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8.4 Point and Interval Estimation of a Population Proportion 167
“occurrence or non-occurrence” or “existence or non-existence” in population.
Variables whose responses were “positive and negative”, “dead and alive”, “healthy
and sick”, “failure and success”, “yes and good”, “vaccinated and not vaccinated”,
“under 15 years and over 15”, “male and female”, etc. are examples of binomial
variables. For example, what proportion of infants are boys? What is the prevalence
rate of depression in the population? What percentages of patients recover with a
particular treatment? Or what is the immunity of a vaccine? These questions have
different forms and shapes that require another answer.
Tip
When the sample size (n) is large, the critical values of t and z are approximately
the same and the z distribution can be used instead of t.
In fact, the proportion itself is a kind of mean that can be reached by making
changes in the formula of estimating the mean to estimate the proportion in
population. Therefore, the same logic expressed for estimating the mean can be
generalized to estimate the proportion.
Tip
If we assign a value of 1 to all subjects who have a feature or an attribute and
a value of 0 to people who do not have that feature, their mean will be equal
to the proportion.
If a sample of n is selected from the study population and r subjects have the
interested characteristic, the p proportion estimate is the number of individuals who
have the characteristic (r) divided by the total number of samples (n), i.e.:
ˆ
p =
r
n
.
This is the relative frequency or point estimate of the proportion in the population.
But as mentioned about the mean, the proportion estimate is also subject to sampling
variations, and its value varies from sample to sample and its confidence interval
should be calculated.
Like any other estimator, the confidence interval is obtained by the following
equation.
Estimator ±
(
Constant coefficient
)
×
(
Standard error
)
.
The standard error of proportion is equal to:
σ
p
=
p
(
1 − p
)
n
=
r
(
n − r
)
n
3
.
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168 8 Estimation and Statistical Inference
Interval estimates or confidence intervals of the proportion can be obtained.
r
n
± z
1−
α
2
r
(
n − r
)
n
3
or
r
n
± z
1−
α
2
ˆ
p
(
1 −
ˆ
p
)
n
.
This range means that the above interval with a probability of %
(
1 − α
)
× 100
includes the true value of the proportion in the population.
Example 6 In the only maternity hospital in a city, 84 babies were born in one month,
of which 37 were girls. Calculate the point estimate and 95% confidence interval of
the proportion of girl birth rate in this city.
Solution: Point estimate of girl birth rate is:
ˆ
p =
r
n
=
37
84
= 0.44.
The 95% confidence interval or interval estimate for girl birth rate is:
r
n
± z
1−
α
2
r
(
n − r
)
n
3
=
37
84
± 1.96
37
(
84 − 37
)
84
3
= 0.44 ± 0.106
=
(
0.334 to 0.546
)
.
Therefore, the 95% interval estimate for girl birth proportion in this city is between
0.334 and 0.546, and this means that with 95% confidence, the above interval covers
the true proportion of girl birth in this city.
Example 7
In an area of 800 patients with Covid-19 have been hospitalized and 32
out of 800 are dead. Find 90% confidence interval for the mortality rate in this area.
Solution: Point estimate of mortality rate is:
ˆ
p =
r
n
=
32
800
= 0.04.
The 90% confidence interval or interval estimate for mortality rate is:
r
n
± z
1−
α
2
r
(
n − r
)
n
3
=
32
800
± 1.645
32
(
800 − 32
)
800
3
= 0.04 ± 0.011
=
(
0.029 to 0.051
)
.
Therefore, the 90% interval estimate for mortality rate in this area is between
0.029 to 0.051, and this means that with 90% confidence, the above interval covers
the true rate of mortality by Covid-19 in this area.
If n is large enough such that both n.p and n
(
1 − p
)
are greater than 10, or both
n and
(
n − r
)
are greater than 10, the distribution of
ˆ
p =
r
n
can be assumed as normal
distribution, i.e.:
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8.4 Point and Interval Estimation of a Population Proportion 169
ˆ
p =
r
n
∼ N
ˆ
p,
ˆ
p
(
1 −
ˆ
p
)
n
≈ N
r
n
,
r
(
n − r
)
n
3
.
In a sample of n, the probability of the sample proportion is greater than p
1
is:
p
r
n
> p
1
=
p
r
n
− p
σ
p
>
(
p
1
− p
)
σ
p
= p
z >
p
1
− p
σ
p
.
Similarly, the probability of the sample proportion is less than p
2
is:
p
r
n
< p
2
= p
z <
p
2
− p
σ
p
.
