1.12: The Digital Divide and the lack Financial Literacy among First Generation
Imagine walking into a bank and seeing no bank tellers around to take deposit money. A few years ago, that is what happened to me. I walked into a private bank to deposit money into my mom’s account with no bank tellers’ insight. I later realized that the bank had done away with the bank tellers, including the drive-through teller, and instead moved basic financial transactions online. Imagine a young person with no knowledge of online banking and no understanding of financial literacy terminology to figure out online banking on their own. Now throw in that the young person may also have no access to a computer and uses an old cell phone where it only works with public WiFi.
As a former assistant branch manager for a private bank and Certified Financial Education Instructor with the National Financial Educators Council, I know the frustration that people go through when they don’t understand how online banking works and the frustration of basic financial literacy terminology. The future of banking going digital has me more worried about first-generation college students, especially among minority groups, because I fall into the two groups. Studies show how the digital divide and lack of financial literacy education among first-generation college students can be set back compared to non-first generation peers.
As a first-generation college student myself, I had to learn financial literacy on my own through life experiences, and I did not have the latest computer technology growing up until I was in high school. A first-generation college student is defined as an individual whose family comes from a low or medium-income background or high school education without attending college (College Board, 2016). When it comes to financial literacy, studies consistently show that a large number of people lack the financial literacy skills necessary to make critical financial decisions such as taking out a mortgage or understanding the stock market (Mandell & Klein, 2009). First-generation college students are vulnerable to falling into debt, ruining their credit score, or lacking the knowledge of building a budget (Lusardi, Mitchell & Curto 2010). A recent study by the TIAA-CREF Institute stated that 12% of college-educated Hispanics demonstrated advanced financial knowledge, and only 32% displayed basic financial literacy (De Bassa, Lusardi, & Yakoboski, 2015). This kind of experience makes it difficult for Hispanics to be financial stable, comprehend financial literature, and build wealth for the future. I don’t see any easy solution to solving financial literacy and the digital divide at the same time.
The solution I have for financial literacy is that every university across the United States requires its students to complete a three-credit financial literacy course. Financial literacy education can play a pivotal role in students’ lives. Ineffective money management can result in individuals making poor financial decisions that may prevent them from reaching full financial relief and possibly living paycheck to paycheck. I have yet to figure out how to solve the digital divide among first-generation college students because of their complexity. For example, the digital divide includes unreliable internet access, older devices that don’t work, inadequate cell phone data places, and computers.
To reach a larger audience for financial literacy education the best way are free tools such as YouTube and social media like Instagram. I can teach an individual the basic concepts of online banking and the terminology associated by publishing online courses on YouTube, which I have done. The tools that I use to create financial literacy education in the digital age is Camtasia to record the video and publish on YouTube and Vimeo for easy access to anyone that is interested. Another digital tool I use to teach financial literacy education is Instagram. I published short videos on Instagram TV for individuals to learn about saving money, creating a financial budget, and developing financial goals.
No matter how many courses I publish on YouTube the individual still needs to access the internet. I can’t teach a person how to obtain a computer or a smartphone capable of opening up a bank account. How do you solve the digital divide and lack of financial literacy among first-generation college students?
References
College Board. (2016). Retrieved January 27, 2016, from https://professionals.collegeboard.com/guidance/prepare/first-generation
De Bassa Scheresberg, C., Lusardi, A., & Yakoboski, P. J. (2015, May). Hispanic Personal Finances: Financial Literacy and Decision-making Among College-Educated Hispanics (Publication). Retrieved January 27, 2016, from TIAA-CREF Institute website: https://www.tiaa-crefinstitute.org/p...ncial-literacy
Lusardi, A., Mitchell, O. S., & Curto, V. (2010). Financial literacy among the young. Journal of consumer affairs , 44 (2), 358-380.
Mandell, L. (2009, January). The impact of financial education in high school and college on financial literacy and subsequent financial decision making. In American Economic Association Meetings, San Francisco, CA .


2.2: Chemical Bonds
Learning Objectives
- Explain the relationship between molecules and compounds
- Distinguish between ions, cations, and anions
- Identify the key difference between ionic and covalent bonds
- Distinguish between nonpolar and polar covalent bonds
- Explain how water molecules link via hydrogen bonds
Atoms separated by a great distance cannot link; rather, they must come close enough for the electrons in their valence shells to interact. But do atoms ever actually touch one another? Most physicists would say no, because the negatively charged electrons in their valence shells repel one another. No force within the human body—or anywhere in the natural world—is strong enough to overcome this electrical repulsion. So when you read about atoms linking together or colliding, bear in mind that the atoms are not merging in a physical sense.
Instead, atoms link by forming a chemical bond. A bond is a weak or strong electrical attraction that holds atoms in the same vicinity. The new grouping is typically more stable—less likely to react again—than its component atoms were when they were separate. A more or less stable grouping of two or more atoms held together by chemical bonds is called a molecule . The bonded atoms may be of the same element, as in the case of H 2 , which is called molecular hydrogen or hydrogen gas. When a molecule is made up of two or more atoms of different elements, it is called a chemical compound . Thus, a unit of water, or H 2 O, is a compound, as is a single molecule of the gas methane, or CH 4 .
Three types of chemical bonds are important in human physiology, because they hold together substances that are used by the body for critical aspects of homeostasis, signaling, and energy production, to name just a few important processes. These are ionic bonds, covalent bonds, and hydrogen bonds.
Ions and Ionic Bonds
Recall that an atom typically has the same number of positively charged protons and negatively charged electrons. As long as this situation remains, the atom is electrically neutral. But when an atom participates in a chemical reaction that results in the donation or acceptance of one or more electrons, the atom will then become positively or negatively charged. This happens frequently for most atoms in order to have a full valence shell, as described previously. This can happen either by gaining electrons to fill a shell that is more than half-full, or by giving away electrons to empty a shell than is less than half-full, thereby leaving the next smaller electron shell as the new, full, valence shell. An atom that has an electrical charge—whether positive or negative—is an ion .
Potassium (K), for instance, is an important element in all body cells. Its atomic number is 19. It has just one electron in its valence shell. This characteristic makes potassium highly likely to participate in chemical reactions in which it donates one electron. (It is easier for potassium to donate one electron than to gain seven electrons.) The loss will cause the positive charge of potassium’s protons to be more influential than the negative charge of potassium’s electrons. In other words, the resulting potassium ion will be slightly positive. A potassium ion is written K + , indicating that it has lost a single electron. A positively charged ion is known as a cation .
Now consider fluorine (F), a component of bones and teeth. Its atomic number is nine, and it has seven electrons in its valence shell. Thus, it is highly likely to bond with other atoms in such a way that fluorine accepts one electron (it is easier for fluorine to gain one electron than to donate seven electrons). When it does, its electrons will outnumber its protons by one, and it will have an overall negative charge. The ionized form of fluorine is called fluoride, and is written as F – . A negatively charged ion is known as an anion .
Atoms that have more than one electron to donate or accept will end up with stronger positive or negative charges. A cation that has donated two electrons has a net charge of +2. Using magnesium (Mg) as an example, this can be written Mg ++ or Mg 2 + . An anion that has accepted two electrons has a net charge of –2. The ionic form of selenium (Se), for example, is typically written Se 2 – .
The opposite charges of cations and anions exert a moderately strong mutual attraction that keeps the atoms in close proximity forming an ionic bond. An ionic bond is an ongoing, close association between ions of opposite charge. The table salt you sprinkle on your food owes its existence to ionic bonding. As shown in Figure \(\PageIndex{1}\), sodium commonly donates an electron to chlorine, becoming the cation Na + . When chlorine accepts the electron, it becomes the chloride anion, Cl – . With their opposing charges, these two ions strongly attract each other.
Water is an essential component of life because it is able to break the ionic bonds in salts to free the ions. In fact, in biological fluids, most individual atoms exist as ions. These dissolved ions produce electrical charges within the body. The behavior of these ions produces the tracings of heart and brain function observed as waves on an electrocardiogram (EKG or ECG) or an electroencephalogram (EEG). The electrical activity that derives from the interactions of the charged ions is why they are also called electrolytes.
Covalent Bonds
Unlike ionic bonds formed by the attraction between a cation’s positive charge and an anion’s negative charge, molecules formed by a covalent bond share electrons in a mutually stabilizing relationship. Like next-door neighbors whose kids hang out first at one home and then at the other, the atoms do not lose or gain electrons permanently. Instead, the electrons move back and forth between the elements. Because of the close sharing of pairs of electrons (one electron from each of two atoms), covalent bonds are stronger than ionic bonds.
Nonpolar Covalent Bonds
Figure \(\PageIndex{2}\) shows several common types of covalent bonds. Notice that the two covalently bonded atoms typically share just one or two electron pairs, though larger sharings are possible. The important concept to take from this is that in covalent bonds, electrons in the outermost valence shell are shared to fill the valence shells of both atoms, ultimately stabilizing both of the atoms involved. In a single covalent bond, a single electron is shared between two atoms, while in a double covalent bond, two pairs of electrons are shared between two atoms. There even are triple covalent bonds, where three atoms are shared.
