Download Comparing Two Means

Comparing Two Proportions
Example: caries incidence
Clinical trial with caries intervention on infants
developed caries
N
by age two
controls
36
27.8%
intervention 68
8.8%
Is this strong evidence of effectiveness of
experimental intervention?
Comparison of two proportions
- two independent samples
These are called “two-sample” tests.
Our goal is usually to estimate p1 – p2, the
corresponding confidence intervals, and to perform
hypothesis tests on:
H0: p1 – p2 = 0.
The obvious statistic to compare the two population
proportions is p̂1 - p̂2 . Where p̂i = number of
successes in group i divided by sample size in group
i.
Probability theory tells us that:
1. p̂1 - p̂2 is the best estimate of p1 – p2
2. the standard error is
p1 (1  p1 ) n1  p2 (1  p2 ) n2
3. If n1p1(1-p1) > 5 and n2p2(1-p2) > 5

pˆ1  pˆ 2 ~ N p1  p2 , p1 (1  p1 ) n1  p2 (1  p2 ) n2

Large-sample confidence interval for p1 – p2
ˆ1  p
ˆ 2  Z1 / 2 
p
ˆ 1 (1  p
ˆ 1 ) n1  p
ˆ 2 (1  p
ˆ 2 ) n2
p
Large-sample Z-test of
H0: p1 – p2 = 0 vs. H1: p1 – p2 ≠ 0
pˆ1  pˆ 2
Z

Test statistic:
SEH 0 ( pˆ1  pˆ 2 )
Where SE H ( pˆ 1  pˆ 2 ) denotes the standard error
estimates using the null hypothesis, p1 = p2.
0
Estimate the common p using
pˆ 
x1  x2
n1  n2 ,
where x1 and x2 are the number of successes in
groups 1 and 2, respectively.
Then
SEH 0 ( pˆ1  pˆ 2 ) 
pˆ (1  pˆ )1 n1  1 n2 
Compare Z to a standard Normal distribution.
Example: Caries incidence
caries by age two
controls
N
36
intervention
68
Number
percent
10
27.8
6
8.8
95% confidence interval:
p̂1 - p̂2 = .278 - .088 = 0.19
SE  .278 (1  .278) 36  .088(1  .088) 68  .082
So 95% confidence interval is
0.19  1.96  0.082  0.029, 0.351
Test: H0: p1 – p2 = 0 vs. H1: p1 – p2 ≠ 0
pˆ  10  6 36  68  .154
SEH 0 ( pˆ1  pˆ 2 )  .154(1  .154)1 36  1 68  .074
Z
0.19
 2.57
.074
Reject at α=.05 level. P-value = 2×P(Z > 2.57) = .0102
Chi-squared Test ( χ2 test)
Chi-square test generalizes two-sample Z-test to
situation with more than two proportions.
Example: perio by gender (NHANES I data):
Evaluate whether periodontitis is independent of
gender by seeing if the proportion of males in each
group defined by periodontal status is the same.
χ2 test utilizes “contingency” tables
Observed Data
Co unt
per iodon tal status
GE NDE R
T o tal
healthy
gin givitis
per io
T o tal
male
11 43
92 9
93 7
30 09
fem ale
26 07
14 90
92 1
50 18
37 50
24 19
18 58
80 27
The null hypothesis is that all proportions are equal
H0: p1 = p2 = p3.
Expected frequencies (under assumption of equal proportions)
male
healthy
periodontal status
gingivitis
perio
Total
3750 ×
(3009/8027)
= 1405.7
2419 ×
(3009/8027)
= 906.8
1858 ×
(3009/8027)
= 696.5
3,009
3750 ×
(5018/8027)
= 2344.3
2419 ×
(5018/8027)
= 1512.2
1858 ×
(5018/8027)
= 1161.5
5,018
3,750
2,419
1,858
8,027
female
Total
Chi-squared statistic:
X2 =
Σ
(observed - expected)2
expected
(1143  1405.7) 2 (929  906.8) 2 (937  696.5) 2



1405.7
906.8
696.5
(2607  2344.3) 2 (1490  1512.2) 2 (921  1161.5) 2



2344.3
1512.2
1161.5
= 212.3
Large (positive) values of X2 indicate evidence
against the null hypothesis.

If H0 is true, then a χ2 statistic from a contingency
table with R rows and C columns should have a
Chi-square distribution with (R-1) × (C-1)
degrees of freedom.

The P-value is the probability that a χ2(R-1) × (C-1)
distribution is greater than the observed statistic.

Note that all the probability in the p-value (and
rejection region) is on one side, since only large
values of X2 would contradict H0.

Our statistic, 212.3, was larger than 15.20, the
99.95th percentile of a χ22 dist’n, so p < 0.0005.

Table 6 in the coursepack has χ2 percentiles.
SPSS output for Chi-square test
GENDER * periodontal status Crosstabulation
per iodon tal status
GENDER
male
Co unt
Ex pected Count
fem ale
Co unt
Ex pected Count
To tal
Co unt
Ex pected Count
healthy
gin givitis
per io
To tal
11 43
92 9
93 7
30 09
14 05.7
90 6.8
69 6.5
30 09.0
26 07
14 90
92 1
50 18
23 44.3
15 12.2
11 61.5
50 18.0
37 50
24 19
18 58
80 27
37 50.0
24 19.0
18 58.0
80 27.0
Chi-Square Tests
Value
Asymp. Sig.
(2- sided)
df
Pearson Chi- Squar e
21 2.271 a
2
.00 0
Lik eliho od Ratio
21 0.264
2
.00 0
Lin ear-by-Linear Associatio n
20 9.324
1
.00 0
N o f Valid Cases
80 27
a. 0 cells (. 0%) have expected count less than 5. The mini mum
expected count is 696.49.
Notes on Chi-squared test:
1. Chi-square test p-values rely on Normal
approximations, so they not valid for small
samples (any expected frequencies < 5).
2. The rejection region for a Chi-square test with
significance level α is the region above the
100(1- α)th percentile of the Chi-square
distribution (i.e. not α/2).
3. The null hypothesis for the Chi-square test can
be equivalently formulated as “X1 is
independent of X2”, where X1 and X2 are the
two categorical variables being compared
(gender and perio status in our example).
4. When comparing two proportions the Chisquare test is equivalent to Z-test.