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6.2 The Standard Normal Distribution 125
Fig. 6.9 Right (positive) and left (negative) skewnesses in distributions
Example 6
The first, second, and third quartiles of albumin level of 185 female in
Table
3.2 are Q
1
= 38.3, Q
2
= 41.2 and Q
3
= 44.3, respectively. Find the skewness
coefficient of albumin level.
Solution: Skewness coefficient for this dataset is:
SK =
Q
3
− 2Q
2
+ Q
1
Q
3
− Q
1
=
44.3 − 2 × 41.2 + 38.3
44.3 − 38.3
=
0.2
6.0
= 0.033.
The value of this coefficient is very small and close to zero. So, the distribution
of albumin levels can be assumed approximately normal in terms of skewness.
https://t.me/med1917
126 6 Normal Distribution
Kurtosis: Some normal distributions may be taller than others. In other words, the
data density around the mean is more in one distribution a is more in the tails of
another distribution. The kurtosis is a measure of the thickness of the tails of a
distribution and the sharpness of its peak. In other words, measure of the combined
weight of a distribution’s tails is relative to the center of the distribution. To measure
the kurtosis of a distribution, the fourth moment is suggested by Pearson:
K =
1
n
×
(
x
i
− x
)
4
σ
4
− 3.
According to this measure, the kurtosis of any normal distribution is zero and each
distribution is compared to the normal distribution. This measure can take positive
or negative values. Positive values (Sk > 0) indicate the elongation of the distribution
and negative values (Sk < 0) indicate the flatness of the distribution. In general if:
•
K = 0, the distribution is assumed to be mesokurtic (middle: neither high nor
low).
•
Less than 0 (K < 0) is considered to be platykurtic (broad or flat).
•
Greater than 0 (K > 0), the distribution is considered to be leptokurtic (skinny:
that are thin and tall).
It should be noted that statistical software such as SPSS and SAS calculates
skewness for the sample, whose formula is slightly different (Fig.
6.10).
Skewness and kurtosis in SPSS can be calculated by choosing the path:
Analyze\Descriptive statistics\Frequencies. After selecting the variable (s), click
on the statistics option and then select the skewness and kurtosis options to
calculate the skewness and kurtosis of the data.
These measures are also obtained by the path: Analyze\Descriptive
statistics\descriptive and then choosing option.
Fig. 6.10 Kurtosis and flatness of a normal distribution
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6.2 The Standard Normal Distribution 127
Example 7 Intraocular pressure (IOP) of the right eye of 25 people is (17–16–19–
18–10–18–15–13–14–11–13–12–18–8–9–20–22–13–16–13–10–16–15–12–12).
Calculate the kurtosis and skewness IOP.
Solution: The mean of this data is 14.4 and the sample standard deviation with
fraction denominator (n − 1) is 3.594.
The skewness and kurtosis coefficients for the formulas used in the reference
books are calculated as follows:
G =
1
n
×
(
x
i
− x
)
3
σ
3
=
200.4
25 × 3.594
3
=
200.4
1160.562
= 0.173,
K =
1
n
×
(
x
i
− x
)
4
σ
4
− 3 =
8863.12
25 × 3.594
4
− 3 =
8863.12
4171.116
− 3 =−0.875.
But their calculations are based on formulas used in software such as SPSS as
follows:
G =
n
(
n − 1
)(
n − 2
)
×
(
x
i
− x
)
3
σ
3
=
25
24 × 23
×
200.4
3.594
3
=
5010
25625.555
= 0.196,
K =
n
(
n + 1
)
(
n − 1
)(
n − 2
)(
n − 3
)
×
(
x
i
− x
)
4
σ
4
− 3
(
n − 1
)
2
(
n − 2
)(
n − 3
)
=
25 × 26
24 × 23 × 22
×
8863.12
3.594
4
− 3
(
24
)
2
23 × 22
=−0.572.
6.2.2 Assessing Normality of a Distribution
There are various methods to ensure that a particular distribution follows an
approximately normal distribution, which are described in order of simplicity
below:
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128 6 Normal Distribution
•
Histogram: Draw a histogram for the data. If the histogram looks symmetrical
and bell-shaped in terms of the number of data, its distribution can be assumed as
normal.
