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156 7 Sampling and Sample Size
=
1.282
√
0.12232 + 1.96
√
0.12329
2
(
0.044
)
2
=
1.292
0.001936
= 667.26
≈ 668
n =
z
1−
α
2
+ z
1−β
2
× 2 × p
(
1 − p
)
(
p
1
− p
2
)
2
=
(
1.96 + 1.28
)
2
× 2 × 0.066 × 0.934
(
0.088 − 0.044
)
2
= 668.51 ≈ 669
n =
z
1−
α
2
+ z
1−β
2
×
p
1
(
1 − p
1
)
+ p
2
(
1 − p
2
)
(
p
1
− p
2
)
2
=
(
1.96 + 1.28
)
2
×
(
0.088 × 0.912 + 0.044 × 0.956
)
(
0.088 − 0.044
)
2
=
10.51 × 0.12232
0.001936
= 664.04 ≈ 665.
Therefore, at least 669 samples in each group should participate in order to be able
to achieve an actual difference in mortality rate between male and female patients
with a 90% probability and a significant level of 0.05.
7.5 Exercises
1. Based on previous studies, the variance of height in men and women is 160 and
140 cm respectively. We want to estimate the height of people in population
with an error of 4 cm and at a significance level of 5%:
A. How many samples are needed to estimate women’s height?
B. How many samples are needed to estimate omen’s height?
C. Explain the reason for the difference in the sample sizes required for men
and women.
2. Should the relation α + β = 1 be considered in estimating the sample size?
Why?
3. To compare the height of men and women in the population with 90% probability
and 95% confidence, how many samples are needed to find out the difference in
height of 10 cm between men and women. However, we know that the variance
of height in these two groups is 155 and 165 cm, respectively.
4. To make X-ray tubes, a tungsten disk known as anode disk is used. The mean
and standard deviations of the radius of this disk are 26 and 4 cm, respectively.
To prevent disk damage, its radius should not exceed 34 cm. How many samples
are needed so that with a probability of 90% the disk radius is not more than
the standard.
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7.5 Exercises 157
5. The researcher wants to estimate the proportion of pregnant women seeking
prenatal care in the first trimester of pregnancy. How many samples should
be taken to make this ratio at the level of significance of 5% and the actual
difference between the proportion of applicants and subsequent pregnancies is
20%. Previous similar studies show that this proportion is not more than 40%.
6. In the design of dental radiography devices, the aluminum thickness in the
radiographic device is lower than the panorex device. If the average thickness
difference between the filters of the two devices is 2.5 mm, what is the minimum
required sample in each group to reveal the above thickness difference with 95%
confidence and 90% power?
7. Assuming the heights of students in a college campus are normally distributed
withaSd = 5 inch, find the required minimum sample size to construct a 95%
confidence interval for mean with a maximum error = 0.5 inch.
8. We are interesting to compare cholesterol concentration between male and
female students in one health sciences faculty. Since cholesterol level is
numerical kind of data basically, you will be comparing their mean between
two groups of people (here is gender, either male or female). Search for
standard deviation and mean of the two groups from reliable sources and find
the minimum sample size required for this comparison at the significance
levels of 5% and 10%.
9. To detect a statistically significant mean difference between the treatment and
placebo groups (1:1 allocation), how many samples are needed in each group?
We assume that the antiviral effect was 87% and 93%, respectively.
10. Here you are going to find smoking habit (yes/no) in one university population.
Smoking habit is categorical data, and in a pilot study, 10 out of 30 university
students experienced smoking. How many samples are needed in this university
campus?
https://t.me/med1917
Chapter 8
Estimation and Statistical Inference
8.1 Introduction
In the preceding chapter, we discussed how sampling serves the purpose of
obtaining results from a random sample of variables under study, enabling us to
generalize those results to the larger population. The fundamental principle
underlying all probability (random) sampling methods is that data is collected from
a random sample of subjects, and this information is then used to draw inferences
about the entire population. However, it’s important to note that different samples
may yield different conclusions. Each sample obtained is not an exact replica of the
original population, leading to deviations from the true population values.
