9.9 Final Remarks

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9.8.2 Factorial ANOVA

Factorial ANOVA is an extension of the one-way ANOVA, assuming the same assumptions, but considering two or more factors. Factorial ANOVA presumes that the quantitative dependent variable is influenced by more than one qualitative explanatory variable (factor). It also tests the possible interactions between the factors, through the resulting effect of the combination of factor A’s level i and factor B’s level j, as discussed by Pestana and Gageiro (2008), Fávero et al. (2009), and Maroco (2014).

For Pestana and Gageiro (2008) and Fávero et al. (2009), the main objective of the factorial ANOVA is to determine whether the means for each factor level are the same (an isolated effect of the factors on the dependent variable), and to verify the interaction between the factors (the joint effect of the factors on the dependent variable).

For educational purposes, the factorial ANOVA will be described for the two-way model.

9.8.2.1 Two-Way ANOVA

According to Fávero et al. (2009) and Maroco (2014), the observations of the two-way ANOVA can be represented, in general, as shown in Table 9.4. For each cell, we can see the values of the dependent variable in the factors A and B that are being studied.

Table 9.4

Observations of the Two-Way ANOVA
		Factor B
		1	2	…	b
Factor A	1	Y₁₁₁	Y₁₂₁	…	Y_ab1
		Y₁₁₂	Y₁₂₂		Y_ab2
		⋮	⋮		⋮
		Y_11n	Y_12n		Y_abn
	2	Y₂₁₁	Y₂₂₁	…	Y_2b1
		Y₂₁₂	Y₂₂₂		Y_2b2
		⋮	⋮		⋮
		Y_21n	Y_22n		Y_2bn
	⋮	⋮	⋮	⋮	⋮
	a	Y_a11	Y_a21	…	Y_ab1
		Y_a12	Y_a22		Y_ab2
		⋮	⋮		⋮
		Y_a1n	Y_a2n		Y_abn

Table 9.4

Source: Fávero, L.P., Belfiore, P., Silva, F.L., Chan, B.L., 2009. Análise de dados: modelagem multivariada para tomada de decisões. Campus Elsevier, Rio de Janeiro; Maroco, J., 2014. Análise estatística com o SPSS Statistics, sixth ed. Edições Sílabo, Lisboa.

where Y_ijk represents observation k (k = 1, …, n) of factor A’s level i (i = 1, …, a) and of factor B’s level j (j = 1, …, b).

First, in order to check the isolated effects of factors A and B, we must test the following hypotheses (Fávero et al., 2009; Maroco, 2014):

$\begin{array}{l} H_{0}^{A} : μ_{1} = μ_{2} = \dots = μ_{a} \\ H_{1}^{A} : \exists_{(i, j)} μ_{i} \neq μ_{j}, i \neq j (i j = 1 \dots a) \end{array}$ $\begin{array}{l} H_{0}^{A} : μ_{1} = μ_{2} = \dots = μ_{a} \\ H_{1}^{A} : \exists_{(i, j)} μ_{i} \neq μ_{j}, i \neq j (i j = 1 \dots a) \end{array}$

si109_e (9.37)

and

$\begin{array}{l} H_{0}^{B} : μ_{1} = μ_{2} = \dots = μ_{b} \\ H_{1}^{B} : \exists_{(i, j)} μ_{i} \neq μ_{j}, i \neq j (i j = 1 \dots b) \end{array}$ $\begin{array}{l} H_{0}^{B} : μ_{1} = μ_{2} = \dots = μ_{b} \\ H_{1}^{B} : \exists_{(i, j)} μ_{i} \neq μ_{j}, i \neq j (i j = 1 \dots b) \end{array}$

si110_e (9.38)

Now, in order to verify the joint effect of the factors on the dependent variable, we must test the following hypotheses (Fávero et al., 2009; Maroco, 2014):

$\begin{array}{l} H_{0} : γ_{ij} = 0, for i \neq j (there is no interaction between the factors A and B) \\ H_{1} : γ_{ij} \neq 0, for i \neq j (there is interaction between the factors A and B) \end{array}$ $\begin{array}{l} H_{0} : γ_{ij} = 0, for i \neq j (there is no interaction between the factors A and B) \\ H_{1} : γ_{ij} \neq 0, for i \neq j (there is interaction between the factors A and B) \end{array}$

si111_e (9.39)

