1. Chi-square test or c2 test

2. What if we are interested in seeing if my “crazy” dice are considered “fair”?
What can I do?

3. Chi-square test Used to test the counts of categorical data
Three types
Goodness of fit (univariate)
Independence (bivariate)
Homogeneity (univariate with two samples)

4. c2 distribution –
df=3
df=5
df=10

5. c2 distribution Different df have different curves Skewed right
As df increases, curve shifts toward right & becomes more like a normal curve

6. c2 assumptions SRS – reasonably random sample
Have counts of categorical data & we expect each category to happen at least once
Sample 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.

7. c2 formula

8. df = number of categories - 1
c2 Goodness of fit test
Based on df –
df = number of categories - 1
Uses univariate data
Want to see how well the observed counts “fit” what we expect the counts to be
Use c2GOF-Test function to do entire calculation
Use c2cdf function on the calculator to find just p-values

9. Hypotheses – written in words
H0: the observed counts equal the expected counts
Ha: the observed counts are not equal to the expected counts
Be sure to write in context!

10. Let’s test our dice!

11. I would expect CEOs to be equally born under all signs.
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 23 Libra 18 Leo 20
Taurus 20 Scorpio 21 Virgo 19
Gemini 18 Sagittarius 19 Aquarius 24
Cancer 23 Capricorn 22 Pisces 29
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

12. 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.)
H0: The number of CEO’s born under each sign is the same.
Ha: The number of CEO’s born under each sign is the different.
P-value = c2cdf(5.102, 10^99, 11) = a = .05
Since p-value > a, I fail to reject H0. There is not sufficient evidence to suggest that the CEOs are born under one sign more than another.

13. Because we do NOT have counts of the type of nuts.
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?
Because we do NOT have counts of the type of nuts.
We could count the number of each type of nut and then perform a c2 test.

14. Since there are 4 categories,
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 a random 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)
Since there are 4 categories,
df = 4 – 1 = 3
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)

15. 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
H0: The distribution of fruit flies is the same as the theoretical model.
Ha: The distribution of fruit flies is not the same as the theoretical model.
P-value = c2cdf(5.671, 10^99, 3) = a = .05
Since p-value > a, I fail to reject H0. There is not sufficient evidence to suggest that the distribution of fruit flies is not the same as the theoretical model.

16. c2 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)

17. Assumptions & formula remain the same!

18. Hypotheses – written in words
H0: two variables are independent
Ha: two variables are dependent
Be sure to write in context!

19. 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 from the South. Also, 150 prefer cut A, 275 prefer cut B, and 75 prefer cut C.

20. Expected Counts Assuming H0 is true,
Must calculate for each cell of table
For example, cell 1,1
To understand the origin of this equation, read pages 653 and 654 of text book

21. If beef preference is independent of geographic region, how would we expect this table to be filled in?
North
South
Total
Cut A
150
Cut B
275
Cut C 75
300
200
500 90 60
165
110 45 30

22. Degrees of freedom
Or cover up one row & one column & count the number of cells remaining! If there are Totals cells, don’t count those.

23. 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?)

24. Using the Calculator Enter the Observed data in Matrix A
Enter the Expected data in Matrix B
Use the STAT ->TESTS->

25. Have a random sample of people All expected counts are greater than 5.
Assumptions:
Have a random sample of people
All expected counts are greater than 5.
H0: geographic region and beef preference are independent Ha: geographic region and beef preference are dependent
P-value = df = 2 a = .05
Since p-value < a, I reject H0. There is sufficient evidence to suggest that geographic region and beef preference are dependent.
Expected Counts:
N S A B C

26. c2 Test for Homogeneity
Used with a single categorical variable from two (or more) independent samples
Used to see if the two populations are the same (homogeneous)

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

28. Hypotheses – written in words
H0: the two (or more) distributions are the same
Ha: the distributions are different
Be sure to write in context!

29. 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)

30. Have 2 random sample of students
Expected Counts:
M F L M H
Assumptions:
Have 2 random sample of students
All expected counts are greater than 5.
H0: drinking behavior is the same for female & male students Ha: drinking behavior is not the same for female & male students
P-value = .000 df = 3 a = .05
Since p-value < a, I reject H0. There is sufficient evidence to suggest that drinking behavior is not the same for female & male students.

31. Telling Chi Squared Tests Apart
Goodness-of-Fit: 1 variable in 1 population
ex: take 1 sample and ask 1 question
Homogeneity of Proportions: 1 variable in 2 or more populations or treatment
ex: take 2 or more samples or treatments and ask 1 question
Independence:2 variables in 1 population
ex: take 1 sample and ask 2 questions