32.3: Lab Report
Modeling a Food Web
-
Draw a food web.
-
What is the ultimate source of energy for your food web?
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What happens to this energy as it moves through the food web? Where does it end up?
Nutrient Cycling in Ecosystems
-
In the space below, diagram the flow of water through the ecosystem around you.
What is the role of plants in the global cycling of water?
Mycorrhizal Networks
-
The fungus obtains its sugars from the living plant tissue. What terms could you apply to this organism?
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The plant synthesizes its own sugars during photosynthesis. What terms could you apply to this organism?
Somewhat recently, a researcher named Suzanne Simard began tracking the exchange of nutrients through mycorrhizae. Not just from tree to fungus, but from tree to tree via connections to the same mycorrhizal partner. By providing certain trees with carbon dioxide containing a heavier isotope of carbon (C14), Dr. Simard could track the sugars formed from those heavy carbon atoms as they moved from tree to tree. What she found was that the connections extended beyond just one tree to another, but that almost all of the trees within her plot were connected through a network of different mycorrhizal fungi. Trees of different species and fungi of different species could exchange nutrients with each other. Further study revealed that the exchange was not limited to nutrients and water, but also included transfer of plant defense compounds that worked like raising an alarm, cuing the other plants to start producing defensive chemicals.
-
How might this information change the way we study ecosystems?
- Sketch a few of the above ground plants that you can see (or imagine) and connect them below ground via a mycorrhizal network. Use arrows to indicate the flow of sugar through this network. Add in some disturbance that would set off a plant’s immune system (attack of the caterpillars, fungal infection, etc…) and depict the movement of the signal throughout the plant community.
- Add a parasitic plant into your ecosystem above and use arrows to depict the movement of sugars.
- Are these parasitic plants autotrophic or heterotrophic? How would you classify them, using ecology terminology? Explain your reasoning.
- What is the difference between a community and an ecosystem?
Community Interactions
- Is the relationship between the fig and the wasp a parasitism or mutualism? Explain your reasoning.
Ecosystem & Eutrophication - Silver Springs
| Table 1: Colorimeter data of Daphnia feeding on Chorella | |
|---|---|
| Absorbance value | |
| Before feeding (time zero) | |
| After feeding (time 30 minutes) |
Questions
-
How did the absorbance change from time 0 to time 30 minutes? Did the absorbance increase, decrease, or stay the same?
-
What does the absorbance value change tell you about the concentration of the
Chorella
? Did it increase, decrease, or stay the same?
-
What does the absorbance value chance tell you about the behavior of the
Daphnia
? Did they consume
Chorella?
How do you know?
-
In a real aquatic ecosystem, do you think zooplankton could decrease the impact of cultural eutrophication and algae blooms? Explain why or why not.
Part 2
Questions
Based on your findings with the model, answer the following questions:
-
What is the maximum number of producers the ecosystem can support before higher trophic levels begin to decline?
-
What happens to community respiration in the simulated algae bloom? Does it increase, decrease, or stay the same?
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What happens to the decomposer population in the simulated algae bloom? Does it increase, decrease, or stay the same? Why?
-
Which group of carnivores (which trophic level) is more greatly impacted by the algae bloom? Why do you think this is the case?
Activity II
QUESTIONS
Based on your findings with the model, answer the following questions:
-
Is an elimination of bread feasible? Why or why not?
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How would the system respond to increasing levels of tourism, assuming bread levels also increased? What trophic levels are most impacted by increased levels of bread?
-
What is the role of bread in the system, and how does its presence or absence impact the ecology of Silver Springs?
-
What would be your suggestion to the managing board of Silver Springs? Should they completely eliminate the tourists feeding the ducks? Why or why not?
LICENSES AND ATTRIBUTIONS
CC LICENSED CONTENT, ORIGINAL
- Biology 102 Labs. Authored by : Lynette Hauser. Provided by : Tidewater Community College. Located at : [www.tcc.edu] . License : CC BY: Attribution
CC LICENSED CONTENT, SHARED PREVIOUSLY
- Trophic Levels. Authored by : Thompsma. Located at : [commons.wikimedia.org] . License : CC BY-SA: Attribution-ShareAlike
PUBLIC DOMAIN CONTENT
- Silver Spring Model. Authored by : Sholto Maud. Provided by : Wikipedia. Located at : [commons.wikimedia.org] . License : Public Domain: No Known Copyright
Ecology Game:
Islands have always provided scientists unique locations to learn about the natural world. Many of Charles Darwin’s key observations that led to his understanding of natural selection came from species living on the many islands that make up the Galapagos archipelago. For example, Darwin observed different species of finches were found on different islands and that the shape of the finches’ beaks corresponded to the type of food they consumed. Further research on the finches of the Galapagos by Dr. Peter Grant and Dr. Rosemary Grant found that even within an island, changes in environmental conditions (i.e. drought, disease) could influence the adaptations observed in the finches.
Because islands are isolated from the mainland and each other by water and often long distances, they can be colonized by new species that have intentionally or unintentionally dispersed to an island. In biology, we call this the founder effect , which is when a new population is established. A founding population is often small and has a different allele frequency than the original population. It is important to remember that every species that finds itself in a new habitat may not be ideally suited for the environmental conditions. Is the habitat, food, climate appropriate for the new species needs? Is there competition for any of these resources with species already living on the island? Large islands that are close to the mainland tend to support more species than small islands that are far from the mainland. As we discuss in class, we know that no two species can occupy the same ecological niche —basically, two species cannot co-exist if they are competing for the same resource. While niches can overlap significantly between two species, they cannot be exactly alike. Any habitat has a limited number of species that can be supported by the available resources. We know that the most successful species are those with adaptations best suited to the environment in which they find themselves and those that can best utilize the available resources. These individuals are able to survive and reproduce on the island and a new population is established there. Occasionally, if there is a mismatch in the individual’s traits and the environment in which they are located or the environmental conditions change unfavorably the individual may no longer succeed in that environment. This can lead to local extinction which is also known as extirpation . An extirpated species can recolonize if environmental conditions become favorable.
Today, you will be playing the role of an individual of one of the six different species on an ancient continent to the East in a game we call the Ecology Game. The objective of the game is to cross the ocean, colonizing islands as you go. However, every species in the game has its own ecological needs and each island along the way has specific environmental conditions and the ability to support different numbers of species. If you are a successful colonizer, a population of your species will exist on the island even after you have moved on. Please note that in order to colonize and establish a new population in real life, you would need a sexually mature male and female or a pregnant female. However, in this game, we are simplifying this concept by using a model of a single individual. Every species will not be able to colonize every island successfully and you have to compete with the other species who are also trying to make it to the new continent. The islands are of varying sizes and successful colonization may cover a small area or a large area depending on the size of the island. The winner is determined by a combination of the order that you reach the new continent and the island area (not how many islands, but how much island space) you have successfully colonized. Remember to think strategically! The Ecology Game is not always about speed, but thinking about your species, its ecological niche, and what we know about colonization and adaptation. Just because you are the first species to reach the new continent that does not mean you win the game. Think of today like an ecological game of chess!
Set up (per group—groups should have no more than 6 students):
- 1 board map
- 2 six-sided dice
- 1 cup of 6 species sticks
- 4 bags of colored game pieces (if groups are larger than 4 and additional game pieces are required use coins or see Manal for additional colored pieces)
- 6 species cards
- 24 island cards
- Environmental factor cards
To Begin:
- Each person in the group should draw a species stick out of the cup. You will play the role of an individual of this species for the entire game.
- Read the associated species card. This will tell you about your ecological niche and any movement limitations you will have in the game.
- Each player can place their colored game piece at any location on the mainland to start the game.
- Roll the dice. The player with the highest roll will go first and play will move counterclockwise around the group.
To Play:
1. Player 1 will roll both die. The sum of the numbers shown on the roll indicates how many squares the player can move.
Note
You cannot move diagonally but you can move vertically and horizontally and you can move forwards or backwards. Some species have limitations on the maximum number of moves they can make. If your roll exceeds the maximum number of moves for your species, you must stop at your maximum number. Every species can stop in the ocean.
2. With each roll, player 1 will also draw an environmental factor card which may influence your movement along the way. This card may affect all players, a subset of players, or just the player drawing the card.
3. If you land on any part of an island, choose the island card and determine if your species can successfully colonize the island. If so, fill that player’s name/species name in the appropriate slot on the card (remember, some islands can only hold a few species) and place a colored game piece on the island to indicate that it has been colonized by you.


5.4: Vitamins, Minerals, and Complementary and Alternative Therapies
By the end of this section, you should be able to:
- 5.4.1 Discuss water-soluble and fat-soluble vitamins and their importance in the body.
- 5.4.2 Identify minerals to treat deficiencies.
- 5.4.3 Discuss chelating agents to remove excess copper, iron, and lead from the body.
- 5.4.4 Identify food source and deficiency conditions associated with vitamins and minerals.
- 5.4.5 Describe the use of common complementary and alternative therapies.
- 5.4.6 Explain how complementary and alternative therapies may potentiate, negate, or cause toxicity with prescribed drugs.
- 5.4.7 Discuss the nursing implications related to complementary and alternative therapies.
- 5.4.8 Explain the client education related to complementary and alternative therapies.
Vitamins and minerals are essential nutrients that the body requires for various physiological functions, including growth, development, and maintenance. Complementary and alternative therapies such as acupuncture, aromatherapy, and herbal supplements are becoming increasingly popular as nontraditional treatments for various medical conditions. This section of the chapter will discuss vitamins, minerals, and common complementary and alternative therapies.
Vitamins
Vitamins are essential organic compounds required for normal metabolic function, growth, and healing. The body requires only small amounts of vitamins daily, which are typically obtained through a well-balanced diet. Vitamins are usually prescribed for clients who have the inability to metabolize and absorb vitamins, such as with celiac disease; increased vitamin losses, such as with diarrhea and crash diets; and when there is an increased vitamin need, such as with pregnancy. There are two types of vitamins, water-soluble and fat-soluble, discussed in the following sections.
Water-Soluble
Water-soluble vitamins dissolve in water. They are excreted by the kidneys in urine and are minimally stored in the body. Water-soluble vitamins need to be consumed regularly in the diet or through supplements to maintain adequate levels in the body. They play a critical role in various bodily functions, such as energy, metabolism, nerve function, and collagen synthesis. Water-soluble vitamins, their functions, food sources, and deficiency conditions are also discussed in Table 5.2.
