15.7B: Dental Caries
- Explain how dental caries can be prevented
Dental caries, also known as tooth decay or a cavity, is an infection, usually bacterial in origin, that causes demineralization of the hard tissues (enamel, dentin, and cementum ) and destruction of the organic matter of the tooth, usually by production of acid by hydrolysis of the food debris accumulated on the tooth surface. If demineralization exceeds saliva and other remineralization factors such as from calcium and fluoridated toothpastes, these tissues progressively break down, producing dental caries (cavities, holes in the teeth). The two bacteria most commonly responsible for dental cavities are Streptococcus mutans and Lactobacillus. If left untreated, the disease can lead to pain, tooth loss, and infection. Today, caries remain one of the most common diseases throughout the world.
Caries can be classified by location, etiology, rate of progression, and affected hard tissues. These forms of classification can be used to characterize a particular case of tooth decay in order to more accurately represent the condition to others and also indicate the severity of tooth destruction.
Tooth decay disease is caused by specific types of bacteria that produce acid in the presence of fermentable carbohydrates such as sucrose, fructose, and glucose. The mineral content of teeth is sensitive to increases in acidity from the production of lactic acid. To be specific, a tooth (which is primarily mineral in content) is in a constant state of back-and-forth demineralization and remineralization between the tooth and surrounding saliva. For people with little saliva, especially due to radiation therapies that may destroy the salivary glands, there also exists remineralization gel. These patients are particularly susceptible to dental caries. When the pH at the surface of the tooth drops below 5.5, demineralization proceeds faster than remineralization (meaning that there is a net loss of mineral structure on the tooth’s surface). Most foods are in this acidic range and without remineralization result in the ensuing decay.
As the enamel and dentin are destroyed, the cavity becomes more noticeable. The affected areas of the tooth change color and become soft to the touch. Once the decay passes through enamel, the dentinal tubules, which have passages to the nerve of the tooth, become exposed, causing a toothache. The pain may worsen with exposure to heat, cold, or sweet foods and drinks. Dental caries can also cause bad breath and foul tastes. In highly progressed cases, infection can spread from the tooth to the surrounding soft tissues. Complications such as cavernous sinus thrombosis and Ludwig’s angina can be life-threatening.
There are four main criteria required for caries formation: a tooth surface (enamel or dentin) caries-causing bacteria, fermentable carbohydrates (such as sucrose), and time. The caries process does not have an inevitable outcome, and different individuals will be susceptible to different degrees depending on the shape of their teeth, oral hygiene habits, and the buffering capacity of their saliva. Dental caries can occur on any surface of a tooth that is exposed to the oral cavity, but not the structures that are retained within the bone. All caries occur from acid demineralization that exceeds saliva and fluoride remineralization, and almost all acid demineralization occurs where food (containing carbohydrate like sugar) is left on teeth. Though most trapped food is left between teeth, over 80% of cavities occur inside pits and fissures on chewing surfaces where brushing, fluoride, and saliva cannot reach to remineralize the tooth as they do on easy-to-reach surfaces that develop few cavities.
In most people, disorders or diseases affecting teeth are not the primary cause of dental caries. Ninety-six percent of tooth enamel is composed of minerals. These minerals, especially hydroxyapatite, will become soluble when exposed to acidic environments. Enamel begins to demineralize at a pH of 5.5. Dentin and cementum are more susceptible to caries than enamel because they have lower mineral content. Thus, when root surfaces of teeth are exposed from gingival recession or periodontal disease, caries can develop more readily. Even in a healthy oral environment, however, the tooth is susceptible to dental caries.
Bacteria in a person’s mouth convert glucose, fructose, and most commonly sucrose (table sugar) into acids such as lactic acid through a glycolytic process called fermentation. If left in contact with the tooth, these acids may cause demineralization, which is the dissolution of its mineral content. The process is dynamic, however, as remineralization can also occur if the acid is neutralized by saliva or mouthwash.
At times, pit and fissure caries may be difficult to detect. Bacteria can penetrate the enamel to reach dentin, but then the outer surface may remineralize, especially if fluoride is present. These caries, sometimes referred to as “hidden caries”, will still be visible on x-ray radiographs, but visual examination of the tooth would show the enamel intact or minimally perforated.
Key Points
- Dentin is produced continuously throughout life by odontoblasts.
- Caries can be classified by location, etiology, rate of progression, and affected hard tissues.
- Enamel is a highly mineralized acellular tissue. Enamel rods, which are the basic unit of the enamel structure, run perpendicularly from the surface of the tooth to the dentin.
- There are four main criteria required for caries formation: a tooth surface (enamel or dentin), caries-causing bacteria, fermentable carbohydrates (such as sucrose), and time.
- Dental caries can occur on any surface of a tooth that is exposed to the oral cavity, but not the structures that are retained within the bone.
- The two bacteria most commonly responsible for dental cavities are Streptococcus mutans and Lactobacillus.
- The use of dental sealants is a means of prevention.
- Professional hygiene care consists of regular dental examinations and cleanings.
- An extraction can also serve as treatment for dental caries.
- In general, early treatment is less painful and less expensive than treatment of extensive decay.
- Recurrent caries, also described as secondary, are caries that appears at a location with a previous history of caries.
- Dentin is produced continuously throughout life by odontoblasts.
- Personal hygiene care consists of proper brushing and flossing daily. The purpose of oral hygiene is to minimize any etiologic agents of disease in the mouth.
- Primary diagnosis involves inspection of all visible tooth surfaces using a good light source, dental mirror and explorer. Dental radiographs (X-rays) may show dental caries before it is otherwise visible, in particular caries between the teeth.
- Caries can be classified by location, etiology, rate of progression, and affected hard tissues.
- In dentin from the deepest layer to the enamel, the distinct areas affected by caries are the advancing front, the zone of bacterial penetration, and the zone of destruction.
- Intrauterine and neonatal lead exposure promote tooth decay. Other divalent cations, such as cadmium, mimic the calcium ion and therefore exposure may promote tooth decay.
- From the deepest layer of the enamel to the enamel surface, the identified areas are the: translucent zone, dark zones, body of the lesion, and surface zone.
- Enamel is a highly mineralized acellular tissue. Enamel rods, which are the basic unit of the enamel structure, run perpendicularly from the surface of the tooth to the dentin.
- Some brands of smokeless tobacco contain high sugar content, increasing susceptibility to caries. Tobacco use is a significant risk factor for periodontal disease, which can cause the gingiva to recede.
- Reduced saliva is associated with increased caries.
- There are four main criteria required for caries formation: a tooth surface (enamel or dentin); caries-causing bacteria; fermentable carbohydrates (such as sucrose); and time.
- If demineralization exceeds saliva and other remineralization factors such as from calcium and fluoridated toothpastes, these tissues progressively break down, producing dental caries (cavities, holes in the teeth).
- The presentation of caries is highly variable.
- Bacteria collect around the teeth and gums in a sticky, creamy-coloured mass called plaque, which serves as a biofilm.
- Dental caries can occur on any surface of a tooth that is exposed to the oral cavity, but not the structures that are retained within the bone.
- The mineral content of teeth is sensitive to increases in acidity from the production of lactic acid.
- Radiographs are used for less visible areas of teeth and to judge the extent of destruction.
- Depending on the extent of tooth destruction, various treatments can be used to restore teeth to proper form, function, and aesthetics, but there is no known method to regenerate large amounts of tooth structure. Instead, dental health organizations advocate preventive and prophylactic measures, such as regular oral hygiene and dietary modifications, to avoid dental caries.
- Tooth decay disease is caused by specific types of bacteria that produce acid in the presence of fermentable carbohydrates such as sucrose, fructose, and glucose.
- Before the cavity forms, the process is reversible, but once a cavity forms, the lost tooth structure cannot be regenerated.
- The two bacteria most commonly responsible for dental cavities are Streptococcus mutans and Lactobacillus.
Key Terms
- cementum : A bony substance that covers the root of a tooth; cement.
- enamel : The hard covering on the exposed part of a tooth.
- dentin : The hard, dense calcareous material that makes up the bulk of a tooth


4.2: Algebraic Expressions
Algebraic Expressions
An algebraic expression is a number, a letter, or a collection of numbers and letters along with meaningful signs of operation.
Algebraic expressions are often referred to simply as expressions , as in the following examples:
\(x + 4\) is an expression
\(7y\) is an expression
\(\dfrac{x-3x^2y}{7+9x}\) is an expression.
The number \(8\) is an expression. \(8\) can be written with explicit signs of operation by writing it as \(8 + 0\) or \(8 \cdot 1\).
\(3x^2 + 6 = 4x - 1\) is not an expression, it is an equation . We will study equations in the next section.
Terms and Factors
In an algebraic expression, the quantities joined by "\(+\)" signs are called terms.
In some expressions it will appear that terms are joined by "\(-\)" signs. We must keep in mind that subtraction is addition of the negative, that is \(a - b = a + (-b)\).
An important concept that all students of algebra must be aware of is the difference between terms and factors .
Any numbers or symbols that are multiplied together are factors of their product.
Terms are parts of
sums
and are therefore joined by addition (or subtraction) signs.
Factors are parts of
products
and are therefore joined by multiplication signs.
Sample Set A
Identify the terms in the following expressions.
\(3x^4 + 6x^2 + 5x + 8\)
The expression has four terms: \(3x^4, 6x^2, 5x, 8\).
\(15y^8\)
In this expression there is only one term. The term is \(15y^8\).
\(14x^5y + (a+3)^2\).
In this expression there are two terms: the terms are \(14x^5y\) and \((a+3)^2\). Notice that the term \((a+3)^2\) is itself composed of two like factors, each of which is composed of the two terms, \(a\) and \(3\).
\(m^3 - 3\)
Using our definition of subtraction, this expression can be written in the form \(m^3+(−3)\). Now we can see that the terms are \(m^3\) and \(−3\).
Rather than rewriting the expression when a subtraction occurs, we can identify terms more quickly by associating the \(+\) or \(−\) sign with the individual quantity.
\(p^4-7p^3-2p-11\).
Associating the sign with the individual quantities we see that the terms of this expression are \(p^4\), \(−7p^3\), \(−2p\), and \(−11\).
Practice Set A
Let’s say it again. The difference between terms and factors is that terms are joined by signs and factors are joined by signs.
- Answer
-
addition, multiplication
\(4x^2 - 8x + 7\)
- Answer
-
\(4x^2, -8x, 7\)
\(2xy + 6x^2 + (x-y)^4\)
- Answer
-
\(2xy, 6x^2, (x-y)^4\)
\(5x^2 + 3x - 3xy^7 + (x-y)(x^3-6)\)
- Answer
-
\(5x^2, 3x, -3xy^7, (x-y)(x^3-6)\)
Sample Set B
Identify the factors in each term.