Consequently, we cannot assert the absolute correctness of our conclusions.
Instead, the conclusion derived from a sample represents our perception and
inference of the true population value, albeit with a certain degree of uncertainty.
This process is known as statistical inference or inferential statistics. Statistical
inference employs the language of probability to quantify the reliability of our
conclusions. Essentially, statistical inference involves using statistical methods to
draw conclusions about unknown aspects of a population (population parameters)
based on a random sample drawn from that population. The outcomes derived from
this random sample are referred to as “estimations”. The two primary approaches to
statistical inference are estimation and hypothesis testing, which we will delve into
in the subsequent chapters.
© 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_8
159
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160 8 Estimation and Statistical Inference
8.2 Estimation
When evaluating a new treatment on a random sample of individuals with a specific
disease within a population, it’s improbable to obtain identical results when
replicating the study with another random sample from the same population. Even
if multiple random samples are drawn from the population of individuals with the
disease and the same test statistic (such as the mean) is calculated for each sample,
it’s highly unlikely that all sample means will match precisely. Variations will
inevitably occur, even in the absence of errors, due to chance factors. Collating all
the means obtained from different samples forms a sampling distribution of the
mean. The standard deviation of this sampling distribution is referred to as the
standard error, which represents the variability in the mean obtained from all
random samples taken from the population. The standard error aids in estimating
the true population parameter.
To better understand the sampling variations, we consider 7 numbers (8–7–6–
5–4–3–2) as the reference population. The mean and variance of this hypothetical
population are 5.00 and 5.60, respectively.
We now estimate the mean of this population by drawing a random sample of 3.
Using a simple random sampling method, the first sample consists of numbers 4,
6 and 1 with the mean of 3.67. Another sample of 3 includes numbers 2, 3, and 8
with the mean of 4.33 will be obtained. If sampling is repeated up to 20 times, the
extracted samples 3 from this population and their means are given in Table
8.1.
The estimated means in the above table show the sample variations of how the
estimation of a parameter will vary from sample to sample. As can be seen, none
of these means are exactly equal to the mean of the reference population. For this
reason, calculating interval estimate is necessary.
If we extract all 210 possible samples (3 out of 7, without replacement) from this
population and calculate their means, these means themselves will have a distribution
named the “sample distribution of the mean”.
The mean and variance of these 20-sample means are 4.4335 and 0.7177
respectively. The mean of the reference population, which is 5.00, is close to the
mean of the samples, but the variance of the reference population, which is 5.60, is
significantly larger than the variance of the sample means. As mentioned, if all 210
Table 8.1 Extracted samples of 3 from a population of 7 elements
x
i
x x
i
x x
i
x x
i
x
Samples 1–6–4 3.67 2–7–8 5.67 6–8–2 5.33 3–7–5 5.00
8–2–3 4.33 7–8–3 6.00 3–2–4 3.00 6–2–4 4.00
2–6–7 5.00 6–5–2 4.33 3–6–4 4.33 7–8–4 6.33
4–8–5 5.67 7–2–8 5.67 2–8–5 5.00 3–6–5 4.67
2–6–5 4.33 4–7–3 4.67 6–1–4 3.67 2–3–7 4.00
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8.2 Estimation 161
possible samples are taken and their means are calculated, it is proved that the
average of these means will be exactly equal to 5.00 and their variance will be
equal to 1.867 which is exactly proportional to the inverse of the sample size (3).
That is, its variance is equal to one-third of the variance of the reference population,
and it can be said that the sample mean is an unbiased estimate of the population
mean, i.e.:
μ
x
= μ
x
.
If we plot the sample means and the reference population mean, it is observed that
the central location of both is the same. But the dispersion of the reference population
will be greater than the dispersion of the samples, because:
σ
2
x
=
σ
2
n
.
The square root of variance of the sample means is called the standard error and
is denoted by Se, i.e.:
Se
x
=
σ
√
n
.
Usually, because t he population standard deviation is not known, an estimate of
S =
(
x−
x
)
2
n−1
is used. So:
Se =
S
√
n
.
