Добавил:
Sekretar
kiopkiopkiop18@yandex.ru
t.me/Prokururor I Вовсе не секретарь, но почту проверяю
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз:
Предмет:
Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_5320_Библиотеки_им_академика_М_И_Перельмана
.pdf
DF
=−=−=k
12
DF
=−=−=Nk1831
MS
between
MS
F
20
https://t.me/med1917
Pharmaceutical Dosage Forms and Drug Delivery
5. Degrees of freedom (DF):
Degrees of freedom between groups (DF
between
Degrees of freedom for the error term (DF
error
between
error
):
13
):
5
Mean squares (MS) of variation:
Mean square between groups (MS
between
Mean square for the error term (MS
):
between
SS
between
===
DF
):
error
SS
== =
error
DF
error
error
1304 2
352 8
2
15
176 4..
86 9..
7. F- ratio:
MS
between
===
MS
error
176 4
86 9
.
.
.
8. Determine the critical F- ratio at the chosen Pα value. Determine the critical F- ratio at (2,
15) degrees of freedom for α = 0.05 is 3.7.
9. Test the hypothesis: Since the obtained F- value is lower than the critical F- value, the null hypoth-
different when reviewed without statistical analysis, the high random error in the observations
An alternate means to test the hypothesis is to use the standard tables to determine the p- value associated
with the observed F- value. If the observed p- value is less than the chosen Pα value (e.g., 0.05), the null
hypothesis is rejected. For example, in the above calculations, the p- value associated with the observed
F- ratio is 0.17. Since this is higher than 0.05, the null hypothesis cannot be rejected.
5.7.9.1.7 Calculations Using Microsoft Excel
calculations. As an illustration, when Microsoft Excel’s data analysis add- in function is utilized for
single- factor ANOVA calculations, the software provides a tabular output of calculated values illustrated
in Table 5.10.
This tabular output of results summarizes statistical parameters associated with the data, followed by
a summary of calculated results in a tabular format. The critical F- value and the p- value associated with
the calculated F- value are indicated to facilitate hypothesis testing.
y
=++++
µτ βγ ε
https://t.me/med1917
Pharmacy Math and Statistics
TABLE 5.10
107
Statistical Results for a Hypothetical Example of a One- Way ANOVA Experiment Using Microsoft Excel
Summary
Groups Count Sum Average Variance
Dose = 0 99 71.1
Dose = 50 134 22.33333
Dose = 100 27.33333
Anova
Source of
variation SS DF MS F p- value F- crit
Between groups 352.7778 2 2.028754
Within groups 15
Total 17
5.7.9.2 Two- Way ANOVA: Design of Experiments
Two- way ANOVA deals with investigation of effects of two variables in a set of experiments. ANOVA
with two or more variables (also called treatments or factors) is most commonly utilized in the design of
experiments.
5.7.9.2.1 Factorial Experiments
When the effects of more than one factor are studied at one or more levels, the factorial experiment is
F- factorial experiment. For example, 3 factors evaluated at two different levels would
be a 23- factorial experiment, and 2 factors evaluated at 3 different levels would be a 32 experiment.
An example of such studies is the effect of temperature and pressure on the progress of a reaction. If
an experiment is run at 2 temperature and pressure values, it is a 22 factorial experiment, with the total
number of runs = 2 × 2 = 4. If the experiment were run at 3 levels of temperature and pressure, it would
be a 32 factorial experiment, with the total number of experimental runs = 3 × 3 = 9. Conversely, if 3
factors (e.g., temperature, pressure, and reactant concentration) were studied at 2 levels each, it would be
a 23 factorial experiment, with 2 × 2 2 = 8 experimental runs. The experiments could be full- factorial or
partial- factorial.
• A full- factorial experiment is one in which all combinations of all factors and levels are studied.
For example, a full- factorial 4- factor, 2- level study would involve 24mental runs. Full- factorial experiments provide information on both the main effects of various
factors and the effects of their interactions. The design and interpretation of a two- factor, two- level
experiment are illustrated in the two- way ANOVA model.
