FPP 28

1
Chi-square test

• FPP 28

2
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
• One categorical variable with more than two
  categories
• Chi-squared goodness of fit test
• Test if two nominal variables are independent
• Two categorical variables with at least one
  having more than two categories
• 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
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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, wed 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.
  
• Lets build a hypothesis test

6
Goodness of fit test

• Hypothesis
• Claim (alternative hyp.) is admission
  probabilities change according to 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

8
Goodness of fit test Test statistic
Cell Obs Exp Dif Dif2 Dif2/Exp
In 7
8-30
31-90
91
9
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.

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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
12
JMP output admissions
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Goodness of fit test Judging p-value

• The .70 is a large p-value, indicating that the
  difference between the observed and expected
  counts 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 change
  according to days from birthday.

14
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
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Test of independence

• Hypotheses
• Want to show 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(delinque
  nt)
• 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

18
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

20
Test of independence

• Calculate the p-value
• X 2 has a chi-squared distribution with degrees
  of freedomdf (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 wed
  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.

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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.