 They are absorbed along with dietary fat and can be stored in the body for extended periods. Excessive intake of fat-soluble vitamins can lead to toxicity because they are not readily excreted. Fat-soluble vitamins are essential to various bodily functions such as vision, immune function, blood clotting, and bone health. Fat-soluble vitamins, their functions, food sources, and deficiency conditions are also discussed in Table 5.2.
| Water-Soluble Vitamins | |||
| Vitamin | Function | Food Source | Deficit Conditions |
| B 1 (Thiamine) | Promotes carbohydrate metabolism and nerve function | Whole-grain and enriched breads and cereals, nuts, liver, black beans, and meats such as fish (richest source), liver, and pork | Metabolic disorders, wet and dry beriberi, Wernicke ’ s encephalopathy, and Korsakoff ’ s psychosis |
| B 2 (Riboflavin) | Promotes use of carbohydrates, fats, and proteins necessary for red blood cell function; biggest role is with protein to promote wound healing | Whole grains, enriched grains, organ meats, and dairy products | Angular stomatitis, cheilosis, glossitis, itchy dermatitis of scrotum or vulva, microcytic anemia, and alcohol substance use disorders |
| B 3 (Niacin) |
Essential for metabolism, glycogenolysis, and nerve function;
participates in the synthesis of steroid hormones and fatty acids |
Fish, poultry, peanuts, mushrooms, and whole grains | Pellagra, hepatotoxicity, and anorexia |
| B 9 (Folic acid) | Necessary for DNA and heme synthesis and intestinal functioning | Leafy vegetables such as spinach and broccoli, yellow vegetables such as squash, liver, dried beans, and peas | Megaloblastic anemias, intestinal disturbances, fetal neural tube defects (caused by deficiency during pregnancy) |
| B 12 (Cobalamin) | Synthesis and maintenance of myelin ; involved in protein synthesis and an essential factor in the synthesis of DNA; helps form red blood cells and assist in nerve function | Fish, poultry, liver, eggs, and dairy products | Poor growth, interruption or delay of nerve impulse transmission, and pernicious anemia |
| C (Ascorbic acid) | Essential for tissue repair and wound healing, necessary for formation of collagen; facilitates absorption of iron | Citrus fruits and juices, strawberries, kiwi, potatoes, tomatoes, and cantaloupe | Poor wound healing, bleeding gums, and scurvy |
| Fat-Soluble Vitamins | |||
| Vitamin | Function | Food Source | Deficit Conditions |
| A (Retinol and beta carotene) | Necessary for immune function and healthy eyes, skin, and hair | Carrots, potatoes, pumpkins, spinach, beef, and eggs | Night blindness, skin lesions, xerophthalmia, and brittle hair |
|
D3 (Cholecalciferol)
D2 (Ergocalciferol) |
Facilitates equilibrium of phosphorus and calcium, crucial for maintaining strong teeth and bones | Milk and other dairy products, such as yogurt and cheese | Rickets, osteomalacia, and osteoporosis |
| E (Alpha-tocopherol) | Promotes functioning of red blood cells, muscles, and tissues and is essential in anti-inflammatory processes, platelet aggregation, and immune response | Fortified cereals, whole grain products, vegetable oils, wheat germ, seeds, and nuts | Retinopathy, peripheral neuropathy, and ataxia |
| K (Phytonadione) | Necessary for blood clotting | Dark green leafy vegetables, turnips, and beets | Increased clotting time, bleeding, or hemorrhage |
Minerals
Minerals are inorganic substances that are essential for certain physiological functions. Minerals are necessary for the formation of strong bones and teeth, regulation of fluid balance, muscle function, and the production of hormones and enzymes. They aid in the transport of oxygen and nutrients throughout the body and support immune function. Table 5.3 lists the heavy metal minerals iron, copper, lead, and zinc and their functions, food sources, chelating agents for toxicity, and deficiency conditions. Table 5.4 lists the functions, food sources, and deficiency conditions for zinc, chromium, and selenium.
| Mineral | Function | Food Source | Chelating Agents for Toxicity | Deficiency Conditions |
|---|---|---|---|---|
| Iron | Component of hemoglobin, helps transport oxygen, supports immune function, and assists in the production of energy | Red meat, poultry, fish, beans, lentils, tofu, fortified cereals, spinach, and other leafy greens | Deferoxamine , deferasirox , and deferiprone bind to excess iron in the body, forming a complex that can then be excreted in the urine or feces. They are used to treat iron overload conditions such as h emochromatosis or thalassemia . | Iron deficiency anemia (microcytic, hypochromic), pregnancy (due to an increased need for RBCs as a result of increased blood volume and the fetus), gastrointestinal bleeding |
| Copper | Helps form red blood cells, supports immune function, and assists in the production of energy and collagen; required for melanin production | Shellfish, nuts, seeds, whole grains, beans, and dark leafy greens | Penicillamine and trientine bind to excess copper and form a complex that can be excreted in the urine. They are used to treat Wilson’s disease , a genetic disorder that causes copper to accumulate in the liver and brain. | Celiac disease, excessive zinc intake (which can interfere with copper absorption and utilization), and Menkes’ disease (a hereditary abnormality that blocks absorption of copper) |
| Lead | Is a toxic heavy metal and has no known essential functions within the body but through exposure can lead to a variety of health problems | Found in contaminated water and food supplies, especially in areas of high environmental pollution and in older homes with lead-based paint or pipes | Dimercaprol, EDTA , and succimer bind to lead in the body, forming a compound that is excreted in the urine or feces; they are used to treat lead poisoning and reduce the body burden of lead. | No deficiency conditions |
| Mineral | Function | Food Source | Deficiency Conditions |
|---|---|---|---|
| Zinc | Supports immune function, helps with wound healing, assists in the production of proteins and DNA synthesis, involved in taste acuity | Oysters, beef, pork, chicken, beans, nuts, seeds, and whole grains | Crohn’s disease, celiac disease, pregnancy, growth retardation, delayed sexual maturation, alopecia, loss of taste sensation, poor wound healing, impaired immunity |
| Chromium | Helps regulate blood sugar levels by potentiating the action of insulin; supports healthy metabolism | Broccoli, grape juice, whole grains, and nuts | Chromium deficiency is rare because the body requires only small amounts; however, severe deficiencies can cause glucose intolerance and weight loss. |
| Selenium | Acts as an antioxidant and helps protect cells from damage, supports immune function, necessary for iodine metabolism to assist in the production of thyroid hormones | Brazil nuts, seafood, poultry, beef, eggs, and whole grains | Malabsorption syndrome |
Complementary and Alternative Therapies
Complementary and alternative therapies , also known as complementary and alternative medicine (CAM) , is a broad term used to describe health practices and treatments that are used in addition to or instead of conventional medical therapies. Examples of CAM include herbal supplements, acupuncture, chiropractic care, massage therapy, and mind-body practices such as meditation and yoga. Some people choose CAM therapies because they believe they are more natural or holistic than conventional medicine or because they have had negative experiences with conventional medicine (Jones et al., 2019; National Institutes of Health, n.d.-b). However, it is important to note that not all CAM therapies are safe and effective, and some may even be harmful.
Dietary Supplement and Health Education Act
The Dietary Supplement and Health Education Act (DSHEA) of 1994 is a federal law that regulates the manufacturing and labeling of dietary supplements. Under DSHEA, dietary supplements are classified as a food, not a drug, and are not subject to the same strict regulations as drugs. However, DSHEA does require that manufacturers of dietary supplements meet certain labeling requirements and notify the Food and Drug Administration (FDA) of any new ingredients they plan to use. DSHEA also allows manufacturers to make certain health claims about their products as long as they are truthful and not misleading (National Institutes of Health, n.d.-a).
Good Manufacturing Practices
Good Manufacturing Practices (GMPs) are a set of guidelines established by the FDA that ensures quality, purity, and strength of dietary supplements. GMPs require manufacturers to use standardized manufacturing processes, perform regular testing of raw materials and finished products, and maintain detailed records of all production and testing. GMPs also require manufacturers to have a system in place for handling complaints and reporting adverse events (National Institutes of Health, n.d.-a).
Commonly Used Herbal Remedies
There are numerous herbal remedies that are used by the general public. Table 5.5 presents common herbal remedies, their use, and special considerations.
| Herbal Remedy | Use | Special Considerations |
|---|---|---|
| Astragalus | To increase stamina and energy and to improve immune function | May increase effects of anti-hypertensive drugs (Han et al., 2019); caution should be used when administering if the client has a fever because it can mask infection |
| Chamomile | Used topically to treat wounds and conjunctivitis; used orally for migraines and anxiety | Clients should not take if pregnant or breastfeeding because it can harm the fetus/infant, can interact with anticoagulants and increase bleeding times, and may cause depression |
| Echinacea | To enhance the immune system and treat colds and influenza; suppresses inflammation | Usage for longer than 12 weeks may cause liver toxicity; do not use with antifungals or hepatotoxic drugs; usage in immunocompromised clients is discouraged |
| Garlic | To treat colds; for cardiovascular health (reduces levels of triglycerides and LDL and raises levels of HDL); helps to reduce blood pressure and suppress platelet aggregation and lower blood glucose levels |
Has anticoagulant effects and should be used cautiously in clients on anticoagulants; lowers blood glucose levels and may cause hypoglycemia in clients already taking medications for their diabetes;
reduces triglycerides and LDL and increases HDL levels—may need a dosage adjustment if client is taking medications for lipids |
| Ginger | To treat nausea, motion sickness, and postoperative nausea; also has anti-inflammatory and analgesic properties | Affects blood clotting and is contraindicated for clients on anticoagulants |
| Ginseng | To treat hypertension; also is a mood elevator and decreases cholesterol levels | May cause irritability if taken with caffeine; may cause interactions with phenelzine and monoamine oxidase inhibitors (MAOIs); may interfere with the effectiveness of digoxin |
| Hawthorn | To treat blood pressure problems and lower cholesterol levels | Use cautiously with digoxin, ACE inhibitors, and central nervous system (CNS) depressants because it may potentiate their effects |
| Licorice root | To treat cough and stomach ulcers | Contraindicated with renal or liver disease, hypertension, coronary artery disease, and pregnancy and in clients who are taking digoxin and thyroid drugs |
| Milk thistle | To treat fatty liver and hepatitis | Can potentiate antihypertensive drugs and immunosuppressant drugs |
| Saw palmetto | To treat urinary symptoms related to prostate issues | May cause orthostatic hypotension; do not use with finasteride because toxicity may occur (Mount Sinai, n.d.) |
| St. John’s wort | To treat depression | Take on an empty stomach to enhance absorption; may interact with other antidepressants as well as theophylline, digoxin, hormonal contraceptives, and certain anticancer drugs; client should not take if pregnant or breastfeeding because it may affect the fetus/infant |
| Turmeric | To treat inflammation | May cause gastrointestinal upset and may increase bleeding times if used with oral anticoagulants |
| Valerian | To treat insomnia and anxiety | May cause severe liver damage; should not be used with CNS depressants because it may cause severe sedation |
Common Complementary Therapies
Complementary therapies are used to promote overall health and well-being. They are often used to alleviate symptoms and side effects associated with medical conditions and treatments, but they are not intended to replace or serve as a substitute for traditional medical care. Table 5.6 describes common complementary therapies, their uses, and special considerations.
Special Considerations
Asian Americans and CAM
Asian American s are more likely to use complementary and alternative medicine (CAM) as compared to other races or ethnicities. Many continue cultural traditions of using Eastern alternatives such as acupuncture, Ayurveda, and yoga.
(Source: Felicilda-Reynaldo et al., 2020.)