• A partial- factorial experiment is one in which half the combinations of levels of all factors are
studied. For example, a partial- factorial 4- factor, 2- level study would involve 2
experimental runs. Partial- factorial experiments provide information on the main effects of various
factors but not on the interaction effects. Design and interpretation of partial- factorial experiments
are beyond the scope of this chapter.
5.7.9.2.2 Model Equation
If there are two variables or treatments being studied in the experiment, the value of each data point is
explained as follows:
ijk ijij ijk
µ
∑∑∑
b
N
S
=−
()
===
ikjnB
b
2
https://t.me/med1917
108
where y
represents the jth observation of the ikth treatment
ijk
μ
β
is the j
j
Pharmaceutical Dosage Forms and Drug Delivery
τ
is the ith treatment effect of the
i
ε
represents random error.
ijk
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
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 two treatments or variables being studied (individual or main
effects,
τ
and
i
β
and effects arising from interaction of these variables,
j
γ
) and random variation (
ij
ε
).
ij
The variables in this experiment are commonly termed factors, and the experiment is termed a factorial
experiment. This equation represents a two- way ANOVA model.
5.7.9.2.3 Null and Alternate Hypotheses
The null hypotheses (H0) for a two- way ANOVA experiment studying factors A and B could be the
following:
• No difference between the population means of samples treated with different levels of factor A.
The alternate hypothesis (H1) would be that the means of underlying populations are not equal.
• No difference between the population means of samples treated with different levels of factor B.
The alternate hypothesis (H1) would be that the means of underlying populations are not equal.
• No difference between the population means of samples treated with different combinations of
different levels of factors A and B (For example, if both factors A and B had two levels— high
and low— the combinations could be high [A] with low [B] versus low [A] with high [B]. A study
of this interaction reveals whether the effect of factor A is different when factor B is low versus
high or not.). The alternate hypothesis (H1) would be that there is an interaction between factors
A and B.
5.7.9.2.4 Calculations
The calculations for a two- way ANOVA experiment are similar to the one- way ANOVA, with the inclusion of the case of a second variable B at levels 1 through b. The equations for the one- way ANOVA in
as below for the inclusion of the effect of variable B.
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.
y
where, y
represents the jth observation of the ith level of treatment of variable A and bth level
ijk
=
== =
ikjnB
11 1
ijB
of treatment of variable B, there being a total of k treatments (i = 1, 2, 3, … k) and n samples
per treatment level (j = 1, 2, 3, …. n) for variable A and b treatments (B = 1, 2, 3, … b) and n
samples per treatment level (j = 1, 2, 3, …. n) for variable BN is the total sample size, including
all treatments and levels.
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).
S
T
∑∑∑
111
y
µ
ijB
S
×−
2
S
×−
S
S
SS
=− −
,,iB
DF
i
k=−1
DF
B
b=−1
DF
=−×Nkb
DF
=−
()
−
()
kb11
MS
between
,
i
https://t.me/med1917
Pharmacy Math and Statistics
3. The sum of squares for the factor is the sum of squares between the columns (SS
between
109
) 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
b
∑
i
S
between,i
S
between, B
nb
=×
nk
=×
∑∑
jnB
==
11
∑∑
jni
==
11
k
b
k
∑
B
y
j
=
1
µ
2
y
B
=
1
µ
b
The sum of squares for interaction between factors A and B is determined by:
2
b
S
between,,iB
=×
b
∑∑
ikB
==
11
n
y
∑
B
=
1
nbynk
×
4. Sum of squares for the random error (SS
k
∑
iB
−
i
i
=
1
−
×
) is the difference between the total sum of squares
error
∑
b
y
BB
=
1
nb
×
∑∑∑
ikjnB
B
−
== =
11 1
nkb
××
2
b
y
ijB
and the sum of squares between and within the columns.
SSSS
errortotal between between
S
5. Degrees of freedom are calculated as follows:
Degrees of freedom between groups (DF
Degrees of freedom for the error term (DF
error
Degrees of freedom for the interaction term (DF
interaction
Mean squares for the random error (MS
):
between
between,
between,
error
):
):
interaction
) and the factor studied (MS
error
) are calculated by
between
dividing their respective sum of squares by their degrees of freedom.