You can see that the covalent bonds shown in Figure \(\PageIndex{2}\) are balanced. The sharing of the negative electrons is relatively equal, as is the electrical pull of the positive protons in the nucleus of the atoms involved. This is why covalently bonded molecules that are electrically balanced in this way are described as nonpolar; that is, no region of the molecule is either more positive or more negative than any other.
Polar Covalent Bonds
Groups of legislators with completely opposite views on a particular issue are often described as “polarized” by news writers. In chemistry, a polar molecule is a molecule that contains regions that have opposite electrical charges. Polar molecules occur when atoms share electrons unequally, in polar covalent bonds.
The most familiar example of a polar molecule is water (Figure \(\PageIndex{3}\)). The molecule has three parts: one atom of oxygen, the nucleus of which contains eight protons, and two hydrogen atoms, whose nuclei each contain only one proton. Because every proton exerts an identical positive charge, a nucleus that contains eight protons exerts a charge eight times greater than a nucleus that contains one proton. This means that the negatively charged electrons present in the water molecule are more strongly attracted to the oxygen nucleus than to the hydrogen nuclei. Each hydrogen atom’s single negative electron therefore migrates toward the oxygen atom, making the oxygen end of their bond slightly more negative than the hydrogen end of their bond.
What is true for the bonds is true for the water molecule as a whole; that is, the oxygen region has a slightly negative charge and the regions of the hydrogen atoms have a slightly positive charge. These charges are often referred to as “partial charges” because the strength of the charge is less than one full electron, as would occur in an ionic bond. As shown in Figure \(\PageIndex{3.a}\), regions of weak polarity are indicated with the Greek letter delta (∂) and a plus (+) or minus (–) sign.
Even though a single water molecule is unimaginably tiny, it has mass, and the opposing electrical charges on the molecule pull that mass in such a way that it creates a shape somewhat like a triangular tent (see Figure \(\PageIndex{3.b}\) ). This dipole, with the positive charges at one end formed by the hydrogen atoms at the “bottom” of the tent and the negative charge at the opposite end (the oxygen atom at the “top” of the tent) makes the charged regions highly likely to interact with charged regions of other polar molecules. For human physiology, the resulting bond is one of the most important formed by water—the hydrogen bond.
Hydrogen Bonds
A hydrogen bond is formed when a weakly positive hydrogen atom already bonded to one electronegative atom (for example, the oxygen in the water molecule) is attracted to another electronegative atom from another molecule. In other words, hydrogen bonds always include hydrogen that is already part of a polar molecule.
The most common example of hydrogen bonding in the natural world occurs between molecules of water. It happens before your eyes whenever two raindrops merge into a larger bead, or a creek spills into a river. Hydrogen bonding occurs because the weakly negative oxygen atom in one water molecule is attracted to the weakly positive hydrogen atoms of two other water molecules (Figure \(\PageIndex{4}\)).
Water molecules also strongly attract other types of charged molecules as well as ions. This explains why “table salt,” for example, actually is a molecule called a “salt” in chemistry, which consists of equal numbers of positively-charged sodium (Na + ) and negatively-charged chloride (Cl – ), dissolves so readily in water, in this case forming dipole-ion bonds between the water and the electrically-charged ions (electrolytes). Water molecules also repel molecules with nonpolar covalent bonds, like fats, lipids, and oils. You can demonstrate this with a simple kitchen experiment: pour a teaspoon of vegetable oil, a compound formed by nonpolar covalent bonds, into a glass of water. Instead of instantly dissolving in the water, the oil forms a distinct bead because the polar water molecules repel the nonpolar oil.
Chapter Review
Each moment of life, atoms of oxygen, carbon, hydrogen, and the other elements of the human body are making and breaking chemical bonds. Ions are charged atoms that form when an atom donates or accepts one or more negatively charged electrons. Cations (ions with a positive charge) are attracted to anions (ions with a negative charge). This attraction is called an ionic bond. In covalent bonds, the participating atoms do not lose or gain electrons, but rather share them. Molecules with nonpolar covalent bonds are electrically balanced, and have a linear three-dimensional shape. Molecules with polar covalent bonds have “poles”—regions of weakly positive and negative charge—and have a triangular three-dimensional shape. An atom of oxygen and two atoms of hydrogen form water molecules by means of polar covalent bonds. Hydrogen bonds link hydrogen atoms already participating in polar covalent bonds to anions or electronegative regions of other polar molecules. Hydrogen bonds link water molecules, resulting in the properties of water that are important to living things.
Interactive Link Questions
Visit this website to learn about electrical energy and the attraction/repulsion of charges. What happens to the charged electroscope when a conductor is moved between its plastic sheets, and why?
Answer: The plastic sheets jump to the nail (the conductor), because the conductor takes on electrons from the electroscope, reducing the repellant force of the two sheets.
Review Questions
Q. Which of the following is a molecule, but not a compound?
A. H 2 O
B. 2 H
C. H 2
D. H +
Answer: C
Q. A molecule of ammonia contains one atom of nitrogen and three atoms of hydrogen. These are linked with ________.
A. ionic bonds
B. nonpolar covalent bonds
C. polar covalent bonds
D. hydrogen bonds
Answer: C
Q. When an atom donates an electron to another atom, it becomes
A. an ion
B. an anion
C. nonpolar
D. all of the above
Answer: A
Q. A substance formed of crystals of equal numbers of cations and anions held together by ionic bonds is called a(n) ________.
A. noble gas
B. salt
C. electrolyte
D. dipole
Answer: B
Q. Which of the following statements about chemical bonds is true?
A. Covalent bonds are stronger than ionic bonds.
B. Hydrogen bonds occur between two atoms of hydrogen.
C. Bonding readily occurs between nonpolar and polar molecules.
D. A molecule of water is unlikely to bond with an ion.
Answer: A
Critical Thinking Questions
Q. Explain why CH 4 is one of the most common molecules found in nature. Are the bonds between the atoms ionic or covalent?
A. A carbon atom has four electrons in its valence shell. According to the octet rule, it will readily participate in chemical reactions that result in its valence shell having eight electrons. Hydrogen, with one electron, will complete its valence shell with two. Electron sharing between an atom of carbon and four atoms of hydrogen meets the requirements of all atoms. The bonds are covalent because the electrons are shared: although hydrogen often participates in ionic bonds, carbon does not because it is highly unlikely to donate or accept four electrons.
Q. In a hurry one day, you merely rinse your lunch dishes with water. As you are drying your salad bowl, you notice that it still has an oily film. Why was the water alone not effective in cleaning the bowl?
A. Water is a polar molecule. It has a region of weakly positive charge and a region of weakly negative charge. These regions are attracted to ions as well as to other polar molecules. Oils are nonpolar, and are repelled by water.
Q. Could two atoms of oxygen engage in ionic bonding? Why or why not?
A. Identical atoms have identical electronegativity and cannot form ionic bonds. Oxygen, for example, has six electrons in its valence shell. Neither donating nor accepting the valence shell electrons of the other will result in the oxygen atoms completing their valence shells. Two atoms of the same element always form covalent bonds.
Glossary
- anion
- atom with a negative charge
- bond
- electrical force linking atoms
- cation
- atom with a positive charge
- covalent bond
- chemical bond in which two atoms share electrons, thereby completing their valence shells
- hydrogen bond
- dipole-dipole bond in which a hydrogen atom covalently bonded to an electronegative atom is weakly attracted to a second electronegative atom
- ion
- atom with an overall positive or negative charge
- ionic bond
- attraction between an anion and a cation
- molecule
- two or more atoms covalently bonded together
- polar molecule
- molecule with regions that have opposite charges resulting from uneven numbers of electrons in the nuclei of the atoms participating in the covalent bond


18.1B: Blood Vessel Function
Blood vessels carry nutrients and oxygen throughout the body and aid in gas exchange.
- List the functions of blood vessels
Key Points
- Systemic and pulmonary circulatory systems efficiently deliver oxygen to the tissues of the body and remove waste products such as carbon dioxide. Arterial blood (except in the pulmonary artery ) is highly saturated with oxygen and supplies oxygen to the body’s tissues.
- Venous blood (except in the pulmonary vein ) is deoxygenated and returns to the heart to be pumped into the lungs for reoxygenation.
- Nutrients carried in the blood are released to tissues via the permeable endothelium of blood vessels.
- Immune cells move throughout the circulatory system and are able to rapidly permeate the walls of blood vessels to attend sites of injury or infection.
- Blood vessels can increase or decrease blood flow near the surface of the body, either increasing or reducing the amount of heat lost as a means of regulating body temperature.