Trending Today
Cannabis has been used for centuries as a natural remedy to alleviate symptoms of various health conditions, such as chronic pain, anxiety, and insomnia. Medical cannabis is a new trend that is legal in many states and countries, allowing clients access to cannabis-based treatments under the supervision of licensed health care providers. With the increasing use of medical cannabis, there is a growing need for health care professionals to be trained in its use and to be able to provide accurate information to clients. Dr. Josh Axe, DNM, DC, CNS, is a doctor of natural medicine, chiropractor, and clinical nutritionist. In the video podcast “Medicinal Benefits of Cannabis Compounds,” Dr. Axe interviews Dr. Bonni Goldstein, MD, who speaks about the different types of cannabis and its medicinal benefits.
| Complementary Therapy | Definition | Use | Special Considerations |
|---|---|---|---|
| Acupuncture | A traditional Chinese medical practice that involves inserting thin needles into specific points on the body to promote healing and pain relief | Treatment of chronic pain, headaches, and musculoskeletal problems | Risk for infection and injury from the needles; not appropriate for people who are taking immunosuppressant or anticoagulant drugs; contraindicated in clients who have diabetes, stroke, or other neurological conditions |
| Aromatherapy | The use of essential oils to promote physical and emotional well-being | Stress reduction, improvement of mood, and promoting relaxation | May cause skin irritation or an allergic reaction |
| Ayurveda | A traditional Indian medical practice that seeks to balance the body, mind, and spirit through diet, lifestyle changes, and herbal remedies | Treatment of digestive disorders, anxiety, and skin problems | Herbal remedies may interact with prescription drugs and should not be used without speaking to the health care provider; some products may contain heavy metals or other toxins |
| Cannabis (marijuana) | A plant that contains compounds that have psychoactive effects | Treatment of chronic pain, muscle spasms, seizures, and nausea and vomiting associated with chemotherapy, and to increase appetite | May cause impaired coordination and memory, increased heart rate, and temporary cognitive impairment; may be addictive |
| Touch therapy | A form of healing that involves the use of touch to promote relaxation and reduce stress | Treatment of anxiety, depression, and chronic pain | Contraindicated in skin conditions where touch therapy may exacerbate the condition, areas of recent injuries or surgery, or areas where cancer treatment such as radiation therapy is being administered |


0.2: Why Study Spanish?
There are many reasons to learn another language. In this module, you will recognize the the prevalence of Spanish in your community, state, and country, the benefits of language learning, and the increasing demand for Spanish skills in the workplace.
Spanish in United States
Click on the link to view the Hispanic population of the United States in 2022 by state. What is the population where you live? Does this fact surprise you?
Duffin, Erin. “Hispanic population of the United States in 2021, by state.” 14 Nov. 2022. Statista, www.statista.com/statistics/259850/hispanic-population-of-the-us-by-state/ . Accessed 4 May 2023.
Now look at the projected forecast of the Hispanic population of the United States. Does the forecast surprise you?
Duffin, Erin. “Forecast of the Hispanic population of the United States from 2016 to 2060.” 5 Oct. 2022. Statista, www.statista.com/statistics/251238/hispanic-population-of-the-us/ . Accessed 4 May 2023.
Spanish in Kansas
Adobe Stock Image
- Kansas is 12.3% Hispanic/Latino.
- Kansas is the 18th most Hispanic/Latino state in the United States.
- The 10 Cities In Kansas With The Largest Hispanic Population For 2023:
City - Percentage (%) of Population that is Hispanic
-
- Liberal - 65.08%
- Dodge City - 64.58%
- Garden City - 53.09%
- Ulysses - 52.24%
- Kansas City - 31.19%
- Emporia - 27.72%
- Great Bend - 23.59%
- Arkansas City - 19.26%
- Newton - 18.8%
- Wichita - 17.63%
Kolmar, Chris. "The 10 Kansas Cities With The Largest Latino Population For 2023." HomeSnacks , 5 Oct. 2023, www.homesnacks.com/ most-hispanic-cities-in-kansas/ . Accessed 23 Dec. 2023.
Spanish at Butler Community College
Did you know that nearly 20% of Butler's student population identifies as Hispanic/Latino?
In the spring of 2023, the Foreign Language Department at Butler Community College conducted a survey of the Hispanic faculty, staff, and students in our very own Butler community. We compiled their stories and insights into a project called ¨Voces de Butler¨ (Butler Voices). When asked why the Spanish language was important to them, we received the following words and phrases to show the impact their native language has in their lives:
- conexiones - connections
- herramientas - tools
- países hispanohablantes - Hispanic countries
- lengua materna - native language
- aprender - to learn
- ayudar - to help
- viajar - to travel
- hablar - to talk
- amistades - friendships
- aventura - adventure
- relaciones - relationships
- conseguir trabajo - getting a job
- abrir la mente - opening your mind
- evitar prejuicios - avoiding prejudices
- cultura - culture


11.7: Considering Multiple Views and Avoiding Bias
Considering Multiple Views
Are multiple views being considered through depth and breadth?
When writing or speech is deep, it covers the complexity of a topic. It doesn’t skim the surface. It dives deeply into the profound knowledge and substantial understanding of a topic. This detailed explanation along with examples is depth , which also includes attention to rhetorical devices (pathos, ethos, logos, and kairos).
When analyzing the depth of any essay, including their own essays and those of their peers, students can ask these types of questions:
- How deeply does this essay go into its topic?
- Is it detailed enough?
- Did it go far enough into the research and reviews of other texts to demonstrate a deep knowledge about the subject? ( Digging into what others have said about your topic, a particular text you’re working with, or the people involved is often where you can often find opposing views ).
- How thoroughly have specific subtopics within a major been researched?
Breadth is how broad or wide a topic has been discussed in writing or in speech. For example, to attain breadth in a persuasive essay, a writer must consider not only one point of view, but all the multiple major perspectives about an issue. Breadth also entails considering multiple contexts of an issue and multiple analytical approaches to solving a problem.
Breadth means reading more than a handful of articles supporting one side of an issue; it means reading more articles supporting various perspectives so the writer can truly understand all viewpoints about the issue and can discuss the issue with breadth that builds a deeper understanding and fairness. When analyzing the breadth of any essay, students and instructors ask questions such as the following:
- Is the content of an essay sufficiently comprehensive enough to cover a wide range of perspectives and angles on a given topic?
- Is anything missing that should be included in the scope of the topic and which would help the essay achieve enough breadth?
- Has the opposing view (i.e., the “naysayer’s” perspective) been explored so as to strengthen the writer’s own argument? (This consideration is particularly key in the development of a fully supported and wisely composed persuasive or argumentative essay.)
- What has not yet been considered to make this idea or essay complete?
Avoiding Bias
Fairness: Is It Objective and Judicious?
The word “fair” is often used synonymously with “just” or “judicious” and is related to “justice.” Especially in essays that are meant to persuade through logical argumentation, topics and points of view (POVs or “perspectives”) need to be treated fairly and diplomatically. A fair, even-handed treatment doesn’t necessarily mean agreeing to opposing (or “naysayer’s”) POVs, but strong, college-level writing must acknowledge the opposing POVs, then must either accommodate or refute them. For example, an essay may state, “The opponents have valid points regarding X and Y. They are right about this and that. However, they are inaccurate about this specific point about X, and their argument doesn’t negate A and B, which remains the most accurate ideas and still strongly support this argument.”
When analyzing the fairness of any work, students and instructors ask questions such as the following:
- Does the writer of this essay exhibit the ability to fairly assess the viewpoints of others, even opposing viewpoints?
- Are there any fallacies, such as ad hominems that unfairly label opponents rather than speak directly and precisely about the opposing argument or POV itself? (Note: The term “ad hominem” is short for “argumentum ad hominem” and is a fallacious argumentative strategy whereby genuine discussion of the topic at hand is avoided by instead attacking the character, motive, or other attribute of the person making the argument, or persons associated with the argument, rather than attacking the substance of the argument itself.)
- Does the writer or speaker have a conflict of interest? Does that conflict of interest appear as bias in the text?
When analyzing our own work (discussions and writings), these questions are crucial:
- What about our own biases? Are we aware of and addressing our own biases? How can we move beyond those?
Check out this video on implicit vs. explicit bias:
Biased Language - OER Series - Writing Center | WC OER from Trefny Center on Vimeo .
English Composition by Contributing Authors is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.


1.1: Introduction
Your alarm goes off and, after hitting “snooze” once or twice, you pry yourself out of bed. You make a cup of coffee to help you get going, and then you shower, get dressed, eat breakfast, and check your phone for messages. On your way to school, you stop to fill your car’s gas tank, almost making you late for the first day of chemistry class. As you find a seat in the classroom, you read the question projected on the screen: “Welcome to class! Why should we study chemistry?”
Do you have an answer? You may be studying chemistry because it fulfills an academic requirement, but if you consider your daily activities, you might find chemistry interesting for other reasons. Most everything you do and encounter during your day involves chemistry. Making coffee, cooking eggs, and toasting bread involve chemistry. The products you use—like soap and shampoo, the fabrics you wear, the electronics that keep you connected to your world, the gasoline that propels your car—all of these and more involve chemical substances and processes. Whether you are aware or not, chemistry is part of your everyday world. In this course, you will learn many of the essential principles underlying the chemistry of modern-day life.


2.8: "Unless"
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The English term “unless” can be tricky to translate. For example,
The Reds will win unless their starting pitcher is injured.
If we use the constant “R” to stand for the atomic proposition, “the Reds will win” and “S” to stand for the atomic proposition, “the Reds’ starting pitcher is injured,” how would we translate this sentence using truth-functional connectives? Think about what the sentence is saying (think carefully). Is the sentence asserting that the Reds will win? No; it is only saying that
The Reds will win as long as their starting pitcher isn’t injured.
“As long as” denotes a conditional statement. In particular, what follows the “as long as” phrase is a sufficient condition, and as we have seen, a sufficient condition is always the antecedent of a conditional. But notice that the sufficient condition also contains a negation. Thus, the correct translation of this sentence is:
~S ⊃ R
One simple trick you can use to translate sentences which use the term “unless” is just substitute the phrase “if it’s not the case that” for the “unless.” But another trick is just to substitute an “or” for the “unless.” Although it may sound strange in English, a disjunction will always capture the truth functional meaning of “unless.” Thus, we could also correctly translate the sentence like this:
S v R
In the next section we will show how we can prove that these two sentences are equivalent using a truth table.


6.1: Soil Nutrient Management
Learning Objectives
- Learn to interpret information contained in a soil test report.
- Make fertilizer recommendations based on soil test results.
- Become familiar with K-State resources as well as publications available to farmers and gardeners.
Proper soil fertility is essential for achieving optimum yields and plant health. It is also critical for minimizing the costs of inputs like fertilizer or irrigation water, as well as for reducing the environmental impacts of production agriculture or manicured landscapes. The key to proper soil fertility is soil testing. We will first tour the K-State Soil Testing Lab. The staff will walk the class through the process, include sample submittal, costs, pH analysis, and nutrient analyses. This will be followed by a problem set in which students will make fertilizer recommendations using soil test results and publically available K-State extension bulletins and resources.
