin is balanced?

9. Testing for goodness of fit (Con’t) These types of tests are called goodness of fit tests. The words expected and observed frequencies are important. Two hypotheses are: H 0 : observed frequencies are the same as expected frequencies H 1 : Observed frequencies are not the same as expected frequencies

10. The test statistic is: 2 2 (O - E ) X = ∑--------------- E Where O j are observed frequencies E j are expected frequencies Decision Rules (Always of this form): 2 Reject H 0 if X-Statistic is > Critical x 2

11. The test statistic (con’t) The degrees of freedom is number of rows minus 1. Refer to table 12-2 on page 314. The degrees of freedom is 6-1=5. Example problem 12-4, page 315-316. Refer to table 12-3 on page 315 The hypotheses are: H 0 : voters today have the same education attainment ads those 20 years ago H 1 : Voters today do not have the same educational attainment as those 20 years ago

12. The test statistic X 2 -statistic = 19.25 Critical x 2 with 4 degrees of freedom and alpha of 0.05 is 9.48773. Decision: reject H 0 Interpretation: Voters today do not have the same education attainment as those 20 years ago.

13. Exercises from book Problem 6 on page 316-317 The 2 hypotheses are: H 0 : Saturday & Sunday each accounts for 25% and each of the 5 remaining days accounts for 10% of all fatal accidents H 1 : Sat and … does not account …accidents

14. Exercises Use alpha = 0.025 N=7 so degree of freedom = 7-1 = 6 Decision rule: reject H 0 if X 2 critical x 2 2 X = 2.4 Critical x 2 from appendix F =14.4494 Decision: Do not reject H 0 Interpretation: Sat & Sun each accounts for 25% & each of the 5 weekdays accounts for 10% of all fatal accidents of the LA freeways Problems for you to do: Prob. #5, p. 316, #9, p.317