Key Terms
- thermoregulation : The maintenance of a constant internal temperature of an organism independent of the temperature of the environment
Blood plays many critical roles within the body: delivering nutrients and chemicals to tissues, removing waste products, and maintaining homeostasis and health. The circulatory system is transports blood through the body to perform these actions, facilitated by the extensive network of blood vessels.
Gas Transfer
The circulatory system can be split into two sections, systemic and pulmonary. In the systemic circulatory system, highly oxygenated blood (95-100%) is pumped from the left ventricle of the heart and into the arteries of the body. Upon reaching the capillary networks, gas exchange between tissue and blood can occur, facilitated by the narrow walls of the capillaries. Oxygen is released from the blood into the tissues and carbon dioxide, a waste product of respiration, is absorbed. The capillaries merge into venules and then veins, carrying the deoxygenated blood (~75%) back to the right atrium of the heart at the end of the systemic circulatory system.
The much smaller pulmonary system reoxygenates the blood and facilitates the removal of carbon dioxide. After leaving the heart through the right ventricle, the blood passes through the pulmonary artery, the only artery in the body that contains deoxygenated blood, and into the capillary network within the lungs. The close association of the thin-walled alveolae with the equally thin-walled capillaries allows for rapid release of carbon dioxide and uptake of oxygen. After leaving the lungs through the pulmonary vein, the only vein which carries oxygenated blood, the blood enters the left atrium. This completes the pulmonary circulatory system.
The Circulatory System : This simplified diagram of the human circulatory system (anterior view) shows arteries in red and veins in blue.
Additional Functions
Blood vessels also facilitate the rapid distribution and efficient transport of factors such as glucose, amino acids, or lipids into the tissues and the removal of waste products for processing elsewhere, such as lactic acid to the liver or urea to the kidneys. Additionally, blood vessels provide the ideal network for immune system surveillance and distribution. Numerous white blood cells circulate around the body, sensing for infection or injury. Once an injury is detected, they rapidly leave the circulatory system by passing through gaps in vessel walls to reach the affected area while signalling for a larger targeted immune response.
Mechanically the blood vessels, especially those near the skin, play a key role in thermoregulation. Blood vessels can swell to allow greater blood flow, allowing for greater radiant heat loss. Conversely, blood flow through these vessels can be lessened to reduce heat loss in colder climates.
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4.4: Sterile Fields
A sterile field is established whenever a patient’s skin is intentionally punctured or incised, during procedures involving entry into a body cavity, or when contact with nonintact skin is possible (e.g., surgery or trauma). Surgical asepsis requires adherence to strict principles and intentional actions to prevent contamination and to maintain the sterility of specific parts of a sterile field during invasive procedures. Creating and maintaining a sterile field is foundational to aseptic technique and encompasses practice standards that are performed immediately prior to and during a procedure to reduce the risk of infection, including the following:
- Handwashing
- Using sterile barriers, including drapes and appropriate personal protective equipment
- Preparing the patient using an approved antimicrobial product
- Maintaining a sterile field
- Using aseptically safe techniques
There are basic principles of asepsis that are critical to understand and follow when creating and maintaining a sterile field. The most basic principle is to allow only sterile supplies within the sterile field once it is established. This means that prior to using any supplies, exterior packaging must be checked for any signs of damage, such as previous exposure to moisture, holes, or tears. Packages should not be used if they are expired or if sterilization indicators are not the appropriate color. Sterile contents inside packages are dispensed onto the sterile field using the methods outlined below. See Figure \(\PageIndex{1}\) [1] for an image of a nurse dispensing sterile supplies from packaging onto an established sterile field.
When establishing and maintaining a sterile field, there are other important principles to strictly follow:
- Disinfect any work surfaces and allow to them thoroughly dry before placing any sterile supplies on the surface.
-
Be aware of areas of sterile fields that are considered contaminated:
- Any part of the field within 1 inch from the edge.
- Any part of the field that extends below the planar surface (i.e., a drape hanging down below the tray tabletop).
- Any part of the field below waist level or above shoulder level.
- Any supplies or field that you have not directly monitored (i.e., turned away from the sterile field or walked out of the room).
- Within 1 inch of any visible holes, tears, or moisture wicked from an unsterile area.
-
When handling sterile kits and trays:
- Sterile kits and trays generally have an outer protective wrapper and four inner flaps that must be opened aseptically.
- Open sterile kits away from your body first, touching only the very edge of the opening flap.
- Using the same technique, open each of the side flaps one at a time using only one hand, being careful not to allow your body or arms to be directly above the opened drape. Take care not to allow already-opened corners to flip back into the sterile area again.
- Open the final flap toward you, being careful to not allow any part of your body to be directly over the field. See Figure \(\PageIndex{2}\) [2] for images of opening the flaps of a sterile kit.
- If you must remove parts of the sterile kit (i.e., sterile gloves), reach into the sterile field with the elbow raised, using only the tips of the fingers before extracting. Pay close attention to where your body and clothing are in relationship to the sterile field to avoid inadvertent contamination.
-
When dispensing sterile supplies onto a sterile field:
- Before dispensing sterile supplies to a sterile field, do not allow sterile items to touch any part of the outer packaging once it is opened, including the former package seal.
- Heavy or irregular items should be opened and held, allowing a second person with sterile gloves to transfer them to the sterile field.
- Wrapped sterile items should be opened similarly to a sterile kit. Tuck each flap securely within your palm, and then open the flap away from your body first. Then open each side flap; secure the flap in the palm one at a time and open. Finally, open the flap (closest to you) toward you, while also protecting the other opened flaps from springing back onto the wrapped item.
- Peel pouches (i.e., gloves, gauze, syringes, etc.) can be opened by firmly grasping each side of the sealed edge with the thumb side of each hand parallel to the seal and pulling carefully apart.
- Drop items from six inches away from the sterile field.
- Sterile solutions should be poured into a sterile bowl or tray from the side of the sterile field and not directly over it. Use only sealed, sterile, unexpired solutions when pouring onto a sterile field. Solution should be held six inches away from the field as it is being poured. Avoid splashing solutions because this allows wicking and transfer of microbes. After pouring of solution stops, it should not be restarted because the edge is considered contaminated. See Figure \(\PageIndex{3}\) [3] of an image of a nurse pouring sterile solution into a receptacle in a sterile field before the procedure begins.
- Don sterile gloves away from the sterile field to avoid contaminating the sterile field.
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5.4: A Population Proportion
During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within three percentage points (if the sample is large enough). Often, election polls are calculated with 95% confidence, so, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43: (0.40 – 0.03,0.40 + 0.03).
Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers.
The procedure to find the confidence interval, the sample size, the error bound, and the confidence level for a proportion is similar to that for the population mean, but the formulas are different. How do you know you are dealing with a proportion problem? First, the underlying distribution is a binomial distribution. (There is no mention of a mean or average.) If \(X\) is a binomial random variable, then
\[X \sim B(n, p)\nonumber \]
where \(n\) is the number of trials and \(p\) is the probability of a success.
To form a proportion, take \(X\), the random variable for the number of successes and divide it by \(n\), the number of trials (or the sample size). The random variable \(P′ \) (read "P prime") is that proportion,
\[P' = \dfrac{X}{n}\nonumber \]
(Sometimes the random variable is denoted as \(\hat{P}\), read "P hat".)
When \(n\) is large and \(p\) is not close to zero or one, we can use the normal distribution to approximate the binomial.
\[X \sim N(np, \sqrt{npq})\nonumber \]
If we divide the random variable, the mean, and the standard deviation by \(n\), we get a normal distribution of proportions with \(P′ \), called the estimated proportion, as the random variable. (Recall that a proportion as the number of successes divided by \(n\).)
\[\dfrac{X}{n} = P' - N\left(\dfrac{np}{n}, \dfrac{\sqrt{npq}}{n}\right)\nonumber \]
Using algebra to simplify:
\[\dfrac{\sqrt{npq}}{n} = \sqrt{\dfrac{pq}{n}}\nonumber \]
P′ follows a normal distribution for proportions:
\[\dfrac{X}{n} = P' - N\left(\dfrac{np}{n}, \dfrac{\sqrt{npq}}{n}\right)\nonumber \]
The confidence interval has the form
\[(p′ – EBP, p′ + EBP).\nonumber \] or \[(p′ – E, p′ + E).\nonumber \]
where
- \(EBP\) or \(E\) is error bound for the proportion.
- \(p′ = \dfrac{x}{n}\)
- \(p′ =\) the estimated proportion of successes ( p′ is a point estimate for p , the true proportion.)
- \(x =\) the number of successes
- \(n =\) the size of the sample
The error bound (EBP or E) for a proportion is
\[EBP = E = \left(z_{\frac{\alpha}{2}}\right)\left(\sqrt{\dfrac{p'q'}{n}}\right)\nonumber \]
where \(q\ = 1 - p'\).