•
Equality of mean, mean, and mode: In the normal distribution, mean, mean, and
mode are equal. Calculate their values. If these three measures are close together
in terms of the number of data, there is a property of normal distribution.
•
Interquartile range: In the normal distribution, the interquartile range (IQR) is
approximately 1.35 times of the standard deviation (
IQR = 1.35Sd). If this
relationship is almost established, it would be another property of a normal
distribution. In the standard normal distribution (with a standard deviation of 1),
the first and third quartiles are equal to −0.674, +0.674, and
IQR =+0.674 −
(
−0.674
)
= 1.35.
•
Standard deviations around the mean: Based on the characteristics of the
normal distribution, if around 68% of the data values lie within 1 standard
deviation of the mean, approximately 95% of the data values are within 2
standard deviations of the mean, and roughly 99.7% of the data values fall
within 3 standard deviations of the mean (based on the sample size), the
distribution can be deemed normal.
•
Distribution skewness and kurtosis: We calculate the coefficients of skewness
and kurtosis. As mentioned in the previous section, the normality of the
distribution can be checked.
•
Kolmogorov–Smirnov test: The more common option is normal plot,
Kolmogorov–Smirnov test, and Shapiro–Wilk’s test, which is easily
implemented in statistical software such as SPSS.
To do this, choose the path: Analyze\Descriptive Statistics\Explore to open a
window. Then click the Plots option to open another window. In this window, select
Normality plots with tests. Its output program provides normal plot and Kolmogorov–
Smirnov tests as well as distribution skewness and kurtosis.
Example 8
Examine the normality of data in Example 7 (intraocular pressure (IOP)
of the right eye).
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6.2 The Standard Normal Distribution 129
Solution: The skewness coefficient and kurtosis coefficient of this data were
calculated in the previous example. To facilitate the calculation of other
procedures, the mean, standard deviation, median, mode and the first and third
quartiles of this data are as are equal to:
x = 14.40, Med = 14.0, Mod = 13.0, Sd = 3.594,
Q
1
= 12.0, Q
2
= 14.0 and Q
3
= 17.5.
A: The proximity of the median (14), mode (13), and mean (14.4) to each other
indicates an almost normal distribution of data.
B: The interquartile is equal to IQR = Q
3
− Q
1
= 17.5 − 12 = 5.5 and 1.5 times
the standard deviation, which is a sign of an almost normal distribution of data (in
perfectly normal distributions, it is 1.35 times the standard deviation).
C: The range of one standard deviation, two standard deviations, and three standard
deviations to the mean should include about 68%, 95%, and 99.7% of the data,
respectively.
x ± Sd = 14.4 ± 3.594 =
(
10.81 − 17.59
)
,
x ± 2Sd = 14.4 ± 2 × 3.594 =
(
7.21 − 21.59
)
,
x ± 3Sd = 14.4 ± 3 × 3.594 =
(
3.62 − 25.18
)
.
It is observed that the interval of one standard deviation to the mean includes 17
observations (68%), the interval of two standard deviations to the mean includes 24
observations (96%), and the interval of three standard deviations to the mean includes
25 observations (100%), which testifies to the almost normal distribution of data.
D: The histogram diagram also shows the nearly normal distribution of data.
E: The p value (Sig column) in both Kolmogorov–Smirnov test and Shapiro–Wilk
test is more than 0.05 which does not reject the normality of the data distribution
Fig.
6.11.
Tests of normality
Kolmogorov–Smirnov
a
Shapiro–Wilk
Statistic df Sig. Statistic df Sig.
IOP 0.132 25 0.200
*
0.980 25 0.890
a
Lilliefors Significance Correction
*
This is a lower bound of the true significance
https://t.me/med1917
130 6 Normal Distribution
Fig. 6.11 Histogram with approximately normal curve
6.3 t Distribution
Many of statistical analyzes are based on the assumption of normality distribution
of the variable under study and that the variance of the sample population is known.