As a result, the probability of the sample proportion is between p
1
and p
2
is:
p
p
1
<
r
n
< p
2
= p
p
1
− p
σ
p
< z <
p
2
− p
σ
p
.
The corresponding value of z can be obtained from the standard normal
distribution table.
Example 8 The proportion of girl birth in Example 6 was 0.44; calculate the
following probabilities in a sample of 75.
The probability of girl birth is less than a third.
The probability of girl birth is greater than two-thirds.
The probability of girl birth is between 0.50 and 0.65.
Solution: According to the provided information in the example, proportion estimate
and estimation of its standard deviation are:
p = 0.44 and σ
p
=
p
(
1 − p
)
n
=
0.44
(
1 − 0.44
)
75
= 0.0573.
The probability of girl birth less than a third is:
p
r
n
< p
2
= p
z <
1
3
− 0.45
0.0573
= p
(
z < −2.033
)
.
According to the standard normal distribution table, we will have:
p
(
z < −2.033
)
= 0.5 − 0.4778 = 0.0212.
The probability of girl birth greater than two-thirds is:
p
r
n
> p
1
= p
z >
2
3
− 0.45
0.0573
= p
(
z > 0.217
)
.
https://t.me/med1917
170 8 Estimation and Statistical Inference
According to the standard normal distribution table, we will have:
p(z > 0.217) = 0.5 − 0.086 = 0.414 .
The probability of girl birth between 0.50 and 0.65 is:
p
p
1
<
r
n
< p
2
= p
p
1
− p
σ
p
< z <
p
2
− p
σ
p
= p
0.50 − 0.45
0.0573
< z <
0.65 − 0.45
0.0573
= p
(
0.871 < z < 3.484
)
.
According to the standard normal distribution table, we will have:
p
(
0.871 < z < 3.484
)
= p
(
z < 3.484
)
− p
(
z < 0.871
)
= 0.4988 − 0.308 = 0.1908.
8.5 Central Limit Theorem
The central limit theorem (CLT) is a fundamental theorem in probability and
statistics, with broad applications across various fields such as medicine, health,
epidemiology, psychology, and social sciences. Specifically, the CLT concerning
the distribution of means enables us to validate the assumption of normal
distribution, thereby facilitating the utilization of statistical formulas that
necessitate normality in our datasets. Furthermore, the CLT permits the application
of a normal distribution in numerous significant and impactful scenarios.
Additionally, the grand mean and standard deviations tend to approximate the
population mean and standard deviations, further highlighting the utility of the CLT
in statistical analysis.
Tips
Whatever the population distribution is (normal or non-normal) with mean μ
and standard deviation σ , the sampling distribution of x for any large sample
(
n ≥ 30), is approximately normal with the same mean μ and same standard
deviation
σ
√
n
.
This means that a sufficiently large sample size can predict the
characteristics of a population accurately.
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8.5 Central Limit Theorem 171
Scores
1
Frequency
2
4
6
8
Fig. 8.1 Uniform distribution of rolling a tetrahedral die
Central limit theorem (CLT) states that the distribution of sample means (x)
from a population with any distribution (uniform, skewed, whatever), regardless of
the distribution s hape in a main population, can be approximated by a normal
distribution with mean
μ and standard deviation
σ
√
n
, given a sufficiently large
sample size (
n ≥ 30). The theorem is a key concept in probability theory because it
shows that probabilistic and statistical methods that work for normal distributions
can be applied to many problems related to other types of distributions. The larger
the sample, the better the approximation will be.
Based on the central limit theorem, the probability that the sample mean x is
between the values a and b is obtained by the following formula:
p
(
a < x < b
)
= p
a − μ
σ
x
<
x − μ
σ
x
<
b − μ
σ
x
= p
(
z
1
< z < z
2
)
.
Example 9 Suppose a tetrahedral four-sided dice, which is shaped like a pyramid,
is used in role-playing games. It has four triangular faces corresponding to numbers
of 2, 4, 6, and 8. So, when the player tosses a tetrahedral die, it comes up with one
of the numbers 2, 4, 6, or 8, which constructs our population. Tossing a tetrahedral
die is like sampling from the above population.
The graph of the original distribution is shown in Fig. 8.1. This is called a uniform
distribution.
The mean, variance, and standard deviation of the hypothetical population are:
Population mean: μ =
2 + 4 + 6 + 8
4
=
20
4
= 5.
Population variance: σ
2
=
(
2−5
)
2
+
(
4−5
)
2
+
(
6−5
)
2
+
(
8−5
)
2
4
=
9+1+1+9
4
= 5.
