Transcript Ch 14

Chapter 14 Tests of Hypotheses
Based on Count Data
 14.2 Tests concerning proportions (large
samples)
 14.3 Differences between proportions
 14.4 The analysis of an r x c table
14.2 Tests concerning proportions
(large samples)
 np>5; n(1-p)>5
 n independent trials;
 X=# of successes
 p=probability of a success
X
 Estimate: pˆ  n
Tests of Hypotheses
 Null H0: p=p0
 Possible Alternatives:
HA: p<p0
HA: p>p0
HA: pp0
Test Statistics
 Under H0, p=p0, and
 pˆ 
 Statistic:
z
p(1  p)

n
pˆ  p0
 pˆ

p0 (1  p0 )
n
pˆ  p0
p0 (1  p0 )
n
is approximately standard normal under H0 .
Reject H0 if z is too far from 0 in either direction.
Rejection Regions
Alternative
Hypotheses
HA: p>p0
HA: p<p0
HA: pp0
Rejection
Regions
z>z
z<-z
z>z/2 or
z<-z/2
Equivalent Form:
z
pˆ  p0
X  np0
n
* 
p0 (1  p0 ) n
np0 (1  p0 )
n
Example 14.1
 H0: p=0.75 vs HA: p0.75
 =0.05
 n=300
 x=206
 Reject H0 if z<-1.96 or z>1.96
ˆ 
p
X
206

 0.68667
n
300
Observed z value
0.68667  0.75
z
 2.5
0.75(1  0.75)
300
or
z
206  225
 2.5
300(0.75)(1  0.75)
 Conclusion: reject H0 since z<-1.96
 P(z<-2.5 or z>2.5)=0.0124< reject H0.
Example 14.2
 Toss a coin 100 times and you get 45 heads
 Estimate p=probability of getting a head
Is the coin balanced one? =0.05
Solution:
H0: p=0.50 vs HA: p0.50
45
pˆ 
 0.45
100
Enough Evidence to Reject H0?
 Critical value z0.025=1.96
 Reject H0 if z>1.96 or z<-1.96
45  100(0.50)
5
z

 1
5
100(0.50)(1  0.50)
 Conclusion: accept H0
Another example
 The following table is for a certain screening test
Truth = surgical biopsy
FNA
status
Result
Positive
Results
Negative
Total
sensitivity 
Cancer
Present
140
Cancer
Absent
80
10
910
150
990
True positive
140

 0.93
True Positives  False Negatives 150
Total
220
920
1140
 Test to see if the sensitivity of the screening
test is less than 97%.
 Hypothesis H : p  p  .97
0
0
Ha : p  p0  .97
 Test statistic
estimated proportion-prestated proportion
z
standard error of the estimated proportion
pˆ  p0
pˆ  p0
140 150  .97



 2.6325
SE pˆ
p0 (1  p0 )
.97  (1  .97)
150
n
What is the conclusion?
 Check p-value when z=-2.6325, p-value =
0.004
 Conclusion: we can reject the null hypothesis
at level 0.05.
One word of caution about sample size:
 If we decrease the sample size by a factor of 10,
Truth = surgical biopsy
FNA
status
Cancer
Present
Cancer
Absent
Total
Result
Positive
14
8
22
Results
Negative
1
91
92
Total
sensitivity 
15
99
114
True positive
14

 0.93
True Positives  False Negatives 15
And if we try to use the z-test,
z

estimated proportion-prestated proportion
standard error of the estimated proportion
pˆ  p0

SE pˆ
pˆ  p0
p0 (1  p0 )
n

14 15  .97
 0.8324
.97  (1  .97)
15
P-value is greater than 0.05 for sure (p=0.2026). So
we cannot reach the same conclusion.
And this is wrong!
So for test concerning proportions
We want
np>5; n(1-p)>5
14.3 Differences Between
Proportions
 Two drugs (two treatments)
 p1 =percentage of patients recovered after
taking drug 1
 p2 =percentage of patients recovered after
taking drug 2
 Compare effectiveness of two drugs
Tests of Hypotheses
 Null H0: p1=p2 (p1-p2 =0)
 Possible Alternatives:
HA: p1<p2
HA: p1>p2
HA: p1p2
Compare Two Proportions