According to the central limit theorem, when the sample is large enough and its
variance is known (regardless of the population distribution), the sample means are
assumed to be normal. But in practice, this assumption is not true, and usually, the
number of samples is small and the variance of the population is unknown. In these
samples, instead of the variance of the population, its sample estimation should be
used, i.e.,
S
2
=
(
x
i
−x
)
2
n−1
. Under such conditions, the statistic z =
x−μ
σ
√
n
will become
the statistic
t =
x−μ
S
√
n
, which has a distribution t.
The shape of this distribution is not exactly normal, but it is very similar to the
standard normal distribution and has similar properties.
•
It is bell-shaped and symmetrical.
•
The mean of this distribution is zero and its variance is greater than one. But the
larger the sample size, the closer the variance gets to one.
•
It is flattered compared to normal distribution and its tails are ticker. That is, its
variance is greater than the normal distribution.
•
The tails of distribution are asymptotic. That is, they can vary from −∞ to +∞.
This distribution has a parameter called degree of freedom (df = n − 1). The
greater the degree of freedom, the closer the t distribution to the normal distribution.
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6.3 t Distribution 131
Fig. 6.12 Student t distributions for n = 10 and n = 20 compared to standard normal distribution
Figure 6.12 shows the distribution of t compared to the standard normal
distribution.
Tips
Similarity of t Distribution to the Standard Normal Distribution:
(1) It is bell-shaped and symmetric about the mean.
(2) The mean, median, and mode are equal to 0 and are located at the center
of the distribution.
(3)
The curve is asymptotic and never touches the X-axis.
Dissimilarity of t Distribution from the Standard Normal Distribution:
(1) Its variance is greater than 1.
(2) The t distribution curves are actually different based on the degrees of
freedom (n − 1).
(3)
As the sample size increases, the t distribution approaches the standard
normal distribution.
One of the applications of t distribution is to test the hypothesis and construct the
confidence interval of the mean sample, which we will discuss in Chaps.
7 and 8.
In this section, it is enough to present the confidence interval formula and solve a
couple of example.
x ± t
(
1−
α
2
, n−1
)
×
S
√
n
.
https://t.me/med1917
132 6 Normal Distribution
Example 9 The accommodation (ACC) of right eye for 25 patients is equal to
(15–14–10–9.5–11–12–11–10–10.5–9–11.5–9–8.5–7–6.5–7.5–8–5–6–5.5–5–2.5–
3.5–1–11.5). These patients were randomly selected from a large population.
Construct the 95% confidence interval for the population mean.
Solution: In order to construct the confidence interval for the mean, we must first
estimate the variance and the standard error and find the critical t value with df = 25–1
= 24 from the student t-table distribution and then substitute them on the formula.
S
2
=
(x
i
− x)
2
n − 1
=
289.5
25 − 1
= 12.06,
Se
(
x
)
= S
x
=
S
√
n
=
√
12.06
√
25
= 0.695,
x =
x
i
n
=
210
25
= 8.4,
t
(
0.025,24
)
= 2.064,
x ± t
(
1−
α
2
,n−1
)
×
S
√
n
= 8.4 ± 2.064 × 0.695
= 8.4 ± 1.43 =
(
6.97 to 9.83
)
.
In Chap. 7, we will interpret the confidence interval and its details in more detail.
Example 10 In Example 9, the mean and variance of accommodation (ACC) of
right eye calculated from a sample of 25 subjects were 8.4 and 12.06, respectively.
If the same mean and variance are obtained from a sample of 81, construct the 95%
confidence interval for the population mean.
Solution: After calculating the standard error, we find the critical t value with df
= 81–1 = 80 from the student t-table distribution and then substitute them on the
formula.
x = 8.4, S
2
= 12.06 and df = 81 − 1 = 80,
t
(
0.025,80
)
= 1.990,
Se
(
x
)
= S
x
=
S
√
n
=
√
12.06
√
81
= 0.386,
x ± t
(
1−
α
2
,n−1
)
×
S
√
n
= 8.4 ± 1.990 × 0.386
= 8.4 ± 0.768 =
(
7.63 to 9.17
)
.
https://t.me/med1917
6.4 Chi-Square Distribution 133
As can be seen, as the sample size increases, the standard error and the critical
value of t become smaller and the confidence interval for the mean becomes narrower.
