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Pharmaceutical Dosage Forms and Drug Delivery
•
population.
•
viz., median and ranks of the data values. They do not make the assumption that the underlying
distribution of the population is known.
Parametric tests are more powerful (with a less probability of type II error, described later) than the
nonparametric tests, since they use more information about the samples. They are frequently used to
provide information, such as the interaction between two variables in a factorial design of experiments.
However, they are also more sensitive to skewness in the distribution of data and the presence of outliers
in the samples. Therefore, nonparametric tests may be preferred for skewed distributions.
t- test, chi- square test, and analysis of variance (ANOVA).
-
metric tests will be described in more detail in the following sections.
5.7.2 Null and Alternate Hypothesis
example, to test the hypothesis that (a) a sampled data set comes from a single population, or that (b) two
sampled data sets come from a single population. A statistical hypothesis represents an assumption about
a population parameter. This assumption may or may not be true and is sought to be tested using the
hypothesis that a given variation within or among data sets occurred purely by chance, it would be termed
the null hypothesis. In this case, therefore, the null hypothesis is the hypothesis of no difference. If the
alternate
hypothesis is assumed to hold true. The alternate hypothesis indicates that the sample observations are
5.7.3 Steps of Hypothesis Testing
The process of testing a hypothesis involves the following general steps:
1. Ask the question (for a practical situation) that can be addressed using one of the statistical tests
2.
assumptions.
3. State null and alternate hypothesis.
4. α = 0.01, 0.05, or 0.1, which indicates 1%, 5%, or 10% probability
the chance of not detecting the differences when they actually do exist.
5.
actually do exist.
Compute the test statistics.
7. Identify the probability (p) of obtaining a test statistic as extreme as the calculated test statistic for
the calculated degrees of freedom using standard probability distribution tables.
8. p
hypothesis is rejected. If p
p, the null hypothesis cannot be rejected.
sample at α
sample at α
< p, the null
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Pharmacy Math and Statistics
5.7.4 One- Tailed and Two- Tailed Hypothesis Tests
97
The null and alternate hypotheses can be stated such that the null hypothesis is rejected when the test stat-
latter is termed two- tailed hypothesis. For example, if
and H0 represents the null hypothesis (H0:
since H0 would be rejected when (
μ
1
μ
μ
μ
1
< d) and (
2
is a two- tailed hypothesis, since the null hypothesis would be rejected in both cases of (
(
μ
μ
> d).
1
2
μ
and
1
d) or (H0:
2
μ
μ
1
μ
represent the means of two populations
2
μ
μ
d) would be one- tailed hypothesis,
1
> d), respectively. However, (H0:
2
2
μ
1
μ
μ
1
2
μ
< d) and
2
= d)
The appropriate statement of null hypothesis depends on the practical situation being addressed. For
example,
• If a sample of tablets were collected during a production run of tableting unit operation and tested
for average tablet weight, the question could be asked whether the average tablet weight is the
target tablet weight. In this case (H0: weight
sample
= 0) or (H0: weight
target
sample
= weight
target
)
would be a two- tailed hypothesis test, since the null hypothesis would be rejected when the sample
weight is higher than or lower than the target weight.
• If a sample of tablets were collected during a production run of the coating unit operation
and tested for coating weight build- up on the tablets, the question could be asked whether
the coating weight build- up has reached the target weight build- up of 3% w/ w. In this case
(H0: weight
sample
target
would be rejected only if the sample weight is less than the target weight.
5.7.5 Regions of Acceptance and Rejection
The regions of acceptance and rejection of a hypothesis refer to regions in the probability distribution of
the sample’s test statistic. Assuming that the null hypothesis is true, a sample’s test statistic is normally
example, Figure 5.8a shows the normal distribution of a test statistic, with a vertical line to the right
indicating the value of the test statistic associated with a probability of occurrence (α) of 0.05, or 5%, by
random chance, or Pαα
For a one- tailed hypothesis test (Figure 5.8a), the region of rejection lies on one (right) side of this
distribution. Suppose the test statistic value obtained for the sample in question is higher than Pα. In that
case, the test statistic in the sample is assumed to lie in the region of rejection, and the null hypothesis
α
to P).
For a two- tailed hypothesis test (Figure 5.8b), the region of rejection lies on either side of the distribution. If the test statistic value obtained for the sample in question is higher than PαPα, the
αPα to Pα).