\(9a^2 - 6a - 12\) contains three terms. Some of the factors in each term are
first term: \(9\) and \(a^2\), or \(9\) and \(a\) and \(a\)
second term: \(-6\) and \(a\)
third term: \(−12\) and \(1\), or, \(12\) and \(−1\)
\(14x^5y + (a+3)^2\) contains two terms. Some of the factors of these terms are
first term: \(14, x^5, y\)
second term: \((a+3)\) and \((a+3)\)
Practice Set B
In the expression \(8x^2 - 5x + 6\), list the factors of the:
first term:
second term:
third term:
- Answer
-
\(8, x, x\)
\(-5, x\)
\(6 \text{ and } 1\) or \(3 \text { and } 2\)
In the expression \(10 + 2(b + 6)(b-18)^2\), list the factors of the:
first term:
second term:
third term:
- Answer
-
\(10 \text{ and } 1\) or \(5 \text{ and } 2\)
\(2, b+6, b−18, b−18\)
Common Factors
Common Factors
Sometimes, when we observe an expression carefully, we will notice that some particular factor appears in every term. When we observe this, we say we are observing common factors . We use the phrase common factors since the particular factor we observe is common to all the terms in the expression. The factor appears in each and every term in the expression.
Sample Set C
Name the common factors in each expression.
\(5x^3 - 7x^3 + 14x^3\)
The factor x^3 appears in each and every term. The expression x^3 is a common factor.
\(4x^2 + 7x\)
The factor \(x\) appears in each term. The term \(4x^2\) is actually \(4xx\). Thus, \(x\) is a common factor.
\(12xy^2 - 9xy + 15\)
The only factor common to all three terms is the number 3. (Notice that \(12=3\cdot4, 9=3\cdot3, 15=3\cdot5\).
\(3(x+5) - 8(x+5)\).
The factor \((x+5)\) appears in each term. So, \((x+5)\) is a common factor.
\(45x^3(x-7)^2 + 15x^2(x-7) - 20x^2(x-7)^5\).
The number \(5\), the \(x^2\), and the \((x-7)\) appear in each term. Also, \(5x^2(x-7)\) is also a factor (since each of the individual quantities is joined by a multiplication sign). Thus, \(5x^2(x-7)\) is a common factor.
\(10x^2+9x-4\)
There is no factor that appears in each and every term. Hence, there are no common factors in this expression.
Practice Set C
List, if any appear, the common factors in the following expressions.
\(x^2 + 5x^2 - 9x^2\)
- Answer
-
\(x^2\)
\(4x^2 - 8x^3 + 16x^4 - 24x^5\)
- Answer
-
\(4x^2\)
\(4(a+1)^3 + 10(a+1)\)
- Answer
-
\(2(a+1)\)
\(9ab(a-8) - 15a(a-8)^2\)
- Answer
-
\(3a(a-8)\)
\(14a^2b^2c(c-7)(2c+5) + 28c(2c+5)\)
- Answer
-
\(14c(2c+5)\)
\(6(x^2-y^2) + 19x(x^2+y^2)\)
- Answer
-
No common factor.
Coefficients
Coefficient
In algebra, as we now know, a letter is often used to represent some quantity. Suppose we represent some quantity by the letter \(x\). The notation \(5x\) means \(x+x+x+x+x\). We can now see that we have five of these quantities. In the expression \(5x\), the number \(5\) is called the numerical coefficient of the quantity \(x\). Often, the numerical coefficient is just called the coefficient. The coefficient of a quantity records how many of that quantity there are.
Sample Set D
\(12x\) means that there are \(12x\)'s.
\(4ab\) means there are four \(ab\)'s.
\(10(x-3)\) means there are ten \(x-3)\)'s.
\(1y\) means there is one \(y\). We usually write just \(y\) rather than \(1y\) since it is clear just be looking that there is only one \(y\).
\(7a^3\) means there are seven \(a^3\)'s.
\(5ax\) means there are five \(ax\)'s. It could also mean there are \(5ax\)'s. This example shows us that it is important for us to be very clear as to which quantity we are working with. When we see the expression 5ax we must ask ourselves "Are we working with the quantity \(ax\) or the quantity \(x\)?".
\(6x^2y^9\) means there are six \(x^2y^9\)'s. It could also mean there are \(6x^2y^9\)'s. It could even mean there are \(6y^9x^2\)'s.
\(5x^3(y-7)\) means there are five \(x^3(y-7)\)'s. It could also mean there are \(5x^3(x-7)\)'s. It could also mean there are \(5(x-7)x^3\)'s.
Practice Set D
What does the coefficient of a quantity tell us?
The Difference Between Coefficients and Exponents It is important to keep in mind the difference between coefficients and exponents.
Coefficients record the number of like terms in an algebraic expression. \(\underbrace{x+x+x+x}_{\text {4 terms }}=\underbrace{4 x}_{\text {coefficient is } 4}\)
Exponents record the number of like factors in a term. \(\underbrace{x \cdot x \cdot x \cdot x}_{\text {4 factors }}=\underbrace{x^{4}}_{\text { exponent is } 4}\)
In a term, the coefficient of a particular group of factors is the remaining group of factors.
- Answer
-
How many of that quantity there are.
Sample Set E
\(3x\)
The coefficient of \(x\) is \(3\).
\(6a^3\)
The coefficient of \(a^3\) is \(6\).
\(9(4-a)\)
The coefficient of \((4-a)\) is \(9\).
\(\dfrac{3}{8}xy^4\).
The coefficient of \(xy^4\) is \(\dfrac{3}{8}\)
\(3x^2y\).
The coefficient of \(x^2y\) is \(3\); the coefficient of \(y\) is \(3x^2\); and the coefficient of \(3\) is \(x^2y\).
\(4(x+y)^2\).
The coefficient of \((x+y)^2\) is \(4\); the coefficient of \(4\) is \((x+y)^2\); and the coefficient of \((x+y)\) is \(4(x+y)\) since \(4(x+y)^2\) can be written as \(4(x+y)(x+y)\).
Practice Set E
Determine the coefficients.
In the term \(6x^3\), the coefficient of:
(a) \(x^3\) is.
(b) \(6\) is.
- Answer
-
(a) \(6\)
(b) \(x^3\)
In the term \(3x(y-1)\), the coefficient of
(a) \(x(y-1)\) is.
(b) \((y-1)\) is.
(c) \(3(y-1)\) is.
(d) \(x\) is.
(e) \(3\) is .
(f) The numerical coefficient is .
- Answer
-
(a) \(3\)
(b) \(3x\)
(c) \(x\)
(d) \(3(y-1)\)
(e) \(x(y-1)\)
(f) \(3\)
In the term \(10ab^4\), the coefficient of
(a) \(ab^4\) is .
(b) \(b^4\) is .
(c) \(a\) is .
(d) \(10\) is .
(e) \(10ab^3\) is .
- Answer
-
(a) \(10\)
(b) \(10a\)
(c) \(10b^4\)
(d) \(ab^4\)
(e) \(b\)
Exercises
For the expressions in the following problems, write the number of terms that appear and then list the terms.
What is an algebraic expression?
- Answer
-
An algebraic expression is a number, a letter, or a collection of numbers and letters along with meaningful signs of operation.
Why is the number 14 considered to be an expression?
Why is the number \(x\) considered to be an expression?
- Answer
-
\(x\) is an expression because it is a letter (see the definition).
For the following problems, list, if any should appear, the common factors in the expressions.
\(2x+1\)
\(6x-10\)
- Answer
-
two
\(6x\), \(-10\)
\(2x^3+x-15\)
\(5x^2+6x-2\)
- Answer
-
three
\(5x^2, 6x, -2\)
\(3x\)
\(5cz\)
- Answer
-
one
\(5cz\)
\(2\)
\(61\)
- Answer
-
one
\(61\)
\(x\)
\(4y^3\)
- Answer
-
one
\(4y^3\)
\(17ab^2\)
\(a+1\)
- Answer
-
two
\(a, 1\)
\((a+1)\)
\(2x + x + 7\)
- Answer
-
three
\(2x, x, 7\)
\(2x + (x+7)\)
\((a+1) + (a-1)\)
- Answer
-
two
\((a+b), (a-1)\)
\(a + 1 + (a-1)\)
For the following problems, list, if any should appear, the common factors in the expressions.
\(x^2 + 5x^2 - 2x^2\)
- Answer
-
\(x^2\)
\(11y^3 - 33y^3\)
\(45ab^2 + 9b^2\)
- Answer
-
\(9b^2\)
\(6x^2y^3 + 18x^2\)
\(2(a+b) - 3(a+b)\)
- Answer
-
\((a+b)\)
\(14ab^2c^2(c+8) + 12ab^2c^2\)
- Answer
-
\(2ab^2c^2\)
\(4x^2y + 5a^2b\)
\(9a(a-3)^2 + 10b(a-3)\)
- Answer
-
\((a-3)\)
\(15x^2 - 30xy^2\)
\(12a^3b^2c - 7(b+1)(c-a)\)
- Answer
-
no common factors
\(0.06ab^2 + 0.03a\)
\(5.2(a+7)^2+17.1(a+7)\)
- Answer
-
\((a+7)\)
\(\dfrac{3}{4}x^2y^2z^2 + \dfrac{3}{8}x^2z^2\)
\(\dfrac{9}{16}(a^2-b^2) + \dfrac{9}{32}(b^2-a^2)\)
- Answer
-
\(\dfrac{9}{32}\)
For the following problems, note how many:
\(a\)'s in \(4a\)?
\(z\)'s in \(12z\)?
- Answer
-
\(12\)
\(x^2\)'s in \(5x^2\)?
\(y^3\)'s in \(6y^3\)?
- Answer
-
\(6\)
\(xy\)'s in \(9xy\)?
\(a^2b\)'s in \(10a^2b\)?
- Answer
-
\(10\)
\((a+1)\)'s in \(4(a+1)\)?
\((9+y)\)'s in \(8(9+y)\)?
- Answer
-
\(8\)
\(y^2\)'s in \(3x^3y^2\)?
\(12x\)'s in \(12x^2y^5\)?
- Answer
-
\(xy^5\)
\((a+5)\)'s in \(2(a+5)\)?
\((x-y)\)'s in \(5x(x-y)\)?
- Answer
-
\(5x\)
\((x+1)\)'s in \(8(x+1)\)?
\(2\)'s in \(2x^2(x-7)\)?
- Answer
-
\(x^2(x-7)\)
\(3(a+8)\)'s in \(6x^2(a+8)^3(a-8)\)?
For the following problems, a term will be given followed by a group of its factors. List the coefficient of the given group of factors.
\(7y; y\)
- Answer
-
\(7\)
\(10x; x\)
\(5a; 5\)
- Answer
-
\(a\)
\(12a^2b^3c^2r^7; a^2c^2r^7\)
\(6x^2b^2(c-1); c-1\)
- Answer
-
\(6x^2b^2\)
\(10x(x+7)^2; 10(x+7)\)
\(9a^2b^5; 3ab^3\)
- Answer
-
\(3ab\)
\(15x^4y^4(z+9a)^3; (z+9a)\)
\((-4)a^6b^2; ab\)
- Answer
-
\((-4)a^5b\)
\((-11a)(a+8)^3(a-1); (a+8)^2\)
Exercises for Review
Simplify \([\dfrac{2x^8(x-1)^5}{x^4(x-1)^2}]^4\)
- Answer
-
\(16x^{16}(x-1)^{12}\)
Supply the missing phrase. Absolute value speaks to the question of and not "which way."
