Chi-square testChi-square testor

2 test

What if we are interested in seeing if my “crazycrazy” dice are considered “fair”?

What can I do?

Let’s test our dice!Let’s test our dice!

Chi-square testChi-square test•Used to test the countscounts of

categorical data•ThreeThree types

–Goodness of fit (univariate)– Independence (bivariate)–Homogeneity (univariate with two samples)

22 distribution – distribution –

df=3

df=5

df=10

22 distribution distribution

•Different df have different curves

•Skewed right•As df increases, curve shifts

toward right & becomes more like a normal curvenormal curve

2 2 assumptionsassumptions• SRS SRS – reasonably random sample• Have countscounts 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.

2 2 formulaformula

exp

expobs 22

2 2 Goodness of fit testGoodness of fit test

•Uses univariate data•Want to see how well the observed counts “fit” what we expect the counts to be

•Use 22cdf functioncdf function on the calculator to find p-valuesp-values

Based on df –Based on df –

df = number of df = number of categoriescategories - 1 - 1

Hypotheses – written in Hypotheses – written in wordswords

H0: the observed counts equal the expected countsHa: the observed counts are not

equal to the expected counts

Be sure to write in context!

Let’s test our dice!

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 Leo20

Taurus 20 Scorpio 21 Virgo 19

Gemini 18 Sagittarius19 Aquarius24

Cancer 23 Capricorn 22 Pisces29

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 = 21.333333Since 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 21.33 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 = 2cdf(5.094, 10^99, 11) = .9265 = .05

Since p-value > , I fail to reject H0. There is not

sufficient evidence to suggest that the CEOs are born under some signs than others.

094.5

3.21

3.2129...

3.21

3.2120

3.21

3.2123222

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 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 countscounts

of the type of nuts.We could countcount the

number of each type of nut and then perform a

2 test.

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 = 56.25Y & S = 18.75E & N = 18.75E & S = 6.25We 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

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 = 2cdf(5.671, 10^99, 3) = .129 = .05

Since p-value > , 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.

671.5

25.625.610

...75.18

75.182025.56

25.5659 2222

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)

Assumptions & Assumptions & formula remain the formula remain the

same!same!

Hypotheses – written in Hypotheses – written in wordswords

H0: two variables are independentHa: two variables are dependent

Be sure to write in context!

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.

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

Total 300 200 500

90 60

165

110

45 30

Expected Counts Expected Counts

•Assuming H0 is true,

totaltable

alcolumn tot totalrow counts expected

Degrees of freedomDegrees of freedom

)1c)(1(r df

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

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.

H0: geographic region and beef preference are

independent Ha: geographic region and beef

preference are dependent

P-value = .0226 df = 2 = .05

Since p-value < , I reject H0. There is sufficient evidence to suggest that geographic region and beef preference are dependent.

576.7...

606050

9090100 22

2

Expected Counts:

N S

A 90 60

B 165 110

C 45 30

22 test for homogeneity test for homogeneity

•Used with a single single categoricalcategorical variable from two (or more) independent two (or more) independent samplessamples

•Used to see if the two populations are the same (homogeneous)

Assumptions & formula remain the same!

Expected counts & df are found the same way as test for independence.

OnlyOnly change is the hypotheses!

Hypotheses – written in Hypotheses – written in wordswords

H0: the two (or more) distributions are the sameHa: the distributions are different

Be sure to write in context!

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)

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 = .05

Since p-value < , I reject H0. There is sufficient evidence to suggest that drinking behavior is not the same for female & male students.

53.96...

4.1674.167186

6.1586.158140 22

2

Expected Counts:

M F

0 158.6 167.4

L 554.0 585.0

M 230.1 243.0

H 38.4 40.6