Tosun IMPORTANCE OF EVIDENCE BASED MEDICINE The Study Objective To determine the quality of health recommendations and claims made on popular medical talk shows Sources Internationally syndicated medical television talk shows that air daily The ID: 918056
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Slide1
Chi Square Tests
PhD
Özgür
Tosun
Slide2IMPORTANCE OF EVIDENCE BASED MEDICINE
Slide3Slide4The
Study
Objective
:
To determine the quality of health recommendations and
claims
made on popular medical talk shows.
Sources
:
Internationally syndicated medical television talk shows that air daily (The
Dr
Oz Show and The Doctors).
Interventions
:
Investigators randomly selected 40 episodes of each of The
Dr
Oz Show and The Doctors from early 2013 and identified and evaluated all recommendations made on each program. A group of experienced evidence reviewers independently searched for, and evaluated as a team, evidence to support 80 randomly selected recommendations from each show.
Main outcomes
measures:
Percentage of recommendations that are supported by evidence as determined by a team of experienced evidence reviewers.
Slide5Slide6Results
On average, The
Dr
Oz Show had 12 recommendations per episode and The Doctors 11.
At
least a case study or better evidence to support 54% (95% confidence interval 47% to 62%) of the 160 recommendations (80 from each show).
For
recommendations in The
Dr
Oz Show, evidence supported 46%, contradicted 15%, and was not found for 39%.
For
recommendations in The Doctors, evidence supported 63%, contradicted 14%, and was not found for 24%.
The
most common recommendation category on The
Dr
Oz Show was dietary advice (39%) and on The Doctors was to consult a healthcare provider (18%).
The
magnitude of benefit was described for 17% of the recommendations on The
Dr
Oz Show and 11% on The
Doctors
Slide7Conclusions
Recommendations
made on medical talk shows often lack adequate information on specific benefits or the magnitude of the effects of these benefits.
Approximately
half of the recommendations have either no evidence or are contradicted by the best available evidence
.
The
public should be skeptical about recommendations made on medical talk shows.
Slide8A Fictional
Answer
for
a
R
andom
Dr.
Oz’s
Recommendation
Dr
Oz:
"
Saturated fat is solid at room temperature, so that means it's solid inside your body."
Patient:
Thanks
, Dr. Oz. You give the best advice.
Carrots are very hard and dense, so they'll
petrify (transform into stone)
your body, turning you into an orange
statue. Am
I doing it right?
Slide9Slide10Pull
down
that
bread
,
kiddo
!!!
Slide11CATEGORICAL
ONE SAMPLE
TWO SAMPLES
>2 SAMPLES
Slide12CATEGORICAL
ONE SAMPLE
TWO SAMPLES
>2 SAMPLES
Independent
Paired
Independent
Slide13CATEGORICAL
ONE SAMPLE
TWO SAMPLES
>2 SAMPLES
Independent
Paired
Independent
2 x 2
Chi
Square
Test
Mc
Nemar
Test
N x M
Chi
Square
Test
One
Sample
Chi
Square
Test
Fisher’s
Exact
Test
Nonparametric
One
sample
difference
of
proportions
test
Parametric
Slide14Cross Table (
Contingency
Table
)
enables showing two or more variables simultaneously in table format
a table of counts cross-classified according to categorical
variables
best way to include sub-group descriptive statistics
simplest contingency table is a 2 x 2 table
is good for demonstrating possible relationships among variables
Slide15Cross Table (Contingency
Table
)
An r
X
c contingency table shows the observed frequencies for two variables.
The
observed frequencies are arranged in r rows and c columns.
The
intersection of a row and a column is called a cell
Slide16Slide17Misreading the Table
it is important to correctly read the information given in a table
although the original data do not change at all, tables can be arranged in several different views
looking at the table does not necessarily show the reader about possible relationships among variables
in order to decide on the existence of relationship, «
statistical hypothesis testing
» is required
Slide18Slide19Slide20Slide21Observed
versus
Expected
In
a
cross
tabulation
,
the
actual
numbers
in
the
cells
of the table are
called the observed valuesObserved Frequencies are obtained empirically through direct observationTheoretical, or Expected Frequencies
are developed on the basis of some hypothesis
Slide22Expected
Frequencies
Assuming the two variables are independent, you can use the contingency table to find the expected frequency for each cell.
Finding the Expected Frequency for Contingency Table Cells
The expected frequency for a cell
E
r,c
in a contingency table is
Slide23Example
:
Find the expected frequency for each “Male” cell in the contingency table for the sample of 321
individuals
.
Assume that the variables, age and gender, are independent.
105
6
10
21
33
22
13
Female
16
10
61 and older
321
38
64
85
73
45
Total
216
28
43
52
51
32
Male
Total
51 – 60
41 – 50
31 – 40
21 – 30
16 – 20
Gender
Age
Slide24Expected Frequency
Example continued
:
105
6
10
21
33
22
13
Female
16
10
61 and older
321
38
64
85
73
45
Total
216
28
43
52
51
32
Male
Total
51 – 60
41 – 50
31 – 40
21 – 30
16 – 20
Gender
Age
Slide25Chi-Square Independence Test
A chi-square independence test is used to test the independence of two variables.