Drug 1: n1 patients, x1 recovered
Drug 2: n2 patients, x2 recovered
x1
x2
ˆ
ˆ
p

;
p

2
Estimates: 1
n1
n2
Statistic for test:
z
ˆ1  p
ˆ2
p
 pˆ
1
ˆ2
p
 If we did this study over and over and drew a histogram of
the resulting values of pˆ1  pˆ 2 , that histogram or
distribution would have standard deviation  pˆ  pˆ
1
2
Estimating the Standard Error
 Under H0, p1=p2=p. So
 p2ˆ  pˆ   p2ˆ   p2ˆ 
1
2
1
2
p1 (1 p1 )
n1

1
1
 p(1  p)( 
)
n1 n2
 Estimate the common p by
n1 pˆ 1  n2 pˆ 2 x1  x2
pˆ 

n1  n2
n1  n2
p2 (1 p2 )
n2
So put them together
z
pˆ1  pˆ 2
1 1
pˆ (1  pˆ )(  )
n1 n2
Example 12.3
 Two sided test:
 H0: p1=p2 vs HA: p1p2
 n1=80, x1=56
 n2=80, x2=38
pˆ1  0.7
pˆ 2  0.475
pˆ 
56  38
 0.5875
80  80
Two Tailed Test
 Observed z-value:
z
0.7  0.475
0.225

 2.88
0.078
1
1
0.5875  0.4125(  )
80 80
 Critical value for two-tailed test: 1.96
 Conclusion: Reject H0 since |z|>1.96
Rejection Regions
Alternative
Hypotheses
HA: p1>p2
HA: p1<p2
HA: p1p2
Rejection
Regions
z>z
z<-z
z>z/2 or
z<-z/2
P-value of the previous example
 P-value=P(z<-2.88)+P(z>2.88)=2*0.004
 So not only we can reject H0 at 0.05 level, we
can also reject at 0.01 level.
14.4 The analysis of an r x c table
Recall Example 12.3
 Two sided test;
H0: p1=p2 vs HA: p1p2
 n1=80, x1=56
n2=80, x2=38
We can put this into a 2x2 table and the question now becomes is there a
relationship between treatment and outcome? We will come back to
this example after we introduce 2x2 tables and chi-square test.
trt1
recover
Not recover
56
24
80
trt2
38
42
80
94
66
2x2 Contingency Table
 The table shows the data from a study of 91 patients who had a
myocardial infarction (Snow 1965). One variable is treatment
(propranolol versus a placebo), and the other is outcome
(survival for at least 28 days versus death within 28 days).
OUTCOME
Survival for at
Death
least 28 days
Treatment
Propranolol 38
Total
7
45
Placebo
29
17
46
Total
67
24
91
Hypotheses for Two-way Tables
The hypotheses for two-way tables are very “broad stroke”.
• The null hypothesis H0 is simply that there is no association
between the row and column variable.
• The alternative hypothesis Ha is that there is an association
between the two variables. It doesn’t specify a particular
direction and can’t really be described as one-sided or twosided.
Hypothesis statement in Our
Example
 Null hypothesis: the method of treating the
myocardial infarction patients did not influence the
proportion of patients who survived for at least 28
days.
 The alternative hypothesis is that the outcome
(survival or death) depended on the treatment,
meaning that the outcomes was the dependent
variable and the treatment was the independent
variable.
Calculation of Expected Cell
Count

To test the null hypothesis, we compare the observed cell
counts (or frequencies) to the expected cell counts (also
called the expected frequencies)
Row1 Total  Column1 Total
E1,1 
Study Total

