fpp 28 chi-square test. more types of inference for nominal variables nominal data is categorical...
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FPP 28
Chi-square test
More types of inference for nominal variablesNominal data is categorical with more than two
categories
Compare observed frequencies of nominal variable to hypothesized probabilitiesOne categorical variable with more than two
categoriesChi-squared goodness of fit test
Test if two nominal variables are independentTwo categorical variables with at least one
having more than two categoriesChi-squared test of independence
Goodness of fit testDo 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 testAssume 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+”
Goodness of fit testIf 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 testHypothesis
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 .
Goodness of fit test: Test statisticChi-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+
Goodness of fit test: Calculate p-valueX2 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-valueTo 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
Chi-squared table
JMP output admissions
31 - 90 8 - 30 91+ Within 7
31 - 90 8 - 30 91+ Within 7
31 - 90
8 - 30
91+
Within 7
Total
Level
69
24
96
11
200
Count
0.34500
0.12000
0.48000
0.05500
1.00000
Prob
4 Levels
Frequencies
31 - 90
8 - 30
91+
Within 7
Level
0.34500
0.12000
0.48000
0.05500
Estim Prob
0.32900
0.12600
0.50400
0.04100
Hypoth Prob
Likelihood Ratio
Pearson
Test
1.3063
1.3974
ChiSquare
3
3
DF
0.7276
0.7061
Prob>Chisq
Method: Fix hypothesized values, rescale omitted
Test Probabilities
Days
Distributions
Goodness of fit test: Judging p-valueThe .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.
Independence testIs 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
Sample of conditional frequencies% Delinquent for each
birth order statusBased 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 independenceHypotheses
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.
Implications of independenceExpected 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 independenceExpected 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
Use the X2 statistic as the test statistic:
245.4205.84
)05.8470(
95.8
)95.823(
34.222
)34.222211(
66.23
)66.2335(
2.308
)2.308312(
80.32
)80.3229(
41.428
)41.428450(
49.45
)59.4524(
2222
22222
X
Test of independence:Calculate the p-valueX 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.
JMP output for chi-squared test
Freq: Column 3
Birth
Order
Eldest
In Betwe
Only Child
Youngest
450
38.99
43.14
94.94
428.407
1.0883
24
2.08
21.62
5.06
45.5927
10.2263
312
27.04
29.91
91.50
308.2
0.0468
29
2.51
26.13
8.50
32.7998
0.4402
70
6.07
6.71
75.27
84.0546
2.3500
23
1.99
20.72
24.73
8.94541
22.0819
211
18.28
20.23
85.77
222.338
0.5782
35
3.03
31.53
14.23
23.662
5.4327
474
41.07
341
29.55
93
8.06
246
21.32
1043
90.38
111
9.62
1154
Delinquent?
Count
Total %
Col %
Row %
Expected
Cell Chi 2
No Yes
Contingency Table
Model
Error
C. Total
N
Source
3
1150
1153
1154
DF
18.54974
346.83395
365.38369
-LogLike
0.0508
RSquare (U)
Likelihood Ratio
Pearson
Test
37.099
42.245
ChiSquare
<.0001
<.0001
Prob>ChiSq
Tests
Contingency Analysis of Delinquent? By Birth Order
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.
Chi-squared test detailsRequires simple random samples.Works best when expected frequencies in
each cell are at least 5.Should not have zero countsHow 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.
Chi-squared testfor collapsed data for
baldness exampleBased on p-value,
baldness and heart disease are not independent.
We see that increasing baldness is associated with increased incidence of heart disease.
Freq: Count
Bal
dnes
s
Little
Much+Extreme
None
Some
221
15.40
28.63
57.25
165
11.50
24.89
42.75
35
2.44
4.53
40.23
52
3.62
7.84
59.77
331
23.07
42.88
56.87
251
17.49
37.86
43.13
185
12.89
23.96
48.68
195
13.59
29.41
51.32
386
26.90
87
6.06
582
40.56
380
26.48
772
53.80
663
46.20
1435
HeartDisease
Count
Total %
Col %
Row %
No Yes
Contingency Table
Model
Error
C. Total
N
Source
3
1431
1434
1435
DF
7.25181
983.27068
990.52249
-LogLike
0.0073
RSquare (U)
Likelihood Ratio
Pearson
Test
14.504
14.510
ChiSquare
0.0023
0.0023
Prob>ChiSq
Tests
Contingency Analysis of HeartDisease By Baldness