Here, we will explain another example:
Example 1 The mean and standard deviations of diastolic blood pressure of 250
nurses are
σ = 9.4 mm/HG,μ = 78.2 mm/HG. We assume these 250 nurses as the
reference population and write their blood pressure measurements on similar papers
and put them in an urn. Bring this urn into a class of 30 students and ask them to draw
10 samples each one. After recording the blood pressure, return it to the urn. The next
student, after mixing thoroughly, takes 10 samples and records the blood pressure.
All 30 students do this and calculate their sample mean of 10 (
x). Finally, we have
30 different means obtained from 10 different samples. Each of these is an estimate
of the reference population mean of 250 flight attendants. But if we take the average
of these entire 30 sample means, the value of
μ = 78.23 mm/HG is obtained, which
is very close to the mean of the reference population µ. The standard deviation of
these 30 means is 3.01 which is close to
σ
√
n
=
9.4
√
10
= 2.97. Now if students take 20
samples instead of 10 samples, and we repeat the same cycle, then the mean of the
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162 8 Estimation and Statistical Inference
20s is μ = 78.14 mm/HG, which is very close to the mean of the main population
μ = 78.20. The standard deviation of the means of 20 is equal to 2.07, which has a
good agreement with σ /
√
n = 9.4/
√
20 = 2.10.
For all the measures and indicators presented in the previous chapters, two types
of estimation can be calculated, point estimate and interval estimate.
Tips
Any estimator should have two important properties:
•
Unbiasedness: The unbiasedness property of the estimators means that, if
we have many samples for the random variable and we calculate the
estimated value corresponding to each sample, the average of these
estimated values approaches the unknown parameter. In other words, an
expected value equal to a population parameter is estimated.
•
Minimum variance: It means that it has lower variance than any other
unbiased estimator for all possible values of the parameter.
8.3 Estimation of the Mean
8.3.1 Point Estimate of the Mean
In statistics, the process of finding an approximate value of the parameter in a
population (such as the mean or proportion), from random samples taken of the
population is called point estimation. In other words, a point estimate is a single
numerical value that estimates of an unknown parameter. The best point estimate of
the population mean
μ is the sample mean (x), and the best estimate of proportion
population (p) is the sample proportion (
ˆ
p
), which are estimated by the following
formulae:
ˆ
p =
r
n
,
x =
n
i=1
f
i
.x
i
n
.
In Table 6.2
, the values 3.67, 3.00, 6.00, 4.33, and… each are point estimates of
the mean. This estimate varies by changing the sample size and even drawing another
sample. Sometimes, this deviation is too high. As can be seen in Table
6.2
, the means
of the 3 from the same population varies from sample to sample. In this example,
it fluctuated between 3.00 and 6.33. These fluctuations are due to sampling changes
and are always present. This fluctuation will be less in larger samples and more in
smaller samples. As can be seen, none of the means of these samples is exactly equal
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8.3 Estimation of the Mean 163
to the reference population mean. However, some are closer and some are farther
away. There is no guarantee that the mean of subsequent samples will fully match to
the mean of the reference population. Therefore, the mean of the population will not
be exactly equal to its estimate. But as the sample size increases, the probability of
the two becoming closer increases. In terms of probabilities, the chances of matching
these two values with each other are very low.
8.3.2 Interval Estimate of the Mean
The goal of statistical analysis is to estimate a true population parameter, which is a
constant but unknown value, using sample data. However, a point estimate derived
from sample data represents the most probable value of the true population
parameter, yet it tends to be imprecise. Interval estimation serves as a method to
address and enhance this imprecision by furnishing a range of values within which
the true population parameter is likely to lie. For instance, the 95% interval
estimation includes the range within which we are 95% certain that the true
population parameter will lie. On the other words, an interval estimate displays the
probability that a population parameter will fall between a pair of values around the
estimated value and measures the degree of uncertainty in a sampling method.
An interval estimate is obtained by adding and subtracting a value to the point
estimate of the mean (or proportion). So, using standard error, we create a range
called “interval estimate” or “confidence interval”. This range is expanded on both
sides of the mean (or proportion) and is a multiplication of the standard error and a
certain probability includes the true value of the population parameter.
