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9.8 Odds Ratio (OR) 207
Example 16 In a study of 96 low birth weight infants, mothers of 60 infants smoked
during pregnancy, and out of 904 low birth weight infants, 440 mothers smoked.
Estimate the odds ratio of low birth weight by smoking during pregnancy.
Solution: Since the study is through the mothers’ history, the study can be considered
retrospective. To calculate the odds ratio, we fill the following Table
9.24.
In the above table: a = 60, b = 440, c = 360, and d = 464.
The odds of low birth weight among smoking mothers are
60
440
and the odds of
low birth weight among non-smoking mothers are
36
464
. By dividing these two odds,
we have:
OR =
60
440
36
464
=
60 × 464
440 × 36
= 1.76.
That is, the odds of low birth weight among mothers who smoked during
pregnancy are 1.76 times the odds of low birth weight for mothers whose mothers
did not smoke during pregnancy.
Confidence interval for relative risk.
Like relative risk (RR) we can calculate confidence interval for OR. Similar to RR,
we first find the standard error of
log
e
OR using the following formula:
Se
log
e
OR
=
1
a
+
1
b
+
1
c
+
1
d
.
So, the lower and upper boundaries of
(
1 − α
)
%
confidence intervals for log
e
OR
are:
log
e
OR − Z
1−
α
2
× Se
log
e
OR
to log
e
OR + Z
1−
α
2
× Se
log
e
OR
,
where Z
1−
α
2
is corresponding value obtained from standard normal distribution.
Finally, using back transformation (antilogarithm) from these boundaries, the
confidence interval for odds ratio (OR) is obtained.
Example 17 Calculate the 95% confidence interval for the odds ratio in the above
example.
Table 9.24 Low birth weight infants in terms of smoking habit of mothers
Low birth weight
Yes No Total
Smoking habit Yes 60 440 500
No 36 464 500
Total 96 904 1000
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208 9 Health and Disease Measures
Solution: In this example, the values are a = 60, b = 440, c = 36, d = 464, OR =
1.76, and
Z
0.957
= 1.96.
Se
log
e
OR
=
1
a
+
1
b
+
1
c
+
1
d
=
1
60
+
1
440
+
1
36
+
1
464
=
√
0.489 = 0.221.
By replacing the above formula, we have:
(
1.76 − 1.96 × 0.221 to 1.76 + 1.96 × 0.221
)
=
(
1.76 − 0.433 to 1.76 − 0.433
)
=
(
1.33 to 2.19
)
.
Due to the large sample size, the calculated confidence interval of the odds ratio
from 1.33 to 2.19 is narrow.
Tips
Risk difference provides a measure of the impact of the risk factor, and focuses
on the number of cases that could potentially be prevented by eliminating the
risk factor.
While Relative risk and odds ratios, provide a measure of the strength of
the association between a factor and a disease or outcome.
If the probability of occurrence of the disease in one group is p
1
and in another
group is
p
2
, it is obvious that the probability of non-occurrence of the disease in those
two groups are
1 − p
1
and 1 − p
2
, respectively. The odds ratio can then be estimated
as follows:
OR =
p
1
1−p
1
p
2
1−p
2
=
p
1
(
1 − p
2
)
p
2
(
1 − p
1
)
.
Calculating the standard error requires knowing the sample size of each group.
Example 18 The probability of a disease in the two groups is p
1
= 0.4 and p
2
= 0.
3, respectively. So 1 − p
1
= 1 − 0.4 = 0. 6 and 1 − p
2
= 1 − 0.3 = 0. and the odds
ratio is:
OR =
p
1
(
1 − p
2
)
p
2
(
1 − p
1
)
=
0.4 × 0.7
0.3 × 0.6
=
0.28
0.18
= 1.56.
That is, this disease is 1.56 times more common in the first population than in the
second population.
https://t.me/med1917
9.9 Risk Difference or Excess Risk 209
Note that RR and OR should not be interpreted equally. Because as it was said
OR =
p
1
1−p
1
p
2
1−p
2
and RR =
p
1
p
2
where p
1
= p(D|E) and p
2
= p(D|E).If p
1
= p(E|D) and
p
2
= p(E|D) are defined, the prevalence ratio for exposure is obtained.