SS
between
,
=
between
,
i
DF
i
MS
B
MS
error
F
i
error,,
F
error
B
,
https://t.me/med1917
110
between
Pharmaceutical Dosage Forms and Drug Delivery
SS
between
,
between
error
B
,
=
,
B
DF
SS
=
error
DF
7. An F- ratio is computed as the ratio of mean squares of factor effect to the mean square of error
effect.
MS
between
=
between
i
MS
MS
=
B
MS
and DF
8. Determine critical F- ratio at (DF
i
between
,
B
error
9. Test the hypothesis: The F- ratio is compared to the Pα value for the F- test at designated degrees
would indicate that the contribution of the factor’s or variable’s effect on the observations is sig-
5.7.9.2.5 Calculations Using Microsoft Excel
As an illustration of two- way ANOVA calculations using Microsoft Excel’s data analysis add- in
tool, the example summarized in Table 5.11 provides a tabular output of calculated values listed in
Table 5.12.
This tabular output of results summarizes statistical parameters associated with the data, followed by a
summary of calculated results in a tabular format. The critical F- value and the p- value associated with the
calculated F- value are indicated to facilitate hypothesis testing. Two- way ANOVA results provide infor
the contribution of columns (pressure) to variation has a p- value of 0.20, while the contribution of rows
(temperature) has a p-
cant, while that of pressure is not.
TABLE 5.11
A Hypothetical Example of a Two- Way ANOVA Experiment.
Temperature (°C) Pressure: 1 atm Pressure: 2 atm
40 95.4 95.8
91.9 92.1
Note: Yield of a chemical synthesis reaction was studied as a
function of temperature and pressure in a 22 full- factorial study
without replication. The data, in terms of percentage yield, are
summarized in the table.
https://t.me/med1917
Pharmacy Math and Statistics
TABLE 5.12
111
Statistical Results for a Hypothetical Example of a Two- Way ANOVA Experiment Using Microsoft Excel
ANOVA: Two- Factor Without Replication
Summary Count Sum Average Variance
Row 1 2 191.2 0.08
Row 2 2 184 92 0.02
Column 1 2 187.3
Column 2 2 187.9 93.95
ANOVA
Source of
variation
Rows 1
Columns 0.09 1 0.09 9 0.204833
Error 0.01 1 0.01
Total 3
Review Questions
SS DF MS F p- value F crit
5.1 Amoxicillin suspension.
A How much water would need to be added to a bottle containing 12.5 g of dry powder for
reconstitution into a 250 mg/ 5 mL suspension? Hint: Use ratio and proportion and remember
to use the same units.
B How many milliliters of amoxicillin suspension containing 250 mg/ 5 mL must be administered
to a patient in need of a 400- mg dose of amoxicillin? Hint: Use ratio and proportion.
C If each 5 mL of a 250 mg/ 5 mL reconstituted amoxicillin suspension contains 0.15 mEq of
sodium, how much sodium does it represent in mg? Hint: Use the atomic weight of sodium.
D Given your answers to (a) and (b) above, how much sodium would the patient be taking per
day if the patient is dosed 400 mg t.i.d.? Hint: Use ratio and proportion.
5.2 Cyclophosphamide tablets.
A Cyclophosphamide is available as 50- mg tablets and has a recommended dose of 5 mg/ kg o.d.
What would be the daily dose for a 175- lb patient? Hint: Use proportion after converting every
quantity to the same units.
B How many tablets should be dispensed for a dosage regimen of 10 days? Hint: Calculate the
5.3 Dosage for Children. For a drug with an adult dose of 100 mg/ kg, what would be the dose for a
4- feet- tall 8- year- old child weighing 80 lbs? Calculate using the nomogram, Fried’s rule, Young’s
rule, and Clark’s rule.
5.4 Tonicity adjustment.
A Calculate the NaCl equivalents (E value) for the following three drugs, given that NaCl has
a molecular weight of 58.5 and dissociates into 2 ions, with a dissociation constant (i) of 1.8.
B For the prescription noted below, calculate the NaCl equivalents present in the formulation.