Materials
- Soil Nutrient Management Problem Set
- Fertilizing Kansas Lawns (Fagerness, 2001)
- Fertilizer Recommendation Program (Kansas State University Soil Testing Laboratory, Kansas State University Department of Agronomy, Manhattan, KS, U.S.)
- Soil Test Interpretations and Fertilizer Recommendations (Leikam et al., 2003)
- Fertilizing Gardens in Kansas (Upham and Marr, 2011)
Recommended Reading
- Soil Analysis (Kansas State University Department of Agronomy, 2020)
Prelab Assignment
Using the recommended reading and the introduction to this lab, consider the questions listed below. These definitions/questions will provide a concise summary of the major concepts to be addressed in the lab. They are also useful as study notes for exams.
- Explain why it is important to collect a soil sample from a representative area of soil and why it is common practice to mix many samples into one composite sample.
- Explain the statement that “soil testing laboratories do not determine the concentration of plant-available nutrients in soils”. What do they actually determine?
- Explain why applying fertilizer to achieve maximum yield may not achieve maximum profit.
- In addition to soil testing, what other diagnostic tools are used to assess fertility?
Activity & Assignment: Problem Set
A problem set will be provided to you at the beginning of the lab. It will familiarize you with how fertilizer recommendations are made based on soil tests. Materials required to complete the problem set are available as printed copies in the lab or as PDFs and Excel Spreadsheets online. You will complete the problem set during lab.


23: Quantitative real time PCR
Summary
Quantitative real time PCR (qRT-PCR) is used to determine the amount of a specific DNA sequence in a DNA mixture.
Also known as
qRT-PCR, qPCR, real time PCR
Note that the acronym RT-PCR can be confusing, as both real time and reverse transcription PCR can be abbreviated “RT,” so care should be taken to avoid confusion.
Samples needed
Sample of DNA, with a mixture of sequences
Method
qRT-PCR is a quantitative variant of typical PCR . It is used to determine how much of a particular DNA sequence is present in a mixture. This can be used to determine gene expression levels (following RT-PCR of cellular RNAs), gene copy number, and for other applications.
With a pair of primers that amplify the target DNA, the amount of the target should double in each PCR cycle. The principle of qRT-PCR is that the less abundant the target DNA sequence in the sample, the more cycles it will take for the abundance of the target to reach a certain threshold. The number of cycles required to reach the threshold is known as the Ct value. Therefore, in two samples, the one with a lower Ct value indicates that the target DNA is more abundant in that sample.
In order to track the production of target DNA in real time, qRT-PCR uses a special machine that can detect fluorescence over the course of the PCR program. In addition to the usual PCR reagents, qRT-PCR requires a method for tracking production of the target. One method uses dyes that only fluoresce when they bind to dsDNA. This is the case for the popular reagent SYBR green. The second method uses probes that are specific to the target DNA sequence and which only fluoresce when they are cleaved by Taq DNA polymerase. These are called Taqman probes. Each of these methods has benefits and drawbacks. In both cases, a certain fluorescence signal is chosen as the threshold value.
qRT-PCR data is either compared to a standard curve or to an internal reference sample to determine absolute or relative DNA quantities, respectively.
Controls
As with any PCR protocol, primer specificity should be considered. If using intercalating dyes like SYBR green, a melt curve can be used to determine how many PCR products are produced in each reaction. After the normal PCR program runs, the temperature of the mixture is raised, and the change in fluorescence is monitored. When a product’s melting temperature is reached, there is a peak in the ΔF signal. More than one peak in the melt curve indicates that the primers are not entirely specific.
When using qRT-PCR to measure gene expression, an endogenous control sequence is often amplified in addition to the sequence of interest. The endogenous control should be a gene whose expression is known to be very stable and is not affected by experimental treatments. This can be used to normalize the signal from the sequence of interest (to account for small differences in total cDNA added). Other controls relating to the reverse transcription step should also be used.
Furthermore, sometimes a no template control (water instead of DNA sample) is used to ensure that none of the other reagents are contaminated with DNA.
Interpretation
In this experiment, researchers performed a CRISPR -generated knockout screen to ask what host cell genes are involved in human cytomegalovius (HCMV) infection. Two knockouts are shown: RRM2 in the top row & ARL6IP6 in the bottom row. In panels g & j (left, red), the plaque-forming units (PFU) generated in each type of knockout cell are shown. PFU correlates with the number of viral particles that can infect new cells. Both knockouts significantly reduce the amount of functional viral particles formed. Panels h & k (right, blue) show the results of a qRT-PCR experiment to quantify the relative number of HCMV genome copies in each knockout cell compared to controls. While RRM2 knockout almost completely abrogates the ability of HCMV to copy its genome, ARL6IP6 does not significantly affect HCMV viral copy number. Therefore, ARL6IP6 mainly acts on a step other than viral genome copying.
Image Descriptions
Figure 1 image description:
Four column graphs. Panels g and j measure PFU in various samples. Panel g shows ~4.3 x 104 PFU per ml in control cells and almost 0 in RRM2 knockout cells with p = 0.003. Panel j shows ~580 x 104 PFU per ml in control cells and ~270 x 104 PFU per ml in ARL6IP6 knockout cells with p - 0.005. Panels h and k show relative genome copies compared to control cells. Panel h shows nearly 0 genome copies in RRM2 knockout cells with p = 0.001, and panel k shows ~0. 85 genome copies in ARL6IP6 knockout cells with p = 0.044. ↵
Thumbnail
"Qpcr-cycling.png" ↗ by Zuzanna K. Filutowska is licensed under CC BY-SA 3.0 ↗.
Description: qPCR amplification plot.
Author
Katherine Mattaini, Tufts University
-
Finkel, Y., A. Nachshon, E. Aharon, T. Arazi, E. Simonovsky, M. Dobešová, Z. Saud, A. Gluck, T. Fisher, R. J. Stanton, M. Schwartz, and N. Stern-Ginossar. 2024. A virally encoded high-resolution screen of cytomegalovirus dependencies. Nature 630:712–719. ↵


6.6: Order Statistics
Descriptive Theory
Recall again the basic model of statistics: we have a population of objects of interest, and we have various measurements (variables) that we make on these objects. We select objects from the population and record the variables for the objects in the sample; these become our data. Our first discussion is from a purely descriptive point of view. That is, we do not assume that the data are generated by an underlying probability distribution. But as always, remember that the data themselves define a probability distribution, namely the empirical distribution .
Order Statistics
Suppose that \(x\) is a real-valued variable for a population and that \(\bs{x} = (x_1, x_2, \ldots, x_n)\) are the observed values of a sample of size \(n\) corresponding to this variable. The order statistic of rank \(k\) is the \(k\)th smallest value in the data set, and is usually denoted \(x_{(k)}\). To emphasize the dependence on the sample size, another common notation is \(x_{n:k}\). Thus, \[ x_{(1)} \le x_{(2)} \le \cdots \le x_{(n-1)} \le x_{(n)} \] Naturally, the underlying variable \(x\) should be at least at the ordinal level of measurement. The order statistics have the same physical units as \(x\). One of the first steps in exploratory data analysis is to order the data, so order statistics occur naturally. In particular, note that the extreme order statistics are \[ x_{(1)} = \min\{x_1, x_2 \ldots, x_n\}, \quad x_{(n)} = \max\{x_1, x_2, \ldots, x_n\} \] The sample range is \(r = x_{(n)} - x_{(1)}\) and the sample midrange is \(\frac{r}{2} = \frac{1}{2}\left[x_{(n)} - x_{(1)}\right]\). These statistics have the same physical units as \(x\) and are measures of the dispersion of the data set.
The Sample Median
If \(n\) is odd, the sample median is the middle of the ordered observations, namely \(x_{(k)}\) where \(k = \frac{n + 1}{2}\). If \(n\) is even, there is not a single middle observation, but rather two middle observations. Thus, the median interval is \(\left[x_{(k)}, x_{(k + 1)}\right]\) where \(k = \frac{n}{2}\). In this case, the sample median is defined to be the midpoint of the median interval, namely \(\frac{1}{2}\left[x_{(k)} + x_{(k+1)}\right]\) where \(k = \frac{n}{2}\). In a sense, this definition is a bit arbitrary because there is no compelling reason to prefer one point in the median interval over another. For more on this issue, see the discussion of error functions in the section on Sample Variance. In any event, sample median is a natural statistic that gives a measure of the center of the data set.
Sample Quantiles
We can generalize the sample median discussed above to other sample quantiles. Thus, suppose that \(p \in [0, 1]\). Our goal is to find the value that is the fraction \(p\) of the way through the (ordered) data set. We define the rank of the value that we are looking for as \((n - 1)p + 1\). Note that the rank is a linear function of \(p\), and that the rank is 1 when \(p = 0\) and \(n\) when \(p = 1\). But of course, the rank will not be an integer in general, so we let \(k = \lfloor (n - 1)p + 1 \rfloor\), the integer part of the desired rank, and we let \(t = [(n - 1)p + 1] - k\), the fractional part of the desired rank. Thus, \((n - 1)p + 1 = k + t\) where \(k \in \{1, 2, \ldots, n\}\) and \(t \in [0, 1)\). So, using linear interpolation , we define the sample quantile of order \(p\) to be \[ x_{[p]} = x_{(k)} + t \left[x_{(k+1)}-x_{(k)}\right] = (1 - t) x_{(k)} + t x_{(k+1)} \] Sample quantiles have the same physical units as the underlying variable \(x\). The algorithm really does generalize the results for sample medians .
The sample quantile of order \(p = \frac{1}{2}\) is the median as defined earlier, in both cases where \(n\) is odd and where \(n\) is even.
The sample quantile of order \(\frac{1}{4}\) is known as the first quartile and is frequently denoted \(q_1\). The the sample quantile of order \(\frac{3}{4}\) is known as the third quartile and is frequently denoted \(q_3\). The sample median which is the quartile of order \(\frac{1}{2}\) is sometimes denoted \(q_2\). The interquartile range is defined to be \(\iqr = q_3 - q_1\). Note that \(\iqr\) is a statistic that measures the spread of the distribution about the median, but of course this number gives less information than the interval \([q_1, q_3]\).
The statistic \( q_1 - \frac{3}{2} \iqr \) is called the lower fence and the statistic \( q_3 + \frac{3}{2} \iqr \) is called the upper fence . Sometimes lower limit and upper limit are used instead of lower fence and upper fence. Values in the data set that are below the lower fence or above the upper fence are potential outliers , that is, values that don't seem to fit the overall pattern of the data. An outlier can be due to a measurement error, or may be a valid but rather extreme value. In any event, outliers usually deserve additional study.