ˆ
p − d to
ˆ
p + d
d = Se × z
1−
α
2
.
Interval estimates, also known as confidence intervals (CIs), offer a range of
plausible values for a population parameter, providing insight into the precision of
the measured variable. Additionally, CIs may offer valuable information regarding
the clinical significance of results. With the interval estimate, the probability of
matching this interval with the population parameter (proportion or mean) increases.
In terms of probabilities, the chances of a point being at an interval also increase.
https://t.me/med1917
164 8 Estimation and Statistical Inference
Tips
If a measure is taken from entire population, calculating the confidence
interval or a statistical test has no advantage. Because the confidence interval
and hypothesis test are to provide uncertainty due to sampling from
population. Therefore, if the entire population is measured, there is no need
to worry about sampling error and there is no need to calculate the confidence
interval and statistical tests.
This argument has another side. Others argue that even if the measures
are taken from entire population, that population represents only a sample
of subjects from a hypothetical super-population. In other words, the study
population, even if enumerated completely without any sampling, represents
merely a biologic sample of a larger set of subjects; therefore, a confidence
interval and statistical tests are justified.
The validity of each argument depends on the context. If consider voter
priorities, it is the actual population they are interested in, and the first
argument is sound. But for biologic phenomena, what happens in an actual
population may be of less interest than the biologic norm that describes the
super-population. Thus, for biologic phenomena, the second argument is
more convincing.
The interval estimate is influenced by the following 3 factors:
Standard deviation of the variable: The more dispersed the variable, the less
homogeneity the data and the larger the standard deviation and the interval estimate
will be wider and broader.
The confidence level or probability of containing the true mean: The more likely
that the true parameter (e.g., mean) will fall within the interval estimate, the estimated
interval should be wider. For instance, 99% intervals, which are from
−3×Sd to μ+
3 × Sd
, are wider than 95% intervals, which are from μ − 2 × Sd to μ + 2 × Sd,
and are wider than 90% intervals, which are from
μ − 1 × Sd to μ + 1 × Sd. If you
are willing to be only 50% sure that an interval contains the true value, then it can
be much narrower.
The sample size of the study: The larger the sample size, the narrower and closer
the interval estimates will be.
Therefore, d can be defined as follows:
d = z
1−
α
2
σ
√
n
= z
1−
α
2
× Se,
where Se =
σ
√
n
is the standard error or standard deviation of the mean. By adding
and subtracting the above value from the sample point estimated mean, an interval
estimate or confidence interval is obtained.
x − z
1−
α
2
σ
√
n
and
x + z
1−
α
2
σ
√
n
.
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8.3 Estimation of the Mean 165
The confidence interval of x + z
1−
α
2
σ
√
n
to x − z
1−
α
2
σ
√
n
is interpreted that this
interval contains the true mean of the population with probability of (1 − α) × 100%.
That is (1 − α) × 100% we are sure that the true mean of the population (μ)isat
this interval.
The smaller value is named the lower limit of the estimate, the larger value upper
limit of the estimate, and both are called “confidence limits”.
In epidemiological studies, interval estimates of the mean are usually calculated
with 95% confidence (sometimes 90 or 99%). Their corresponding values of
z
1−
α
2
are 2.58 for 99%, 1.96 for 95%, and 1.64 for 90%. If another confidence level is
considered by the researcher, the corresponding value can be obtained from the
standard normal distribution table.
Example 2 Male diastolic blood pressure in a population has an approximately
normal distribution with a variance of
σ
2
= 10.4. Based on a sample of 25, the mean
diastolic blood pressure of men was estimated to be
x = 8.5:
Calculate the 90% interval estimate of the mean.
Obtain 95% confidence interval of the mean.
Determine limits that can contain the true mean with the 99% probability.