This formula is similar to the error bound formula for a mean, except that the "appropriate standard deviation" is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is \(\dfrac{\sigma}{\sqrt{n}}\). For a proportion, the appropriate standard deviation is
\[\sqrt{\dfrac{pq}{n}}.\nonumber \]
However, in the error bound formula, we use
\[\sqrt{\dfrac{p'q'}{n}}\nonumber \]
as the standard deviation, instead of
\[\sqrt{\dfrac{pq}{n}}.\nonumber \]
In the error bound formula, the sample proportions p′ and q′ are estimates of the unknown population proportions p and q . The estimated proportions \(p′\) and \(q′\) are used because \(p\) and \(q\) are not known. The sample proportions \(p′\) and \(q′\) are calculated from the data: \(p′\) is the estimated proportion of successes, and \(q′\) is the estimated proportion of failures.
The confidence interval can be used only if the number of successes \(np′\) and the number of failures \(nq′\) are both greater than five.
For the normal distribution of proportions, the \(z\)-score formula is as follows.
If
\[P' - N\left(p, \sqrt{\dfrac{pq}{n}}\right)\]
then the \(z\)-score formula is
\[z = \dfrac{p'-p}{\sqrt{\dfrac{pq}{n}}} \]
Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes - they own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones.
Solution A
- The first solution is step-by-step (Solution A).
- The second solution uses a function of the TI-83, 83+ or 84 calculators (Solution B).
Let \(X =\) the number of people in the sample who have cell phones. \(X\) is binomial.
\[X \sim B(500,\dfrac{421}{500}).\nonumber \]
To calculate the confidence interval, you must find \(p′\), \(q′\), and \(EBP\).
- \(n = 500\)
- \(x =\) the number of successes \(= 421\)
\[p′ = \dfrac{x}{n} = \dfrac{421}{500} = 0.842\nonumber \]
- \(p′ = 0.842\) is the sample proportion; this is the point estimate of the population proportion.
\[q′ = 1 – p′ = 1 – 0.842 = 0.158\nonumber \]
Since \(CL = 0.95\), then
\[\alpha = 1 – CL = 1 – 0.95 = 0.05\left(\dfrac{\alpha}{2}\right) = 0.025.\nonumber \]
Then
\[z_{\dfrac{\alpha}{2}} = z_{0.025 = 1.96}\nonumber \]
Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find \(z_{0.025}\). Remember that the area to the right of \(z_{0.025}\) is \(0.025\) and the area to the left of \(z_{0.025}\) is \(0.975\). This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.
\[EBP = E = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = (1.96)\sqrt{\dfrac{(0.842)(0.158)}{500}} = 0.032\nonumber \]
\[p' – E = 0.842 – 0.032 = 0.81\nonumber \]
\[p′ + E = 0.842 + 0.032 = 0.874\nonumber \]
The confidence interval for the true binomial population proportion is \((p′ – E, p′ + E) = (0.810, 0.874)\).
Interpretation
We estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have cell phones.
Explanation of 95% Confidence Level
Ninety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones.
Solution B
Press
STAT
and arrow over to
TESTS
.
Arrow down to
A:1-PropZint
. Press
ENTER
.
Arrow down to xx and enter 421.
Arrow down to nn and enter 500.
Arrow down to
C-Level
and enter .95.
Arrow down to
Calculate
and press
ENTER
.
The confidence interval is (0.81003, 0.87397).
Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets.
- Answer
-
(0.3315, 0.4525)
For a class project, a political science student at a large university wants to estimate the percent of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students who are registered voters, and interpret the confidence interval.
Answer
- The first solution is step-by-step (Solution A).
- The second solution uses a function of the TI-83, 83+, or 84 calculators (Solution B).
Solution A
- \(x = 300\) and
- \(n = 500\)
\[p' = \dfrac{x}{n} = \dfrac{300}{500} = 0.600\nonumber \]
\[q′ = 1 − p′ = 1 − 0.600 = 0.400\nonumber \]
Since \(CL = 0.90\), then
\[\alpha = 1 – CL = 1 – 0.90 = 0.10\left(\dfrac{\alpha}{2}\right) = 0.05\]
\[z_{\dfrac{\alpha}{2}} = z_{0.05} = 1.645\nonumber \]
Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find \(z_{0.05}\). Remember that the area to the right of \(z_{0.05}\) is 0.05 and the area to the left of \(z_{0.05}\) is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table.
\[EBP = E = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = (1.645)\sqrt{\dfrac{(0.60)(0.40)}{500}} = 0.036\nonumber \]
\[p′ – E = 0.60 − 0.036 = 0.564\nonumber \]
\[p′ + E = 0.60 + 0.036 = 0.636\nonumber \]
The confidence interval for the true binomial population proportion is \((p′ – E, p′ + E) = (0.564,0.636)\).
Interpretation
- We estimate with 90% confidence that the true percent of all students that are registered voters is between 56.4% and 63.6%.
- Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL students are registered voters.
Explanation of 90% Confidence Level
Ninety percent of all confidence intervals constructed in this way contain the true value for the population percent of students that are registered voters.
Solution B
Press
STAT
and arrow over to
TESTS
.
Arrow down to
A:1-PropZint
. Press
ENTER
.
Arrow down to xx and enter 300.
Arrow down to nn and enter 500.
Arrow down to
C-Level
and enter 0.90.
Arrow down to
Calculate
and press
ENTER
.
The confidence interval is (0.564, 0.636).
A student polls his school to see if students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation.
- Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval.
- In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone.
- Answer a
-
(0.7731, 0.8269); We estimate with 90% confidence that the true percent of all students in the district who are against the new legislation is between 77.31% and 82.69%.
- Answer b
-
Sixty-eight percent (68%) of students own an iPod and a smart phone.
\[p′ = 0.68\nonumber \]
\[q′ = 1–p′ = 1 – 0.68 = 0.32\nonumber \]
Since \(CL = 0.97\), we know
\[\alpha = 1 – 0.97 = 0.03\nonumber \]
and
\[\dfrac{\alpha}{2} = 0.015.\nonumber \]
The area to the left of \(z_{0.05}\) is 0.015, and the area to the right of \(z_{0.05}\) is 1 – 0.015 = 0.985.
Using the TI 83, 83+, or 84+ calculator function InvNorm(0.985,0,1),
\[z_{0.05} = 2.17\nonumber \]
\[EPB = E = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = 2.17\sqrt{\dfrac{0.68(0.32)}{300}} \approx 0.0269\nonumber \]
\[p′ – E = 0.68 – 0.0269 = 0.6531\nonumber \]
\[p′ + E = 0.68 + 0.0269 = 0.7069\nonumber \]
We are 97% confident that the true proportion of all students who own an iPod and a smart phone is between 0.6531 and 0.7069.
Calculator
Press
STAT
and arrow over toTESTS
.Arrow down to
A:1-PropZint
. PressENTER
.
Arrow down to x and enter 300*0.68.
Arrow down to n and enter 300.
Arrow down to C-Level and enter 0.97.
Arrow down to Calculate and pressENTER
.The confidence interval is (0.6531, 0.7069).
"Plus Four" Confidence Interval for \(p\)
There is a certain amount of error introduced into the process of calculating a confidence interval for a proportion. Because we do not know the true proportion for the population, we are forced to use point estimates to calculate the appropriate standard deviation of the sampling distribution. Studies have shown that the resulting estimation of the standard deviation can be flawed.
Fortunately, there is a simple adjustment that allows us to produce more accurate confidence intervals. We simply pretend that we have four additional observations. Two of these observations are successes and two are failures. The new sample size, then, is \(n + 4\), and the new count of successes is \(x + 2\). Computer studies have demonstrated the effectiveness of this method. It should be used when the confidence level desired is at least 90% and the sample size is at least ten.
A random sample of 25 statistics students was asked: “Have you smoked a cigarette in the past week?” Six students reported smoking within the past week. Use the “plus-four” method to find a 95% confidence interval for the true proportion of statistics students who smoke.
Solution A
Six students out of 25 reported smoking within the past week, so \(x = 6\) and \(n = 25\). Because we are using the “plus-four” method, we will use \(x = 6 + 2 = 8\) and \(n = 25 + 4 = 29\).
\[p' = \dfrac{x}{n} = \dfrac{8}{29} \approx 0.276\nonumber \]
\[q′ = 1 – p′ = 1 – 0.276 = 0.724\nonumber \]
Since \(CL = 0.95\), we know \(\alpha = 1 – 0.95 = 0.05\) and \(\dfrac{\alpha}{2} = 0.025\).
\[z_{0.025} = 1.96\nonumber \]
\(EPB = E = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = (1.96)\sqrt{\dfrac{0.276(0.724)}{29}} \approx 0.163\)
\[p′ – E = 0.276 – 0.163 = 0.113\nonumber \]
\[p′ + E = 0.276 + 0.163 = 0.439\nonumber \]
We are 95% confident that the true proportion of all statistics students who smoke cigarettes is between 0.113 and 0.439.