The model presented by Pestana and Gageiro (2008) can be described as:

$Y_{ijk} = μ + α_{i} + β_{j} + γ_{ij} + ɛ_{ijk}$ $Y_{ijk} = μ + α_{i} + β_{j} + γ_{ij} + ɛ_{ijk}$

(9.40)

where:

μ is the population’s global mean;
α_i is the effect of factor A’s level i, given by μ_i − μ;
β_i is the effect of factor B’s level j, given by μ_j − μ;
γ_ij is the interaction between the factors;
ɛ_ijk is the random error that follows a normal distribution with a mean equal to zero and a constant variance.

To standardize the effects of the levels chosen of both factors, we must assume that:

$\sum_{i = 1}^{a} α_{i} = \sum_{j = 1}^{b} β_{j} = \sum_{i = 1}^{a} γ_{ij} = \sum_{i = 1}^{b} γ_{ij} = 0$ $\sum_{i = 1}^{a} α_{i} = \sum_{j = 1}^{b} β_{j} = \sum_{i = 1}^{a} γ_{ij} = \sum_{i = 1}^{b} γ_{ij} = 0$

si113_e (9.41)

Let’s consider $\bar{Y}$ $\bar{Y}$ , ${\bar{Y}}_{ij}$ ${\bar{Y}}_{ij}$ , ${\bar{Y}}_{i}$ ${\bar{Y}}_{i}$ , and ${\bar{Y}}_{j}$ ${\bar{Y}}_{j}$ the general mean of the global sample, the mean per sample, the mean of factor A’s level i, and the mean of factor B’s level j, respectively.

We can describe the residual sum of squares (RSS) as:

$RSS = \sum_{i = 1}^{a} \sum_{j = 1}^{b} \sum_{k = 1}^{n} {(Y_{ijk} - {\bar{Y}}_{ij})}^{2}$ $RSS = \sum_{i = 1}^{a} \sum_{j = 1}^{b} \sum_{k = 1}^{n} {(Y_{ijk} - {\bar{Y}}_{ij})}^{2}$

si118_e (9.42)

On the other hand, the sum of squares of factor A (SSF_A), the sum of squares of factor B (SSF_B), and the sum of squares of the interaction (SSF_AB) are represented below in Expressions (9.43)–(9.45), respectively:

${SSF}_{A} = b \cdot n \cdot \sum_{i = 1}^{a} {({\bar{Y}}_{i} - \bar{Y})}^{2}$ ${SSF}_{A} = b \cdot n \cdot \sum_{i = 1}^{a} {({\bar{Y}}_{i} - \bar{Y})}^{2}$

si119_e (9.43)

${SSF}_{B} = a \cdot n \cdot \sum_{j = 1}^{b} {({\bar{Y}}_{j} - \bar{Y})}^{2}$ ${SSF}_{B} = a \cdot n \cdot \sum_{j = 1}^{b} {({\bar{Y}}_{j} - \bar{Y})}^{2}$

si120_e (9.44)

${SSF}_{AB} = n \cdot \sum_{i = 1}^{a} \sum_{j = 1}^{b} {({\bar{Y}}_{ij} - {\bar{Y}}_{i} - {\bar{Y}}_{j} + \bar{Y})}^{2}$ ${SSF}_{AB} = n \cdot \sum_{i = 1}^{a} \sum_{j = 1}^{b} {({\bar{Y}}_{ij} - {\bar{Y}}_{i} - {\bar{Y}}_{j} + \bar{Y})}^{2}$

si121_e (9.45)

Therefore, the sum of total squares can be written as follows:

$TSS = RSS + {SSF}_{A} + {SSF}_{B} + {SSF}_{AB} = \sum_{i = 1}^{a} \sum_{j = 1}^{b} \sum_{k = 1}^{n} {(Y_{ijk} - \bar{Y})}^{2}$ $TSS = RSS + {SSF}_{A} + {SSF}_{B} + {SSF}_{AB} = \sum_{i = 1}^{a} \sum_{j = 1}^{b} \sum_{k = 1}^{n} {(Y_{ijk} - \bar{Y})}^{2}$

si122_e (9.46)

Thus, the ANOVA statistic for factor A is given by:

$F_{A} = \frac{\frac{{SSF}_{A}}{(a - 1)}}{\frac{RSS}{(n - 1) \cdot ab}} = \frac{{MSF}_{A}}{MSR}$ $F_{A} = \frac{\frac{{SSF}_{A}}{(a - 1)}}{\frac{RSS}{(n - 1) \cdot ab}} = \frac{{MSF}_{A}}{MSR}$

si123_e (9.47)

where:

MSF_A is the mean square of factor A;
MSR is the mean square of the errors.