Solution: The three terms “interval estimate”, “confidence interval”, and “limits
determination” are synonymous, and similar methods are used to calculate. Based
on given information,
x = 8.5 and Se =
σ
√
n
=
√
10.4
√
25
=
3.22
5
= 0.645. Assuming
two-tailed estimation,
z
0.90
= 1.64, z
0.95
= 1.96, z
0.99
= 2.58
8.5 ± 1.64 × 0.645 = 8.5 ± 1.06 =
(
7.44 to 9.56
)
8.5 ± 1.96 × 0.645 = 8.5 ± 1.26 =
(
7.24 to 9.76
)
8.5 ± 2.58 × 0.645 = 8.5 ± 1.66 =
(
6.84 to 10.16
)
.
It can be seen that with increasing confidence level, the confidence interval has
become wider.
Example 3
If in the Example 2, the mean is based on a sample of 256. Recalculate
the mentioned estimates.
Solution: According to the given information: x = 8.5 and Se =
σ
√
n
=
√
10.4
√
256
=
3.22
16
= 0.201. Assuming two-tailed estimation, z
0.90
= 1.64, z
0.95
= 1.96, z
0.99
=
2.58
8.5 ± 1.64 × 0.201 = 8.5 ± 0.33 =
(
8.17 to 8.83
)
8.5 ± 1.96 × 0.201 = 8.5 ± 0.39 =
(
8.11 to 8.89
)
8.5 ± 2.58 × 0.201 = 8.5 ± 0.52 =
(
7.98 to 9.02
)
.
It is observed that with increasing the sample size, the standard error becomes
smaller and the confidence interval becomes narrower and closer.
Example 4 If in the Example 2 the mean is based on a sample of 25, but the
population variance is
σ
2
= 15.8. Recalculate the mentioned estimates.
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166 8 Estimation and Statistical Inference
Solution: According to the given information: x = 8.5 and Se =
σ
√
n
=
√
15.8
√
25
=
3.97
5
= 0.795. Assuming two-tailed estimation, z
0.90
= 1.64, z
0.95
= 1.96, z
0.99
=
2.58
8.5 ± 1.64 × 0.795 = 8.5 ± 1.30 =
(
7.20 to 9.80
)
8.5 ± 1.96 × 0.795 = 8.5 ± 1.56 =
(
6.94 to 10.06
)
8.5 ± 2.58 × 0.795 = 8.5 ± 2.05 =
(
6.45 to 10.55
)
.
It can be seen that when the dispersion of the population is greater, the standard
error is larger and the confidence interval widens.
By comparing the results of these three examples, the role of the above three
components in estimating the confidence interval can be realized.
It should be noted that the true value of variance of the population is usually
unknown to the researchers and they should use the estimated variance of S
2
instead.
In this case, if the sample size is small (less than 30), another coefficient called t is
used instead of
z
1−
α
2
.
d = t
(
df ,1−
α
2
)
×
S
√
n
.
The degree of freedom (df = n − 1) plays a role in this coefficient, the value of
which is obtained from the t distribution table at the end of statistical books.
Therefore, when the variance of the population is unknown and estimated from
the sample, the interval estimate of the mean is calculated as follows:
x + t
(
df , 1−
α
2
)
×
S
√
n
to
x − t
(
df , 1−
α
2
)
×
S
√
n
.
Example 5 The pain relief time of a new drug for 6 patients with arthritis is: 4.3–
3.7–2.5–4.9–2.4–2–2 h. Calculate a 95% confidence interval of mean.
Solution: In this example, the mean is x =
4.3+3.7+···+2.2
6
= 3.3, standard deviation
Sd = 1.13, n = 6 and df = 1–6 = 5, and standard error
Se =
Sd
√
n
=
1.13
√
6
= 0.46.The
critical value of t withadf = 5 is equal to t
(
5, 1−
0.05
2
)
= t
(
5, 0.975
)
=2.57
. Therefore, the
95% confidence interval is calculated as follows:
3.3 ± 2.57 × 0.46 = 3.3 ± 1.18 =
(
2.12 to 4.48
)
h.
8.4 Point and Interval Estimation of a Population
Proportion
In epidemiology and health sciences, we are usually dealing with the proportion of
subjects with a feature or the rate of a disease in the population. Because many
attributes and variables are used in binomial scale (yes, no). That is, they indicate
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