Find the value of \(−[−6(−4−2)+7(−3+5)]\).
- Answer
-
\(-50\)
Find the value of \(\dfrac{2^5-4^2}{3^{-2}}\)
Express 0.0000152 using scientific notation.
- Answer
-
\(1.52 \times 10^{-5}\)


11.5: Applications
The Five-Step Method
When solving practical problems, it is often more convenient to introduce two variables rather than only one. Two variables should be introduced only when two relationships can be found within the problem. Each relationship will produce an equation, and a system of two equations in two variables will result.
We will use the five-step method to solve these problems.
- Introduce two variables, one for each unknown quantity.
- Look for two relationships within the problem. Translate the verbal phrases into mathematical expressions to form two equations.
- Solve the resulting system of equations.
- Check the solution.
- Write a conclusion.
Sample Set A (Number Problems)
The sum of two numbers is \(37\). One number is \(5\) larger than the other. What are the numbers?
Step 1:
Let \(x = \) smaller number.
Let \(y = \) larger number.
Step 2: There are two relationships.
a) The sum is \(37\)
\(x + y = 37\)
b) One is \(5\) larger than the other.
Step 3:
\(\left\{\begin{array}{r}
x + y = 37\\
y = x + 5
\end{array}\right.\)
We can easily solve this system by substitution. Substitute \(x + 5\) for \(y\) in equation 1.
\(\begin{aligned}
x + (x + 5) & = 37\\
x + x + 5 &= 37\\
2x + 5 &= 37\\
2x &= 32\\
x &= 16 & \text{ Then, } y = 16 + 5 = 21\\
x &= 16
\end{aligned}
Step 4: The sum if \(37\).
\(\begin{aligned}
x + y &= 37\\
16 + 21 &= 37 \text{ Is this correct? }\\
37 &= 37 \text{ Yes, this is correct. }
\end{aligned}\)
One is \(5\) larger than the other.
\(\begin{aligned}
y &= x + 5\\
21 &= 16 + 5\\
21 &= 21 \text{ Yes, this is correct. }
\end{aligned}\)
Step 5: The two numbers are \(16\) and \(21\).
Practice Set A
The difference of two numbers is 9, and the sum of the same two numbers is 19. What are the two numbers?
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
- Answer
-
The two numbers are 14 and 5.
Value And Rate Problems: Coin Problems Problems And Mixture Problems
The problems in Sample Sets B and C are value problems . They are referred to as value problems because one of the equations of the system used in solving them is generated by considering a value, or rate, or amount times a quantity .
Sample Set B (Coin Problems)
A parking meter contains 27 coins consisting only of dimes and quarters. If the meter contains $4.35, how many of each type of coin is there?
Step 1:
Let \(D = \) number of dimes.
Let \(Q = \) number of quarters.
Step 2: There are two relationships.
a) There are 27 coins. \(D + Q = 27\).
b) Contribution due to dimes =
Contribution due to quarters =
\(10D + 25Q = 435\)
Step 3:
\(\left\{\begin{array}{r}
D + Q = 27\\
10D + 25Q = 435
\end{array}\right.\)
We can solve this system using elimination by addition. Multiply both sides of equation (1) by \(-10\) and add.
\(\begin{array}{c}
-10 D-10 Q=-270 \\
10 D+25 Q=435 \\
\hline 15 Q=165 \\
Q=11
\end{array}\)
Then, \(D + 11 = 27\\
D = 16\)
\(D = 16, Q = 11\)
Step 4: 16 dimes and 11 quarters is 27 coins.
\(\begin{aligned}
10(16) + 11(25) &= 435 \text{ Is this correct? }\\
160 + 275 &= 435 \text{ Is this correct? }\\
435 &= 435 \text{ Yes, this is correct. }
\end{aligned}\)
The solution checks.
Step 5: There are 11 quarters and 16 dimes.
Practice Set B
A bag contains only nickels and dimes. The value of the collection is $2. If there are 26 coins in all, how many of each coin are there?
- Answer
-
There are 14 dimes and 12 nickels.
Sample Set C (Mixture Problems)
A chemistry student needs 40 milliliters (ml) of a 14% acid solution. She had two acid solutions, A and B, to mix together to form the 40 ml acid solution. Acid solution A is 10% acid and acid solution B is 20% acid. How much of each solution should be used?
Step 1:
Let \(x = \) number of ml is solution A.
Let \(y = \) number of ml of solution B.
Step 2: There are two relationships
a) The sum of the number of ml of the two solutions is 40.
\(x + y = 40\).
b) To determine the second equation, draw a picture of the situation
The equation follows directly from the drawing if we use the idea of amount times quantity.
Step 3: \(.10x + .20y = .14(40)\)
\(\left\{\begin{array}{r}
x + y = 40\\
.10x + .20y = .14(40)
\end{array}\right.\)
\(\left\{\begin{array}{r}
x + y = 40\\
.10x + .20y = .14(40)
\end{array}\right.\)
Solve this system by addition. First, eliminate decimals in equation 2 by multiplying both sides by 100.
\(\left\{\begin{array}{r}
x + y = 40\\
10x + 20y = 14(40)
\end{array}\right.\)
Eliminate \(x\) by multiplying equation 1 by \(-10\) and then adding.
\(\begin{array}{c}
-10 x-10 y=-400 \\
10 x+20 y=560 \\
\hline 10 y=160 \\
y=16
\end{array}\)
Then, \(x + 16 = 40\\
x = 24\)
\(x = 24, y = 16\)
Step 4: 24 ml and 16ml to add to 40ml.
\(\begin{aligned}
10(24) + 20(16) &= 560 \text{ Is this correct? }\\
240 + 320 &= 560 \text{ Is this correct? }\\
560 &= 560 \text{ Yes, this is correct. }
\end{aligned}\)
The solution checks.
Step 5: The student should use 24 ml of acid solution A and 16 ml of acid solution B.
Practice Set C
A chemistry student needs 60 ml of a 26% salt solution. He has two salt solutions, A and B, to mix together to form the 60 ml solution. Salt solution A is 30% salt and salt solution B is 20% salt. How much of each solution should be used?
- Answer
-
The student should use 36 ml of salt solution A and 24 ml of salt solution B.
Exercises
The sum of two numbers is 22. One number is 6 more than the other. What are the numbers?
- Answer
-
The two numbers are 14 and 8.
The sum of two numbers is 32. One number is 8 more than the other. What are the numbers?
The difference of two numbers is 12 and one number is three times as large as the other. Whatare the numbers?
- Answer
-
The two numbers are 18 and 6.
The difference of two numbers is 9 and one number is 10 times larger than the other. Whatare the numbers?
Half the sum of two numbers is 14 and half the difference is 2. What are the numbers?
- Answer
-
The numbers are 16 and 12.
One third of the sum of two numbers is 6 and one fifth of the difference is 2. What are the numbers?
A 14 pound mixture of grapes sells for $3.10. Type 1 grape sells for 25¢ a pound and type 2 grape sells for 20¢ a pound. How many pounds of each type of grape were used?
- Answer
-
6 pounds of Type 1 and 8 pounds of Type 2.
The cost of 80 liters of a blended cleaning solution is $28. Type 1 solution costs 20¢ a liter and type 2 solution costs 40¢ a liter. How many liters of each solution were used to form the blended solution?
The cost of 42 grams of a certain chemical compound is $14.40. Type 1 chemical costs 45¢ a gram and type 2 chemical costs 30¢ a gram. How many grams of each chemical were used to form the compound?
- Answer
-
12 grams of Type 1 and 30 grams of Type 2.
A play was attended by 342 people, some adults and some children. Admission for adults was $1.50 and for children 75¢. How many adults and how many children attended the play?
200 tickets were sold to a college’s annual musical performance. Tickets for students were $2.50 and for nonstudents $3.50. The total amount collected was $537. How many nonstudents purchased tickets for the performance?
- Answer
-
37 non-student tickets.
A chemistry student needs 22 ml of a 38% acid solution. She has two acid solutions, A and B, to mix together to form the solution. Acid solution A is 40% acid and acid solution B is 30% acid. How much of each solution should be used?
A chemistry student needs 50 ml of a 72% salt solution. He has two salt solutions, A and B, to mix together to form the solution. Salt solution A is 60% salt and salt solution B is 80% salt. How much of each solution should be used?
- Answer
-
30 ml of 80% solution; 20 ml of 60% solution.
A chemist needs 2 liters of an 18% acid solution. He has two solutions, A and B, to mix together to form the solution. Acid solution A is 10% acid and acid solution B is 15% acid. Can the chemist form the needed 18% acid solution? (Verify by calculation.) If the chemist locates a 20% acid solution, how much would have to be mixed with the 10% solution to obtain the needed 2-liter 18% solution?
A chemist needs 3 liters of a 12% acid solution. She has two acid solutions, A and B, to mix together to form the solution. Acid solution A is 14% acid and acid solution B is 20% acid. Can the chemist form the needed 12% solution? (Verify by calculation.) If the chemist locates a 4% acid solution, how much would have to be mixed with the 14% acid solution to obtain the needed 3-liter 12% solution?
- Answer
-
a) Cannot form 12% solution;
b) \(\dfrac{3}{5}\) liter of 4% solution.
A chemistry student needs 100 ml of a 16% acid solution. He has a bottle of 20% acid solution. How much pure water and how much of the 20% solution should be mixed to dilute the 20% acid solution to a 16% acid solution?
A chemistry student needs 1 liter of a 78% salt solution. She has a bottle of 80% salt solution. How much pure water and how much of the 80% salt solution should be mixed to dilute the 80% salt solution to a 78% salt solution?
- Answer
-
25 ml of pure water; 975 ml of 80% salt solution.
25 ml of pure water; 975 ml of 80% salt solution.
A child’s bank contains 78 coins. The coins are only pennies and nickels. If the value of the coins is $1.50, how many of each coin are there?
- Answer
-
18 nickels; 60 pennies.
Exercises For Review
Simplify \(\dfrac{x^{n + 6}}{x^2}\)
Find the product: \((3x - 5)^2\)
- Answer
-
\(9x^2 - 30x + 25\)
Find the difference: \(\dfrac{2x - 1}{x^2 + x - 6} - \dfrac{x + 2}{x^2 + 5x + 6}\)
Use substitution method to solve
\(\left\{\begin{array}{r}
4x - y = 3\\
2x - 5 = 15
\end{array}\right.\)
- Answer
-
\((0,−3)\)
Use the addition method to solve
\(\left\{\begin{array}{r}
3x + 4y = -7\\
4x - 5y = 1
\end{array}\right.\)


24.4: The Modern Period
Japanese Art in the Meiji Period
The art of the Meiji period (1868–1912) was marked by a division between European and traditional Japanese styles.