9.9 Risk Difference or Excess Risk
The risk difference (RD), excess risk,or attributable risk is the difference between
the risk of an outcome (disease) in the exposed group and the unexposed group.
Suppose we fill the corresponding numbers of outcome and exposure situation in the
table as following format then
p
e
is the incidence (proportion) of an outcome in the
exposed group and
p
u
is the incidence (proportion) of an outcome in the unexposed
group (Table
9.25).
The point estimation of the risk difference is:
RD = p
e
− p
u
,
where p
e
=
a
a+b
, p
u
=
c
c+d
and RD = p
e
− p
u
=
a
a+b
−
c
c+d
.
If n
e
and n
u
are the sample sizes of exposed and unexposed groups, the sampling
distribution of RD is approximately normal with standard error as follows:
Se
(
RD
)
=
p
e
(
1 − p
e
)
n
e
+
p
u
(
1 − p
u
)
n
u
=
a × b
(
a + b
)
3
+
c × d
(
c + d
)
3
.
Confidence interval (1 − α)% is obtained as:
CI
1−α
= p
e
− p
u
± z
1−
α
2
× SE
(
RD
)
.
Example 19 To illustrate the risk difference (RD), we use data from a prospective
cohort study (from literature) in which was examined the association between
Table 9.25 Disease status and exposure
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210 9 Health and Disease Measures
smoking and lung cancer. They classified anyone who smoked regularly as a
“smoker” and was shown in Table
9.26.
Solution: In this example, a = 40, b = 20, c = 10, and d = 130. The proportion of
lung cancer in the exposed and unexposed groups and risk difference (RD) is:
p
e
=
a
a + b
=
40
40 + 20
= 0.667 p
u
=
c
c + d
=
10
10 + 130
= 0.071
RD = p
e
− p
u
= 0.667 − 0.071 = 0.596.
To estimate the 95% confidence interval of the risk difference, we must first
calculate the RD standard error as follows.
SE
(
RD
)
=
p
e
(
1 − p
e
)
n
e
+
p
u
(
1 − p
u
)
n
u
=
0.667
(
1 − 0.667
)
60
+
0.071
(
1 − 0.071
)
140
=
√
0.00418 = 0.0646
or
SE
(
RD
)
=
a × b
(
a + b
)
3
+
c × d
(
c + d
)
3
=
40 × 20
(
40 + 20
)
3
+
10 × 130
(
10 + 130
)
3
=
√
0.00418 = 0.0646.
Assuming 95% confidence level, z
1−
α
2
= 1.96 and confidence interval is estimated
as:
CI
1−α
= p
e
− p
u
± z
1−
α
2
× SE
(
RD
)
= 0.667 − 0.071 ± 1.96 × 0.0646
= 0.596 ± 0.127 =
(
0.469 to 0.723
)
.
Table 9.26 Incidence of lung
cancer in smokers and
non-smokers
Lung cancer No lung cancer Total
Smokers 40 20 60
Non-smokers 10 130 140
Total 50 140 200
https://t.me/med1917
9.10 Exercises 211
The risk difference means that in the smoking group, compared to the non-smoking
group during the study period, there were 596 excess (additional) lung cancer cases
per 1000 subjects. According to this study, out of 100 sampling times, 95 times the
true risk difference lies between 469 and 723 (in 1000 cases) excess (additional)
cases of lung cancer.
As a rule of thumb, to interpret the 95% confidence interval for the risk difference,
we should use the following wording: “Based on this sample, out of 100 sampling
times, 95 times the true risk difference lies between the estimated [upper limit] and
the [lower limit] excess/fewer cases of [disease]”.
The attributable fraction among the exposed (AF
e
) is the proportion of incidents
in the exposed group that are attributable to the risk factor and are usually expressed
as percentage.
AF
e
=
p
e
− p
u
p
e
× 100.
The proportion of the cases can be attributed to a particular exposure.
Example 20 For the example 19, AF
e
is equal to:
AF
e
=
p
e
− p
u
p
e
=
0.667 − 0.071
0.667
× 100 = 89.4% .