The five statistics \(\left(x_{(1)}, q_1, q_2, q_3, x_{(n)}\right)\) are often referred to as the
five-number summary
. Together, these statistics give a great deal of information about the data set in terms of the center, spread, and skewness. The five numbers roughly separate the data set into four intervals each of which contains approximately 25% of the data. Graphically, the five numbers, and the outliers, are often displayed as a
boxplot
, sometimes called a
box and whisker plot
. A boxplot consists of an axis that extends across the range of the data. A line is drawn from smallest value that is not an outlier (of course this may be the minimum \(x_{(1)}\)) to the largest value that is not an outlier (of course, this may be the maximum \(x_{(n)}\)). Vertical marks (
whiskers
) are drawn at the ends of this line. A rectangular box extends from the first quartile \(q_1\) to the third quartile \(q_3\) and with an additional whisker at the median \(q_2\). Finally, the outliers are denoted as points (beyond the extreme whiskers). All statistical packages will compute the quartiles and most will draw boxplots. The picture below shows a boxplot with 3 outliers.
Alternate Definitions
The algorithm given above is not the only reasonable way to define sample quantiles, and indeed there are lots of alternatives. One natural method would be to first compute the empirical distribution function \[ F(x) = \frac{1}{n} \sum_{i=1}^n \bs{1}(x_i \le x), \quad x \in \R \] Recall that \(F\) has the mathematical properties of a distribution function, and in fact \(F\) is the distribution function of the empirical distribution of the data. Recall that this is the distribution that places probability \(\frac{1}{n}\) at each data value \(x_i\) (so this is the discrete uniform distribution on \(\{x_1, x_2, \ldots, x_n\}\) if the data values are distinct). Thus, \(F(x) = \frac{k}{n}\) for \(x \in [x_{(k)}, x_{(k+1)})\). Then, we could define the quantile function to be the inverse of the distribution function, as we usually do for probability distributions: \[ F^{-1}(p) = \min\{x \in \R: F(x) \ge p\}, \quad p \in (0, 1) \] It's easy to see that with this definition, the quantile of order \(p \in (0, 1)\) is simply \(x_{(k)}\) where \(k = \lceil n p \rceil\).
Another method is to compute the rank of the quantile of order \(p \in (0, 1)\) as \((n + 1)p\), rather than \((n - 1)p + 1\), and then use linear interpolation just as we have done. To understand the reasoning behind this method, suppose that the underlying variable \(x\) takes value in an interval \((a, b)\). Then the \(n\) points in the data set \(\bs{x}\) separate this interval into \(n + 1\) subintervals, so it's reasonable to think of \(x_{(k)}\) as the quantile of order \(\frac{k}{n + 1}\). This method also reduces to the standard calculation for the median when \(p = \frac{1}{2}\). However, the method will fail if \(p\) is so small that \((n + 1) p \lt 1\) or so large that \( (n + 1) p > n \).
The primary definition that we give above is the one that is most commonly used in statistical software and spreadsheets. Moreover, when the sample size \(n\) is large, it doesn't matter very much which of these competing quantile definitions is used. All will give similar results.
Transformations
Suppose again that \(\bs{x} = (x_1, x_2, \ldots, x_n)\) is a sample of size \(n\) from a population variable \(x\), but now suppose also that \(y = a + b x\) is a new variable, where \(a \in \R\) and \(b \in (0, \infty)\). Recall that transformations of this type are location-scale transformations and often correspond to changes in units. For example, if \(x\) is the length of an object in inches, then \(y = 2.54 x\) is the length of the object in centimeters. If \(x\) is the temperature of an object in degrees Fahrenheit, then \(y = \frac{5}{9}(x - 32)\) is the temperature of the object in degrees Celsius. Let \(\bs{y} = \bs{a} + b \bs{x}\) denote the sample from the variable \(y\).
Order statistics and quantiles are preserved under location-scale transformations:
- \(y_{(i)} = a + b x_{(i)}\) for \(i \in \{1, 2, \ldots, n\}\)
- \(y_{[p]} = a + b x_{[p]}\) for \(p \in [0, 1]\)
Proof
Part (a) follows easily from the fact that the location-scale transformation is strictly increasing and hence preserves order: \(x_i \lt x_j\) if and only if \(a + b x_i \lt a + b x_j\). For part (b), let \(p \in [0, 1]\) and let \(k \in \{1, 2, \ldots,n\}\) and \(t \in [0, 1)\) be as above in the definition of the sample quantile or order \(p\). Then
\[ y_{[p]} = y_{(k)} + t[y_{(k+1)} - y_{(k)}] = a + b x_{(k)} + t[a + b x_{(k+1)} - (a + b x_{(k)})] = a + b\left(x_{(k)} + t [x_{(k+1)}- x_{(k)}]\right) = a + b x_{[p]} \]Like standard deviation (our most important measure of spread), range and interquartile range are not affected by the location parameter, but are scaled by the scale parameter.
The range and interquartile range of \(\bs{y}\) are
- \(r(\bs{y}) = b \, r(\bs{x})\)
- \(\iqr(\bs{y}) = b \, \iqr(\bs{x})\)
Proof
These results follow immediately from the previous result.
More generally, suppose \(y = g(x)\) where \(g\) is a strictly increasing real-valued function on the set of possible values of \(x\). Let \(\bs{y} = \left(g(x_1), g(x_2), \ldots, g(x_n)\right)\) denote the sample corresponding to the variable \(y\). Then (as in the proof of Theorem 2), the order statistics are preserved so \(y_{(i)} = g(x_{(i)})\). However, if \(g\) is nonlinear, the quantiles are not preserved (because the quantiles involve linear interpolation). That is, \(y_{[p]}\) and \(g(x_{[p]})\) are not usually the same. When \(g\) is convex or concave we can at least give an inequality for the sample quantiles.
Suppose that \(y = g(x)\) where \(g\) is strictly increasing. Then
- \(y_{(i)} = g\left(x_{(i)}\right)\) for \(i \in \{1, 2, \ldots, n\}\)
- If \(g\) is convex then \(y_{[p]} \ge g\left(x_{[p]}\right)\) for \(p \in [0, 1]\)
- If \(g\) is concave then \(y_{[p]} \le g\left(x_{[p]}\right)\) for \(p \in [0, 1]\)
Proof
As noted, part (a) follows since \(g\) is strictly increasing and hence preserves order. Part (b) follows from the definition of convexity. For \(p \in [0, 1]\), and \(k \in \{1, 2, \ldots, n\}\) and \(t \in [0, 1)\) as in the definition of the sample quantile of order \(p\), we have \[ y_{[p]} = (1 - t) y_{(k)} + t y_{(k+1)} = (1 - t) g\left(x_{(k)}\right) + t g\left(x_{(k+1)}\right) \ge g\left[(1 - t) x_{(k)} + t x_{(k+1)}\right] = g\left(x_{[p]}\right) \] Part (c) follows by the same argument.
Stem and Leaf Plots
A stem and leaf plot is a graphical display of the order statistics \(\left(x_{(1)}, x_{(2)}, \ldots, x_{(n)}\right)\). It has the benefit of showing the data in a graphical way, like a histogram, and at the same time, preserving the ordered data. First we assume that the data have a fixed number format: a fixed number of digits, then perhaps a decimal point and another fixed number of digits. A stem and leaf plot is constructed by using an initial part of this string as the stem , and the remaining parts as the leaves. There are lots of variations in how to do this, so rather than give an exhaustive, complicated definition, we will just look at a couple of examples in the exercise below.
Probability Theory
We continue our discussion of order statistics except that now we assume that the variables are random variables. Specifically, suppose that we have a basic random experiment, and that \(X\) is a real-valued random variable for the experiment with distribution function \(F\). We perform \(n\) independent replications of the basic experiment to generate a random sample \(\bs{X} = (X_1, X_2, \ldots, X_n)\) of size \(n\) from the distribution of \(X\). Recall that this is a sequence of independent random variables, each with the distribution of \(X\). All of the statistics defined in the previous section make sense, but now of course, they are random variables. We use the notation established previously, except that we follow our usual convention of denoting random variables with capital letters. Thus, for \(k \in \{1, 2, \ldots, n\}\), \(X_{(k)}\) is the \(k\)th order statistic , that is, the \(k\) smallest of \((X_1, X_2, \ldots, X_n)\). Our interest now is on the distribution of the order statistics and statistics derived from them.
Distribution of the \(k\)th order statistic
Finding the distribution function of an order statistic is a nice application of Bernoulli trials and the binomial distribution.
The distribution function \( F_k \) of \(X_{(k)}\) is given by \[ F_k(x) = \sum_{j=k}^n \binom{n}{j} \left[F(x)\right]^j \left[1 - F(x)\right]^{n - j}, \quad x \in \R \]
Proof
For \( x \in \R \), let \[ N_x = \sum_{i=1}^n \bs{1}(X_i \le x) \] so that \( N_x \) is the number of sample variables that fall in the interval \( (-\infty, x] \). The indicator variables in the sum are independent, and each takes the value 1 with probability \( F(x) \). Thus, \( N_x \) has the binomial distribution with parameters \( n \) and \( F(x) \). Next note that \(X_{(k)} \le x\) if and only if \(N_x \ge k\) for \(x \in \R\) and \(k \in \{1, 2, \ldots, n\}\), since both events mean that there are at least \( k \) sample variables in the interval \( (-\infty x] \). Hence \[ \P\left(X_{(k)} \le x\right) = \P\left(N_x \ge k\right) = \sum_{j=k}^n \binom{n}{j} \left[F(x)\right]^j \left[1 - F(x)\right]^{n - j} \]
As always, the extreme order statistics are particularly interesting.
The distribution functions \( F_1 \) of \( X_{(1)} \) and \( F_n \) of \( X_{(n)} \) are given by
- \(F_1(x) = 1 - \left[1 - F(x)\right]^n\) for \(x \in \R\)
- \(F_n(x) = \left[F(x)\right]^n\) for \(x \in \R\)
The quantile functions \( F_1^{-1} \) and \( F_n^{-1} \) of \( X_{(1)} \) and \( X_{(n)} \) are given by
- \( F_1^{-1}(p) = F^{-1}\left[1 - (1 - p)^{1/n}\right] \) for \( p \in (0, 1) \)
- \( F_n^{-1}(p) = F^{-1}\left(p^{1/n}\right) \) for \( p \in (0, 1) \)
Proof
The formulas follow from the previous theorem and simple algebra. Recall that if \( G \) is a distribution function, then the corresponding quantile function is given by \( G^{-1}(p) = \min\{x \in \R: G(x) \ge p\} \) for \( p \in (0, 1) \).
When the underlying distribution is continuous, we can give a simple formula for the probability density function of an order statistic.