Explain how the conflict caused by Europeanization and modernization during the Meiji Period was reflected in the artwork of the time
Key Points
- The Meiji period (September 1868 through July 1912) represents the first half of the Empire of Japan, during which Japanese society moved from being an isolated feudalism to its modern form.
- During this period, western style painting (Yōga) was officially promoted by the government, which sent promising young artists abroad for studies and hired foreign artists to establish an art curriculum at Japanese schools.
- After an initial burst of western style art, there was a revival of appreciation for traditional Japanese styles (Nihonga) led by art critic Okakura Kakuzo and educator Ernest Fenollosa.
- In the 1880s, western style art was banned from official exhibitions and was severely criticized by critics. Supported by Okakura and Fenollosa, the Nihonga style evolved with influences from the European pre-Raphaelite movement and European romanticism.
- In 1907, with the establishment of the Bunten exhibitions, both competing groups—Yōga and Nihonga—found mutual recognition and co-existence and even began the process toward mutual synthesis.
Key Terms
- pre-Raphaelite movement : An art movement founded by a group of English painters, poets, and critics with the intention of reforming art by rejecting what they considered to be the mechanistic approach first adopted by the Mannerist artists who succeeded Raphael and Michelangelo.
- feudalism : A social system based on personal ownership of resources, personal fealty of a lord by a subject, and a hierarchical social structure reinforced by religion.
- Romanticism : An artistic, literary, and intellectual movement that originated in Europe toward the end of the 18th century and in most areas was at its peak from 1800 to 1840; partly a reaction to the Industrial Revolution, it was also a revolt against aristocratic social and political norms of the Age of Enlightenment and a reaction against the scientific rationalization of nature.
Overview: The Meiji Period
The Meiji period (明 Meiji-jidai) was an era in Japanese history that extended from September 1868 through July 1912. This period represents the first half of Japan’s time as an imperial power. Fundamental changes affected Japan’s social structure, internal politics, economy, military, and foreign relations. Japanese society moved from being an isolated feudalism to its modern form. In art, this period was marked by the division into competing European and traditional indigenous styles. In 1907, with the establishment of the Bunten exhibition under the aegis of the Ministry of Education, both competing groups found mutual recognition and co-existence and even began the process towards mutual synthesis.
The Yōga Style
During the Meiji period, Japan underwent a tremendous political and social change in the course of the Europeanization and modernization campaign organized by the Meiji government. Western style painting (Yōga) was officially promoted by the government, which sent promising young artists abroad for studies. The Yōga style painters formed the Meiji Bijutsukai (Meiji Fine Arts Society) to hold its own exhibitions and to promote a renewed interest in western art. Foreign artists were also hired to come to Japan to establish an art curriculum in Japanese schools. The Yōga style encompassed oil painting, watercolors, pastels, ink sketches, lithography, etching, and other techniques developed in western culture.
The Nihonga Style
After an initial burst of western style art, however, the pendulum swung in the opposite direction. Led by art critic Okakura Kakuzo and educator Ernest Fenollosa, there was a revival of appreciation for traditional Japanese styles (Nihonga). In the 1880s, western style art was banned from official exhibitions and was severely criticized by critics. Supported by Okakura and Fenollosa, the Nihonga style evolved with influences from the European pre-Raphaelite movement and European romanticism. Paintings of this style were made in accordance with traditional Japanese artistic conventions, techniques, and materials based on traditions over a thousand years old.
Japanese Art in the Showa Period
During the Shōwa period, Japan shifted toward totalitarianism until its defeat in World War II, when it led an economic and cultural recovery.
Create a timeline describing the upheaval, occupation, democratic reforms, and economic boom of the pre- and post-war Shōwa period
Key Points
- The Shōwa period in Japanese history corresponds to the reign of the Shōwa Emperor, Hirohito, from December 25, 1926 through January 7, 1989.
- Japanese painting in the pre-war Shōwa period was largely dominated by the work of Yasui Sōtarō (1888–1955) and Umehara Ryūzaburō (1888–1986).
- During World War II, government controls and censorship meant that only patriotic themes could be expressed, and many artists were recruited into the government propaganda effort.
- After the end of World War II in 1945, many artists began working in art forms derived from the international scene, moving away from local artistic developments into the mainstream of world art.
Key Terms
- Treaty of San Francisco : A treaty between Japan and part of the Allied Powers, officially signed by 48 nations on September 8, 1951 and coming into force on April 28, 1952; representing the official conclusion of World War II, it ended Japan’s position as an imperial power and allocated compensation to Allied civilians and former prisoners of war who had suffered Japanese war crimes.
- Surrealism : An artistic movement and an aesthetic philosophy, pre-dating abstract expressionism, that aims for the liberation of the mind by emphasizing the critical and imaginative powers of the subconscious.
- fascism : A political regime having totalitarian aspirations, ideologically based on a relationship between business and the centralized government, business-and-government control of the market place, repression of criticism or opposition, a leader cult, and exalting the state and/or religion above individual rights.
Overview: The Shōwa Period
The Shōwa period in Japanese history corresponds to the reign of the Shōwa Emperor, Hirohito (裕), from December 25, 1926 through January 7, 1989. The Shōwa period was longer than the reign of any previous Japanese emperor. During the pre-1945 period, Japan moved toward political totalitarianism, ultra-nationalism, and fascism, culminating in Japan’s invasion of China in 1937. This was part of an overall global period of social upheavals and conflicts such as the Great Depression and the Second World War.
Defeat in the Second World War brought radical change to Japan. For the first and only time in its history, Japan was occupied by foreign powers. This occupation by the United States on behalf of the Allied Forces (which included the Soviet Union, the United Kingdom, and China) lasted seven years. Allied occupation brought forth sweeping democratic reforms, leading to the end of the emperor’s status as a living god and the transformation of Japan into a democracy with a constitutional monarch. In 1952, with the Treaty of San Francisco, Japan became a sovereign nation once more and underwent an economic revitalization. In these ways, the pre-1945 and post-war periods regard completely different states: the pre-1945 Shōwa period (1926–1945) concerns the Empire of Japan, while the post-1945 Shōwa period (1945–1989) was a part of the State of Japan.
Art in the Pre-War Shōwa Period
Japanese painting in the pre-war Shōwa period was largely dominated by Yasui Sōtarō (1888–1955) and Umehara Ryūzaburō (1888–1986). These artists introduced the concepts of pure art and abstract painting to the Nihonga tradition (a style based on traditional Japanese art forms) and thus created a more interpretative version of that genre. Yasui Sōtarō was strongly influenced by the realistic styles of the French artists Jean-François Millet, Pierre-Auguste Renoir, and Paul Cézanne; he incorporated clear outlines and vibrant colors in his portraits and landscapes, combining western realism with the softer touches of traditional Nihonga techniques. This trend was further developed by Leonard Foujita (also known as Fujita Tsuguharu) and the Nika Society to encompass surrealism. To promote these trends, the Independent Art Association was formed in 1930.
By the early 20th century, European art forms were also introduced into Japanese architecture. Their marriage with traditional Japanese styles of architecture produced notable buildings like the Tokyo Train Station and the National Diet Building that still exist today.
Art in the Post-War Shōwa Period
During World War II, government controls and censorship meant that only patriotic themes could be expressed, and many artists were recruited into the government propaganda effort. After the end of World War II in 1945, many artists began working in art forms derived from the international scene, moving away from local artistic developments into the mainstream of world art. Traditional Japanese conceptions endured, however, particularly in the use of modular space in architecture, certain spacing intervals in music and dance, a propensity for certain color combinations, and characteristic literary forms.
Japanese Art after World War II
After World War II, Japanese artists became preoccupied with the mechanisms of urban life and moved from abstraction to anime-influenced art.
Describe the flourishing of painting, calligraphy, and printmaking after World War II
Key Points
- In the post-World War II period of Japanese history, the government-sponsored Japan Art Academy (Nihon Geijutsuin) was formed in 1947, containing both nihonga and yōga divisions.
- After World War II, painters, calligraphers, and printmakers flourished in the big cities, particularly Tokyo, and became preoccupied with the mechanisms of urban life, reflected in the flickering lights, neon colors, and frenetic pace of their abstractions.
- After the abstractions of the 1960s, the 1970s saw a return to realism strongly flavored by the “op” and “pop” art movements, embodied in the 1980s in the explosive works of Ushio Shinohara.
- By the late 1970s, the search for Japanese qualities and a national style caused many artists to reevaluate their artistic ideology and turn away from what some felt were the empty formulas of the West. Contemporary paintings began to make conscious use of traditional Japanese art.
- There are also a number of contemporary painters in Japan whose work is largely inspired by anime subcultures and other aspects of popular and youth culture, such as the work of Takashi Murakami.
Key Terms
- Rinpa school : One of the major historical schools of Japanese painting, created in 17th-century Kyoto by Hon’ami Kōetsu (1558–1637) and Tawaraya Sōtatsu (d. c. 1643).
- Nitten : The annual Japan Art Academy Awards and the premier art exhibition in Japan.
Japanese Art After World War II
Welcoming the new post-World War II period of Japanese history, the government-sponsored Japan Art Academy (Nihon Geijutsuin) was formed in 1947. The Academy contained both nihonga (traditional Japanese) and yōga (European-influenced) divisions. Government sponsorship of art exhibitions had ended, but they were replaced by private exhibitions, such as the Nitten, on an even larger scale. Although the Nitten was initially the exhibition of the Japan Art Academy, since 1958 it has been run by a separate private corporation. Participation in the Nitten became almost a prerequisite for nomination to the Japan Art Academy.
The arts of the Edo and prewar periods (1603–1945) had been supported by merchants and urban people, but they were not as popular as the arts of the postwar period. After World War II, painters, calligraphers, and printmakers flourished in the big cities—particularly Tokyo—and became preoccupied with the mechanisms of urban life, reflected in the flickering lights, neon colors, and frenetic pace of their abstractions. Styles of the New York-Paris art world were fervently embraced. After the abstractions of the 1960s, the 1970s saw a return to realism strongly flavored by the “op” and “pop” art movements, embodied in the 1980s in the explosive works of Ushio Shinohara.
Many such outstanding avant-garde artists worked both in Japan and abroad, winning international prizes. Some of these artists felt more identified with the international school of art rather than anything specifically Japanese. By the late 1970s, the search for Japanese qualities and a national style caused many artists to reevaluate their artistic ideology and turn away from what some felt were the empty formulas of the West. Contemporary paintings within the modern idiom began to make conscious use of traditional Japanese art forms, devices, and ideologies.
Mono-ha
Mono-ha is the name given to group of 20th century Japanese artists. The mono-ha artists explored the encounter between natural and industrial materials, such as stone, steel plates, glass, light bulbs, cotton, sponge, paper, wood, wire, rope, leather, oil, and water, arranging them in mostly unaltered, ephemeral states. The works focus as much on the interdependency of these various elements and the surrounding space as on the materials themselves. A number of mono-ha artists turned to painting to recapture traditional nuances in spatial arrangements, color harmonies, and lyricism.