Solution B
Press
STAT
and arrow over to
TESTS
.
Arrow down to
A:1-PropZint
. Press
ENTER
.
REMINDER
Remember that the plus-four method assume an additional four trials: two successes and two failures. You do not need to change the process for calculating the confidence interval; simply update the values of x and n to reflect these additional trials.
Arrow down to
x
and enter eight.
Arrow down to
n
and enter 29.
Arrow down to
C-Level
and enter 0.95.
Arrow down to
Calculate
and press
ENTER
.
The confidence interval is (0.113, 0.439).
Out of a random sample of 65 freshmen at State University, 31 students have declared a major. Use the “plus-four” method to find a 96% confidence interval for the true proportion of freshmen at State University who have declared a major.
Solution A
Using “plus four,” we have \(x = 31 + 2 = 33\) and \(n = 65 + 4 = 69\).
\[p′ = 3369 \approx 0.478\nonumber \]
\[q′ = 1 – p′ = 1 – 0.478 = 0.522\nonumber \]
Since \(CL = 0.96\), we know \(\alpha = 1 – 0.96 = 0.04\) and \(\dfrac{\alpha}{2} = 0.02\).
\[z_{0.02} = 2.054\nonumber \]
\[EPB = E = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = (2.054)\left(\sqrt{\dfrac{(0.478)(0.522)}{69}}\right) - 0.124\nonumber \]
\[p′ – E = 0.478 – 0.124 = 0.354\nonumber \]
\[p′ + E = 0.478 + 0.124 = 0.602\nonumber \]
We are 96% confident that between 35.4% and 60.2% of all freshmen at State U have declared a major.
Solution B
Press
STAT
and arrow over to
TESTS
.
Arrow down to
A:1-PropZint
. Press
ENTER
.
Arrow down to
x
and enter 33.
Arrow down to
n
and enter 69.
Arrow down to
C-Level
and enter 0.96.
Arrow down to
Calculate
and press
ENTER
.
The confidence interval is (0.355, 0.602).
The Berkman Center for Internet & Society at Harvard recently conducted a study analyzing the privacy management habits of teen internet users. In a group of 50 teens, 13 reported having more than 500 friends on Facebook. Use the “plus four” method to find a 90% confidence interval for the true proportion of teens who would report having more than 500 Facebook friends.
Solution A
Using “plus-four,” we have \(x = 13 + 2 = 15\) and \(n = 50 + 4 = 54\).
\[p′ = 1554 \approx 0.278\nonumber \]
\[q′ = 1 – p′ = 1 − 0.241 = 0.722\nonumber \]
Since \(CL = 0.90\), we know \(\alpha = 1 – 0.90 = 0.10\) and \(\dfrac{\alpha}{2} = 0.05\).
\[z_{0.05} = 1.645\nonumber \]
\[EPB = E = \left(z_{\dfrac{\alpha}{2}}\right)\left(\sqrt{\dfrac{p'q'}{n}}\right) = (1.645)\left(\sqrt{\dfrac{(0.278)(0.722)}{54}}\right) \approx 0.100\nonumber \]
\[p′ – E = 0.278 – 0.100 = 0.178\nonumber \]
\[p′ + E = 0.278 + 0.100 = 0.378\nonumber \]
We are 90% confident that between 17.8% and 37.8% of all teens would report having more than 500 friends on Facebook.
Solution B
Press
STAT
and arrow over to
TESTS
.
Arrow down to
A:1-PropZint
. Press
ENTER
.
Arrow down to
x
and enter 15.
Arrow down to
n
and enter 54.
Arrow down to
C-Level
and enter 0.90.
Arrow down to
Calculate
and press
ENTER
.
The confidence interval is (0.178, 0.378).
The Berkman Center Study referenced in Example talked to teens in smaller focus groups, but also interviewed additional teens over the phone. When the study was complete, 588 teens had answered the question about their Facebook friends with 159 saying that they have more than 500 friends. Use the “plus-four” method to find a 90% confidence interval for the true proportion of teens that would report having more than 500 Facebook friends based on this larger sample. Compare the results to those in Example.
Answer
Solution A
Using “plus-four,” we have \(x = 159 + 2 = 161\) and \(n = 588 + 4 = 592\).
\[p′ = 161592 \approx 0.272\nonumber \]
\[q′ = 1 – p′ = 1 – 0.272 = 0.728\nonumber \]
Since CL = 0.90, we know \(\alpha = 1 – 0.90 = 0.10\) and \(\dfrac{\alpha}{2} = 0.05\)
\[EPB = E = \left(z_{\dfrac{\alpha}{2}}\right)\left(\sqrt{\dfrac{p'q'}{n}}\right) = (1.645)\left(\sqrt{\dfrac{(0.272)(0.728)}{592}}\right) \approx 0.030\nonumber \]
\[p′ – E = 0.272 – 0.030 = 0.242\nonumber \]
\[p′ + E = 0.272 + 0.030 = 0.302\nonumber \]
We are 90% confident that between 24.2% and 30.2% of all teens would report having more than 500 friends on Facebook.
Solution B
-
Press
STAT
and arrow over toTESTS
. -
Arrow down to
A:1-PropZint
. PressENTER
. -
Arrow down to
x
and enter 161. -
Arrow down to
n
and enter 592. -
Arrow down to
C-Level
and enter 0.90. -
Arrow down to
Calculate
and pressENTER
. - The confidence interval is (0.242, 0.302).
Conclusion: The confidence interval for the larger sample is narrower than the interval from Example. Larger samples will always yield more precise confidence intervals than smaller samples. The “plus four” method has a greater impact on the smaller sample. It shifts the point estimate from 0.26 (13/50) to 0.278 (15/54). It has a smaller impact on the EPB (or E) , changing it from 0.102 to 0.100. In the larger sample, the point estimate undergoes a smaller shift: from 0.270 (159/588) to 0.272 (161/592). It is easy to see that the plus-four method has the greatest impact on smaller samples.
Calculating the Sample Size \(n\)
If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. The error bound formula for a population proportion is
\[EBP = E = \left(z_{\frac{\alpha}{2}}\right)\left(\sqrt{\dfrac{p'q'}{n}}\right)\nonumber \]
Solving for \(n\) gives you an equation for the sample size.
\[n = \dfrac{\left(z_{\frac{\alpha}{2}}\right)^{2}(p'q')}{E^{2}}\nonumber \]
Suppose a mobile phone company wants to determine the current percentage of customers aged 50+ who use text messaging on their cell phones. How many customers aged 50+ should the company survey in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of customers aged 50+ who use text messaging on their cell phones.
Answer
From the problem, we know that \(\bf{E = 0.03}\) (3%=0.03) and \(z_{\dfrac{\alpha}{2}} z_{0.05} = 1.645\) because the confidence level is 90%.
However, in order to find \(n\), we need to know the estimated (sample) proportion \(p′\). Remember that \(q′ = 1 – p′\). But, we do not know \(p′\) yet. Since we multiply \(p′\) and \(q′\) together, we make them both equal to 0.5 because \(p′q′ = (0.5)(0.5) = 0.25\) results in the largest possible product. (Try other products: \((0.6)(0.4) = 0.24\); \((0.3)(0.7) = 0.21\); \((0.2)(0.8) = 0.16\) and so on). The largest possible product gives us the largest \(n\). This gives us a large enough sample so that we can be 90% confident that we are within three percentage points of the true population proportion. To calculate the sample size \(n\), use the formula and make the substitutions.
\[n = \dfrac{z^{2}p'q'}{E^{2}}\nonumber \]
gives
\[n = \dfrac{1.645^{2}(0.5)(0.5)}{0.03^{2}} = 751.7\nonumber \]
Round the answer to the next higher value. The sample size should be 752 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of all customers aged 50+ who use text messaging on their cell phones.
Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 90% confident that the estimated proportion is within five percentage points of the true population proportion of customers who click on ads on their smartphones?
- Answer
-
271 customers should be surveyed. Check the Real Estate section in your local
Glossary
- Binomial Distribution
- a discrete random variable (RV) which arises from Bernoulli trials; there are a fixed number, \(n\), of independent trials. “Independent” means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV \(X\) is defined as the number of successes in \(n\) trials. The notation is: \(X \sim B(\mathbf{n},\mathbf{p})\). The mean is \(\mu = np\) and the standard deviation is \(\sigma = \sqrt{npq}\). The probability of exactly \(x\) successes in \(n\) trials is \(P(X = x = \left(\binom{n}{x}\right))p^{x}q^{n-x}\).
- Error Bound for a Population Proportion (\(EBP\) or \(E\)
- the margin of error; depends on the confidence level, the sample size, and the estimated (from the sample) proportion of successes.