On the other hand, the ANOVA statistic for factor B is given by:

$F_{B} = \frac{\frac{{SSF}_{B}}{(b - 1)}}{\frac{RSS}{(n - 1) \cdot ab}} = \frac{{MSF}_{B}}{MSR}$ $F_{B} = \frac{\frac{{SSF}_{B}}{(b - 1)}}{\frac{RSS}{(n - 1) \cdot ab}} = \frac{{MSF}_{B}}{MSR}$

si124_e (9.48)

where:

MSF_B is the mean square of factor B.

And the ANOVA statistic for the interaction is represented by:

$F_{AB} = \frac{\frac{{SSF}_{AB}}{(a - 1) \cdot (b - 1)}}{\frac{RSS}{(n - 1) \cdot ab}} = \frac{{MSF}_{AB}}{MSR}$ $F_{AB} = \frac{\frac{{SSF}_{AB}}{(a - 1) \cdot (b - 1)}}{\frac{RSS}{(n - 1) \cdot ab}} = \frac{{MSF}_{AB}}{MSR}$

si125_e (9.49)

where:

MSF_AB is the mean square of the interaction.

The calculations of the two-way ANOVA are summarized in Table 9.5.

Table 9.5

Calculations of the Two-Way ANOVA
Source of Variation	Sum of Squares	Degrees of Freedom	Mean Squares	F
Factor A	${SSF}_{A} = b \cdot n \cdot \sum_{i = 1}^{a} {({\bar{Y}}_{i} - \bar{Y})}^{2}$ ${SSF}_{A} = b \cdot n \cdot \sum_{i = 1}^{a} {({\bar{Y}}_{i} - \bar{Y})}^{2}$	a − 1	${MSF}_{A} = \frac{{SSF}_{A}}{(a - 1)}$ ${MSF}_{A} = \frac{{SSF}_{A}}{(a - 1)}$	$F_{A} = \frac{{MSF}_{A}}{MSR}$ $F_{A} = \frac{{MSF}_{A}}{MSR}$
Factor B	${SSF}_{B} = a \cdot n \cdot \sum_{j = 1}^{b} {({\bar{Y}}_{j} - \bar{Y})}^{2}$ ${SSF}_{B} = a \cdot n \cdot \sum_{j = 1}^{b} {({\bar{Y}}_{j} - \bar{Y})}^{2}$	b − 1	${MSF}_{B} = \frac{{SSF}_{B}}{(b - 1)}$ ${MSF}_{B} = \frac{{SSF}_{B}}{(b - 1)}$	$F_{B} = \frac{{MSF}_{B}}{MSR}$ $F_{B} = \frac{{MSF}_{B}}{MSR}$
Interaction	${SSF}_{AB} = n \cdot \sum_{i = 1}^{a} \sum_{j = 1}^{b} {({\bar{Y}}_{ij} - {\bar{Y}}_{i} - {\bar{Y}}_{j} + \bar{Y})}^{2}$ ${SSF}_{AB} = n \cdot \sum_{i = 1}^{a} \sum_{j = 1}^{b} {({\bar{Y}}_{ij} - {\bar{Y}}_{i} - {\bar{Y}}_{j} + \bar{Y})}^{2}$	(a − 1). (b − 1)	${MSF}_{AB} = \frac{{SSF}_{AB}}{(a - 1) \cdot (b - 1)}$ ${MSF}_{AB} = \frac{{SSF}_{AB}}{(a - 1) \cdot (b - 1)}$	$F_{AB} = \frac{{MSF}_{AB}}{MSR}$ $F_{AB} = \frac{{MSF}_{AB}}{MSR}$
Error	$RSS = \sum_{i = 1}^{a} \sum_{j = 1}^{b} \sum_{k = 1}^{n} {(Y_{ijk} - {\bar{Y}}_{ij})}^{2}$ $RSS = \sum_{i = 1}^{a} \sum_{j = 1}^{b} \sum_{k = 1}^{n} {(Y_{ijk} - {\bar{Y}}_{ij})}^{2}$	(n − 1) ⋅ ab	$MSR = \frac{RSS}{(n - 1) \cdot ab}$ $MSR = \frac{RSS}{(n - 1) \cdot ab}$
Total	$TSS = \sum_{i = 1}^{a} \sum_{j = 1}^{b} \sum_{k = 1}^{n} {(Y_{ijk} - \bar{Y})}^{2}$ $TSS = \sum_{i = 1}^{a} \sum_{j = 1}^{b} \sum_{k = 1}^{n} {(Y_{ijk} - \bar{Y})}^{2}$	N − 1