Nihonga, Rinpa, and Kano Influence
Japanese-style, or nihonga painting continued in a pre-war fashion, updating traditional expressions while retaining their intrinsic character. Some artists within this style still painted on silk or paper with traditional colors and ink, while others used new materials, such as acrylics.
Many other older schools of art were still practiced, most notably those of the Edo and pre-war periods. For example, the decorative naturalism of the Rinpa school, characterized by brilliant, pure colors and bleeding washes, was reflected in the work of many artists of the postwar period in the 1980s art of Hikosaka Naoyoshi. The realism of Maruyama Ōkyo’s School and the calligraphic and spontaneous Japanese style of the gentlemen-scholars were both widely practiced in the 1980s. At times, all of these schools (along with older ones, such as the Kano School ink traditions) were drawn on by contemporary artists in the Japanese style and in the modern idiom. Many Japanese-style painters were honored with awards and prizes as a result of renewed popular demand for Japanese-style art beginning in the 1970s. More and more, the international modern painters also drew on the Japanese schools as they turned away from Western styles in the 1980s. The tendency had been to synthesize East and West, and some artists such as Shinoda Toko had already leapt the gap between the two. Shinoda’s bold sumi ink abstractions were inspired by traditional calligraphy but were realized as lyrical expressions of modern abstraction.
Anime Influence
There are also a number of contemporary painters in Japan whose work is largely inspired by anime subcultures and other aspects of popular and youth culture. Takashi Murakami is perhaps among the most famous and popular of these, along with the other artists in his Kaikai Kiki studio collective. His work centers on expressing issues and concerns of post-war Japanese society through seemingly innocuous forms. He draws heavily from anime and related styles but produces paintings and sculptures in media more traditionally associated with fine arts, intentionally blurring the lines between commercial, popular, and fine arts.
- Curation and Revision. Provided by : Boundless.com. License : CC BY-SA: Attribution-ShareAlike
- Meiji period. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Meiji_period. License : CC BY-SA: Attribution-ShareAlike
- Japanese painting. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Japanese_painting%23Prewar_period_.281868-1945.29. License : CC BY-SA: Attribution-ShareAlike
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- pre-Raphaelite movement. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/pre-Raphaelite%20movement. License : CC BY-SA: Attribution-ShareAlike
- Kuroki Neko by Hishida Shunso. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Kuroki_Neko_by_Hishida_Shunso.jpg. License : Public Domain: No Known Copyright
- Kuroda Maiko. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Kuroda_Maiko.jpg. License : Public Domain: No Known Copyright
- Treaty of San Francisco. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Treaty%20of%20San%20Francisco. License : CC BY-SA: Attribution-ShareAlike
- Showa period. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Showa_period. License : CC BY-SA: Attribution-ShareAlike
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- fascism. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/fascism. License : CC BY-SA: Attribution-ShareAlike
- Surrealism. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/Surrealism. License : CC BY-SA: Attribution-ShareAlike
- Kuroki Neko by Hishida Shunso. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Kuroki_Neko_by_Hishida_Shunso.jpg. License : Public Domain: No Known Copyright
- Kuroda Maiko. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Kuroda_Maiko.jpg. License : Public Domain: No Known Copyright
- 1280px-Tokyo_station05s3872.jpg. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Japanese_art#/media/File:Tokyo_station05s3872.jpg. License : CC BY-SA: Attribution-ShareAlike
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- Japanese painting. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Japanese_painting%23Prewar_period_.281868-1945.29. License : CC BY-SA: Attribution-ShareAlike
- Nitten. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Nitten. License : CC BY-SA: Attribution-ShareAlike
- rimpa school. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/rimpa%20school. License : CC BY-SA: Attribution-ShareAlike
- Mono-ha. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Mono-ha. License : CC BY-SA: Attribution-ShareAlike
- Kuroki Neko by Hishida Shunso. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Kuroki_Neko_by_Hishida_Shunso.jpg. License : Public Domain: No Known Copyright
- Kuroda Maiko. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Kuroda_Maiko.jpg. License : Public Domain: No Known Copyright
- 1280px-Tokyo_station05s3872.jpg. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Japanese_art#/media/File:Tokyo_station05s3872.jpg. License : CC BY-SA: Attribution-ShareAlike
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2: Working Effectively with Home Care Clients
All of us have unique needs and backgrounds. We look different from each other; we have different cultural backgrounds, religions, ages, social classes, financial statuses, occupations, interests, and families. We may speak different languages. We have different values, or thoughts on what is important to us. What makes each of us the same, each human being the same as the next, is that we have the same basic physical and psychological needs. This module will explore the basic physical and psychological needs of humans. It will also look at how culture influences people’s behavior, values, and health care practices. You will gain some basic information about a variety of cultures and religions with whom you may work if you choose to pursue a career as a Home Health Aide/Personal Care Aide. During this module, you will learn a variety of Communication skills to use in working with patients, and how to best develop a positive working relationship with patients. We will also explore how to document observations and the work done with patients. Finally, you will learn about the different types of abuse , the signs to look for, and how to report suspicions of abuse. We will also discuss the importance of patient rights and how best to support the rights of patients.


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3.1: Bonding in compounds
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Covalent and ionic bonds that hold atoms together in molecules are described. Representing a compound by formula and properties of the compounds are also introduced.
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3.2: Naming binary ionic compounds
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Naming cations, anions, and binary ionic compounds are described. Writing the formulae from the names of the ionic compounds is also described.
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3.3: Polyatomic ions and their compounds
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Polyatomic ions, their nomenclature, the name of the compounds containing polyatomic ions, and the writing formulae of these compounds from their names are described.
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3.4: Naming acids
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The nomenclature of acids that do not contain oxyanions and those containing oxyanions is described.
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3.5: Naming binary covalent compounds
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Nomenclature of binary covalent compounds, i.e., writing names from formulae and writing formulae from the names are described.
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3.6: Lewis structures of molecules
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A systematic procedure for writing Lewis structures of compounds and exceptions to the octet rule are discribed.
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3.7: Molecular shapes –Valence shell electron pair repulsion (VSEPR) theory
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The 3D shapes of molecules are explained on the basis of valence shell electron pair repulsion theory.
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3.8: Polarity of molecules
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The polarity of molecules based on polar bonds and vector addition of the polarity vectors of the bonds in a molecule are described.
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3.9: Intramolecular forces and intermolecular forces
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Intramolecular forces, i.e., ionic, covalent, and metallic bonds, and intermolecular forces based on electrostatic interactions between permanent or transient dipoles, and hydrogen bonding are described.


16.10: Factorial ANOVA 3- Unbalanced Designs
Factorial ANOVA is a very handy thing to know about. It’s been one of the standard tools used to analyse experimental data for many decades, and you’ll find that you can’t read more than two or three papers in psychology without running into an ANOVA in there somewhere. However, there’s one huge difference between the ANOVAs that you’ll see in a lot of real scientific articles and the ANOVA that I’ve just described: in real life, we’re rarely lucky enough to have perfectly balanced designs. For one reason or another, it’s typical to end up with more observations in some cells than in others. Or, to put it another way, we have an unbalanced design .
Unbalanced designs need to be treated with a lot more care than balanced designs, and the statistical theory that underpins them is a lot messier. It might be a consequence of this messiness, or it might be a shortage of time, but my experience has been that undergraduate research methods classes in psychology have a nasty tendency to ignore this issue completely. A lot of stats textbooks tend to gloss over it too. The net result of this, I think, is that a lot of active researchers in the field don’t actually know that there’s several different “types” of unbalanced ANOVAs, and they produce quite different answers. In fact, reading the psychological literature, I’m kind of amazed at the fact that most people who report the results of an unbalanced factorial ANOVA don’t actually give you enough details to reproduce the analysis: I secretly suspect that most people don’t even realise that their statistical software package is making a whole lot of substantive data analysis decisions on their behalf. It’s actually a little terrifying, when you think about it.  The
coffee.Rdata
file contains a hypothetical data set (the
coffee
data frame) that produces an unbalanced 3×2 ANOVA. Suppose we were interested in finding out whether or not the tendency of people to
babble
when they have too much coffee is purely an effect of the coffee itself, or whether there’s some effect of the
milk
and
sugar
that people add to the coffee. Suppose we took 18 people, and gave them some coffee to drink. The amount of coffee / caffeine was held constant, and we varied whether or not milk was added: so
milk
is a binary factor with two levels,
"yes"
and
"no"
. We also varied the kind of sugar involved. The coffee might contain
"real"
sugar, or it might contain
"fake"
sugar (i.e., artificial sweetener), or it might contain
"none"
at all, so the
sugar
variable is a three level factor. Our outcome variable is a continuous variable that presumably refers to some psychologically sensible measure of the extent to which someone is “babbling”. The details don’t really matter for our purpose. To get a sense of what the data look like, we use the
some()
function in the
car
package. The
some()
function randomly picks a few of the observations in the data frame to print out, which is often very handy:
load("./rbook-master/data/coffee.rdata")
some( coffee )
## milk sugar babble
## 1 yes real 4.6
## 5 yes real 5.1
## 6 no real 5.5
## 7 yes none 3.9
## 9 yes none 3.7
## 10 no fake 5.6
## 11 no fake 4.7
## 13 no real 6.0
## 14 no real 5.4
## 17 no none 5.3
If we use the
aggregate()
function to quickly produce a table of means, we get a strong impression that there are differences between the groups:
aggregate( babble ~ milk + sugar, coffee, mean )
## milk sugar babble
## 1 yes none 3.700
## 2 no none 5.550
## 3 yes fake 5.800
## 4 no fake 4.650
## 5 yes real 5.100
## 6 no real 5.875
This is especially true when we compare these means to the standard deviations for the
babble
variable, which you can calculate using
aggregate()
in much the same way. Across groups, this standard deviation varies from .14 to .71, which is fairly small relative to the differences in group means.
246
So far, it’s looking like a straightforward factorial ANOVA, just like we did earlier. The problem arises when we check to see how many observations we have in each group:
xtabs( ~ milk + sugar, coffee )
## sugar
## milk none fake real
## yes 3 2 3
## no 2 4 4
This violates one of our original assumptions, namely that the number of people in each group is the same. We haven’t really discussed how to handle this situation.
“Standard ANOVA” does not exist for unbalanced designs
Unbalanced designs lead us to the somewhat unsettling discovery that there isn’t really any one thing that we might refer to as a standard ANOVA. In fact, it turns out that there are three fundamentally different ways 247 in which you might want to run an ANOVA in an unbalanced design. If you have a balanced design, all three versions produce identical results, with the sums of squares, F-values etc all conforming to the formulas that I gave at the start of the chapter. However, when your design is unbalanced they don’t give the same answers. Furthermore, they are not all equally appropriate to every situation: some methods will be more appropriate to your situation than others. Given all this, it’s important to understand what the different types of ANOVA are and how they differ from one another.