11.2: Supporting Children’s Learning
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Much of children’s development in the creative arts during the preschool years proceeds naturally and needs only fertile soil, along with time, to grow. Children initiate many behaviors and routines when they simply go about their play. They practice many skills along the way, and supportive physical and social–emotional development occurs as children progress from ages three to five. At the same time, their drawings become more mature and expressive, their pretend characters and settings become more complex and social, their musical expression skills grow with their muscular coordination and abilities to discern beat, tone, and melody, and the movements they coordinate with music or simply orchestrate in silence gain in surety and expressive complexity.
A primary responsibility of the preschool teacher is to let such natural developments occur. Child-initiated artistic activity is valuable not only because it is so enmeshed with a host of developments for children, but also because children cherish ownership of much of what they do. Children follow their hearts and minds to what interests them and to areas where they experience increasing mastery. They draw as they will and may not be interested in exactly what thing, animal, or person the creation represents. They may hold firm to their idea of how to draw a tree, behave like a bear, or sing like a bird; it often becomes important for teachers and other adults to avoid critiquing such expressions (except where the child may solicit advice). Teachers would do well to let the child experiment with, and perhaps revise, her expressions as the need occurs or as maturing views of the world and its possible representations take hold.
Along with child-initiated art, a complementary perspective needs reinforcement. This perspective recognizes the ways that teachers can and should support young learners in their development. An element of this scaffolding is creating conditions in the preschool program in which interesting and important connections between the arts and other developments can take place. Capitalizing on language and communication opportunities is another example; placing children in settings where cooperation is important and where cooperative dispositions and skills may grow is yet another. Some art activities can help children become aware of and reflect on differences among people, become exposed to diverse art forms from different cultures, and create a common platform of learning for children between home and school. These considerations will set the stage for children’s growth and interest in the arts. [2]
[1] Image by Kathy Desy is in the public domain.
[2] The California Preschool Curriculum Framework, Volume 2 by the California Department of Education is used with permission


1.4: Aseptic Technique
- Define aseptic technique, sterilization, and contamination.
- Discuss the importance of growth media and keeping it sterile.
- Recognize potential sources of contamination.
- Complete the Aseptic Technique Procedure.
Aseptic Technique and Sterile Environments
The goal of microbiology is to generate and study pure cultures : growth media containing a single microbial species. Pure cultures provide objective, unbiased data that can be used in clinical analysis and research. If unwanted growth of other microbial species is present in a culture, the sample is contaminated. To prevent sample contamination, maintaining a sterile environment through aseptic technique is vital. Sterile environments are free from all unwanted microbes including exogenous fungi, yeasts, endospores, vegetative cells, and viruses. (OpenStax, 2024) Sterilization can be accomplished through chemicals, like 10% bleach solutions, and physical means, like exposur e to high heat and pressure in an autoclave. All steps and procedures that help prevent contamination are a part of aseptic technique, "aseptic" meaning "without contamination". Routine disinfection, media sterilization, equipment sterilization, and proper transfer of samples/microbes are all key aseptic techniques that you will complete in this lab course to maintain a sterile working environment.
Routine Disinfection
Disinfection uses antimicrobial chemicals to kill microbes on surfaces. These chemicals should be fast acting, easy to prepare and use, and shelf stable. (OpenStax, 2024) Natural disinf ectants include vinegar whose high acidity is capable of inactivating most microbes. In this microbiology lab, we will use the chemical disinfectant chlorine bleach. 10% bleach soluti ons will be used to clean your work area (lab bench) daily, before and after experiments. Undiluted Bleach will be used to clean up any spills as described in "1.1 Laboratory Safety". We will also use mild antimicrobial soap to wash hands during the lab. While disinfection does not completely sterilize a work area or hands, some microbes will always remain, it is still a key component of aseptic technique.
Growth Media
Microbes are grown on or in substances composed of nutrients needed for cellular metabolism. These substances are called media or growth media. Growth media can be liquid, solid, or semi-solid and is ma de with different carbon sources, sugar sources, and any other nutrient needed to support microbial growth. Liquid media is also called "broth". Broth cultures can be used to observe metabolic pathways in bacteria, such as urease production, and are a good tool for growing large amounts of a microbe quickly. Both semi-solid and solid media are made using a polysaccharide, called agar, in different amounts to create a solid substance. Agar is naturally solid at room temperature, contains no nutrients, and is not digestible by microbes (specifically bacteria) making it an ideal solidification agent. (Hartline, 2023) Semi-solid agar can be used to test for microbial motility. The semi-solid state of the media allows the microbes to penetrate the media and move out from the area of inoculation, serving as a quick method to determine if motility structures are present. Finally, solid media has the greatest variety of experimental usage. As seen in Figure 1, solid agar can be used to make slants, agar deeps, and petri plates. Typically, slants are used for making stock cultures, agar deeps for metabolism experiments, and plates for observing colony morphologies. Media in test tubes should always be covered with caps to prevent contamination and kept in test tube racks. Petri dishes are kept inverted, see Figure 2. Petri dishes are kept in incubators with the lid-side down and agar-side up to prevent any dirt, dust, condensation, or foreign microbes from falling onto the agar during incubation.
Figure 1. Media can be liquid (broth), semi-solid, or solid depending on the needs of an experiment.
Figure 2. Petri dishes should be inverted in incubators to decrease the chance of contamination.
Microbial Transfer Procedures and Media Inoculation
Growth media needs to be sterile prior to the experiment and remain uncontaminated during lab use. Growth media is sterilized using physical means, specifcally, an autoclave. Autoclaves use high heat (steam) and pressure to inactivate or terminate microbes including vegetative cells and spores. The autoclave functions with a pressure of 15-20lbs/ in 2 and reaches temperatures between 121 o C and 132 o C. Media and other BSL1 liquids, used petri dishes, and used perishable items are autoclaved for 15 to 20 minutes depending on the volume of material present. (OpenStax, 2024 and Hartline, 2023) To keep growth media uncontaminated during an experiment, specific transfer procedures are used to minimize microbial transmission. A key step in microbial transfer is equipment sterilization. Sterile swabs, inoculation loops, and inoculation needles can be used to move microbes between different medias. Inoculation loops and needles must be sterilized before and after direct contact with a microbe. Either a Bunsen burner or table-top "mini" incinerator should be used to sterilize loops and needles. The wire must be kept in the Bunsen burner flame or incinerator body for at least 10 seconds until the wire has turned red/orange. Never leave the wire unattended in either heat source, the wire will become hot and may cause burns or melt. The inoculation loop or needle should be allowed to cool for 10 to 15 seconds before picking up any microbe. If the wire is still hot, it can burn, damage, or kill the microbe being worked with. Once a sterile swab, loop, or needle is ready to use, microbes should be moved/transferred in the following order:
- Stock Broth to Sample Broth Tube
- Sample Broth Tube to Stock Plate (Petri Dish)
-
Stock Plate to Sample Plate
- In this lab, you will be provided with pre-made stock plates to use for sample plate preparation.
The process of transferring microbes to d ifferent medias and spreading them across a media's surface is called inoculation. Before transferring and inoculating microbes, test tubes and petri dishes must be properly labeled. Test tubes should be labeled using labeling tape and include: student initials, date, course section, and sample type. Petri dishes can be written on directly using a sharpie. To label a petri dish, write in small letters around the edge of the bottom plate and include: student initials, date, course section, and sample type. Do not write on the lid, the lid will not be visible in the incubator.
Figure 3. Example for how to label the bottom of a petri-dish. Write in small letters around the edge of the bottom dish. Never write on the lid of a petri dish. The lid my be damaged, lost, or not visible in an incubator.
Once everything is properly labeled, to collect microbes from a stock plate:
- Place the plate lid-side up and agar-side down on the work bench.
- Using a non-dominant hand, lift the lid slightly. Never take the lid fully off, always keep it partially covering the plate to shield the agar.
- Touch a sterile swab, loop, or needle to an area with no microbial growth to test that the metal has cooled down.
- Gently, touch part of a microbial colony.  There are thousands of cells in a single colony. A little bacteria goes a long way!
- Be careful to not press too hard on the agar, it can break.
- After microbes have been collected, replace the lid on the stock plate and prepare to transfer the microbes to a secondary media: broths, slants, deeps, or sample plates.
Figure 4. This image shows the proper technique for opening petri dishes. Always keep the lid over the exposed media to prevent contamination.
To inoculate a sample plate:
-
In this lab we will complete quad streak inoculations. A more in-depth lesson on completing a quad streak can be found in "2.1: Introduction to Streak Plate Techniques". This chapter will present a brief description of proper microbial streaking.
- We will also discuss slant, semi-solid, and deep agar inoculations in "Biochemical Characterization Part I and II".
-
With a non-dominant hand, lift the lid slightly on the sample plate.

- With the swab, loop, or needle containing microbes swipe the bacteria in a zig-zag pattern across one region/quadrant of the plate.
- Now, sterilize the wire and let it cool. When the wire or swab is ready, drag a small portion of bacteria from the first quadrant and swipe in a new zig-zag pattern in a new secondary quadrant.