5.7.6 Probability Value and Power of a Test
p- value). The
p- value is the fractional probability of accepting the null hypothesis, assuming that the null hypothesis is
true. In other words, lower the p- value of the test, expressed as a fractional probability (e.g., 0.01, 0.05, or
0.1, representing 1%, 5%, or 10% probability, respectively), the greater the chance of accepting the null
hypothesis and not detecting differences between two samples. A lower p- value indicates a greater difference between the two samples. The commonly used probability level for accepting the null hypothesis is
5%, corresponding to the p- value of 0.05.
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98
FIGURE 5.8 An illustration of regions of acceptance and rejection in a normal probability distribution. Knowing the prob-
α = 0.05) can quantify a cut- off point, indicated by a vertical line in the plot. This vertical
α line has
a lower than 5% chance of occurrence and is said to fail in the region of rejection. This is one- tailed hypothesis, since data
values on only one side of the mean are being considered for hypothesis testing. This side could be the positive side, as
indicated in (a), or the negative side, which would be indicated by the α line on the left of the mean. In a two- tailed hypothesis
testing (b), data values on both positive and negative sides of the mean are considered. Data values that are more extreme than
the α line are said to fall in the region of rejection. All other data values are considered in the region of acceptance.
Pharmaceutical Dosage Forms and Drug Delivery
null hypothesis is not true. In other words, the higher the power of the test, expressed in %, greater the
chance that true differences between two different sample sets would be detected. The power of a test can
be increased by increasing the sample size. The commonly accepted power of a test is 80%.
5.7.7 Types of Error
test statistic:
• -
esis. This is the error of rejecting a null hypothesis when it is actually true. In other words, type
different. The probability of type I error is denoted by α.
• α, is higher.
Therefore, using lower α tends to reduce the probability of a type I error.
•
null hypothesis. This is the error of not rejecting a null hypothesis when it is actually not true. In
The probability of type II error is higher when the chosen power of the test, β, is lower. Therefore,
using higher β tends to reduce the probability of a type II error.
y
ij iij
=++
µτ ε
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Pharmacy Math and Statistics
5.7.8 Questions Addressed by Tests of Significance
99
and a probability distribution of the test statistic. For example, the differences between means are tested
using a t- test, the differences between proportions are tested using a z- test, and the differences in the
frequency of a categorical variable are tested using the
2
χ
example situation, underlying assumptions of tests, statement of null hypothesis, and calculations of the
test statistic are summarized in .
• The calculation of a test statistic, which represents the difference between the expected and the
observed values, or the values of two samples. It also takes into account the variability in the
sample through the incorporation of standard error. The calculation of test statistics involves quantifying the extent of observed differences vis- à- vis the variability.
•
(Pα) for the given degrees of freedom. The degree of freedom is calculated based on the sample size
and sometimes also the number of variables studied. The degrees of freedom affect the distribution
plot of the test statistic and thus the Pα value for a given α.
Having calculated the Pα
is 0.942, and the Pα value at the desired probability of error of 5% is 1.347. In that case, the test statistic
falls in the acceptance region. Hence, the null hypothesis cannot be rejected. On the other hand, if the test
statistic value were higher than 1.347, the test statistic would fall in the region of rejection. Hence, the
null hypothesis would be rejected.
5.7.9 Analysis of Variance
means of different samples. Any number of samples or subgroups may be compared in an ANOVA
experiment. ANOVA is based on the underlying explanation of the variation of sample values from the
population mean as being a linear combination of the variable effect and random error.
The number of variables (also termed treatments or factors) in an ANOVA experiment can be one (oneway ANOVA), two (two- way ANOVA), or more. Each variable or factor can be studied at different levels,
indicating the intensity. For example, a clinical study that evaluates one dose of an experimental drug
is a one- variable one- level experiment. A study that evaluates two doses of an experimental drug would
be a one- variable two- level study. Another study that evaluates three doses of two experimental drugs
would be a two- variable three- level study. The level may be a quantitative number, such as the dose in
the above examples, or it may be a numerical designation of the presence or intensity of an effect, such
as “0” and “1.”