The first kind of ANOVA is conventionally referred to as Type I sum of squares . I’m sure you can guess what they other two are called. The “sum of squares” part of the name was introduced by the SAS statistical software package, and has become standard nomenclature, but it’s a bit misleading in some ways. I think the logic for referring to them as different types of sum of squares is that, when you look at the ANOVA tables that they produce, the key difference in the numbers is the SS values. The degrees of freedom don’t change, the MS values are still defined as SS divided by df, etc. However, what the terminology gets wrong is that it hides the reason why the SS values are different from one another. To that end, it’s a lot more helpful to think of the three different kinds of ANOVA as three different hypothesis testing strategies . These different strategies lead to different SS values, to be sure, but it’s the strategy that is the important thing here, not the SS values themselves. Recall from Section 16.5 and 16.6 that any particular F-test is best thought of as a comparison between two linear models. So when you’re looking at an ANOVA table, it helps to remember that each of those F-tests corresponds to a pair of models that are being compared. Of course, this leads naturally to the question of which pair of models is being compared. This is the fundamental difference between ANOVA Types I, II and III: each one corresponds to a different way of choosing the model pairs for the tests.
Type I sum of squares
The Type I method is sometimes referred to as the “sequential” sum of squares, because it involves a process of adding terms to the model one at a time. Consider the coffee data, for instance. Suppose we want to run the full 3×2 factorial ANOVA, including interaction terms. The full model, as we’ve discussed earlier, is expressed by the R formula
babble ~ sugar + milk + sugar:milk
, though we often shorten it by using the
sugar * milk
notation. The Type I strategy builds this model up sequentially, starting from the simplest possible model and gradually adding terms.
The simplest possible model for the data would be one in which neither milk nor sugar is assumed to have any effect on babbling. The only term that would be included in such a model is the intercept, and in R formula notation we would write it as
babble ~ 1
. This is our initial null hypothesis. The next simplest model for the data would be one in which only one of the two main effects is included. In the coffee data, there are two different possible choices here, because we could choose to add milk first or to add sugar first (pardon the pun). The order actually turns out to matter, as we’ll see later, but for now let’s just make a choice arbitrarily, and pick sugar. So the second model in our sequence of models is
babble ~ sugar
, and it forms the alternative hypothesis for our first test. We now have our first hypothesis test:
knitr::kable(tibble::tribble(
~V1, ~V2,
"Null model:", "`babble ~ 1`",
"Alternative model:", "`babble ~ sugar`"
), col.names= c("", ""))
| Null model: |
babble ~ 1
|
| Alternative model: |
babble ~ sugar
|
This comparison forms our hypothesis test of the main effect of
sugar
. The next step in our model building exercise it to add the other main effect term, so the next model in our sequence is
babble ~ sugar + milk
. The second hypothesis test is then formed by comparing the following pair of models:
knitr::kable(tibble::tribble(
~V1, ~V2,
"Null model:", "`babble ~ sugar`",
"Alternative model:", "`babble ~ sugar + milk`"
), col.names= c("", ""))
| Null model: |
babble ~ sugar
|
| Alternative model: |
babble ~ sugar + milk
|
This comparison forms our hypothesis test of the main effect of
milk
. In one sense, this approach is very elegant: the alternative hypothesis from the first test forms the null hypothesis for the second one. It is in this sense that the Type I method is strictly sequential. Every test builds directly on the results of the last one. However, in another sense it’s very inelegant, because there’s a strong asymmetry between the two tests. The test of the main effect of
sugar
(the first test) completely ignores
milk
, whereas the test of the main effect of
milk
(the second test) does take
sugar
into account. In any case, the fourth model in our sequence is now the full model,
babble ~ sugar + milk + sugar:milk
, and the corresponding hypothesis test is
knitr::kable(tibble::tribble(
~V1, ~V2,
"Null model:", "`babble ~ sugar + milk`",
"Alternative model:", "`babble ~ sugar + milk + sugar:milk`"
), col.names= c("", ""))
| Null model: |
babble ~ sugar + milk
|
| Alternative model: |
babble ~ sugar + milk + sugar:milk
|
Type I sum of squares is the default hypothesis testing method used by the
anova()
function, so it’s easy to produce the results from a Type I analysis. We just type in the same commands that we always did. Since we’ve now reached the point that we don’t need to hide the fact that ANOVA and regression are both linear models, I’ll use the
lm()
function to run the analyses:
mod <- lm( babble ~ sugar + milk + sugar:milk, coffee )
anova( mod )
## Analysis of Variance Table
##
## Response: babble
## Df Sum Sq Mean Sq F value Pr(>F)
## sugar 2 3.5575 1.77876 6.7495 0.010863 *
## milk 1 0.9561 0.95611 3.6279 0.081061 .
## sugar:milk 2 5.9439 2.97193 11.2769 0.001754 **
## Residuals 12 3.1625 0.26354
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Leaving aside for one moment the question of how this result should be interpreted, let’s take note of the fact that our three p-values are .0109, .0811 and .0018 respectively. Next, let’s see if we can replicate the analysis using tools that we’re a little more familiar with. First, let’s fit all four models:
mod.1 <- lm( babble ~ 1, coffee )
mod.2 <- lm( babble ~ sugar, coffee )
mod.3 <- lm( babble ~ sugar + milk, coffee )
mod.4 <- lm( babble ~ sugar + milk + sugar:milk, coffee )
To run the first hypothesis test comparing
mod.1
to
mod.2
we can use the command
anova(mod.1, mod.2)
in much the same way that we did in Section 16.5. Similarly, we can use the commands
anova(mod.2, mod.3)
and
anova(mod.3, mod.4)
and to run the second and third hypothesis tests. However, rather than run each of those commands separately, we can enter the full sequence of models like this:
anova( mod.1, mod.2, mod.3, mod.4 )
## Analysis of Variance Table
##
## Model 1: babble ~ 1
## Model 2: babble ~ sugar
## Model 3: babble ~ sugar + milk
## Model 4: babble ~ sugar + milk + sugar:milk
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 17 13.6200
## 2 15 10.0625 2 3.5575 6.7495 0.010863 *
## 3 14 9.1064 1 0.9561 3.6279 0.081061 .
## 4 12 3.1625 2 5.9439 11.2769 0.001754 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
This output is rather more verbose than the last one, but it’s telling essentially the same story. 248
The big problem with using Type I sum of squares is the fact that it really does depend on the order in which you enter the variables. Yet, in many situations the researcher has no reason to prefer one ordering over another. This is presumably the case for our milk and sugar problem. Should we add milk first, or sugar first? It feels exactly as arbitrary as a data analysis question as it does as a coffee-making question. There may in fact be some people with firm opinions about ordering, but it’s hard to imagine a principled answer to the question. Yet, look what happens when we change the ordering:
mod <- lm( babble ~ milk + sugar + sugar:milk, coffee )
anova( mod )
## Analysis of Variance Table
##
## Response: babble
## Df Sum Sq Mean Sq F value Pr(>F)
## milk 1 1.4440 1.44400 5.4792 0.037333 *
## sugar 2 3.0696 1.53482 5.8238 0.017075 *
## milk:sugar 2 5.9439 2.97193 11.2769 0.001754 **
## Residuals 12 3.1625 0.26354
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
The p-values for both main effect terms have changed, and fairly dramatically. Among other things, the effect of
milk
has become significant (though one should avoid drawing any strong conclusions about this, as I’ve mentioned previously). Which of these two ANOVAs should one report? It’s not immediately obvious.
When you look at the hypothesis tests that are used to define the “first” main effect and the “second” one, it’s clear that they’re qualitatively different from one another. In our initial example, we saw that the test for the main effect of
sugar
completely ignores
milk
, whereas the test of the main effect of
milk
does take
sugar
into account. As such, the Type I testing strategy really does treat the first main effect as if it had a kind of theoretical primacy over the second one. In my experience there is very rarely if ever any theoretically primacy of this kind that would justify treating any two main effects asymmetrically.
The consequence of all this is that Type I tests are very rarely of much interest, and so we should move on to discuss Type II tests and Type III tests. However, for the sake of completeness – on the off chance that you ever find yourself needing to run Type I tests – I’ll comment briefly on how R determines the ordering of terms in a Type I test. The key principle in Type I sum of squares is that the hypothesis testing be sequential, with terms added one at a time. However, it does also imply that main effects be added first (e.g., factors
A
,
B
,
C
etc), followed by first order interaction terms (e.g., terms like
A:B
and
B:C
), then second order interactions (e.g.,
A:B:C
) and so on. Within each “block” you can specify whatever order you like. So, for instance, if we specified our model using a command like this,
mod <- lm( outcome ~ A + B + C + B:C + A:B + A:C )
and then used
anova(mod)
to produce sequential hypothesis tests, what we’d see is that the main effect terms would be entered
A
then
B
and then
C
, but then the interactions would be entered in the order
B:C
first, then
A:B
and then finally
A:C
. Reordering the terms within each group will change the ordering, as we saw earlier. However, changing the order of terms across blocks has no effect. For instance, if we tried to move the interaction term
B:C
to the front, like this,
mod <- lm( outcome ~ B:C + A + B + C + A:B + A:C )
it would have no effect. R would still enter the terms in the same order as last time. If for some reason you really, really need an interaction term to be entered first, then you have to do it the long way, creating each model manually using a separate
lm()
command and then using a command like
anova(mod.1, mod.2, mod.3, mod.4)
to force R to enter them in the order that you want.
