11/12/09

1

FPP 28

Chi-square test

More types of inference for nominal variables  Nominal data is categorical with more than two categories

 Compare observed frequencies of nominal variable to hypothesized probabilities

 Chi-squared goodness of fit test

 Test if two nominal variables are independent

 Chi-squared test of independence

11/12/09

2

Goodness of fit test  Do people admit themselves to

hospitals more frequently close to their birthday?

 Data from a random sample of 200 people admitted to hospitals

Days from birthday

Number of admissions

within 7 11

8-30 24

31-90 69

91+ 96

Goodness of fit test  Assume there is no birthday effect, that is, people admit

randomly. Then, Pr (within 7) = = .0411

Pr (8 - 30) = = .1260 Pr (31-90) = = .3288 Pr (91+) = = .5041

 So, in a sample of 200 people, we’d expect

to be in “within 7” to be in “8 - 30” to be in “31 - 90” to be in “91+”

11/12/09

3

Goodness of fit test   If admissions are random, we expect the sample frequencies

and hypothesized probabilities to be similar

 But, as always, the sample frequencies are affected by chance error

 So, we need to see whether the sample frequencies could have been a plausible result from a chance error if the hypothesized probabilities are true.

 Let’s build a hypothesis test

Goodness of fit test  Hypothesis

 Claim (alternative hyp.) is admission probabilities depend on the days since birthday

 Opposite of claim (null hyp.) is probabilities in accordance with random admissions.

 H0 : Pr (within 7) = .0411 Pr (8 - 30) = .1260 Pr (31-90) = .3288 Pr (91+) = .5041

 HA : probabilities different than those in H0 .

11/12/09

4

Goodness of fit test: Test statistic  Chi-squared test statistic

€

X 2 = sum (observed - expected)2

expected

Goodness of fit test: Test statistic

€

X 2 = sum (observed - expected)2

expected

= .94 + .057 + .16 + .23 =1.397

Cell Obs Exp Dif Dif2 Dif2/Exp

In 7

8-30

31-90

91+

11/12/09

5

Goodness of fit test: Calculate p-value  X2 has a chi-squared distribution with degrees of freedom

equal to number of categories minus 1.   In this case, df = 4 – 1 = 3.

Goodness of fit test: Calculate p-value  To get a p-value, calculate the area under the chi-squared

curve to the right of 1.397

 Using JMP, this area is 0.703. If the null hypothesis is true, there is a 70% chance of observing a value of X2 as or more extreme than 1.397

 Using the table the p-value is between 0.9 and 0.70

11/12/09

6

Chi-squared table

JMP output admissions

11/12/09

7

Goodness of fit test: Judging p-value  The .70 is a large p-value, indicating the data could well

occur by random chance when the null hypothesis is true. Therefore, we cannot reject the null hypothesis. There is not enough evidence to conclude that admissions rates are independent of time from birthday.

Independence test   Is birth order related to

delinquency?

 Nye (1958) randomly sampled 1154 high school girls and asked if they had been “delinquent”.

Eldest 24 450

In Between 29 312

Youngest 35 211

Only 23 70

11/12/09

8

Sample of conditional frequencies  % Delinquent for each birth

order status  Based on conditional

frequencies, it appears that youngest are more delinquent

 Could these sample frequencies have plausibly occurred by chance if there is no relationship between birth order and delinqeuncy

Oldest .05

Middle .085

Youngest .14

Only .25

Test of independence  Hypotheses

 Claim is that there is some relationship between birth order and delinquency.

 Opposite is that there is no relationship.

 H0 : birth order and delinquency are independent.

 HA : birth order and delinquency are dependent.

11/12/09

9

Implications of independence   Expected counts

 Under independence,   Pr(oldest and delinquent) = Pr(oldest)*Pr(delinquent)

 Estimate Pr(oldest) as marginal frequency of oldest

 Estimate Pr(delinquent) as marginal frequency of delinquent

 Hence, estimate Pr(oldest and delinquent) as

 The expected number of oldest and delinquent, under independence, equals

 This is repeated for all the other cells in table

Test of independence  Expected counts

 Next we compare the observed counts with the expected to get a test statistic

Oldest 45.59 428.41

In Between 32.80 308.2

Youngest 23.66 222.34

Only 8.95 84.05

11/12/09

10

 Use the X2 statistic as the test statistic:

Test of independence:   Calculate the p-value   X 2 has a chi-squared distribution with degrees of freedom:

df = (number rows – 1) * (number columns – 1)

  In delinquency problem, df = (4 - 1) * (2 - 1) = 3.

  The area under the chi-squared curve to the right of 42.245 is less than .0001. There is only a very small chance of getting an X2 as or more extreme than 42.245.

11/12/09

11

JMP output for chi-squared test   This is a small p-value. It is

unlikely we’d observe data like this if the null hypothesis is true. There does appear to be an association between delinquency and birth order.

11/12/09

12

Chi-squared test details  Requires simple random samples.  Works best when expected frequencies in each cell are at

least 5.

 Should not have zero counts  How one specifies categories can affect results.

Chi-squared test items   What do I do when expected counts are less than 5?   Try to get more data. Barring that, you can collapse categories.

Example: Is baldness related to heart disease? (see JMP for data set)

Baldness Disease Number of people None Yes 251 None No 331 Little Yes 165 Little No 221 Some Yes 195 Some No 185 Combine “extreme” and “much” categories Much Yes 50 Much or extreme Yes 52 Much No 34 Much or extreme No 35 Extreme Yes 2 Extreme No 1

This changes the question slightly, since we have a new category.

11/12/09

13

Chi-squared test   for collapsed data for baldness

example  Based on p-value, baldness and

heart disease are not independent.

 We see that increasing baldness is associated with increased incidence of heart disease.