5.7.9.1 One- Way ANOVA
5.7.9.1.1 Model Equation
When sample sets are treated with a single variable at i different levels (i = 1, 2, 3, …, k), the value of
each data point is explained as:
where, yij represents the jth observation of the iμ is the mean of all
τ
is the i
i
ε
represents random error.
ij
TABLE 5.6
t
d
−
()
−
SE
12
SE
sd
=+
DF =−
()
−
()
norn whichever is smaller
t
dD
SE
−
SE
n
()
()
DFn=−1
z
pp
−
12
S
11
12
P
Pn Pn
×+×
22
12
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Test question
or situation
and the test of
signicancetouse Example
To test difference
between two
means, use two-
sample t- test.
To test difference
between matched
pairs, use matched
pairs t- test.
To test difference
between two
proportions,
use the twoproportion z- test.
Two batches of tablets were
manufactured, with an
average tablet weight of 200
mg. A sample of 100 tablets
each was tested from each
of these batches. Do the
two batches have different
average tablet weight?
Tablet friability test was
conducted on a batch on 10
different occasions. Total
tablet weight was recorded
before and after the
friability test in each case. Is
tablet friability >1%?
Edge- chipping defects in
tablets were counted in
a sample of 400 coated
and 350 uncoated tablets.
22 coated tablets had this
defect. Is edge chipping
more likely for the coated or
the uncoated tablets?
Statement of
hypothesis (for the
example given) Equations and abbreviations Underlying assumptions
H0:
μ
=
μ
1
H1:
2
μ
μ
1
2
OR
H0:
μ
μ
1
H1:
2
μ
μ
1
2
where, d = 0
= d
d
2
mean mean
=
where,
11, H01, alter-
12
1
nsdn
1
2
2
2
and
dsd, standard
n
• Random sampling
• Independent samples
• Population follows a normal
or near- normal distribution
• Population size is at least
tenfold higher than the
sample size
mean
t, test statistic for the
t- distribution.
H0:
μ
> D
d
H1:
μ
D
d
where, D = 1
where,
=
∑−
=
2
dd n
i
−
1/
and
H0, null
1d, difference between the two
• Random sampling
• Data sets not independent
• Population follows a normal
or near- normal distribution
D), standard deviation of differences of matched
di, difference for the matched pair id, mean of difference between
n
d, mean difference between matched pairs.
H0: P1 = P
H1: P1P
2
2
OR
H0: P1P2 = d
H1: P1P2d
where, d = 0
E
=×−
PP
, where
=
SE
11
=
pooled
nn
+
1
(
pooled pooled
×+
)
nn
)
H01P, proportion of observations
P
d
pooled
• Random sampling
• Independent samples
• Sample includes at least ten
events and ten nonevents for
calculating the proportion
• Population size is at least ten-
fold higher than sample size
and SE, standard error of the pooled sample proportion.
100
Pharmaceutical Dosage Forms and Drug Delivery
td=
−
()
−mean mean
SE
12
SE
sdnsd
n
=+
1
2
1
2
2
2
DF =−
()
−
()
norn whichever is smaller
12
11,
t
dD
SE
=−SE
dd n
n
i
=
∑−
()
−
()
2
1/
DFn=−1
d
d
z
pp
=
−
12
SE
SE
pooled pooled
=×−
(
)
×+
PP
nn
1
11
12
)
P
Pn Pn
nn
pooled
=
×+×
+
1122
12
χ
2
()
OE
E
ii
i
E nP
ii
=×
DF =−k 1
χ
()
OE
ir
,,
E
nn
n
ir
×
DFi r=−
()
−
()
11
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newgenrtpdf
To test whether
a categorical
variable follows
a hypothesized
frequency
distribution, use
the chi- square
goodness- of- t
test.
To test whether a
categorical variable follows the
same frequency
distribution in
two or more
populations, use
the chi- square test
of homogeneity.
A controlled- release capsule
formulation uses drug
microspheres encapsulated
in hard gelatin capsules.
Three types of microspheres
are encapsulated: 30% w/
w of immediate release,
35% w/ w of delayed release
delayed release by 4 h. In
an analysis of 20 capsules,
the proportions of these
components were 23.2% w/
w, 37.9% w/ w, and 38.9%
w/ w. Does this sample
represent the targeted
amount for each capsule in
the formulation?
An antihypertensive drug was
tested in 320 male and 290
female human volunteers.