Population standard deviation: σ =
√
σ
2
=
√
5 = 2.236.
Now, we are rolling two tetrahedral dice simultaneously. The resulting sequence
of two observed numbers represents a sample of size 2 drawn with replacement from
the specified population. There are 16 possible samples, and they are all equally
likely. Table
8.2 lists them and provides the value of the sample mean x for each.
A frequency distribution of 16 sample means is shown in Table 8.3:
The histogram of the 16 samples means appears to be approximately normal.
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172 8 Estimation and Statistical Inference
Table 8.2 16 possible samples of size 2 and their sample means
Samples Mean (x)
2, 2 2
2, 4 3
2, 6 4
2, 8 5
4, 2 3
4, 4 4
4, 6 5
4, 8 6
6, 2 4
6, 4 5
6, 6 6
6, 8 7
8, 2 5
8, 4 6
8, 6 7
8, 8 8
Table 8.3 Frequency
distribution of sample means
of rolling tetrahedral dice
together
x f
2 1
3 2
4 3
5 4
6 3
7 2
8 1
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8.5 Central Limit Theorem 173
The mean and standard deviations of the sample means are denoted by μ
x
and σ
x
,
respectively.
μ
¯
x
=
(2 × 1) + (3 × 2) + ··· + (7 × 2) + (8 × 1)
16
=
80
16
= 5
σ
¯
x
=
(
2 − 5
)
2
× 1
+
(
3 − 5
)
2
× 2
+ ··· +
(
7 − 5
)
2
× 1
+
(
8 − 5
)
2
× 1
16
=
40
16
=
√
2.5 = 1.581
If we divided the population standard deviation by
√
n = 2 = 1.414.
1.414 × 1.581 = 2.236.
Conclusion is made that:
μ
x
= μ
x
and σ
x
=
σ
x
√
n
.
In statistics, σ
x
=
σ
x
√
n
is named standard error, which is denoted by Se.
Example 10 In the example of pain relief in patients with arthritis (Example 5),
assume that the pain relieved time in the population has normal distribution with
mean and standard deviations of 3.3 and 1.3 h. In a sample of 6, find the following
probabilities:
(a) The mean is less than 2 h.
(b)
The mean is between 2 to 4 h.
(c) The maximum distance between x and population mean is one.
Solution: In the example, the information of x = 3.3, S = 1.13, n = 6, df =
6 − 1 = 5 and Se =
S
√
n
=
1.13
√
6
= 0.46 is provided.
(a) The probability that the pain relief time is less than 2 h is equal to:
p
(
x < 2
)
= p
x − μ
σ
x
<
b − μ
σ
x
= p
z <
2 − 3.3
0.46
= p(z < −2.83)
According to the normal distribution table, we have:
p
(
x < 2
)
= p
(
z < −2.83
)
= 0.5 − 0.4997 = 0.0023.
(b) The probability that the pain relief time is between 2 and 4 h is equal to:
p
(
2 < x < 4
)
= p
a − μ
σ
x
<
x − μ
σ
x
<
b − μ
σ
x
= p
2 − 3.3
0.46
< z <
4 − 3.3
0.46
https://t.me/med1917
174 8 Estimation and Statistical Inference
= p
(
2.83 < z < 1.52
)
= p
(
0 < z < 2.83
)
+ p
(
0 < z < 1.52
)
.
According to the normal distribution table, we have:
p
(
0 < z < 2.83
)
+ p
(
0 < z < 1.52
)
= 0.4977 + 0.4357 = 0.9334.
(c) The probability that the sample mean of pain relief time to the population mean
of up to one hour is equal to:
p
(
μ − 1 < x <μ + 1
)
= p
μ − 1 − μ
σ
x
<
x − μ
σ
x
<
μ + 1 − μ
σ
x
=
−1
σ
x
< z <
+1
σ
x
=
−1
0.46
< z <
+1
0.46
=
(
−2.17 < z ≤ 2.17
)
= 2 ×
(
0 < z ≤ 2.17
)
.
According to the standard normal distribution table, we have:
2 ×
(
0 < z ≤ 2.17
)
= 2 × 0.4850 = 0.97.
8.6 Estimation of Variance and Standard Deviation
Aside from the population mean, the variance and standard deviation of the
population may also be of interest. Because the standard deviation (often SD) is a
measure of variability and indicates how close the data is to the mean. The variance
is used as an estimate of the variability of the population from which the sample
was drawn. The estimation of variance is important because it can monitor the
status of quantities and is highly indicative of how one clinical trial or experiment
is performing. It can be used to measure the confidence in statistical data and is
used to compare different sets of data.