- Repeat this process until four quadrants of bacteria have been created on one plate.
Inoculated petri dishes should be wrapped in parafilm. Parafilm keeps samples uncontaminated while also protecting the user from direct transfer contamination. Potential sources of contamination include (CDC, 2016):
- People: students, instructors, and anyone else who may enter the lab space.
- Surfaces: tables, counter tops, personal items like bags or books, phones and other electronics, sinks and faucets.
- Equipment: incinerators, inoculation loops or needles, and microscopes.
- Foreign microbes, dust, dirt, and other biological materials.
Sample plates should always be stored in an incubator, a temperature controlled growth environment, to promote growth of the microbe. In this lab, our incubators are set to 37°C as we are working with mesophiles, microbes that prefer to grow in or on human hosts. After 24 to 48 hours, microbial growth should be visible on or in all media types and ready to use for follow-up experiments.
Medical Asepsis
Aseptic technique in a healthcare setting, between patient and healthcare worker, uses a strict set of hygiene practices known as medical asepsis. Healthcare workers closely monitor patient hygiene and mobility to determine what intervention is necessary to treat infections, prevent new infections, and promote patient health. Patient health and hygiene in medical asepsis pays close attention to the skin, hair, nails, oral cavity, and perineum. (Openstax, 2024). These cells and tissues can easily become infected or transmit diseases if aseptic techniques are not used correctly. Here, we will focus on the skin.
A patient’s skin is their largest organ and first defense against external pathogens. Normal, healthy skin will help a patient regulate their body temperature, produce sebum to stop pathogens, and prevent water lose. The skin can become weakened and damaged in healthcare settings. Long-term immobility or even medications like anticoagulants can affect skin health. Changes to skin health put the patient at risk for developing infections like Scabies.
Scabies is a contagious skin infection caused by the Sarcoptes scabiei mite. Scabies outbreaks are common in nursing facilities, childcare facilities, and detention facilities. (CDC, 2023). The mite targets a patient’s weakened skin barrier and burrows into the upper skin layer. It causes intense itching and pain and appears as a red pimple-like rash. Scabies will be found most often in skin folds near the armpits or knees and other locations where patients may have difficulty cleaning themselves. The mite is spread by direct contact and can spread from patient to healthcare worker if proper aseptic technique is not followed. Washing clothing and bedding, wearing proper PPE, and limited physical contact will decrease the risk of transmission. Scabies can be prevented with regular cleaning, exfoliation, and moisturizing of the patient’s skin.
Attributions and Text Citations
"Microbiology Laboratory Manual: Labs, 1.7 Aseptic Technique" by Dr. Rosanna Hartline , LibreTexts: Biology , West Hills College Lemoore is licensed under CC BY-NC 4.0
"Microbiology Textbook: Chapter 9, Microbial Growth" by OpenStax , Digital ISBN 13: 978-1-947172-23-4 is licensed under CC BY 4.0
"Microbiology Textbook: Chapter 13, Control of Microbial Growth" by OpenStax , Digital ISBN 13: 978-1-947172-23-4 is licensed under CC BY 4.0
"Microbiology Labs 1: 2 , Aseptic Transfers " by Jackie Reynolds, LibreText Biology , Dallas College, Richland Campus is licensed under CC BY-NC 4.0
"Microbiology for Allied Health Students: Lab Manual, 3.5 Motility Agar" by Molly Smith and Sara Shelby, LibreText Biology , South Georgia State College via GALILEO Open Learning Materials is licensed under CC BY 4.0
"How Infections Spread" by Centers for Diseases Control and Prevention , Use of CDC Materials Statement is in the Public Domain
"Clinical Nursing Skills Textbook: Chapter 7, Hygiene" by Openstax , Digital ISBN: 13: 978-1-961584-40-2 is licensed under CC BY 4.0
"Venous Thromboembolism (Blood Clots): Testing and Diagnosis" by Centers for Disease Control and Prevention , Use of CDC Materials Statement is in the Public Domain
"Scabies: About Scabies" by Centers for Disease Control and Prevention , Use of CDC Materials Statement is in the Public Domain
"Scabies: Public Health Strategies for Scabies Outbreaks in Institutional Settings" by Centers for Disease Control and Prevention , Use of CDC Materials Statement is in the Public Domain
Image Citations
Figure 1, "Microbiology Laboratory Manual: Labs, 1.7 Aseptic Technique" by Dr. Rosanna Hartline , LibreTexts: Biology , West Hills College Lemoore is licensed under CC BY-NC 4.0
Figure 2, Modified From:
- "Petri Dish or Culture Plate containing agar medium used for stool culture.jpg" by Frankincense Diala, Wikimedia Commons is in the Public Domain, CC0
Figure 3, Modified From:
- "Bio 225 OER Images" by Madison Rost, Associate Professor , HGTC and are in the Public Domain
- "File:Streptomyces mirabilis NRRL B-2400 (Type Strain).jpg" by Agricultural Research Service (NRRL)Culture Collection, Wikimedia Commons is in the Public Domain
Figure 4, "Microbiology Laboratory Manual: Labs, 1.7 Aseptic Technique" by Dr. Rosanna Hartline , LibreTexts: Biology , West Hills College Lemoore is licensed under CC BY-NC 4.0
Video 1, "HOW TO: Streak Bacteria " by jeanhuanglab , YouTube, Creative Commons Attribution license (reuse allowed) is licensed under CC BY 4.0


2.2: Rogerian Argument
The Rogerian argument, inspired by the influential psychologist Carl Rogers, aims to find compromise on a controversial issue.
If you are using the Rogerian approach your introduction to the argument should accomplish three objectives:
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Introduce the author and work
Usually, you will introduce the author and work in the first sentence, as in this example: In Dwight Okita’s “In Response to Executive Order 9066,” the narrator addresses an inevitable by-product of war – racism. The first time you refer to the author, refer to him or her by his or her full name. After that, refer to the author by last name only. Never refer to an author by his or her first name only. -
Provide the audience a short but concise summary of the work to which you are responding
Remember, your audience has already read the work you are responding to. Therefore, you do not need to provide a lengthy summary. Focus on the main points of the work to which you are responding and use direct quotations sparingly. Direct quotations work best when they are powerful and compelling. - State the main issue addressed in the work Your thesis, or claim, will come after you summarize the two sides of the issue.
The Introduction
The following is an example of how the introduction of a Rogerian argument can be written. The topic is racial profiling.
In Dwight Okita’s “In Response to Executive Order 9066,” the narrator — a young Japanese-American — writes a letter to the government, who has ordered her family into a relocation camp after Pearl Harbor. In the letter, the narrator details the people in her life, from her father to her best friend at school. Since the narrator is of Japanese descent, her best friend accuses her of “trying to start a war” (18). The narrator is seemingly too naïve to realize the ignorance of this statement, and tells the government that she asked this friend to plant tomato seeds in her honor. Though Okita’s poem deals specifically with World War II, the issue of race relations during wartime is still relevant. Recently, with the outbreaks of terrorism in the United States, Spain, and England, many are calling for racial profiling to stifle terrorism. The issue has sparked debate, with one side calling it racism and the other calling it common sense.Once you have written your introduction, you must now show the two sides to the debate you are addressing. Though there are always more than two sides to a debate, Rogerian arguments put two in stark opposition to one another. Summarize each side, then provide a middle path. Your summary of the two sides will be your first two body paragraphs. Use quotations from outside sources to effectively illustrate the position of each side.
An outline for a Rogerian argument might look like this:
- Introduction
- Side A
- Side B
- Claim
- Conclusion
The Claim
Since the goal of Rogerian argument is to find a common ground between two opposing positions, you must identify the shared beliefs or assumptions of each side. In the example above, both sides of the racial profiling issue want the U.S. A solid Rogerian argument acknowledges the desires of each side, and tries to accommodate both. Again, using the racial profiling example above, both sides desire a safer society, perhaps a better solution would focus on more objective measures than race; an effective start would be to use more screening technology on public transportation. Once you have a claim that disarms the central dispute, you should support the claim with evidence, and quotations when appropriate.
Quoting Effectively
Remember, you should quote to illustrate a point you are making. You should not, however, quote to simply take up space. Make sure all quotations are compelling and intriguing: Consider the following example. In “The Danger of Political Correctness,” author Richard Stein asserts that, “the desire to not offend has now become more important than protecting national security” (52). This statement sums up the beliefs of those in favor of profiling in public places.