C Calculate the amount of NaCl equivalents that would need to be added to the above formula-
tion to make it isotonic for ophthalmic administration.
D If NaCl were incompatible with one or more of drugs, how much dextrose (molecular
weight = 180) may be used instead.
https://t.me/med1917
112
Drug A
Drug B
Drug C
Drug A 40 mg
Drug B 25 mg
Drug C 100 mg
Water q.s. 10 mL
Molecular weight = 220, ions = 3,i
Molecular weight = 180, ions = 1,i = 1
Molecular weight = 140, ions = 2,i = 1.9
Pharmaceutical Dosage Forms and Drug Delivery
5.5 Volume and weight interconversions
A Glycerin is a highly viscous liquid that may be weighed instead of measured in volume. How
3?
B Ethanol is a low- viscosity liquid that is easier measured in volume than in weight. Given that
its density is 0.78 g/ cm3, how much volume of ethanol is needed to prepare 25 mL of a 5% v/
v solution?
C Ethanol is a low- viscosity liquid that is easier measured in volume than in weight. Given that
its density is 0.78 g/ cm3, how much volume of ethanol is needed to prepare 25 g of a 5% w/
w solution?
Concentration calculations
A What would be the equivalent weight of calcium chloride (CaCl2) if its molecular weight is
111 g/ mol?
B What amount of CaCl2 would be needed to make 50 mL of a 0.5 M solution?
C What amount of CaCl2 would be needed to make 50 mL of a 0.5 N solution?
D A drug product was found to contain 40 ppm of an impurity during analysis. How many
milligrams of this impurity might be ingested by an average 150- lb adult human being if the
drug is to be administered in doses of 5 mg/ kg/ day in four divided doses?
E What is the mole fraction of an isotonic NaCl solution? The molecular weight of NaCl is 58.5
and that of water is 18. Hint: Isotonic NaCl solution has 0.9% w/ v salt concentration.
F
stock solution?
G How much of the 0.1 N HCl solution would be needed to prepare 200 mL of a 2 N solution,
using the 5 N stock solution of HCl?
5.7 Calculate the mean, median, variance, and standard deviation of the following sets of values:
A
B
C
D By reviewing the above results, which of the three data sets has the highest spread around the
central tendency?
E By reviewing the above results, which of the three data sets has the least spread around the
central tendency?
F By reviewing the above results, what are the differences between the means of which two data
G By reviewing the above results, what are the differences between the means of which two data
5.8
of 5 mg and an accepted error rate of 5% is?
A mg
B 10 mg
C 100 mg
D 1000 mg
https://t.me/med1917
Pharmacy Math and Statistics
113
5.9
minimum trituration strength can you use?
A 1:20
B 1:24
C 1:30
D 1:50
REFERENCES
Canal P., Chatelut E., and Guichard S. (1998) Practical treatment guide for dose individualisation in cancer
chemotherapy, Drugs 56(6): 1019.
Chatelut E., Canal P., Brunner V. et al., (1995) Prediction of carboplatin clearance from standard morphological
and biological patient characteristics, J Natl Cancer Inst 87(8): 573.
Dowdy S., Weardon S., and Chilko D. (2004) Statistics for Research, Hoboken, NJ: Wiley- Interscience.
Math Calculations for Pharmacy Technicians: A Worktext, St. Louis,
MO: Saunders.
Hempel G. and Boos J. (2007) Oncologist 12(8): 924.
Hopkins W.A. (2005) APhA’s Complete Math Review for the Pharmacy Technician, Washington, DC: APhA
Publications.
Narang A.S., and Desai D.S. (2009) Anticancer drug development. In Mahato R.I. and Lu Y. (Eds.)
Pharmaceutical Perspectives of Cancer Therapeutics, New York: AAPS- Springer, p. 49.
Po A.L.W. (1998) Statistics for Pharmacists, Oxford: Wiley- Blackwell.
patients with impaired hepatic function, J Natl Cancer Inst 88(12): 817.
https://t.me/med1917
https://t.me/med1917
Part II
Physicochemical Principles
Соседние файлы в папке Библиотека им академика М.И. Перельмана