The process of comparing the observed counts with the
expected counts is called a goodness-of-fit test. (If the chisquare value is small, the fit is good and the null hypothesis
is not rejected.)
Observed cell counts
OUTCOME
Survival for at
Death
least 28 days
Treatment
Propranolol 38
Placebo
Total
29
7
17
67
24
Total
45
46
91
Expected cell counts
OUTCOME
Survival for at
Death
least 28 days
Treatment
Propranolol 33.13
Placebo
Total
33.87
11.87
12.13
67
24
Total
45
46
91
The Chi-Square ( c2) Test Statistic
The chi-square statistic is a measure of how much the observed
cell counts in a two-way table differ from the expected cell counts.
It can be used for tables larger than 2 x 2, if the average of the
expected cell counts is > 5 and the smallest expected cell count is
> 1; and for 2 x 2 tables when all 4 expected cell counts are > 5.
The formula is:
c2 = S(observed count – expected count)2/expected count
Degrees of freedom (df) = (r –1) x (c – 1)
Where “observed” is an observed sample count and “expected” is
the computed expected cell count for the same cell, r is the number
of rows, c is the number of columns, and the sum (S) is over all the
r x c cells in the table (these do not include the total cells).
The Chi-Square ( c2) Test Statistic
Calculatio n of the chi - square ( c 2 ) value
 (O  E ) 2 
c  

E


(38  33.13) 2 (7  11.87) 2 (29  33.87) 2 (17  12.13) 2




33.13
11.87
33.87
12.13
(4.87) 2 (4.87) 2 (4.87) 2 (4.87) 2




33.13
11.87
33.87
12.13
23.72 23.72 23.72 23.72




33.13 11.87 33.87 12.13
 0.72  2  0.7  1.96  5.38
2
Adobe Acrobat 7.0
Document
df  ( R  1)(C  1)  1
p  value  0.025
Interpreta tion :
The results noted in this 2  2 table are statistica lly significan t.
That is, it is highly probable (only 1 chance in about 50 of being wrong)
that the investigat or can reject the null hypothesis of independen ce and
accept the alternativ e hypothesis that propranolo l does affect the outcome
of myocardial infarction .
Example: Patient Compliance w/ Rx
In a study of 100 patients with hypertension, 50 were randomly allocated to a
group prescribed 10 mg lisinopril to be taken once daily, while the other 50
patients were prescribed 5 mg lisinopril to be taken twice daily. At the end of
the 60 day study period the patients returned their remaining medication to the
research pharmacy. The pharmacy then counted the remaining pills and
classified each patient as < 95% or 95%+ compliant with their prescription.
The two-way table for Compliance and Treatment was:
Compliance
Treatment
10 mg Daily
5 mg bid
Total
95%+
46
40
86
< 95%
4
10
14
50
50
100
Total
Example: Patient Compliance w/ Rx
Treatment
Compliance 10 mg Daily
5 mg bid
Total
95%+
460
400
860
< 95%
40
100
140
Total
500
500
1000
The expected cell counts were:
Treatment
Compliance 10 mg Daily
5 mg bid
Total
95%+
430
430
860
< 95%
70
70
140
Total
500
500
1000
c2 = 29.9, df = (2-1)*(2-1) = 1, P-value <0.001
If we use the two sample test for proportion
460
400
860
pˆ1 
, pˆ 2 
, pˆ 
500
500
1000
pˆ1  pˆ 2
z
 5.46812
1 1
pˆ (1  pˆ )(  )
n1 n2
5.468122  29.9
z 2  c 2 (1)
The c2 and z Test Statistics
The comparison of the proportions of “successes” in two
populations leads to a 2 x 2 table, so the population proportions can
be compared either using the c2 test or the two-sample z test . It
really doesn’t matter, because they always give exactly the same
result, because the c2 is equal to the square of the z statistic and the
chi-square with one degree of freedom c2(1) critical values are equal
to the squares of the corresponding z critical values.
• A P-value for the 2 x 2 c2 can be found by calculating the square
root of the chi-square, looking that up in Table for P(Z > z) and
multiplying by 2, because the chi-square always tests the twosided alternative.
• For a 2 x 2 table with a one-sided alternative hypothesis the twosample z statistic would need to be used.
• To test more than two populations the chi-square must be used
• The chi-square is the one most often seen in the literature
Summary: Computations for Two-way Tables
1. create the table, including observed cell counts, column and row
totals.
2. Find the expected cell counts.
• Determine if a c2 test is appropriate
• Calculate the c2 statistic and number of degrees of freedom
3. Find the approximate P-value
• use Table III chi-square table to find the approximate P-value
• or use z-table and find the two-tailed p-value if it is 2 x 2.
4. Draw conclusions about the association between the row and
column variables.
Yates Correction for Continuity
 The chi-square test is based on the normal approximation of the
binomial distribution (discrete), many statisticians believe a correction
for continuity is needed.
2