Table 9.5

The calculated values of the statistics (F_A^cal, F_B^cal, and F_AB^cal) must be compared to the critical values obtained from the F-distribution table (Table A in the Appendix): F_A^c = F_{a − 1, (n − 1)ab, α}, F_B^c = F_{b − 1, (n − 1)ab, α}, and F_AB^c = F_{(a − 1)(b − 1), (n − 1)ab, α}. For each statistic, if the value lies in the critical region (F_A^cal > F_A^c, F_B^cal > F_B^c, F_AB^cal > F_AB^c), we must reject the null hypothesis. Otherwise, we do not reject H₀.

Example 9.13

Using the Two-Way ANOVA

A sample with 24 passengers who travel from Sao Paulo to Campinas in a certain week is collected. The following variables are analyzed (1) travel time in minutes, (2) the bus company chosen, and (3) the day of the week. The main objective is to verify if there is a relationship between the travel time and the bus company, between the travel time and the day of the week, and between the bus company and the day of the week. The levels considered in the variable bus company are Company A (1), Company B (2), and Company C (3). On the other hand, the levels regarding the day of the week are Monday (1), Tuesday (2), Wednesday (3), Thursday (4), Friday (5), Saturday (6), and Sunday (7). The results of the sample are shown in Table 9.E.17 and are available in the file Two_Way_ANOVA.sav as well. Test these hypotheses, considering a 5% significance level.

Table 9.E.17

Data From Example 9.13 (Using the Two-Way ANOVA)
Time (Min)	Company	Day of the Week
90	2	4
100	1	5
72	1	6
76	3	1
85	2	2
95	1	5
79	3	1
100	2	4
70	1	7
80	3	1
85	2	3
90	1	5
77	2	7
80	1	2
85	3	4
74	2	7
72	3	6
92	1	5
84	2	4
80	1	3
79	2	1
70	3	6
88	3	5
84	2	4

9.8.2.1.1 Solving the Two-Way ANOVA Test by Using SPSS Software

The use of the images in this section has been authorized by the International Business Machines Corporation©.

Step 1: In this case, the most suitable test is the two-way ANOVA.

First, we must verify if there is normality in the variable Time (metric) in the model (as shown in Fig. 9.46). According to this figure, we can conclude that variable Time follows a normal distribution, with a 95% confidence level. The hypothesis of variance homogeneity will be verified in Step 4.

Step 2: The null hypothesis H₀ of the two-way ANOVA for this example assumes that the population means of each level of the factor Company and of each level of the factor Day_of_the_week are equal, that is, H₀^A : μ₁ = μ₂ = μ₃ and H₀^B : μ₁ = μ₂ = … = μ₇.

The null hypothesis H₀ also states that there is no interaction between the factor Company and the factor Day_of_the_week, that is, H₀: γ_ij = 0 for i ≠ j.

Step 3: The significance level to be considered is 5%.

Step 4: The F statistics in ANOVA for the factor Company, for the factor Day_of_the_week, and for the interaction Company ⁎ Day_of_the_week will be obtained through the SPSS software, according to the procedure specified below.

In order to do that, let´s click on Analyze → General Linear Model → Univariate …, as shown in Fig. 9.47.

After that, let´s include the variable Time in the box of dependent variables (Dependent Variable) and the variables Company and Day_of_the_week in the box of Fixed Factor(s), as shown in Fig. 9.48.

Fig. 9.48 Selection of the variables to elaborate the two-way ANOVA.

