1. Chi-square test Chi-square test or  2 test

2. Chi-square test countsUsed to test the counts of categorical data ThreeThree types –Goodness of fit (univariate) –Independence (bivariate) –Homogeneity (univariate with two samples)

3.  2 distribution – df=3 df=5 df=10

4.  2 distribution Different df have different curves Skewed right normal curveAs df increases, curve shifts toward right & becomes more like a normal curve

5.  2 assumptions SRSSRS – reasonably random sample countsHave counts of categorical data & we expect each category to happen at least once Sample sizeSample size – to insure that the sample size is large enough we should expect at least five in each category. ***Be sure to list expected counts!! Combine these together: All expected counts are at least 5.

6.  2 formula

7.  2 Goodness of fit test Uses univariate data Want to see how well the observed counts “fit” what we expect the counts to be  2 cdf function p-valuesUse  2 cdf function on the calculator to find p-values Based on df – df = number of categories - 1

8. Hypotheses – written in words H 0 : proportions are equal H a : at least one proportion is not the same Be sure to write in context!

9. Does your zodiac sign determine how successful you will be? Fortune magazine collected the zodiac signs of 256 heads of the largest 400 companies. Is there sufficient evidence to claim that successful people are more likely to be born under some signs than others? Aries 23Libra18Leo20 Taurus20Scorpio21Virgo19 Gemini18Sagittarius19Aquarius24 Cancer23Capricorn22Pisces29 How many would you expect in each sign if there were no difference between them? How many degrees of freedom? I would expect CEOs to be equally born under all signs. So 256/12 = 21.333333 Since there are 12 signs – df = 12 – 1 = 11

10. Assumptions: Have a random sample of CEO’s All expected counts are greater than 5. (I expect 21.33 CEO’s to be born in each sign.) H 0 : The proportions of CEO’s born under each sign are the same. H a : At least one of the proportion of CEO’s born under each sign is different. P-value =  2 cdf(5.094, 10^99, 11) =.9265  =.05 Since p-value > , I fail to reject H 0. There is not sufficient evidence to suggest that the CEOs are born under some signs more than others.

11. A company says its premium mixture of nuts contains 10% Brazil nuts, 20% cashews, 20% almonds, 10% hazelnuts and 40% peanuts. You buy a large can and separate the nuts. Upon weighing them, you find there are 112 g Brazil nuts, 183 g of cashews, 207 g of almonds, 71 g or hazelnuts, and 446 g of peanuts. You wonder whether your mix is significantly different from what the company advertises? Why is the chi-square goodness-of-fit test NOT appropriate here? What might you do instead of weighing the nuts in order to use chi-square? counts Because we do NOT have counts of the type of nuts. count We could count the number of each type of nut and then perform a  2 test.

12. Offspring of certain fruit flies may have yellow or ebony bodies and normal wings or short wings. Genetic theory predicts that these traits will appear in the ratio 9:3:3:1 (yellow & normal, yellow & short, ebony & normal, ebony & short) A researcher checks 100 such flies and finds the distribution of traits to be 59, 20, 11, and 10, respectively. What are the expected counts? df? Are the results consistent with the theoretical distribution predicted by the genetic model? (see next page) Expected counts: Y & N = 56.25 Y & S = 18.75 E & N = 18.75 E & S = 6.25 We expect 9/16 of the 100 flies to have yellow and normal wings. (Y & N) Since there are 4 categories, df = 4 – 1 = 3

13. Assumptions: Have a random sample of fruit flies All expected counts are greater than 5. Expected counts: Y & N = 56.25, Y & S = 18.75, E & N = 18.75, E & S = 6.25 H 0 : The proportions of fruit flies are the same as the theoretical model. H a : At least one of the proportions of fruit flies is not the same as the theoretical model. df = 4 – 1 = 3P-value =  2 cdf(5.671, 10^99, 3) =.129  =.05 Since p-value > , I fail to reject H 0. There is not sufficient evidence to suggest that the distribution of fruit flies is not the same as the theoretical model.

14.  2 test for independence Used with categorical, bivariate data from ONE sample Used to see if the two categorical variables are associated (dependent) or not associated (independent) Key phrasing: independent/ dependent, related to, affected by, etc.

15. Assumptions & formula remain the same!

16. Hypotheses – written in words H 0 : two variables are independent H a : two variables are dependent Be sure to write in context!

17. The following data are based on the results of an (old) survey to examine media preferences. Does the evidence support the claim that the type of media preferred and age group are not independent? TV fansRadio Listeners Newspaper Readers Totals 18 to 34294131101 35 to 49222729 78 50 or older503240 122 Totals 101100 301

18. If preferred media type is independent, how would we expect this table to be filled in? TV fansRadio Listeners Newspaper Readers Totals 18 to 34 33.8933.55 101 35 to 49 26.1725.91 78 50 or older 40.9440.53 122 Totals 101100 301

19. Expected Counts Assuming H 0 is true,

20. Degrees of freedom Or cover up one row & one column & count the number of cells remaining!

21. Assumptions: Have a random sample of people All expected counts are greater than 5. (expected counts listed on the previous slide) H 0 : age & media preference are independent H a : age and & media preference are dependent P-value =.1144df = 4  =.05 Since p-value > , I fail to reject H 0. There is not sufficient evidence to suggest that age and media preference are dependent.

22.  2 test for homogeneity single categorical two (or more) independent samplesUsed with a single categorical variable from two (or more) independent samples Used to see if the two populations are the same (homogeneous) Key phrasing: is different from, a difference between, is the same, etc.

23. Assumptions & formula remain the same! Expected counts & df are found the same way as test for independence. Only Only change is the hypotheses!

24. Hypotheses – written in words H 0 : the proportions for the two (or more) distributions are the same H a : At least one of the proportions for the distributions is different Be sure to write in context!

25. The following data is on drinking behavior for independently chosen random samples of male and female students. Does there appear to be a gender difference with respect to drinking behavior? (Note: low = 1-7 drinks/wk, moderate = 8-24 drinks/wk, high = 25 or more drinks/wk) MenWomenTotal None140186326 Low4786611139 Moderate300173473 High631679 Total98110362017

26. Assumptions: Have 2 random sample of students All expected counts are greater than 5. H 0 : the proportions of drinking behaviors is the same for female & male students H a : at least one of the proportions of drinking behavior is different for female & male students P-value = 8.67 x 10^-21df = 3  =.05 Since p-value < , I reject H 0. There is sufficient evidence to suggest that drinking behavior is not the same for female & male students. Expected Counts: M F 0158.6167.4 L554.0585.0 M230.1243.0 H38.440.6