6.4 Chi-Square Distribution
Many of important probabilistic distributions in statistics are related t o the normal
distribution. One of these important distributions that is widely used in the analysis
of categorical data, goodness of fit of the observed distributions, and the confidence
interval of the variance of the population when the distribution of the main population
is normal is the Chi-square distribution.
Suppose k standard normal random variable (Z
1
, Z
2
, Z
3
, ··· , Z
k
), their sum of
squares (
Z
2
1
+ Z
2
2
+ Z
2
3
+ ··· + Z
2
k
) has the Chi-square distribution with degree of
freedom k and is denoted by the Greek letter
χ
2
. The degree of freedom plays an
important role in this distribution. The greater the degree of freedom of Chi-square
distribution, the more symmetrical the distribution shape will be. In this book, only
Chi-square distribution with the degree of freedom 1 will explain and we denote it
by
χ
2
(
1
)
.
The standardized normal variable z follows the distribution N (1, 0). Its square
(Z
2
) is also a non-negative random variable and can change from 0 to + (skewed to
the right). The distribution of the random variable Z
2
is called the distribution of χ
2
(1)
and its graph is shown in Fig. 5.10.
This distribution was first proposed in 1900 by the famous statistical scientist Karl
Pearson (Fig.
6.13).
The percentages of this distribution (critical values) are given in Table VI of the
appendix for different degrees of freedom. This statistic is always executed as one
tailed. In the distribution of
χ
2
(
1
)
with 1 degree of freedom, one or two points are
important.
Fig. 6.13 Chi-square distribution with different degrees of freedom
https://t.me/med1917
134 6 Normal Distribution
Mean distribution is 1, because E
z
2
=
E
(
x−μ
)
2
σ
2
=
σ
2
σ
2
= 1.
Percentages of this distribution can also be obtained from the standard normal
distribution. For example, the percentile corresponding to the 0.05 probability is +
1.96 or −1.96 and their square is (1.96)
2
= 3.841. The distribution of χ
2
(
1
)
at the
same 5% level equals to 3.841. The same is true for other percentile values of this
distribution. This same relation is true for other percentile points of the distribution.
Due to the equivalency between the standard normal distribution N (1, 0) and the
Chi-square distribution of
χ
2
(
1
)
, many issues of the standardized random variable can
be expressed by both distributions. However, it should be noted that using Z
2
(or
χ
2
(
1
)
) eliminates the z direction. Therefore, if the direction of the deviations from the
mean is important, the N(1, 0) distribution should be used.
Chi-square computational statistic is defined as follows:
χ
2
=
k
i=1
(
O
i
− E
i
)
2
E
i
,
where O
i
is the ith observed value and E
i
is the expected value for the ith subject.
The expected frequency (E
i
) is also obtained by multiplying the “sum of
corresponding column for the cell (
C
i
)” by the “sum of the related row for the same
cell (
R
i
)” divided by the “grand total (N )”:
E
i
=
C
i
× R
i
N
.
Example 11 The last digit of the weight of a sample of 96 adults (in kilograms) is
recorded in Table
6.1. We want to investigate and judge the accuracy of weight in
this study.
Solution: If the weighing and recording of people are accurate and there is no
tendency to round the l ast digits to 0 or 0.5, all the unit’s digits on the right place (0,
1, 2, … and 9) should be repeated equally. Because the probability of observing
each one is equal to 0.1. Therefore, the expected number of each unit’s digit is
obtained by multiplying this probability by the total number of observations (96).
That is, each units’ digit (
96 × 0.1 = 9.6) is repeated, which is given in the third
row (
E
i
) of the table above. The set of all values in this row is called theoretical
distribution.
It should be noted that the sum of the expected frequencies and the sum of the
observed frequencies must always be equal.
Table 6.1 Last digit of the weight of individuals in a sample of 96
Last digit 0 1 2 3 4 5 6 7 8 9 Tot a l
Freq 13 8 10 9 10 14 5 12 11 4 96
E
i
9.6 9.6 9.6 9.6 9.6 9.6 9.6 9.6 9.6 9.6 96
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