This example is based on the one-way ANOVA, in which the factors are fixed. If one of the factors were chosen randomly, it would be inserted into the box Random Factor(s), resulting in a case of a three-way ANOVA. The button Model … defines the variance analysis model to be tested. Through the button Contrasts …, we can assess if the category of one of the factors is significantly different from the other categories of the same factor. Charts can be constructed through the button Plots …, thus allowing the visualization of the existence or nonexistence of interactions between the factors. Button Post Hoc …, on the other hand, allows us to compare multiple means. Finally, from the button Options …, we can obtain descriptive statistics and the result of Levene’s variance homogeneity test, as well as select the appropriate significance level (Fávero et al., 2009; Maroco, 2014).

Therefore, since we want to test variance homogeneity, we must select, in Options …, the option Homogeneity tests, as shown in Fig. 9.49.

Finally, let’s click on Continue and on OK to obtain Levene’s variance homogeneity test and the two-way ANOVA table.

In Fig. 9.50, we can see that the variances between groups are homogeneous (P = 0.451 > 0.05).

Based on Fig. 9.51, we can conclude that there are no significant differences between the travel times of the companies analyzed, that is, the factor Company does not have a significant impact on the variable Time (P = 0.330 > 0.05).

On the other hand, we conclude that there are significant differences between the days of the week, that is, the factor Day_of_the_week has a significant effect on the variable Time (P = 0.003 < 0.05).

We finally conclude that there is no significant interaction, with a 95% confidence level, between the two factors Company and Day_of_the_week, since P = 0.898 > 0.05.

9.8.2.1.2 Solving the Two-Way ANOVA Test by Using Stata Software

The use of the images in this section has been authorized by StataCorp LP©.

The command anova on Stata specifies the dependent variable being analyzed, as well as the respective factors. The interactions are specified using the character # between the factors. Thus, the two-way ANOVA is generated through the following syntax:

anova variable_Y⁎ factor_A⁎ factor_B⁎ factor_A#factor_B

or simply:

anova variable_y⁎ factor_A⁎## factor_B⁎

in which the term variable_y⁎ should be substituted for the quantitative dependent variable and the terms factor_A⁎ and factor_B⁎ for the respective factors.

If we type the syntax anova variable_Y⁎ factor_A⁎ factor_B⁎, only the ANOVA for each factor will be elaborated, and not between the factors.

The data presented in Example 9.13 are available in the file Two_Way_ANOVA.dta. The quantitative dependent variable is called time and the factors correspond to the variables company and day_of_the_week. Thus, we must type the following command:

anova time company##day_of_the_week

The results can be seen in Fig. 9.52 and are similar to those presented on SPSS, which allows us to conclude, with a 95% confidence level, that only the factor day_of_the_week has a significant effect on the variable time (P = 0.003 < 0.05), and that there is no significant interaction between the two factors analyzed (P = 0.898 > 0.05).

9.8.2.2 ANOVA With More Than Two Factors

The two-way ANOVA can be generalized to three or more factors. According to Maroco (2014), the model becomes very complex, since the effect of multiple interactions can make the effect of the factors a bit confusing. The generic model with three factors presented by the author is:

$Y_{ijkl} = μ + α_{i} + β_{j} + γ_{k} + {αβ}_{ij} + {αγ}_{ik} + {βγ}_{jk} + {αβγ}_{ijk} + ɛ_{ijkl}$ $Y_{ijkl} = μ + α_{i} + β_{j} + γ_{k} + {αβ}_{ij} + {αγ}_{ik} + {βγ}_{jk} + {αβγ}_{ijk} + ɛ_{ijkl}$

(9.50)

9.9 Final Remarks

This chapter presented the concepts and objectives of parametric hypotheses tests and the general procedures for constructing each one of them.

We studied the main types of tests and the situations in which each one of them must be used. Moreover, the advantages and disadvantages of each test were established, as well as their assumptions.

We studied the tests for normality (Kolmogorov-Smirnov, Shapiro-Wilk, and Shapiro-Francia), variance homogeneity tests (Bartlett’s χ², Cochran’s C, Hartley’s F_max, and Levene’s F), Student’s t-test for one population mean, for two independent means, and for two paired means, as well as ANOVA and its extensions.