Three effects of this drug
were tracked— reduction in
blood pressure of at least
20 mm Hg, and skin rashes
and nausea as adverse
events. The proportion of
populations showing these
events were 220, 12, and
and 19 for females, respectively. Is it likely that the
drug’s effects are affected
by gender?
H0: Ps = P
H1: PsP
H0: P
H1: P
i,r
i,r
= P
P
h
h
i,r
i,r
for each categorical
variable i in each
population r.
2
=∑
−
where,
and
H01Ps
PhOi, observed proportion of
the iEi, expected proportion of the
inPi, hypothesized
proportion of the i
k, number of categorical variables in the sample (e.g., k = 3
for the example cited in column 2).
2
−
ir ir
2
=∑
where,
E
,
=
ir
,
H01P
in rO
E
, expected proportion of the ith variable in rni,
i,r
, observed proportion of the ith variable in rth popu-
i,r
and
, proportion of ith variable
i,r
total number of observations of the i
nr, total number of observations of the rn, total number of
• Random sampling
Pharmacy Math and Statistics
• Categorical variable
• Population size is at
least tenfold higher than
sample size
• Expected value for each
categorical variable is at
• Random sampling
• Categorical variable
• Population size is at
least tenfold higher than
sample size
• Expected value for each
categorical variable in each
101
σ
µ
2
µ
==
∑∑
n
N
11
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102
Pharmaceutical Dosage Forms and Drug Delivery
Hence, the value of each data point in an experiment is represented in terms of the mean of all samples
and deviations arising from the effect of treatment or variable being studied (
τ
) and random variation (
i
ε
).
ij
This equation represents a one- way ANOVA model.
5.7.9.1.2 Underlying Assumptions
ANOVA is used to test hypotheses regarding means of two or more samples, assuming the following:
• The underlying populations are normally distributed.
• Variances of the underlying populations are approximately equal.
• The errors (
and a variance of
5.7.9.1.3 Fixed- and Random- Effects Model
ε
) are random and have a normal and independent distribution, with a mean of zero
ij
.
yij) from the mean of all data points
(μ) as a combination of random variation (
subgroups of the experimental data points can be subjected to different levels of the treatment,
ε
) and the effect of a known variable or treatment (τ). Different
ij
τ
, where
i
i = 1, 2, 3, … kxed- effects model. On the
other hand, if the levels of the treatment are randomly assigned from several possible levels, the model is
termed a random- effects model.
-
pharmacokinetic study of a given drug at dose levels of 0, 50, and 100 mg. A random effects model would
three different drugs A, B, and C at unknown and variable dose levels (e.g., dose titration by the physician for individualization to the patient). The effects are assumed to be random in the latter case, since
random- effects model.
5.7.9.1.4 Null and Alternate Hypothesis
The null hypothesis (H0) for a one- way ANOVA experiment would be no difference between the population means of samples treated with different levels of the selected factor. The alternate hypothesis (H1)
states that the means of underlying populations are not equal.
5.7.9.1.5 Calculations for Fixed- Effects Model from First Principles
ANOVA is based on the calculation of ratio of variance introduced by the factor and random variations.
Although many software tools are currently available that reduce the requirement for tedious calculations,
1. Mean of all samples in the experiment (μ) is calculated by adding all observations and dividing
by the total number of samples in the experiment.
where, yij represents the jth observation of the ith level of treatment of the variable, there being a
total of k treatments (i = 1, 2, 3, … k) and n samples per treatment level (j = 1, 2, 3, … n), and N
being the total sample size, including all treatments and levels.
y
ikj
=
ij
SS
ikj
n
=−
()
==
S
2
S
between
=−
DF
=−k 1
DF
=−Nk
MS
MS
error
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Pharmacy Math and Statistics
103
2. Total sum of squares (SST) of all observations is calculated by squaring all observations and
subtracting from the mean of all samples in the experiment (μ).
T
∑∑
y
µ
ij
3. The sum of squares for the factor studied is the sum of squares between the columns (SS
between
)
if each level of the factor is arranged in a column. It is calculated by subtracting the mean value
for each column from the mean of all samples, squaring this value, and adding it for all columns.
k
S
between
4. The sum of squares for the random error (SS
n
∑
i
∑
j
j
k
=
1
error
=× −
y
j
=
1
µ
) is the difference between the total sum of
squares and the sum of squares between and within the columns.