(|
O

E
|

0
.
5
)
2
Yates c   

E


 It makes little difference if the numbers in the table are large, but in
tables with small numbers it is worth doing.
 It reduces the size of the chi-square value and so reduces the chance of
finding a statistically significant difference, so that correction for
continuity makes the test more conservative.
What do we do if the expected
values in any of the cells in a 2x2
table is below 5?
For example, a sample of teenagers might be divided into male
and female on the one hand, and those that are and are not
currently dieting on the other. We hypothesize, perhaps, that the
proportion of dieting individuals is higher among the women
than among the men, and we want to test whether any
difference of proportions that we observe is significant. The
data might look like this:
men women total
dieting
1
not dieting 11
totals
12
9
3
12
10
14
24
The question we ask about these data is: knowing that 10 of
these 24 teenagers are dieters, what is the probability that
these 10 dieters would be so unevenly distributed between
the girls and the boys? If we were to choose 10 of the
teenagers at random, what is the probability that 9 of them
would be among the 12 girls, and only 1 from among the 12
boys?
--Hypergeometric distribution!
--Fisher’s exact test uses hypergeometric distribution to
calculate the “exact” probability of obtaining such set of the
values.
Fisher’s exact test
Before we proceed with the Fisher test, we first
introduce some notation. We represent the cells by
the letters a, b, c and d, call the totals across rows
and columns marginal totals, and represent the
grand total by n. So the table now looks like this:
men women total
dieting
a
b
not dieting c
d
totals
a+c b+d
a+b
c+d
n
dieting
not dieting
totals
men
women
total
a
c
a+c
b
d
b+d
a+b
c+d
n
Fisher showed that the probability of obtaining
any such set of values was given by the
hypergeometric distribution:
In our example
As extreme
as observed
men
women
total
dieting
1
9
10
not dieting
11
3
totals
12
12
p
More
extreme than
observed
men
women
total
dieting
0
10
10
14
not dieting
12
2
14
24
totals
12
12
24
10!14!12!12!
 0.00134
24!1!9!11!3!
10!14!12!12!
p
 0.00003
24!0!10!12!2!
Recall that p-value is the probability of observing data as
extreme or more extreme if the null hypothesis is true. So
the p-value is this problem is 0.00137.
The fisher Exact Probability Test
 Used when one or more of the expected counts in a contingency
table is small (<2).
 Fisher's Exact Test is based on exact probabilities from a
specific distribution (the hypergeometric distribution).
 There's really no lower bound on the amount of data that is
needed for Fisher's Exact Test. You can use Fisher's Exact Test
when one of the cells in your table has a zero in it. Fisher's Exact
Test is also very useful for highly imbalanced tables. If one or
two of the cells in a two by two table have numbers in the
thousands and one or two of the other cells has numbers less
than 5, you can still use Fisher's Exact Test.
 Fisher's Exact Test has no formal test statistic and no critical
value, and it only gives you a p-value.