Suppose now that \(X\) has a continuous distribution with probability density function \( f \). Then \(X_{(k)}\) has a continuous distribution with probability density function \( f_k \) given by \[ f_k(x) = \frac{n!}{(k - 1)! (n - k)!} \left[F(x)\right]^{k-1} \left[1 - F(x)\right]^{n-k} f(x), \quad x \in \R\]
Proof
Of course, \(f_k(x) = F_k^\prime(x)\). We take the derivatives term by term and use the product rule on \[ \frac{d}{dx}\left[F(x)\right]^j \left[1 - F(x)\right]^{n-j} = j \left[F(x)\right]^{j-1} f(x) \left[1 - F(x)\right]^{n-j} - (n - j)\left[F(x)\right]^j \left[1 - F(x)\right]^{n-j-1}f(x) \] We use the binomial identities \(j \binom{n}{j} = n \binom{n - 1}{j - 1}\) and \((n - j) \binom{n}{j} = n \binom{n - 1}{j}\). The net effect is \[ f_k(x) = n f(x) \left[ \sum_{j=k}^n \binom{n - 1}{j - 1}[F(x)]^{j-1} [1 - F(x)]^{(n-1)-(j-1)} - \sum_{j=k}^{n-1} \binom{n-1}{j} [F(x)]^j [1 - F(x)]^{(n-1)-j}\right] \] The sums cancel, leaving only the \(j = k\) term in the first sum. Hence \[f_k(x) = n f(x) \binom{n-1}{k-1}[F(x)]^{k-1}[1 - F(x)]^{n-k}\] But \(n \binom{n-1}{k-1} = \frac{n!}{(k-1)!(n-k)!}\).
Heuristic Proof
There is a simple heuristic argument for this result First, \(f_k(x) \, dx\) is the probability that \(X_{(k)}\) is in an infinitesimal interval of size \(dx\) about \(x\). On the other hand, this event means that one of sample variables is in the infinitesimal interval, \(k - 1\) sample variables are less than \(x\), and \(n - k\) sample variables are greater than \(x\). The number of ways of choosing these variables is the multinomial coefficient \[ \binom{n}{k - 1, 1, n - k} = \frac{n!}{(k - 1)! (n - k)!} \] By independence, the probability that the chosen variables are in the specified intervals is \[ \left[F(x)\right]^{k-1} \left[1 - F(x)\right]^{n-k} f(x) \, dx \]
Here are the special cases for the extreme order statistics.
The probability density function \( f_1 \) of \( X_{(1)} \) and \( f_n \) of \( X_{(n)} \) are given by
- \(f_1(x) = n \left[1 - F(x)\right]^{n-1} f(x)\) for \(x \in \R\)
- \(f_n(x) = n \left[F(x)\right]^{n-1} f(x)\) for \(x \in \R\)
Joint Distributions
We assume again that \(X\) has a continuous distribution with distribution function \( F \) and probability density function \( f \).
Suppose that \(j, k \in \{1, 2, \ldots, n\}\) with \(j \lt k\). The joint probability density function \( f_{j,k} \) of \(\left(X_{(j)}, X_{(k)}\right)\) is given by
\[ f_{j,k}(x, y) = \frac{n!}{(j - 1)! (k - j - 1)! (n - k)!} \left[F(x)\right]^{j-1} \left[F(y) - F(x)\right]^{k - j - 1} \left[1 - F(y)\right]^{n-k} f(x) f(y); \quad x, \, y \in \R, x \lt y \]Heuristic Proof
We want to compute the probability that \(X_{(j)}\) is in an infinitesimal interval \(dx\) about \(x\) and \(X_{(k)}\) is in an infinitesimal interval \(dy\) about \(y\). Note that there must be \(j - 1\) sample variables that are less than \(x\), one variable in the infinitesimal interval about \(x\), \(k - j - 1\) sample variables that are between \(x\) and \(y\), one variable in the infinitesimal interval about \(y\), and \(n - k\) sample variables that are greater than \(y\). The number of ways to select the variables is the multinomial coefficient \[ \binom{n}{j-1, 1, k - j - 1, 1, n - k} = \frac{n!}{(j - 1)! (k - j - 1)! (n - k)!} \] By independence, the probability that the chosen variables are in the specified intervals is \[ \left[F(x)\right]^{j-1} f(x) dx \left[F(y) - F(x)\right]^{k - j - 1} f(y) dy \left[1 - F(y)\right]^{n-k} \]
From the joint distribution of two order statistics we can, in principle, find the distribution of various other statistics: the sample range \(R\); sample quantiles \(X_{[p]}\) for \(p \in [0, 1]\), and in particular the sample quartiles \(Q_1\), \(Q_2\), \(Q_3\); and the inter-quartile range IQR. The joint distribution of the extreme order statistics \((X_{(1)}, X_{(n)})\) is a particularly important case.
The joint probability density function \( f_{1,n} \) of \(\left(X_{(1)}, X_{(n)}\right)\) is given by \[ f_{1,n}(x, y) = n (n - 1) \left[F(y) - F(x)\right]^{n-2} f(x) f(y); \quad x, \, y \in \R, x \lt y \]
Proof
This is a corollary of Theorem 7 with \(j = 1\) and \(k = n\).
Arguments similar to the one above can be used to obtain the joint probability density function of any number of the order statistics. Of course, we are particularly interested in the joint probability density function of all of the order statistics. It turns out that this density function has a remarkably simple form.
\(\left(X_{(1)}, X_{(2)}, \ldots, X_{(n)}\right)\) has joint probability density function \( g \) given by \[ g(x_1, x_2, \ldots, x_n) = n! f(x_1) \, f(x_2) \, \cdots \, f(x_n), \quad x_1 \lt x_2 \lt \cdots \lt x_n \]
Proof
For each permutation \(\bs{i} = (i_1, i_2, \ldots, i_n)\) of \((1, 2, \ldots, n)\), let \(S_\bs{i} = \{\bs{x} \in \R^n: x_{i_1} \lt x_{i_2} \lt \cdots \lt x_{i_n}\}\). On \(S_\bs{i}\), the mapping \((x_1, x_2, \ldots, x_n) \mapsto (x_{i_1}, x_{i_2}, \ldots, x_{i_n})\) is one-to-one, has continuous first partial derivatives, and has Jacobian 1. The sets \(S_\bs{i}\) where \(\bs{i}\) ranges over the \(n!\) permutations of \((1, 2, \ldots, n)\) are disjoint. The probability that \((X_1, X_2, \ldots, X_n)\) is not in one of these sets is 0. The result now follows from the multivariate change of variables formula.
Heuristic Proof
Again, there is a simple heuristic argument for this result. For each \(\bs{x} \in \R^n\) with \(x_1 \lt x_2 \lt \cdots \lt x_n\), there are \(n!\) permutations of the coordinates of \(\bs{x}\). The probability density of \((X_1, X_2, \ldots, X_n)\) at each of these points is \(f(x_1) \, f(x_2) \, \cdots \, f(x_n)\). Hence the probability density of \((X_{(1)}, X_{(2)}, \ldots, X_{(n)})\) at \(\bs{x}\) is \(n!\) times this product.
Probability Plots
A probability plot , also called a quantile-quantile plot or a Q-Q plot for short, is an informal, graphical test to determine if observed data come from a specified distribution. Thus, suppose that we observe real-valued data \((x_1, x_2, \ldots, x_n)\) from a random sample of size \(n\). We are interested in the question of whether the data could reasonably have come from a continuous distribution with distribution function \(F\). First, we order that data from smallest to largest; this gives us the sequence of observed values of the order statistics: \(\left(x_{(1)}, x_{(2)}, \ldots, x_{(n)}\right)\).
Note that we can view \(x_{(i)}\) has the sample quantile of order \(\frac{i}{n + 1}\). Of course, by definition, the distribution quantile of order \(\frac{i}{n + 1}\) is \( y_i = F^{-1} \left( \frac{i}{n + 1} \right) \). If the data really do come from the distribution, then we would expect the points \( \left(\left(x_{(1)}, y_1\right), \left(x_{(2)}, y_2\right) \ldots, \left(x_{(n)}, y_n\right)\right) \) to be close to the diagonal line \(y = x\); conversely, strong deviation from this line is evidence that the distribution did not produce the data. The plot of these points is referred to as a probability plot .
Usually however, we are not trying to see if the data come from a particular distribution, but rather from a parametric family of distributions (such as the normal, uniform, or exponential families). We are usually forced into this situation because we don't know the parameters; indeed the next step, after the probability plot, may be to estimate the parameters. Fortunately, the probability plot method has a simple extension for any location-scale family of distributions. Thus, suppose that \(G\) is a given distribution function. Recall that the location-scale family associated with \(G\) has distribution function \( F(x) = G \left( \frac{x - a}{b} \right)\) for, \(x \in \R \), where \(a \in \R\) is the location parameter and \(b \in (0, \infty)\) is the scale parameter. Recall also that for \(p \in (0, 1)\), if \(z_p = G^{-1}(p)\) denote the quantile of order \(p\) for \(G\) and \(y_p = F^{-1}(p)\) the quantile of order \(p\) for \(F\). Then \( y_p = a + b \, z_p \). It follows that if the probability plot constructed with distribution function \(F\) is nearly linear (and in particular, if it is close to the diagonal line), then the probability plot constructed with distribution function \(G\) will be nearly linear. Thus, we can use the distribution function \(G\) without having to know the location and scale parameters.
In the exercises below, you will explore probability plots for the normal, exponential, and uniform distributions. We will study a formal, quantitative procedure, known as the chi-square goodness of fit test in the chapter on Hypothesis Testing.
Exercises and Applications
Basic Properties
Suppose that \(x\) is the temperature (in degrees Fahrenheit) for a certain type of electronic component after 10 hours of operation. A sample of 30 components has five number summary \((84, 102, 113, 120, 135)\).
- Classify \(x\) by type and level of measurement.
- Find the range and interquartile range.
- Find the five number summary, range, and interquartile range if the temperature is converted to degrees Celsius. The transformation is \(y = \frac{5}{9}(x - 32)\).
Answer
- continuous, interval
- 51, 18
- \((28.89, 38.89, 45.00, 48.89, 57.22)\), 28.33, 10
Suppose that \(x\) is the length (in inches) of a machined part in a manufacturing process. A sample of 50 parts has five number summary (9.6, 9.8, 10.0, 10.1, 10.3).
- Classify \(x\) by type and level of measurement.
- Find the range and interquartile range.
- Find the five number summary, range, and interquartile if length is measured in centimeters. The transformation is \(y = 2.54 x\).
Answer
- continuous, ratio
- 0.7, 0.3
- \((24.38, 24.89, 25.40, 25.65, 26.16)\), 1.78, 0.76
Professor Moriarity has a class of 25 students in her section of Stat 101 at Enormous State University (ESU). For the first midterm exam, the five number summary was (16, 52, 64, 72, 81) (out of a possible 100 points). Professor Moriarity thinks the grades are a bit low and is considering various transformations for increasing the grades.
- Find the range and interquartile range.
- Suppose she adds 10 points to each grade. Find the five number summary, range, and interquartile range for the transformed grades.
- Suppose she multiplies each grade by 1.2. Find the five number summary, range, and interquartile range for the transformed grades.
- Suppose she uses the transformation \(w = 10 \sqrt{x}\), which curves the grades greatly at the low end and very little at the high end. Give whatever information you can about the five number summary of the transformed grades.
- Determine whether the low score of 16 is an outlier.
Answer
- 65, 20
- \((26, 62, 74, 82, 91)\), 65, 20
- \((19.2, 62.4, 76.8, 86.4, 97.2)\), 78, 24
- \(y_{(1)} = 40\), \(q_1 \le 72.11\), \(q_2 \le 80\), \(q_3 \le 84.85 \), \(y_{(25)} = 90\)
- The lower fence is 27, so yes 16 is an outlier.