And means that 89.4% of the adverse outcomes in the exposed group can be
attributed to the exposure (AF
e
= 89.4%).
9.10 Exercises
1. Out of 1000 patients (538 males and 462 females) 15–50 years old referred to a
general practitioner, 70 (42 males and 28 females) were diagnosed with asthma
and 930 were diagnosed with other causes. Find the odds ratio of asthma among
men and women.
2.
Out of 200 breast cancer patients who have referred for chemotherapy, 100
received the new method and 100 received the conventional standard method.
In the new method, 85 people and in the conventional method, 70 people have
severe nausea and vomiting. Calculate the relative risk of severe nausea and
vomiting with the new method.
3. Calculate the specificity, sensitivity, positive predictive value, and negative
predictive value of screening tests conducted on 2150 pupils in a city to
evaluate vision impairment as well as standard assessments. Of the 2107
patients whose standard examination was negative (70 positive and 2037
negative), and of the 43 patients whose standard examination was positive (11
positive and 32 negative).
https://t.me/med1917
212 9 Health and Disease Measures
Screening
Positive Negative Total
Standard examination Positive 11 32 43
Negative 70 2037 2107
Total 81 2069 2150
4. Determine the relative risk and risk difference between smokers and women who
have never smoked, as well as the associated 95% confidence interval.
Smoking habit Incidence rate per 100,000
person-years
Observed person-years No. of strokes
Never smoked 17.7 395,594 70
Smoker 49.6 280,141 139
5. The table below shows the findings of a case control research. Subjects are placed
into two groups based on their recent history of consuming meat (yes or no),
and those who do not reach part of their intestinal tissue and have turned black
(necrotic) are compared to those who do not have the sickness. Calculate the odds
ratio for necrotizing illness and its confidence intervals for meat consumption
exposure. 50 of the 66 participants who consumed meat had necrotizing illness,
while 11 of the 52 who did not eat meat recently were unwell.
Recent history of meat consumption
Yes No Total
Necrotizing disease Yes 50 11 61
No 16 41 57
Total 66 52 118
6. The table below displays the fractured bone diagnosis provided by the surgeon
and X-ray testing for 200 patients. Calculate their sensitivity, specificity, positive
predictive value, negative predictive value, and confidence interval for each one.
Diagnosis by the surgeon
Positive Negative Tot a l
X-ray Positive 54 36 90
Negative 123 12 135
Total 177 48 225
https://t.me/med1917
9.10 Exercises 213
7. The results of the questionnaires and interviews conducted with faculty
students regarding their smoking habits are compiled in the table below.
Determine Cohen’s Kappa’s value and its 95% confidence interval.
Smoking habit Interview
Yes No Total
Questionnaire Ye s 100 5 105
No 4 41 45
Total 104 46 150
https://t.me/med1917
Chapter 10
Foreground Toolbox for Hypothesis
Testing
10.1 Introduction
Statistical inference (judgment, decision-making, and generalization of sample
results to the population) includes two stages. The first step is to estimate the
desired parameter from a random sample expressed in the previous chapters. The
second step is to perform a significant test or hypothesis test to determine whether
the observed difference between sample data and specified value on the null
hypothesis is true or due to sampling random variations. Statistical tests depend on
the subject, hypothesis, the nature, and structure of the population (distribution of
the studied property), whether or not the variance is known, and…. is different.
Tip
The “statistical significance” and “clinical importance” are two completely
separate issues. “Non-significant” is not equivalent with “no effect” or “no
difference” between the two groups, because the researcher may not have been
able to formulate a good null hypothesis or the sample size may not have been
large enough.
Therefore, the requirement of these tests is knowledge of some statistical concepts
such as null and alternative hypothesis, test statistic, critical value, significance level,
p-value, one-sided or two-sided test and decision rule, which we will briefly describe
first.
© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2024
S. H. Saneii and H. Doosti, Practical Biostatistics for Medical and Health Sciences,
https://doi.org/10.1007/978-981-97-3083-4_10
215
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216 10 Foreground Toolbox for Hypothesis Testing
10.2 Definitions
Statistical significance
Statistical significance refers to the likelihood that observed differences from an
expected value or between studied groups are substantial, genuine, and not merely a
result of random chance. It indicates that the disparity between groups is attributable
to inherent differences rather than random fluctuations. Statistical hypothesis testing
is employed to ascertain significance, yielding a p-value that quantifies the probability
of obtaining results as extreme as those observed, assuming randomness as the sole
cause.