SSSSS
errortotal
5. Degrees of freedom are calculated as follows:
Degrees of freedom between groups (DF
7. Degrees of freedom for the error term (DF
8. Mean squares for the random error (MS
):
between
between
error
error
) and the factor studied (MS
error
):
) are calculated by
between
dividing their respective sum of squares by their DF.
SS
between
=
between
error
DF
between
SS
error
=
DF
error
9. An F- ratio is computed as the ratio of mean squares of factor effect to the mean square of error
effect.
F =
MS
between
MS
a. Mean
n
n
==
==
∑∑
11
6
6
b
=−
==
∑∑
2
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104
10. Determine critical F- ratio at (DF
between
Pharmaceutical Dosage Forms and Drug Delivery
and DF
) degrees of freedom for α = 0.05.
error
11. Test the hypothesis. The F- ratio is compared to the Pα value for the F- test at designated degrees
results indicates that the contribution of the factor’s or variable’s effect on the observations is
5.7.9.1.6 Example of Calculations for Fixed- Effects Model
tration of two doses of a test antihyperlipidemic compound and a placebo to a set of six patients in each
group. Hypothetical results of this study in terms of reduction of blood cholesterol level are summarized
in Table 5.7. These data can be rephrased in statistical terms, as presented in Table 5.8.
y
i
. SS
of observationsin eachgroup i
n
foreachi
()
between,i
=−
n
()
y
i
1
j
µ
j
2
6
y
1
j
6
y
i
j
22
.11
i
TABLE 5.8
Rephrasing the Data in Statistical Terms for a Hypothetical Example of a One- Way ANOVA Experiment
Subject #
j = 1 y
j= 2 y
j = 3 y
j = 4 y
j = 5 y
jn y
a 22.3 27.3
b 30.9 0.1 27.9
c 185.2 0.5
d 352.8
TABLE 5.7
A Hypothetical Example of a One- Way ANOVA Experiment
Subject # Dose = 0 Dose = 50 mg Dose = 100 mg
1 20 18 28
2 18 25 22
3 14
4 30 28 29
5 5 15 24
12 12 15
Factor A, level 1
i = 1
= 20 y
1,1
= 18 y
1,2
= 14 y
1,3
= 30 y
1,4
= 5 y
1,5
= 12 y
Factor A, level 2
i = 2
= 18 y
2,1
= 25 y
2,2
y
2,3
= 28 y
2,4
= 15 y
2,5
= 12 y
Factor A, level 3
i = 3 (k = 3)
= 28
3,1
= 22
3,2
3,3
= 29
3,4
= 24
3,5
= 15
c. SS
==
∑∑
2
2
d. SSs
()
µ
==
×
∑∑
n
N
63
S
()
==
ikj
n
9
S
=+
S
between
=−
S
=−=
2
https://t.me/med1917
Pharmacy Math and Statistics
n
between,ii
3
=× −
j
∑
between
=
i
1
no.of obsvns
×=×−
j
n
n
y
∑
i
=
j
1
n
y
i
1
j
µ
=×
2
µ
umof allvalues
=
6
yy
1
j
i
6
6
−
22 1
.
c
105
1. Mean of all samples in the experiment (μ):
y
397
==
ikj
ij
11
=
221.
2. Total sum of squares of variation in all data points (SST):
The calculations are illustrated in Table 5.9.
Squaring (yij- μ) values and adding them together,
ST=−
∑∑
11
2
y
=
1656
µ
ij
.
3. The sum of squares of variation coming from the factor studied (SS
between
):
As calculated in Table 5.8.
4. The sum of squares of variation coming from random error (SS
error
):
As calculated in Table 5.8.
TABLE 5.9
SS SS
totalbetween error
SSSSS
errortotal
S
1656 9 352 8 1304
error
.. .
Calculations for a Hypothetical Example of a One- Way ANOVA Experiment
Subject #
j = 1 y
j = 2 y
j = 3 y
j = 4 y
j = 5 y
j y
Factor A, level 1
i = 1
μ y
1,1
μ y
1,2
μ y
1,3
μ y
1,4
μ y
1,5
μ y
Factor A, level 2
i = 2
y
2,1
μ y
2,2
μ y
2,3
μ y
2,4
μ y
2,5
μ y
Factor A, level 3
i = 3
3,1
μ
3,2
μ
3,3
μ
3,4
μ
3,5
μ
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