Computational Exercises
All statistical software packages will compute order statistics and quantiles, draw stem-and-leaf plots and boxplots, and in general perform the numerical and graphical procedures discussed in this section. For real statistical experiments, particularly those with large data sets, the use of statistical software is essential. On the other hand, there is some value in performing the computations by hand, with small, artificial data sets, in order to master the concepts and definitions. In this subsection, do the computations and draw the graphs with minimal technological aids.
Suppose that \(x\) is the number of math courses completed by an ESU student. A sample of 10 ESU students gives the data \(\bs{x} = (3, 1, 2, 0, 2, 4, 3, 2, 1, 2)\).
- Classify \(x\) by type and level of measurement.
- Give the order statistics
- Compute the five number summary and draw the boxplot.
- Compute the range and the interquartile range.
Answer
- discrete, ratio
- \((0, 1, 1, 2, 2, 2, 2, 3, 3, 4)\)
- \((0, 1.25, 2, 2.75, 4)\)
- 4, 1.5
Suppose that a sample of size 12 from a discrete variable \(x\) has empirical density function given by \(f(-2) = 1/12\), \(f(-1) = 1/4\), \(f(0) = 1/3\), \(f(1) = 1/6\), \(f(2) = 1/6\).
- Give the order statistics.
- Compute the five number summary and draw the boxplot.
- Compute the range and the interquartile range.
Answer
- \((-2, -1, -1, -1, 0, 0, 0, 0, 1, 1, 2, 2)\)
- \((-2, -1, 0, 1, 2)\)
- 4, 2
The stem and leaf plot below gives the grades for a 100-point test in a probability course with 38 students. The first digit is the stem and the second digit is the leaf. Thus, the low score was 47 and the high score was 98. The scores in the 6 row are 60, 60, 62, 63, 65, 65, 67, 68.
\[\begin{array}{l|l}
4 & 7 \\
5 & 0346 \\
6 & 00235578 \\
7 &0112346678899 \\
8 &0367889 \\
9 & 1368
\end{array} \nonumber \]
Compute the five number summary and draw the boxplot.
Answer
\((47, 65, 75, 83, 98) \)
App Exercises
In the histogram app, construct a distribution with at least 30 values of each of the types indicated below. Note the five number summary.
- A uniform distribution.
- A symmetric, unimodal distribution.
- A unimodal distribution that is skewed right.
- A unimodal distribution that is skewed left.
- A symmetric bimodal distribution.
- A \(u\)-shaped distribution.
In the error function app, Start with a distribution and add additional points as follows. Note the effect on the five number summary:
- Add a point below \(x_{(1)}\).
- Add a point between \(x_{(1)}\) and \(q_1\).
- Add a point between \(q_1\) and \(q_2\).
- Add a point between \(q_2\) and \(q_3\).
- Add a point between \(q_3\) and \(x_{(n)}\).
- Add a point above \(x_{(n)}\).
In the last problem, you may have noticed that when you add an additional point to the distribution, one or more of the five statistics does not change. In general, quantiles can be relatively insensitive to changes in the data.
The Uniform Distribution
Recall that the standard uniform distribution is the uniform distribution on the interval \([0, 1]\).
Suppose that \(\bs{X}\) is a random sample of size \(n\) from the standard uniform distribution. For \( k \in \{1, 2, \ldots, n\} \), \(X_{(k)}\) has the beta distribution, with left parameter \(k\) and right parameter \(n - k + 1\). The probability density function \( f_k \) is given by \[f_k(x) = \frac{n!}{(k - 1)! (n - k)!} x^{k-1} (1 - x)^{n-k}, \quad 0 \le x \le 1\]
Proof
This follows immediately from the basic theorem above since \(f(x) = 1\) and \(F(x) = x\) for \(0 \le x \le 1\). From the form of \(f_k\) we can identify the distribution as beta with left parameter \(k\) and right parameter \(n - k + 1\).
In the order statistic experiment, select the standard uniform distribution and \(n = 5\). Vary \(k\) from 1 to 5 and note the shape of the probability density function of \(X_{(k)}\). For each value of \(k\), run the simulation 1000 times and compare the empirical density function to the true probability density function.
It's easy to extend the results for the standard uniform distribution to the general uniform distribution on an interval.
Suppose that \( \bs{X} \) is a random sample of size \( n \) from the uniform distribution on the interval \( [a, a + h] \) where \( a \in \R \) and \( h \in (0, \infty) \). For \( k \in \{1, 2, \ldots, n\} \), \( X_{(k)} \) has the beta distribution with left parameter \( k \), right parameter \( n - k + 1 \), location parameter \( a \), and scale parameter \( h \). In particular,
- \( \E\left(X_{(k)}\right) = a + h \frac{k}{n + 1} \)
- \( \var\left(X_{(k)}\right) = h^2 \frac{k (n - k + 1)}{(n + 1)^2 (n + 2)}\)
Proof
Suppose that \( \bs{U} = (U_1, U_2, \ldots, U_n) \) is a random sample of size \( n \) from the standard uniform distribution, and let \( X_i = a + h U_i \) for \( i \in \{1, 2, \ldots, n\} \). Then \( \bs{X} = (X_1, X_2, \ldots, X_n) \) is a random sample of size \( n \) from the uniform distribution on the interval \( [a, a + h] \), and moreover, \( X_{(k)} = a + h U_{(k)} \). So the distribution of \( X_{(k)} \) follows from the previous result. Parts (a) and (b) follow from standard results for the beta distribution.
We return to the standard uniform distribution and consider the range of the random sample.
Suppose that \(\bs{X}\) is a random sample of size \(n\) from the standard uniform distribution. The sample range \( R \) has the beta distribution with left parameter \( n - 1 \) and right parameter 2. The probability density function \( g \) is given by \[ g(r) = n (n - 1) r^{n-2} (1 - r), \quad 0 \le r \le 1 \]
Proof
From the result above , the joint PDF of \((X_{(1)}, X_{(n)})\) is \(f_{1, n}(x, y) = n (n - 1) (y - x)^{n - 2}\) for \(0 \le x \le y \le 1\). Hence, for \(r \in [0, 1]\), \[ \P(R \gt r) = \P(X_{(n)} - X_{(1)} \gt r) = \int_0^{1-r} \int_{x+r}^1 n (n - 1) (y - x)^{n-2} \, dy \, dx = (n - 1) r^n - n r^{n-1} + 1 \] It follows that the CDF of \(R\) is \(G(r) = n r^{n-1} - (n - 1)r^n\) for \(0 \le r \le 1\). Taking the derivative with respect to \(r\) and simplifying gives the PDF \(g(r) = n (n - 1) r^{n-2} (1 - r)\) for \(0 \le r \le 1\). We can tell from the form of \(g\) that the distribution is beta with left parameter \(n - 1\) and right parameter 2.
Once again, it's easy to extend this result to a general uniform distribution.
Suppose that \( \bs{X} = (X_1, X_2, \ldots, X_n) \) is a random sample of size \( n \) from the uniform distribution on \( [a, a + h] \) where \( a \in \R \) and \( h \in (0, \infty) \). The sample range \( R = X_{(n)} - X_{(1)} \) has the beta distribution with left parameter \( n - 1 \), right parameter \( 2 \), and scale parameter \( h \). In particular,
- \( \E(R) = h \frac{n - 1}{n + 1} \)
- \( \var(R) = h^2 \frac{2 (n _ 1)}{(n + 1)^2 (n + 2)} \)
Proof
Suppose again that \( \bs{U} = (U_1, U_2, \ldots, U_n) \) is a random sample of size \( n \) from the standard uniform distribution, and let \( X_i = a + h U_i \) for \( i \in \{1, 2, \ldots, n\} \). Then \( \bs{X} = (X_1, X_2, \ldots, X_n) \) is a random sample of size \( n \) from the uniform distribution on the interval \( [a, a + h] \), and moreover, \( X_{(k)} = a + h U_{(k)} \). Hence \( X_{(n)} - X_{(1)} = h(U_{(n)} - U_{(1)} \) so the distribution of \( R \) follows from the previous result. Parts (a) and (b) follow from standard results for the beta distribution.
The joint distribution of the order statistics for a sample from the uniform distribution is easy to get.
Suppose that \((X_1, X_2, \ldots, X_n)\) is a random sample of size \(n\) from the uniform distribution on the interval \([a, a + h]\), where \(a \in \R\) and \( h \in (0, \infty) \). Then \(\left(X_{(1)}, X_{(2)}, \ldots, X_{(n)}\right)\) is uniformly distributed on \(\left\{\bs{x} \in [a, a + h]^n: a \le x_1 \le x_2 \le \cdots \le x_n \lt a + h\right\}\).
Proof
This follows easily from the fact that \((X_1, X_2, \ldots, X_n)\) is uniformly distributed on \([a, a + h]^n\). From the result above , the joint PDF of the order statistics is \(g(x_1, x_2, \ldots, x_n) = n! / h^n\) for \((x_1, x_2, \ldots, x_n) \in [a, a + h]^n\) with \(a \le x_1 \le x_2 \le \cdots \le x_n \le a + h\).
The Exponential Distribution
Recall that the exponential distribution with rate parameter \(\lambda \gt 0\) has probability density function \[ f(x) = \lambda e^{-\lambda x}, \quad 0 \le x \lt \infty \] The exponential distribution is widely used to model failure times and other random times under certain ideal conditions. In particular, the exponential distribution governs the times between arrivals in the Poisson process.
Suppose that \(\bs{X}\) is a random sample of size \(n\) from the exponential distribution with rate parameter \(\lambda\). The probability density function of the \(k\)th order statistic \(X_{(k)}\) is \[ f_k(x) = \frac{n!}{(k - 1)! (n - k)!} \lambda (1 - e^{-\lambda x})^{k-1} e^{-\lambda(n - k + 1)x}, \quad 0 \le x \lt \infty \] In particular, the minimum of the variables \(X_{(1)}\) also has an exponential distribution, but with rate parameter \(n \lambda\).
Proof
The PDF of \(X_{(k)}\) follows from the theorem above since \(F(x) = 1 - e^{-\lambda x}\) for \(0 \le x \lt \infty\). Substituting \(k = 1\) gives \(f_1(x) = n \lambda e^{-n \lambda x}\) for \(0 \le x \lt \infty\).
In the order statistic experiment, select the standard exponential distribution and \(n = 5\). Vary \(k\) from 1 to 5 and note the shape of the probability density function of \(X_{(k)}\). For each value of \(k\), run the simulation 1000 times and compare the empirical density function to the true probability density function.