Conversely, clinical significance assesses the extent to which an observed
phenomenon manifests in real-world implications. While statistical significance
aims to refute chance occurrences, clinical significance validates the extent to
which an event or effect genuinely occurs in a measurable manner. In essence,
statistical significance underscores the improbability of an event occurring by
chance alone.
Null hypothesis
The null hypothesis, which is denoted by H
0
, states that a population parameter (such
as the mean, the standard deviation, the proportion and so on) is equal to a claimed
value or hypothesized value. When two or more populations are compared, their
parameters (such as mean and proportion) are equal. For this reason, the hypothesis
is called null hypothesis, which considers the difference between the true value of the
parameter with the specified value in the hypothesis as zero (for one population) or
the difference of the parameter in two populations equal to zero. The null hypothesis
is often an initial claim that is based on researchers’ previous analyses or specialized
knowledge. The null hypothesis may be written to one of the following forms.
One Population:
{
H
0
: μ = μ
0
→ μ − μ
0
= 0
H
0
: p = p
0
→ p − p
0
= 0
Two populations:
{
H
0
: μ
1
= μ
2
→ μ
1
− μ
2
= 0
H
0
: p
1
= p
2
→ p
1
− p
2
= 0
Alternative hypothesis
The alternative hypothesis, denoted as H
1
or H
a
, posits that a population parameter
differs from the hypothesized value stated in the null hypothesis. It suggests that
the parameter may be smaller, greater, or simply different from the null hypothesis.
The alternative hypothesis represents what the researcher either believes to be true
or aims to demonstrate.
In contrast, the null and alternative hypotheses are mutually exclusive statements
regarding population parameters. A hypothesis test utilizes sample data to determine
whether to reject the null hypothesis.
https://t.me/med1917
10.2 Definitions 217
It’s important to note that hypothesis tests are not designed to determine whether
to accept the null hypothesis but rather to decide whether to reject it. The alternative
hypothesis can be formulated as either one-tailed or two-tailed.
Alpha and p-value are two critical terms frequently used in statistical analyses,
particularly in hypothesis testing scenarios. They aid in determining whether a test
statistic is considered “significant” or “non-significant” in a research study. These
terms assist researchers in concluding whether data conforms to a specific
distribution, assessing if two processes share the same average or variation,
identifying variables with an impact on the response variable in experimental
designs, and more.
Significance level (Alpha)
The significance level, also known as alpha (α), represents the probability that a
researcher will erroneously reject the null hypothesis when it is actually true. This
alpha value determines the likelihood of a type I error, where we reject a null
hypothesis that is, in reality, true. It is termed a pre-defined threshold value because
researchers establish its value before examining the data.
For instance, if a researcher sets a significance level of 0.05, it means they
anticipate (or desire) to erroneously reject the true null hypothesis approximately
once out of every 20 times. This is because 0.05 equals 1/20, and 20 multiplied by
0.05 equals 1.
Reality of H
0
True Not true
Test result H
0
Don’t reject Reject Don’t reject Reject
1 − α α β 1 − β
The p-value
The p-value is another crucial component of significance tests, derived from statistical
tests. Each test statistic is associated with a corresponding probability or p-value.
Unlike alpha, the p-value arises from a distinct source. This probability represents
the likelihood of observing the statistic solely by chance, under the assumption that
the null hypothesis is true.
Tips
When the p-value is close to the significant level of α (e.g., between 0.045 and
0.055), caution should be exercised, and decision-making and judgment require
more precision. Contradictory conclusions from these two values (0.045 and
0.055) are not logical. Because with a little change in the sample size, the value
of p will increase and decrease and another result will be obtained.
It is recommended that in these cases, the estimated confidence interval
be calculated and presented so that the value of P is interpreted according to
the estimated confidence interval. However, the use of the words “difficult to
reject” or “just reject” is recommended.
https://t.me/med1917
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