Type III sum of squares
Having just finished talking about Type I tests, you might think that the natural thing to do next would be to talk about Type II tests. However, I think it’s actually a bit more natural to discuss Type III tests (which are simple) before talking about Type II tests (which are trickier). The basic idea behind Type III tests is extremely simple: regardless of which term you’re trying to evaluate, run the F-test in which the alternative hypothesis corresponds to the full ANOVA model as specified by the user, and the null model just deletes that one term that you’re testing. For instance, in the coffee example, in which our full model was
babble ~ sugar + milk + sugar:milk
, the test for a main effect of
sugar
would correspond to a comparison between the following two models:
knitr::kable(tibble::tribble(
~V1, ~V2,
"Null model:", "`babble ~ milk + sugar:milk`",
"Alternative model:", "`babble ~ sugar + milk + sugar:milk`"
), col.names= c("", ""))
| Null model: |
babble ~ milk + sugar:milk
|
| Alternative model: |
babble ~ sugar + milk + sugar:milk
|
Similarly the main effect of
milk
is evaluated by testing the full model against a null model that removes the
milk
term, like so:
knitr::kable(tibble::tribble(
~V1, ~V2,
"Null model:", "`babble ~ sugar + sugar:milk`",
"Alternative model:", "`babble ~ sugar + milk + sugar:milk`"
), col.names= c("", ""))
| Null model: |
babble ~ sugar + sugar:milk
|
| Alternative model: |
babble ~ sugar + milk + sugar:milk
|
Finally, the interaction term
sugar:milk
is evaluated in exactly the same way. Once again, we test the full model against a null model that removes the
sugar:milk
interaction term, like so:
knitr::kable(tibble::tribble(
~V1, ~V2,
"Null model:", "`babble ~ sugar + milk`",
"Alternative model:", "`babble ~ sugar + milk + sugar:milk`"
), col.names= c("", ""))
| Null model: |
babble ~ sugar + milk
|
| Alternative model: |
babble ~ sugar + milk + sugar:milk
|
The basic idea generalises to higher order ANOVAs. For instance, suppose that we were trying to run an ANOVA with three factors,
A
,
B
and
C
, and we wanted to consider all possible main effects and all possible interactions, including the three way interaction
A:B:C
. The table below shows you what the Type III tests look like for this situation:
knitr::kable(tibble::tribble(
~V1, ~V2, ~V3,
"`A`", "`B + C + A:B + A:C + B:C + A:B:C`", "`A + B + C + A:B + A:C + B:C + A:B:C`",
"`B`", "`A + C + A:B + A:C + B:C + A:B:C`", "`A + B + C + A:B + A:C + B:C + A:B:C`",
"`C`", "`A + B + A:B + A:C + B:C + A:B:C`", "`A + B + C + A:B + A:C + B:C + A:B:C`",
"`A:B`", "`A + B + C + A:C + B:C + A:B:C`", "`A + B + C + A:B + A:C + B:C + A:B:C`",
"`A:C`", "`A + B + C + A:B + B:C + A:B:C`", "`A + B + C + A:B + A:C + B:C + A:B:C`",
"`B:C`", "`A + B + C + A:B + A:C + A:B:C`", "`A + B + C + A:B + A:C + B:C + A:B:C`",
"`A:B:C`", "`A + B + C + A:B + A:C + B:C`", "`A + B + C + A:B + A:C + B:C + A:B:C`"
), col.names = c( "Term being tested is", "Null model is `outcome ~ ...`", "Alternative model is `outcome ~ ...`"))
| Term being tested is |
Null model is
outcome ~ ...
|
Alternative model is
outcome ~ ...
|
|---|---|---|
A
|
B + C + A:B + A:C + B:C + A:B:C
|
A + B + C + A:B + A:C + B:C + A:B:C
|
B
|
A + C + A:B + A:C + B:C + A:B:C
|
A + B + C + A:B + A:C + B:C + A:B:C
|
C
|
A + B + A:B + A:C + B:C + A:B:C
|
A + B + C + A:B + A:C + B:C + A:B:C
|
A:B
|
A + B + C + A:C + B:C + A:B:C
|
A + B + C + A:B + A:C + B:C + A:B:C
|
A:C
|
A + B + C + A:B + B:C + A:B:C
|
A + B + C + A:B + A:C + B:C + A:B:C
|
B:C
|
A + B + C + A:B + A:C + A:B:C
|
A + B + C + A:B + A:C + B:C + A:B:C
|
A:B:C
|
A + B + C + A:B + A:C + B:C
|
A + B + C + A:B + A:C + B:C + A:B:C
|
As ugly as that table looks, it’s pretty simple. In all cases, the alternative hypothesis corresponds to the full model, which contains three main effect terms (e.g.
A
), three first order interactions (e.g.
A:B
) and one second order interaction (i.e.,
A:B:C
). The null model always contains 6 of thes 7 terms: and the missing one is the one whose significance we’re trying to test.
At first pass, Type III tests seem like a nice idea. Firstly, we’ve removed the asymmetry that caused us to have problems when running Type I tests. And because we’re now treating all terms the same way, the results of the hypothesis tests do not depend on the order in which we specify them. This is definitely a good thing. However, there is a big problem when interpreting the results of the tests, especially for main effect terms. Consider the coffee data. Suppose it turns out that the main effect of
milk
is not significant according to the Type III tests. What this is telling us is that
babble ~ sugar + sugar:milk
is a better model for the data than the full model. But what does that even
mean
? If the interaction term
sugar:milk
was also non-significant, we’d be tempted to conclude that the data are telling us that the only thing that matters is
sugar
. But suppose we have a significant interaction term, but a non-significant main effect of
milk
. In this case, are we to assume that there really is an “effect of sugar”, an “interaction between milk and sugar”, but no “effect of milk”? That seems crazy. The right answer simply
must
be that it’s meaningless
249
to talk about the main effect if the interaction is significant. In general, this seems to be what most statisticians advise us to do, and I think that’s the right advice. But if it really is meaningless to talk about non-significant main effects in the presence of a significant interaction, then it’s not at all obvious why Type III tests should allow the null hypothesis to rely on a model that includes the interaction but omits one of the main effects that make it up. When characterised in this fashion, the null hypotheses really don’t make much sense at all.
Later on, we’ll see that Type III tests can be redeemed in some contexts, but I’d better show you how to actually compute a Type III ANOVA first. The
anova()
function in R does not directly support Type II tests or Type III tests. Technically, you
can
do it by creating the various models that form the null and alternative hypotheses for each test, and then using
anova()
to compare the models to one another. I outlined the gist of how that would be done when talking about Type I tests, but speaking from first hand experience
250
I can tell you that it’s very tedious. In practice, the
anova()
function is only used to produce Type I tests or to compare specific models of particular interest (see Section 16.5). If you want Type II or Type III tests you need to use the
Anova()
function in the
car
package. It’s pretty easy to use, since there’s a
type
argument that you specify. So, to return to our coffee example, our Type III tests are run as follows:
mod <- lm( babble ~ sugar * milk, coffee )
Anova( mod, type=3 )
## Anova Table (Type III tests)
##
## Response: babble
## Sum Sq Df F value Pr(>F)
## (Intercept) 41.070 1 155.839 3.11e-08 ***
## sugar 5.880 2 11.156 0.001830 **
## milk 4.107 1 15.584 0.001936 **
## sugar:milk 5.944 2 11.277 0.001754 **
## Residuals 3.162 12
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
As you can see, I got lazy this time and used
sugar * milk
as a shorthand way of referring to
sugar + milk + sugar:milk
. The important point here is that this is just a regular ANOVA table, and we can see that our Type III tests are significant for all terms, even the intercept.
Except, as usual, it’s not that simple. One of the perverse features of the Type III testing strategy is that the results turn out to depend on the contrasts that you use to encode your factors (see Section 16.7 if you’ve forgotten what the different types of contrasts are). The results that I presented in the ANOVA table above are based on the R default, which is treatment contrasts; and as we’ll see later, this is usually a very poor choice if you want to run Type III tests. So let’s see what happens if switch to Helmert contrasts:
my.contrasts <- list( milk = "contr.Helmert", sugar = "contr.Helmert" )
mod.H <- lm( babble ~ sugar * milk, coffee, contrasts = my.contrasts )
Anova( mod.H, type=3 )
## Anova Table (Type III tests)
##
## Response: babble
## Sum Sq Df F value Pr(>F)
## (Intercept) 434.29 1 1647.8882 3.231e-14 ***
## sugar 2.13 2 4.0446 0.045426 *
## milk 1.00 1 3.8102 0.074672 .
## sugar:milk 5.94 2 11.2769 0.001754 **
## Residuals 3.16 12
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Oh, that’s not good at all. In the case of
milk
in particular, the p-value has changed from .002 to .07. This is a pretty substantial difference, and hopefully it gives you a sense of how important it is that you take care when using Type III tests.
Okay, so if the p-values that come out of Type III analyses are so sensitive to the choice of contrasts, does that mean that Type III tests are essentially arbitrary and not to be trusted? To some extent that’s true, and when we turn to a discussion of Type II tests we’ll see that Type II analyses avoid this arbitrariness entirely, but I think that’s too strong a conclusion. Firstly, it’s important to recognise that some choices of contrasts will always produce the same answers. Of particular importance is the fact that if the columns of our contrast matrix are all constrained to sum to zero, then the Type III analysis will always give the same answers. This means that you’ll get the same answers if you use
contr.Helmert
or
contr.sum
or
contr.poly
, but different answers for
contr.treatment
or
contr.SAS
.
random.contrasts <- matrix( rnorm(6), 3, 2 ) # create a random matrix
random.contrasts[, 1] <- random.contrasts[, 1] - mean( random.contrasts[, 1] ) # contrast 1 sums to 0
random.contrasts[, 2] <- random.contrasts[, 2] - mean( random.contrasts[, 2] ) # contrast 2 sums to 0
random.contrasts # print it to check that we really have an arbitrary contrast matrix...
## [,1] [,2]
## [1,] -1.523759 0.6891740
## [2,] -0.334936 0.9999209
## [3,] 1.858695 -1.6890949
contrasts( coffee$sugar ) <- random.contrasts # random contrasts for sugar
contrasts( coffee$milk ) <- contr.Helmert(2) # Helmert contrasts for the milk factor
mod.R <- lm( babble ~ sugar * milk, coffee ) # R will use the contrasts that we assigned
Anova( mod.R, type = 3 )
## Anova Table (Type III tests)
##
## Response: babble
## Sum Sq Df F value Pr(>F)
## (Intercept) 434.29 1 1647.8882 3.231e-14 ***
## sugar 2.13 2 4.0446 0.045426 *
## milk 1.00 1 3.8102 0.074672 .
## sugar:milk 5.94 2 11.2769 0.001754 **
## Residuals 3.16 12
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Yep, same answers.
Type II sum of squares
Okay, so we’ve seen Type I and III tests now, and both are pretty straightforward: Type I tests are performed by gradually adding terms one at a time, whereas Type III tests are performed by taking the full model and looking to see what happens when you remove each term. However, both have some serious flaws: Type I tests are dependent on the order in which you enter the terms, and Type III tests are dependent on how you code up your contrasts. Because of these flaws, neither one is easy to interpret.
Type II tests are broadly similar to Type III tests: start with a “full” model, and test a particular term by deleting it from that model. However, Type II tests are based on the
marginality principle
which states that you should not omit a lower order term from your model if there are any higher order ones that depend on it. So, for instance, if your model contains the interaction
A:B
(a 2nd order term), then it really ought to contain the main effects
A
and
B
(1st order terms). Similarly, if it contains a three way interaction term
A:B:C
, then the model must also include the main effects
A
,
B
and
C
as well as the simpler interactions
A:B
,
A:C
and
B:C
. Type III tests routinely violate the marginality principle. For instance, consider the test of the main effect of
A
in the context of a three-way ANOVA that includes all possible interaction terms. According to Type III tests, our null and alternative models are:
knitr::kable(tibble::tribble(
~V1, ~V2,
"Null model:", "`outcome ~ B + C + A:B + A:C + B:C + A:B:C`",
"Alternative model:", "`outcome ~ A + B + C + A:B + A:C + B:C + A:B:C`"
), col.names = c("", ""))
| Null model: |
outcome ~ B + C + A:B + A:C + B:C + A:B:C
|
| Alternative model: |
outcome ~ A + B + C + A:B + A:C + B:C + A:B:C
|
Notice that the null hypothesis omits
A
, but includes
A:B
,
A:C
and
A:B:C
as part of the model. This, according to the Type II tests, is not a good choice of null hypothesis. What we should do instead, if we want to test the null hypothesis that
A
is not relevant to our
outcome
, is to specify the null hypothesis that is the most complicated model that does not rely on
A
in any form, even as an interaction. The alternative hypothesis corresponds to this null model plus a main effect term of
A
. This is a lot closer to what most people would intuitively think of as a “main effect of
A
”, and it yields the following as our Type II test of the main effect of
A
.
