AP Statistics Tuesday, 05 April 2016 OBJECTIVE TSW (1) identify the conditions to use a chi-square test; (2) examine the chi- square test for independence; and (3) develop an understanding for a chi-square test for homogeneity.

Chi-square test Chi-square test or  2 test

fuzzy Suppose we are interested in seeing if some “fuzzy” dice are considered “fair”? What can we do?

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

 2 distribution – df=3 df=5 df=10

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

 2 assumptions SRSSRS – reasonably random sample countsHave counts of categorical data & we expect each category to happen at least once Sample sizeSample size – to ensure 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.

 2 formula

 2 Goodness of fit test Uses univariate data We 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

Hypotheses – written in words H 0 : the observed counts equal the expected counts H a : the observed counts are not equal to the expected counts Be sure to write in context!

Let’s test our dice! No, let’s skip this part.

AP Statistics Wednesday, 06 April 2016 OBJECTIVE TSW (1) identify the conditions to use a chi-square goodness of fit test; (2) examine the chi-square test for independence; and (3) develop an understanding for a chi-square test for homogeneity. TODAY’S ASSIGNMENT –WS Chi-Square Goodness of Fit Everyone needs a calculator.

Example 1 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 = Since there are 12 signs – df = 12 – 1 = 11

Assumptions: Have a random sample of CEO’s All expected counts are greater than 5. (I expect CEO’s to be born in each sign.) H 0 : The number of CEO’s born under each sign is the same. H a : The number of CEO’s born under each sign is different. P-value =  2 cdf(5.094, 10^99, 11) =  = 0.05 Since p-value > , I fail to reject H 0. There is not sufficient evidence to suggest that the number of CEOs born under each sign is different.

Example 2 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 of 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.

Example 3 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? Expected counts: Y & N = Y & S = E & N = 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

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 distribution of fruit flies is the same as the theoretical model. H a : The distribution of fruit flies is not the same as the theoretical model. P-value =  2 cdf(5.671, 10^99, 3) =  =.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.

Assignment WS Chi Square Goodness of Fit –Due before you leave today.

 2 Test For Independence

 2 test for independence Used with categorical, bivariate data from ONE sample Used to determine if the two categorical variables are associated (dependent) or not associated (independent)

Assumptions & formula remain the same!

Hypotheses – written in words H 0 : are independent H a : are dependent Be sure to write in context! Do NOT just write: “The two variables are independent / The two variables are dependent.”

Expected Counts Assuming H 0 is true,

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

Example 4 A beef distributor wishes to determine whether there is a relationship between geographic region and cut of meat preferred. If there is no relationship, we will say that beef preference is independent of geographic region. Suppose that, in a random sample of 500 customers, 300 are from the North and 200 are from the South. Also, 150 prefer cut A, 275 prefer cut B, and 75 prefer cut C.

If beef preference is independent of geographic region, how would we expect this table to be filled in? NorthSouthTotal Cut A150 Cut B275 Cut C75 Total

Now suppose that in the actual sample of 500 consumers the observed numbers were as follows: (on your paper) Is there sufficient evidence to suggest that geographic regions and beef preference are not independent? (Is there a difference between the expected and observed counts?)

Assumptions: Have a random sample of people All expected counts are greater than 5.  approximately normal distribution H 0 : geographic region and beef preference are independent H a : geographic region and beef preference are dependent P-value = df = 2  = 0.05 Since p-value < , I reject H 0. There is sufficient evidence to suggest that geographic region and beef preference are dependent. Expected Counts: N S A90 60 B C45 30

Assignment WS Chi-Square Independence –Due on Thursday, 16 April WS Chi-Square Goodness of Fit –Due by the end of the period today.

AP Statistics Monday, 11 April 2016 OBJECTIVE TSW (1) identify the conditions to use a chi-square test; (2) examine the chi-square test for independence; and (3) develop an understanding for a chi- square test for homogeneity. ASSIGNMENT DUE –WS Chi-Square Independence  wire basket QUIZ: Chi-Square is on Wednesday, 13 April –We will have ‘C’ lunch on Wednesday. TEST: Chi-Square Inference is on Monday, 18 April Everyone needs a calculator.

 2 Test For Homogeneity

 2 test for homogeneity single categorical two (or more) independent samplesUsed with a single categorical variable from two (or more) independent samples Used to determine if the two populations are the same (homogeneous)

Assumptions & formula remain the same! Expected counts & df are found the same way as the chi-square test for independence. only The only change is the hypotheses!

Hypotheses – written in words H 0 :The two (or more) distributions of are the same. H a :The distributions of are different. Be sure to write in context!

Example 5 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) Assignment: Get together with the people in your row and complete this example (full write-up) on one piece of notebook paper AND copy this in your notes. Hand to me when your group finishes.

Assumptions ▪ 2 independent random samples ▪ Expected counts: all expected counts ≥ 5  approximately normal distribution H 0 :Drinking behavior is the same for male and female students. H a :Drinking behavior is different for male and female students. χ 2 = p = df = 3 α = 0.05 p < α  Reject H 0. There is sufficient evidence to suggest that drinking behavior is different for male and female students. MenWomen None Low Moderate High

Assignments WS Chi-Square –Due on Wednesday, 13 April WS Mixed Hypothesis Review #2 –Due on Wednesday, 13 April  you will get this tomorrow! WS Mixed Hypothesis Review #3 –Due on Monday, 18 April  you will get this tomorrow! WS AP Review Chi-Square  you will get this on Friday, 04/15/16! QUIZ: Chi-Square is on Wednesday, 13 April TEST: Chi-Square Inference is on Monday, 18 April 2016.