Regardless of the application’s main goal, parametric tests can provide good and interesting research results that will be useful in the decision-making process. From a conscious choice of the modeling software, the correct use of each test must always be made based on the underlying theory, without ever ignoring the researcher’s experience and intuition.

9.10 Exercises

(1) In what situations should parametric tests be applied and what are the assumptions of these tests?

(2) What are the advantages and disadvantages of parametric tests?

(3) What are the main parametric tests to verify the normality of the data? In what situations must we use each oneof them?

(4) What are the main parametric tests to verify the variance homogeneity between groups? In what situations must we use each one of them?

(5) To test a single population mean, we can use z-test and Student’s t-test. In what cases must each one of them be applied?

(6) What are the main mean comparison tests? What are the assumptions of each test?

(7) The monthly aircraft sales data throughout last year can be seen in the table below. Check and see if there is normality in the data. Consider α = 5%.

Jan.	Feb.	Mar.	Apr.	May	Jun.	Jul.	Aug.	Sept.	Oct.	Nov.	Dec.
48	52	50	49	47	50	51	54	39	56	52	55

Unlabelled Table

(8) Test the normality of the temperature data listed (α = 5%):

12.5	14.2	13.4	14.6	12.7	10.9	16.5	14.7	11.2	10.9	12.1	12.8
13.8	13.5	13.2	14.1	15.5	16.2	10.8	14.3	12.8	12.4	11.4	16.2
14.3	14.8	14.6	13.7	13.5	10.8	10.4	11.5	11.9	11.3	14.2	11.2
13.4	16.1	13.5	17.5	16.2	15.0	14.2	13.2	12.4	13.4	12.7	11.2

Unlabelled Table

(9) The table shows the final grades of two students in nine subjects. Check and see if there is variance homogeneity between the students (α = 5%).

Student 1	6.4	5.8	6.9	5.4	7.3	8.2	6.1	5.5	6.0
Student 2	6.5	7.0	7.5	6.5	8.1	9.0	7.5	6.5	6.8

Unlabelled Table

(10) A fat-free yogurt manufacturer states that the number of calories in each cup is 60 cal. In order to check if this information is true, a random sample with 36 cups is collected; and we observed that the average number of calories was 65 cal. with a standard deviation of 3.5. Apply the appropriate test and check if the manufacturer’s statement is true, considering a significance level of 5%.

(11) We would like to compare the average waiting time before being seen by a doctor (in minutes) in two hospitals. In order to do that, we collected a sample with 20 patients from each hospital. The data are available in the tables. Check and see if there are differences between the average waiting times in both hospitals. Consider α = 1%.

Hospital 1

72	58	91	88	70	76	98	101	65	73
79	82	80	91	93	88	97	83	71	74

Unlabelled Table

Hospital 2

66	40	55	70	76	61	53	50	47	61
52	48	60	72	57	70	66	55	46	51

Unlabelled Table

(12) Thirty teenagers whose total cholesterol level is higher than what is advisable underwent treatment that consisted of a diet and physical activities. The tables show the levels of LDL cholesterol (mg/dL) before and after the treatment. Check if the treatment was effective (α = 5%).

Before the treatment

220	212	227	234	204	209	211	245	237	250
208	224	220	218	208	205	227	207	222	213
210	234	240	227	229	224	204	210	215	228

Unlabelled Table

After the treatment

195	180	200	204	180	195	200	210	205	211
175	198	195	200	190	200	222	198	201	194
190	204	230	222	209	198	195	190	201	210

Unlabelled Table

(13) An aerospace company produces civilian and military helicopters at its three factories. The tables show the monthly production of helicopters in the last 12 months at each factory. Check if there is a difference between the population means. Consider α = 5%.

Factory 1

Unlabelled Table

Factory 2

Unlabelled Table

Factory 3

Unlabelled Table

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Table of Contents for 9.9 Final Remarks

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9.8.2 Factorial ANOVA

9.8.2.1 Two-Way ANOVA

9.8.2.1.1 Solving the Two-Way ANOVA Test by Using SPSS Software

9.8.2.1.2 Solving the Two-Way ANOVA Test by Using Stata Software

9.8.2.2 ANOVA With More Than Two Factors

9.9 Final Remarks

9.10 Exercises

Table of Contents for
9.9 Final Remarks