Using
a chi-square test, you can determine whether the occurrence of one variable affects the probability of the occurrence of the other variable.
For the chi-square independence test to be used, the following must be true.
The observed frequencies must be obtained by using a random sample.
Each expected frequency must be greater than or equal to 5.
Slide26Chi-Square Independence Test
We are looking for significant differences between the
observed
frequencies
in a table (
f
o
) and those that would be
expected
by random chance (
f
e
)
Slide272 x 2 Chi
Square
df = (r-1)(c-1)=1
+
-
1
O
11
2
Total
N
First criteria
Total
Second
Criteria
O
12
O
21
O
22
O
.1
O
.2
O
1
.
O
2.
E
ij
should be greater than or equal to 5.
Slide28Is
squi
n
t more
common among children with a positive family history?
Is there an association between squint and family history of squint?
+
-
+
20
3
0
50
-
15
55
70
Total
35
85
120
Squint
(
Şaşılık
)
Total
Family
History
2
(1,0.025)
=5.024 > 4.869. Accept H
0
.
There is no relation between squint and family history
14.58
35.42
20.42
49.58
Slide29Attention
In
2 X 2
contingency
tables
,
i
f
any expected frequencies are less than 5, then alternative procedure to called Fisher’s Exact Test should be performed.
Slide30An Example
A study was conducted to analyze the relation between coronary heart disease (CHD) and smoking. 40 patients with CHD and 50 control subjects were randomly selected from the records and smoking habits of these subjects were examined. Observed values are as follows:
Slide31Observed and
expected
frequencies
+
-
Yes
No
Total
90
Smoking
Total
CHD
30
4
46
14
76
40
50
10
6.2
33.8
7.8
42.2
Slide32df
= (r-1)(c-1)=(2-1)(2-1)=1
2
(1,0.05)
=
3
.
841
Conclusion: There is a relation between CHD and smoking.
2
=
4. 9
5
>
reject
H
0
Slide33Slide34An Example for Fisher’s Exact Test
Research question: does positive BRCA1 gene actually affects the occurrence of breast cancer?
Slide35Since the percentage of the cells which have expected count < 5 is 50%, Fisher’s exact test should be applied.
According to Fisher’s test, p value is 0.070
p>α
Fail to reject H
0
BRCA1 gene has no affect on breast cancer
Slide36McNemar Test
35 patients were evaluated for arrhythmia with two different medical devices. Is there any statistically significant difference between the diagnose of two devices?
Device I
Device II
Total
Arrhythmia
(+)
Arrhythmia
(-)
Arrhythmia
(+)
10
3
13
Arrhythmia
(-)
13
9
22
Total
23
12
35
Slide37The
significance
test
for
the
difference
between
two
dependent
population
/
McNemar
test
H0
: P1=P2 Ha: P1 P
2
Critical z value is ±1.96 Reject H
0
Slide38McNemar
test
approach
:
2
(1,0.05)
=
3.841<5.1 p<0.05
;
reject
H
0
.
Slide39Evaluation of
a
rrhythmia
patients using these two devices will provide significantly different results. Further research is required to understand which one is better for diagnosis.
Slide40A
researcher
wants
to
know
whether
the
mothers
age
is
affecting
the
probability of having
congenital abnormality of neonatals or not. The
collected data is given in the table:
Congenital
abnormality
Total
Present
Absent
Age
groups
≤25
3
22
25
26-35
8
34
42
>35
18
16
34
Total
29
72
101
N
x
M
Chi
Square
Slide41H
0
:
There
is no
relation
between
the
age
of
mother
and
presence of
congenital
abnormality
.Under the
assumption that null hypothesis is
true:(Expected count)
Congenital
abnormality
Present
Absent
Age
groups
≤25
3
(7.2)
22
(17.8)
26-35
8
(12.1)
34
(29.9)
>35
18
(9.8)
16
(24.2)
Slide42Reject
H
0
Slide43Congenital
abnormality
χ
2
Present
Absent
Age
groups
≤25
3
(7.2)
22
(17.8)
3.44
26-35
8
(12.1)
34
(29.9)
1,95
>35
18
(9.8)
16
(24.2)
9,64
Omit
the
>35
age
group
Slide44Congenital
abnormality
Present
Absent
Age
groups
≤25
3
22
26-35
8
34
H
0
is
accepted
Slide45At the end of the analysis, we should conclude that the risk of having a baby with congenital abnormality is significantly higher for >35 age group.
However, risk is not differing significantly between <= 25 age group and 26-35 age group
Slide46Slide47Attention
In
N x M
contingency
tables
,
i
f
the
proportion
of
cells
those
have
expe
cted
frequencies less
than 5 is above 20%, then it is not possible
to perform any statistical analysis
Slide48EXAMPLE
: Researcher wants to know if there is any significant difference among education groups in terms of their alcohol consumption rates
Slide49At the end of the analysis, since the proportion of cells which have expected count <5 is 50%, we must conclude that this hypothesis cannot be tested under this circumstances. The samples size in the study is not high enough.
Calculated p value is not valid.