chi-square test - duke universitygp42/sta101/notes/fpp28_2pp.pdf · 11/12/09 1 fpp 28 chi-square...
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11/12/09
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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
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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+”
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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 .
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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+
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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
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Chi-squared table
JMP output admissions
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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
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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.
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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
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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.
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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.
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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.
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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.
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