On the other hand, uniformity in research work as well as in most production
processes is an important criterion for quality control. That is, it must be ensured that
variations in measurements do not exceed a certain value, because having similar
outputs is important. Therefore, it is important to calculate the confidence intervals
for the standard deviation and variance of a population, assuming that the distribution
of the population is normal.
Point estimate of variance and standard deviation from a sample of size n is
obtained from the following formula:
S
2
= ˆσ
2
=
n
i=1
(
x
i
− x
)
2
n − 1
and S = ˆσ =
n
i=1
(
x
i
− x
)
2
n − 1
.
https://t.me/med1917
8.6 Estimation of Variance and Standard Deviation 175
Because S
2
has a Chi-square distribution with degree of freedom (n − 1), the
interval estimate of the variance is calculated using Chi-square:
χ
2
=
(
n − 1
)
S
2
σ
2
(
n − 1
)
S
2
χ
2
df ,
α
2
to
(
n − 1
)
S
2
χ
2
df , 1−
α
2
.
The (1 − α)% interval estimate of standard deviation is also obtained by square
root from the above limits.
(
n − 1
)
S
2
χ
2
df ,
α
2
,
(
n − 1
)
S
2
χ
2
df , 1−
α
2
.
For ease of 95% confidence interval is equal to:
S
(
n − 1
)
χ
2
df , 0.025
, S
(
n − 1
)
χ
2
df , 0.975
.
Example 11 The standard deviation of pain relief time in patients with arthritis in
a sample of 6 is 1.3 h. Find the 95% confidence interval for variance and standard
deviation of pain relief time.
Solution: The following information is provided, S
2
= 1.69, n = 6, df = n−1 = 5
and from the Chi-square distribution table χ
2
0.025
= 14.449,χ
2
0.975
= 0.831 replacing
on the previous formula, gives interval estimates of variance and standard deviation.
(
6 − 1
)
1.69
14.449
to
(
6 − 1
)
1.69
0.831
=
(
0.585, 10.168
)
√
0.585,
√
10.168
=
(
0.765, 3.189
)
.
Area = 1 −
Area =
Area =
https://t.me/med1917
176 8 Estimation and Statistical Inference
8.7 Exercises
1. Specify the answers of the three examples of interval estimation (diastolic blood
pressure) on the axis and interpret the role of factors affecting confidence interval.
2. In a sample of 225 people with asthma, 40% had a positive reaction to dust.
Calculate a 95% confidence interval for the proportion of people with positive
reactions and interpret it.
3.
Based on a sample of 25 men, the mean and variance of their height are 165
and 160 cm, respectively. Based on another sample of 36 women, the mean and
variance of their height were 155 and 140 cm, respectively.
A. Calculate and interpret the 95% confidence interval for the height of men
and women in population.
B. Is it possible to judge the mean height of men and women based on the
confidence interval? Why and how?
4. As part of the malaria control plan, 10,000 rural homes must be sprayed. For this
purpose, cost estimation is essential. But measuring all houses is not possible. A
sample of 100 is selected and the area to be sprayed is measured. The average
spraying area in these 100 houses is 23.2 square meters with a standard deviation
of 5.9.
A. Find an interval estimation that covers 95% of the total mean.
B. If this probability increases to 99%, what will be the above interval?
C. If the above estimation is from a 20 sample, what will happen in parts A and
B.
5. The distribution of blood types in a study is included in the table below. Can the
results of this table be generalized to the entire population? This means that can
we calculate the proportion of the blood types in the population corresponding
to the percentages of this table? If not, how do you interpret these data?
Blood types A B AB O Unknown Sum
Frequency 30 32 8 34 96 200
Percentage 28.84 30.77 7.69 32.69 . 100
Relative percentage 15.0 16.0 4.0 17.0 48.0 100
6. The mean and standard deviations of diastolic blood pressure in 35-year-old men
in a 64 sample were 105 and 12 mmHg, respectively. Based on this information,
what range of diastolic blood pressure includes the actual blood pressure of the
population.
7.
In a sample of 225 medical diagnostic devices, 40% had piezoelectric transducers.
Calculate and interpret the 95% interval estimate for the proportion of devices
with piezoelectric transducers.
8.
Eighty compound generators are set up simultaneously in a factory. After
checking the output voltage of the generators, it was found that 11 of them were
defective compounds and the rest were additional compounds. If a company
buys 30 generators a year from this factory, what percentage of the companies
are defective.
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