251
knitr::kable(tibble::tribble(
~V1, ~V2,
"Null model:", "`outcome ~ B + C + B:C`",
"Alternative model:", "`outcome ~ A + B + C + B:C`"
), col.names = c("", ""))
| Null model: |
outcome ~ B + C + B:C
|
| Alternative model: |
outcome ~ A + B + C + B:C
|
Anyway, just to give you a sense of how the Type II tests play out, here’s the full table of tests that would be applied in a three-way factorial ANOVA:
knitr::kable(tibble::tribble(
~V1, ~V2, ~V3,
"`A`", "`B + C + B:C`", "`A + B + C + B:C`",
"`B`", "`A + C + A:C`", "`A + B + C + A:C`",
"`C`", "`A + B + A:B`", "`A + B + C + A:B`",
"`A:B`", "`A + B + C + A:C + B:C`", "`A + B + C + A:B + A:C + B:C`",
"`A:C`", "`A + B + C + A:B + B:C`", "`A + B + C + A:B + A:C + B:C`",
"`B:C`", "`A + B + C + A:B + A:C`", "`A + B + C + A:B + A:C + B:C`",
"`A:B:C`", "`A + B + C + A:B + A:C + B:C`", "`A + B + C + A:B + A:C + B:C + A:B:C`"
), col.names = c( "Term being tested is", "Null model is `outcome ~ ...`", "Alternative model is `outcome ~ ...`"))
| Term being tested is |
Null model is
outcome ~ ...
|
Alternative model is
outcome ~ ...
|
|---|---|---|
A
|
B + C + B:C
|
A + B + C + B:C
|
B
|
A + C + A:C
|
A + B + C + A:C
|
C
|
A + B + A:B
|
A + B + C + A:B
|
A:B
|
A + B + C + A:C + B:C
|
A + B + C + A:B + A:C + B:C
|
A:C
|
A + B + C + A:B + B:C
|
A + B + C + A:B + A:C + B:C
|
B:C
|
A + B + C + A:B + A:C
|
A + B + C + A:B + A:C + B:C
|
A:B:C
|
A + B + C + A:B + A:C + B:C
|
A + B + C + A:B + A:C + B:C + A:B:C
|
In the context of the two way ANOVA that we’ve been using in the coffee data, the hypothesis tests are even simpler. The main effect of
sugar
corresponds to an F-test comparing these two models:
knitr::kable(tibble::tribble(
~V1, ~V2,
"Null model:", "`babble ~ milk`",
"Alternative model:", "`babble ~ sugar + milk`"
), col.names = c("", ""))
| Null model: |
babble ~ milk
|
| Alternative model: |
babble ~ sugar + milk
|
The test for the main effect of
milk
is
knitr::kable(tibble::tribble(
~V1, ~V2,
"Null model:", "`babble ~ sugar`",
"Alternative model:", "`babble ~ sugar + milk`"
), col.names = c("", ""))
| Null model: |
babble ~ sugar
|
| Alternative model: |
babble ~ sugar + milk
|
Finally, the test for the interaction
sugar:milk
is:
knitr::kable(tibble::tribble(
~V1, ~V2,
"Null model:", "`babble ~ sugar + milk`",
"Alternative model:", "`babble ~ sugar + milk + sugar:milk`"
), col.names = c("", ""))
| Null model: |
babble ~ sugar + milk
|
| Alternative model: |
babble ~ sugar + milk + sugar:milk
|
Running the tests are again straightforward. We use the
Anova()
function, specifying
type=2
:
mod <- lm( babble ~ sugar*milk, coffee )
Anova( mod, type = 2 )
## Anova Table (Type II tests)
##
## Response: babble
## Sum Sq Df F value Pr(>F)
## sugar 3.0696 2 5.8238 0.017075 *
## milk 0.9561 1 3.6279 0.081061 .
## sugar:milk 5.9439 2 11.2769 0.001754 **
## Residuals 3.1625 12
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Type II tests have some clear advantages over Type I and Type III tests. They don’t depend on the order in which you specify factors (unlike Type I), and they don’t depend on the contrasts that you use to specify your factors (unlike Type III). And although opinions may differ on this last point, and it will definitely depend on what you’re trying to do with your data, I do think that the hypothesis tests that they specify are more likely to correspond to something that you actually care about. As a consequence, I find that it’s usually easier to interpret the results of a Type II test than the results of a Type I or Type III test. For this reason, my tentative advice is that, if you can’t think of any obvious model comparisons that directly map onto your research questions but you still want to run an ANOVA in an unbalanced design, Type II tests are probably a better choice than Type I or Type III. 252
Effect sizes (and non-additive sums of squares)
The
etaSquared()
function in the
lsr
package computes η2 and partial η2 values for unbalanced designs and for different Types of tests. It’s pretty straightforward. All you have to do is indicate which
type
of tests you’re doing,
etaSquared( mod, type=2 )
## eta.sq eta.sq.part
## sugar 0.22537682 0.4925493
## milk 0.07019886 0.2321436
## sugar:milk 0.43640732 0.6527155
and out pops the η
2
and partial η
2
values, as requested. However, when you’ve got an unbalanced design, there’s a bit of extra complexity involved. To see why, let’s expand the output from the
etaSquared()
function so that it displays the full ANOVA table:
es <- etaSquared( mod, type=2, anova=TRUE )
es
## eta.sq eta.sq.part SS df MS F
## sugar 0.22537682 0.4925493 3.0696323 2 1.5348161 5.823808
## milk 0.07019886 0.2321436 0.9561085 1 0.9561085 3.627921
## sugar:milk 0.43640732 0.6527155 5.9438677 2 2.9719339 11.276903
## Residuals 0.23219530 NA 3.1625000 12 0.2635417 NA
## p
## sugar 0.017075099
## milk 0.081060698
## sugar:milk 0.001754333
## Residuals NA
Okay, if you remember back to our very early discussions of ANOVA, one of the key ideas behind the sums of squares calculations is that if we add up all the SS terms associated with the effects in the model, and add that to the residual SS, they’re supposed to add up to the total sum of squares. And, on top of that, the whole idea behind η 2 is that – because you’re dividing one of the SS terms by the total SS value – is that an η 2 value can be interpreted as the proportion of variance accounted for by a particular term.
Now take a look at the output above. Because I’ve included the η 2 value associated with the residuals (i.e., proportion of variance in the outcome attributed to the residuals, rather than to one of the effects), you’d expect all the η 2 values to sum to 1. Because, the whole idea here was that the variance in the outcome variable can be divided up into the variability attributable to the model, and the variability in the residuals. Right? Right? And yet when we add up the η 2 values for our model…
sum( es[,"eta.sq"] )
## [1] 0.9641783
… we discover that for Type II and Type III tests they generally don’t sum to 1. Some of the variability has gone “missing”. It’s not being attributed to the model, and it’s not being attributed to the residuals either. What’s going on here?
Before giving you the answer, I want to push this idea a little further. From a mathematical perspective, it’s easy enough to see that the missing variance is a consequence of the fact that in Types II and III, the individual SS values are not obliged to the total sum of squares, and will only do so if you have balanced data. I’ll explain why this happens and what it means in a second, but first let’s verify that this is the case using the ANOVA table. First, we can calculate the total sum of squares directly from the raw data:
ss.tot <- sum( (coffee$babble - mean(coffee$babble))^2 )
ss.tot
## [1] 13.62
Next, we can read off all the SS values from one of our Type I ANOVA tables, and add them up. As you can see, this gives us the same answer, just like it’s supposed to:
type.I.sum <- 3.5575 + 0.9561 + 5.9439 + 3.1625
type.I.sum
## [1] 13.62
However, when we do the same thing for the Type II ANOVA table, it turns out that the SS values in the table add up to slightly less than the total SS value:
type.II.sum <- 0.9561 + 3.0696 + 5.9439 + 3.1625
type.II.sum
## [1] 13.1321
So, once again, we can see that there’s a little bit of variance that has “disappeared” somewhere.
Okay, time to explain what’s happened. The reason why this happens is that, when you have unbalanced designs, your factors become correlated with one another, and it becomes difficult to tell the difference between the effect of Factor A and the effect of Factor B. In the extreme case, suppose that we’d run a 2×2 design in which the number of participants in each group had been as follows:
knitr::kable(tibble::tribble(
~V1, ~V2, ~V3,
"milk", "100", "0",
"no milk", "0", "100"
), col.names = c( "", "sugar", "no sugar"))
| sugar | no sugar | |
|---|---|---|
| milk | 100 | 0 |
| no milk | 0 | 100 |
Here we have a spectacularly unbalanced design: 100 people have milk and sugar, 100 people have no milk and no sugar, and that’s all. There are 0 people with milk and no sugar, and 0 people with sugar but no milk. Now suppose that, when we collected the data, it turned out there is a large (and statistically significant) difference between the “milk and sugar” group and the “no-milk and no-sugar” group. Is this a main effect of sugar? A main effect of milk? Or an interaction? It’s impossible to tell, because the presence of sugar has a perfect association with the presence of milk. Now suppose the design had been a little more balanced:
knitr::kable(tibble::tribble(
~V1, ~V2, ~V3,
"milk", "100", "5",
"no milk", "5", "100"
), col.names = c( "", "sugar", "no sugar"))
| sugar | no sugar | |
|---|---|---|
| milk | 100 | 5 |
| no milk | 5 | 100 |
This time around, it’s technically possible to distinguish between the effect of milk and the effect of sugar, because we have a few people that have one but not the other. However, it will still be pretty difficult to do so, because the association between sugar and milk is still extremely strong, and there are so few observations in two of the groups. Again, we’re very likely to be in the situation where we know that the predictor variables (milk and sugar) are related to the outcome (babbling), but we don’t know if the nature of that relationship is a main effect of one predictor, or the other predictor or the interaction.
This uncertainty is the reason for the missing variance. The “missing” variance corresponds to variation in the outcome variable that is clearly attributable to the predictors, but we don’t know which of the effects in the model is responsible. When you calculate Type I sum of squares, no variance ever goes missing: the sequentiual nature of Type I sum of squares means that the ANOVA automatically attributes this variance to whichever effects are entered first. However, the Type II and Type III tests are more conservative. Variance that cannot be clearly attributed to a specific effect doesn’t get attributed to any of them, and it goes missing.