The Conclusion
Your conclusion should:
- Bring the essay back to what is discussed in the introduction
- Tie up loose ends
- End on a thought-provoking note
The following is a sample conclusion:
Though the debate over racial profiling is sure to continue, each side desires to make the United States a safer place. With that goal in mind, our society deserves better security measures than merely searching a person who appears a bit dark. We cannot waste time with such subjective matters, especially when we have technology that could more effectively locate potential terrorists. Sure, installing metal detectors and cameras on public transportation is costly, but feeling safe in public is priceless.Contributors and Attributions
- Rogerian Argument. Provided by : Utah State University. Located at : ocw.usu.edu/English/introduction-to-writing-academic-prose/rogerian-argument.html. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
- Image of Carl Rogers. Authored by : Didius. Provided by : Wikimedia. Located at : http://commons.wikimedia.org/wiki/File:Carl_Ransom_Rogers.jpg . License : CC BY: Attribution


1.3: Determining the Equation of a Line
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Learning Objectives
In this section, you will learn to:
- Find an equation of a line if a point and the slope are given.
- Find an equation of a line if two points are given.
So far, we were given an equation of a line and were asked to give information about it. For example, we were asked to find points on the line, find its slope and even find intercepts. Now we are going to reverse the process. That is, we will be given either two points, or a point and the slope of a line, and we will be asked to find its equation.
An equation of a line can be written in three forms, the slope-intercept form , the point-slope form, or the standard form . We will discuss each of them in this section.
A line is completely determined by two points, or by a point and slope. The information we are given about a particular line will influence which form of the equation is most convenient to use. Once we know any form of the equation of a line, it is easy to re-express the equation in the other forms if needed.
THE SLOPE-INTERCEPT FORM OF A LINE: \(y = mx + b\)
In the last section we learned that the equation of a line whose slope = \(m\) and
\(y\)-intercept = \(b\) is \[\mathbf{y=mx+b}. \nonumber \] This is called the
slope-intercept form
of the line and is the most commonly used form.
Example \(\PageIndex{1}\)
Find an equation of a line whose slope is 5, and \(y\)-intercept is 3.
Solution
Since the slope is \(m = 5\), and the \(y\)- intercept is \(b = 3\), the equation is \(y = 5x + 3\).
Example \(\PageIndex{2}\)
Find the equation of the line that passes through the point (\(2, 7\)) and has slope \(3\).
Solution
Since \(m = 3\), the partial equation is \(y = 3x + b\).
Now \(b\) can be determined by substituting the point (\(2, 7\)) in the equation \(y = 3x + b\).
\begin{aligned}
&7=3(2)+b \nonumber \\
&b=1 \nonumber
\end{aligned}
Therefore, the equation is \(y = 3x + 1\).
Example \(\PageIndex{3}\)
Find an equation of the line that passes through the points (-1, 2), and (1, 8).
Solution
\(m=\frac{8-2}{1-(-1)}=\frac{6}{2}=3\). So the partial equation is \(y = 3x + b\).
We can use either of the two points (-1, 2) or (1, 8), to find \(b\). Substituting (-1, 2) gives
\begin{aligned}
&2=3(-1)+b \nonumber \\
&5=b \nonumber
\end{aligned}
So the equation is \(y = 3x +5\).
Example \(\PageIndex{4}\)
Find an equation of the line that has \(x\)-intercept 3, and \(y\)-intercept 4.
Solution
\(x\)-intercept = 3, and \(y\)-intercept = 4 correspond to the points (3, 0), and (0, 4), respectively.
\[ m=\frac{4-0}{0-3} = -\frac{4}{3} \nonumber \]
We are told the \(y\)-intercept is 4; thus \(b\) = 4
Therefore, the equation is \(y = -\frac{4}{3} x + 4\).
THE POINT-SLOPE FORM OF A LINE: \(y - y_1 = m(x - x_1)\)
The point-slope form is useful when we know two points on the line and want to find the equation of the line.
Let \(L\) be a line with slope \(m\), and known to contain a specific point (\(x_1\), \(y_1\)). If (\(x, y\)) is any other point on the line \(L\), then the definition of a slope leads us to the point-slope form or point-slope formula.
The slope is \( \frac{y-y_1}{x-x_1}= m\)
Multiplying both sides by (\(x-x_1\)) gives the point-slope form:
\[\mathbf{y - y_1 = m(x - x_1)} \nonumber \]
Example \(\PageIndex{5}\)
Find the point-slope form of the equation of a line that has slope 1.5 and passes through the point (12,4).
Solution
Substituting the point \((x_1,y_1) = (12,4)\) and \(m= 1.5\) in the point-slope formula, we get
\[\mathbf{y - y_1 = m(x - x_1)} \nonumber \]
\[y - 4 = 1.5(x - 12) \nonumber \]
The student may be tempted to simplify this into the slope intercept form \(y = mx + b\). But since the problem specifically requests point-slope form we will not simplify it.
THE STANDARD FORM OF A LINE: \(Ax + By = C\)
Another useful form of the equation of a line is the standard form.
If we know the equation of a line in point-slope form, \(y - y_1 = m(x - x_1)\), or if we know the equation of the line in slope-intercept form \(y = mx + b\), we can simplify the formula to have all terms for the \(x\) and \(y\) variables on one side of the equation, and the constant on the other side of the equation.
The result is referred to as the standard form of the line: \[\mathbf{Ax + By = C}. \nonumber \]
Example \(\PageIndex{6}\)
Using the point-slope formula, find the standard form of an equation of the line that passes through the point (2, 3) and has slope \(-3/5\).
Solution: Substituting the point (2, 3) and \(m= - 3/5\) in the point-slope formula, we get
\[y - 3 = - 3/5(x - 2) \nonumber \]
Multiplying both sides by 5 gives us
\begin{aligned}
&5(y-3)=-3(x-2)\\
&5 y-15=-3 x+6\\
&3 x+5 y=21 \quad \text { Standard Form }
\end{aligned}
Example \(\PageIndex{7}\)
Find the standard form of the line that passes through the points (1, -2), and (4, 0).
Solution
First we find the slope: \(m = \frac{0-(-2)}{4-1} = \frac{2}{3}\)
Then, the point-slope form is: \(y - (-2) = \frac{2}{3}(x -1)\)
Multiplying both sides by 3 gives us
\begin{aligned}&3(y+2)=2(x-1) \\
&3 y+6=2 x-2\\
&-2 x+3 y=-8\\
&2 x-3 y=8 \quad \text{ Standard Form } \end{aligned}
We should always be able to convert from one form of an equation to another. For example, if we are given a line in the slope-intercept form, we should be able to express it in the standard form, and vice versa.
Example \(\PageIndex{8}\)
Write the equation \(y = -\frac{2}{3}x + 3\) in the standard form.
Solution
Multiplying both sides of the equation by 3, we get
\begin{aligned}
&3y = -2x + 9 \\
&2x + 3y = 9 \quad \text { Standard Form }
\end{aligned}
Example \(\PageIndex{9}\)
Write the equation \(3x - 4y = 10\) in the slope-intercept form.
Solution
Solving for \(y\), we get
\begin{aligned}
&-4y = -3x + 10 \\
&y = \frac{3}{4}x - \frac{5}{2} \quad \text { Standard Form }
\end{aligned}
Finally, we learn a very quick and easy way to write an equation of a line in the standard form. But first we must learn to find the slope of a line in the standard form by inspection.
By solving for \(y\), it can easily be shown that the slope of the line \(Ax + By = C\) is \(-A/B\).
The reader should verify this.
Example \(\PageIndex{10}\)
Find the slope of the following lines, by inspection.
- \(3x-5y=10\)
- \(2x+7y=20\)
- \(4x-3y=8\)
Solution
- \(A=3\), \(B=-5\), therefore, \(m=-\frac{3}{-5}=\frac{3}{5}\)
- \(A=2\), \(B=7\), therefore, \(m=-\frac{2}{7}\)
- \(m=-\frac{4}{-3}=\frac{4}{3}\)
Now that we know how to find the slope of a line in the standard form by inspection, our job in finding the equation of a line is going to be easy.
Example \(\PageIndex{11}\)
Find an equation of the line that passes through (2, 3) and has slope - 4/5.
Solution
Since the slope of the line is - 4/5, we know that the left side of the equation is \(4x + 5y\), and the partial equation is going to be
\[4x + 5y = c \nonumber \]
Of course, \(c\) can easily be found by substituting for \(x\) and \(y\).
\begin{aligned}
&4(2)+5(3)=c\\
&23=c
\end{aligned}
The desired equation is
\[4x + 5y = 23. \nonumber \]
If you use this method often enough, you can do these problems very quickly.
We summarize the forms for equations of a line below:
Slope Intercept form:
\(\mathbf{y = mx + b}\),
where \(m\) = slope, \(b\) = \(y\)-intercept
Point Slope form:
\(\mathbf{y - y_1 = m(x - x_1)}\),
where \(m\) = slope, \((x_1,y_1)\) is a point on the line
Standard form: \(\mathbf{Ax + By = C}\)
Horizontal Line:
\(\mathbf{y = b}\)
where \(b\) = \(y\)-intercept
Vertical Line:
\(\mathbf{x = a}\)
where \(a\) = \(x\)-intercept