Suppose again that \(\bs{X}\) is a random sample of size \(n\) from the exponential distribution with rate parameter \(\lambda\). The sample range \(R\) has the same distribution as the maximum of a random sample of size \(n - 1\) from the exponential distribution. The probability density function is \[ h(t) = (n - 1) \lambda (1 - e^{-\lambda t})^{n - 2} e^{-\lambda t}, \quad 0 \le t \lt \infty \]
Proof
By the result above , \((X_{(1)}, X_{(n)})\) has joint PDF \(f_{1, n}(x, y) = n (n - 1) \lambda^2 (e^{-\lambda x} - e^{-\lambda y})^{n-2} e^{-\lambda x} e^{-\lambda y}\) for \(0 \le x \le y \lt \infty\). Hence for \(0 \le t \lt \infty\), \[ \P(R \le t) = \P(X_{(n)} - X_{(1)} \le t) = \int_0^\infty \int_x^{x + t} n (n - 1) \lambda^2 (e^{-\lambda x} - e^{-\lambda y})^{n-2} e^{-\lambda x} e^{-\lambda y} \, dy \, dx \] Substituting \(u = e^{-\lambda y}\), \(du = -\lambda e^{-\lambda y} \, dy\) into the inside integral and evaluating gives \[ \P(R \le t) = \int_0^\infty n \lambda e^{-n \lambda x} (1 - e^{-\lambda t})^{n-1} \, dx = (1 - e^{-\lambda t})^{n-1}\] Differentiating with respect to \(t\) gives the the PDF. Comparing with our previous result , we see that this is the PDF of the maximum of a sample of size \(n - 1\) from the exponential distribution.
Suppose again that \(\bs{X}\) is a random sample of size \(n\) from the exponential distribution with rate parameter \(\lambda\). The joint probability density function of the order statistics \((X_{(1)}, X_{(2)}, \ldots, X_{(n)})\) is \[ g(x_1, x_2, \ldots, x_n) = n! \lambda^n e^{-\lambda(x_1 + x_2 + \cdots + x_n)}, \quad 0 \le x_1 \le x_2 \cdots \le x_n \lt \infty \]
Proof
This follows from the result above and simple algebra.
Dice
Four fair dice are rolled. Find the probability density function of each of the order statistics.
Answer
| \(x\) | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| \(f_1(x)\) | \(\frac{671}{1296}\) | \(\frac{369}{1296}\) | \(\frac{175}{1296}\) | \(\frac{65}{1296}\) | \(\frac{15}{1296}\) | \(\frac{1}{1296}\) |
| \(f_2(x)\) | \(\frac{171}{1296}\) | \(\frac{357}{1296}\) | \(\frac{363}{1296}\) | \(\frac{261}{1296}\) | \(\frac{123}{1296}\) | \(\frac{21}{1296}\) |
| \(f_3(x)\) | \(\frac{21}{1296}\) | \(\frac{123}{1296}\) | \(\frac{261}{1296}\) | \(\frac{363}{1296}\) | \(\frac{357}{1296}\) | \(\frac{171}{1296}\) |
| \(f_4(x)\) | \(\frac{1}{1296}\) | \(\frac{15}{1296}\) | \(\frac{65}{1296}\) | \(\frac{175}{1296}\) | \(\frac{369}{1296}\) | \(\frac{671}{1296}\) |
In the dice experiment, select the order statistic and die distribution given in parts (a)–(d) below. Increase the number of dice from 1 to 20, noting the shape of the probability density function at each stage. Now with \(n = 4\), run the simulation 1000 times, and note the apparent convergence of the relative frequency function to the probability density function.
- Maximum score with fair dice.
- Minimum score with fair dice.
- Maximum score with ace-six flat dice.
- Minimum score with ace-six flat dice.
Four fair dice are rolled. Find the joint probability density function of the four order statistics.
Answer
The joint probability density function \(g\) is defined on \(\{(x_1, x_2, x_3, x_4) \in \{1, 2, 3, 4, 5, 6\}^4: x_1 \le x_2 \le x_3 \le x_4\}\)
- \(g(x_1, x_2, x_3, x_4) = \frac{1}{1296}\) if the coordinates are all the same (there are 6 such vectors).
- \(g(x_1, x_2, x_3, x_4) = \frac{4}{1296}\) if there are two distinct coordinates, one value occurring 3 times and the other value once (there are 30 such vectors).
- \(g(x_1, x_2, x_3, x_4) = \frac{6}{1296}\) if there are two distinct coordinates in \((x_1, x_2, x_3, x_4)\), each value occurring 2 times (there are 15 such vectors).
- \(g(x_1, x_2, x_3, x_4) = \frac{12}{1296}\) if there are three distinct coordinates, one value occurring twice and the other values once (there are 60 such vectors).
- \(g(x_1, x_2, x_3, x_4) = \frac{24}{1296}\) if the coordinates are distinct (there are 15 such vectors).
Four fair dice are rolled. Find the probability density function of the sample range.
Answer
\(R\) has probability density function \(h\) given by \(h(0) = \frac{6}{1296}, \; h(1) = \frac{70}{1296}, \; h(2) = \frac{300}{1296}, \; h(3) = \frac{300}{1296}, \; h(4) = \frac{318}{1296}, \; h(5) = \frac{302}{1296}\)
Probability Plot Simulations
In the probability plot experiment, set the sampling distribution to normal distribution with mean 5 and standard deviation 2. Set the sample size to \(n = 20\). For each of the following test distributions, run the experiment 50 times and note the geometry of the probability plot:
- Standard normal
- Uniform on the interval \([0, 1]\)
- Exponential with parameter 1
In the probability plot experiment, set the sampling distribution to the uniform distribution on \([4, 10]\). Set the sample size to \(n = 20\). For each of the following test distributions, run the experiment 50 times and note the geometry of the probability plot:
- Standard normal
- Uniform on the interval \([0, 1]\)
- Exponential with parameter 1
In the probability plot experiment, Set the sampling distribution to the exponential distribution with parameter 3. Set the sample size to \(n = 20\). For each of the following test distributions, run the experiment 50 times and note the geometry of the probability plot:
- Standard normal
- Uniform on the interval \([0, 1]\)
- Exponential with parameter 1
Data Analysis Exercises
Statistical software should be used for the problems in this subsection.
Consider the petal length and species variables in Fisher's iris data.
- Classify the variables by type and level of measurement.
- Compute the five number summary and draw the boxplot for petal length.
- Compute the five number summary and draw the boxplot for petal length by species.
- Draw the normal probability plot for petal length.
Answers
- petal length: continuous, ratio. type: discrete, nominal
- \((10, 15, 44, 51, 69)\)
- type 0: \((10, 14, 15, 16, 19)\); type 1: \((45, 51, 55.5, 59, 69)\); type 2: \((30, 40, 44, 47, 56)\)
Consider the erosion variable in the Challenger data set.
- Classify the variable by type and level of measurement.
- Compute the five number summary and draw the boxplot.
- Identify any outliers.
Answer
- continuous, ratio
- \((0, 0, 0, 0, 53)\)
- All of the positive values 28, 40, 48, and 53 are outliers.
A stem and leaf plot of Michelson's velocity of light data is given below. In this example, the last digit (which is always 0) has been left out, for convenience. Also, note that there are two sets of leaves for each stem, one corresponding to leaves from 0 to 4 (so actually from 00 to 40) and the other corresponding to leaves from 5 to 9 (so actually from 50 to 90). Thus, the minimum value is 620 and the numbers in the second 7 row are 750, 760, 760, and so forth.
\[\begin{array}{l|l}
6 & 2 \\
6 & 5 \\
7 & 222444 \\
7 &<PHONE_NUMBER>99 \\
8 &<PHONE_NUMBER>11111223344444444 \\
9 &<PHONE_NUMBER> \\
9 &<PHONE_NUMBER>8 \\
10 & 000 \\
10 & 7
\end{array} \nonumber \]
Classify the variable by type and level of measurement.
- Compute the five number summary and draw the boxplot.
- Compute the five number summary for the velocity in \(\text{km}/\text{hr}\). The transformation is \(y = x + 299\,000\).
- Draw the normal probability plot.
Answer
- continuous, interval
- \((620, 805, 850, 895, 1071)\)
- \((299\,620, 299\,805, 299\,850, 299\,895, 300\,071)\)
Consider Short's paralax of the sun data.
- Classify the variable by type and level of measurement.
- Compute the five number summary and draw the boxplot.
- Compute the five number summary and draw the boxplot if the variable is converted to degrees. There are 3600 seconds in a degree.
- Compute the five number summary and draw the boxplot if the variable is converted to radians. There are \(\pi/180\) radians in a degree.
- Draw the normal probability plot.
Answer
- continuous, ratio
- \((5.76, 8.34, 8.50, 9.02, 10.57)\)
- \((0.00160, 0.00232, 0.00236, 0.00251, 0.00294)\)
- \((0.0000278, 0.0000404, 0.0000412, 0.0000437, 0.0000512)\)
Consider Cavendish's density of the earth data.
- Classify the variable by type and level of measurement.
- Compute the five number summary and draw the boxplot.
- Draw the normal probability plot.
Answer
- continuous, ratio
- \((4.88, 5.30, 5.46, 5.61, 5.85)\)
Consider the M&M data.
- Classify the variables by type and level of measurement.
- Compute the five number summary and draw the boxplot for each color count.
- Construct a stem and leaf plot for the total number of candies.
- Compute the five number summary and draw the boxplot for the total number of candies.
- Compute the five number summary and draw the boxplot for net weight.
Answer
- color counts: discrete ratio. net weight: continuous ratio.
- red: \((3, 5.5, 9, 14, 20)\); green: \((2, 5, 7, 9, 17)\); blue: \((1, 4, 6.5, 10, 19)\); orange: \((0, 3.5, 6, 10.5, 13)\); yellow: \((3, 8, 13.5, 18, 26)\); brown: \((4, 8, 12.5, 18, 20)\)
-
5 0 5 3 5 4 5 5 5 5 5 6 6 6 6 7 7 7 5 8 8 8 8 8 8 8 8 8 9 9 9 6 0 0 1 1 - \((50, 55.5, 58, 60, 61)\)
- \((46.22, 48.28, 49.07, 50.23, 52.06)\)
Consider the body weight, species, and gender variables in the Cicada data.
- Classify the variables by type and level of measurement.
- Compute the five number summary and draw the boxplot for body weight.
- Compute the five number summary and draw the boxplot for body weight by species.
- Compute the five number summary and draw the boxplot for body weight by gender.
Answer
- body weight: continuous, ratio. species: discrete, nominal. gender: discrete, nominal.
- \((0.08, 0.13, 0.17, 0.22, 0.39)\)
- species 0: \((0.08, 0.13, 0.16, 0.21, 0.27)\); species 1: \((0.08, 0. 14, 0.18, 0.23, 0.31)\); species 2: \((0.12, 0.12, 0.215, 0.29, 0.39)\)
- female: \((0.08, 0.17, 0.21, 0.25, 0.31)\); male: \((0.08, 0.12, 0.14, 0.16, 0.39)\)
Consider Pearson's height data.
- Classify the variables by type and level of measurement.
- Compute the five number summary and sketch the boxplot for the height of the father.
- Compute the five number summary and sketch the boxplot for the height of the son.
Answer
- continuous ratio
- \((59.0, 65.8, 67.8, 69.6, 75.4)\)
- \((58.5, 66.9, 68.6, 70.5, 78.4)\)