1.6: Muromachi and Momoyama Periods (1338-1615 CE)
Introduction
Although the Emperor of Japan was the titular head of government, the shogun (samurai military commander) was the authority that governed the country from 1185 until 1868. As the leader of the daimyo (local samurai warlords) and their samurai (warriors), the shogun had an elite army to fight, conquer, control, and manage the masses. Early shogun and daimyo were in constant warfare until the Edo period (1615-1868) when Tokugawa Shogunate unified the country. 1450-1600 cover the Muromachi period (1338-1573) and Japan's Momoyama period (1573-1615). These periods yielded artworks forming the foundation of today's Japanese visual culture: ink washes, Zen paintings, and screen paintings.
Sumi-e (or suibokuga ) is a style of painting originating in China and brought to Japan in the 14 th century by Zen monks. The black and white ink wash painting style – often presented in a vertical, hanging scroll format – became popular during the Muromachi period in the hands of Zen priest-artists. The subject matters of ink wash paintings vary from the representations of Buddhist deities, eccentric Zen masters to evocative landscapes, painted as vehicles of expressions, aids for meditation, and even a path towards enlightenment.
Muromachi Period Ink Wash Paintings
Tōyō (1420-1506) was the most significant Zen priest-artist in the second half of the 15 th century. Comparing his two ink wash paintings demonstrates his deft use of the brush and his versatility in manipulating gradations of ink as his vehicle of expression. In Autumn Landscape (1.6.1), Sesshū's brushstroke control is seen in the variety of lines. Thicker jagged lines form the outlines of rocks and rooftops of the village in the foreground, with shorter and softer strokes depicting vegetations in the mid-ground, under the recessed mountains. While the rock formations and the mountains were outlined with black concentrated ink, the shapes were filled with broader brushstrokes of greys. Beyond the trees, the slight hint of distant mountains is in the far horizon, represented by arcs of sheer diluted ink.
In Haboku Landscape (1.6.2) , Sesshū's careful delineations have given way to a more gestural method of painting. "Haboku" means "splashed ink," however it does not mean the artist simply throws ink onto the surface. Brushes are still used with differing tonalities of ink wash, replacing distinct outlines to render abstract shapes. From the surface of Haboku Landscape, spontaneous impressions of mountains, rock outcroppings, vegetations, and rivers hover between reality and dreams. Sesshu anchored these suggestions with delineations of rooftops and a pole peeking out behind fences; towards the bottom right as two people float on a boat. These concrete representations complement the more abstract shapes and let the viewers connect with the painting through objects they recognize. The wide range of brushstrokes and his command of ink in these two contrasting paintings exemplify Sesshū's mastery as the foremost artist of the Muromachi period.
Momoyama Period (1573-1615)
The sedate religious themes in monochromatic ink wash paintings of the Muromachi period gave way to secular themes in the large-scale opulence of Momoyama paintings. This painting style -- developed by the Kano School in the 16 th century -- became the primary mode of castle interior decorations and continued to be heavily patronized by samurai elites in the next few centuries.
The Momoyama period is a short transitional period full of violent confrontations; daimyo vied for territory and power; the ultimate goal was to rule the country as a shogun . When a daimyo successfully dominated a region, an imposing castle was constructed as his residence. Part testament to daimyo military prowess and part stronghold, a castle was a sprawling complex with many steep walls and moats to protect the central keep. As the last defensible haven, the central keep towered high over the complex. Himeji Castle (1.6.3) in Hyōgo Prefecture is a fine example of daimyo's castle still standing today.
Most paintings in the Momoyama period existed to decorate the interiors of castles. In concert with the imposing nature of a castle, the paintings were equally flamboyant. The subject matters of these paintings ranged from landscapes, vegetation, animals to genre scenes of life in the cities. These were painted on the large-scale surfaces prevalent in the castle interior: fusuma (sliding door panels) and byobu (free-standing folding screens).
Kano School
Kano Eitoku's Chinese Guardians Lions (1.6.4) painted on a 6-panel byobu typifies this visual style. Emerging from a field of gold, a pair of mythical animals with muscular bodies and glaring eyes stride confidently toward the viewers; the bold images are set in a shimmering background on an awe-inspiring, large tableau. The tableau provided a visual backdrop for the daimyo when they received their vassals, entertained their guests, issued their orders, or meted their punishments. The gold glittered and gleamed as the screens reflected candlelight in the castle, impressing the daimyo's audience. As successful military leaders in turbulent times, daimyo sought to perpetuate the image of power, authority, and fortune to bolster their hard-fought status.
Under Kano Eitoku (1543 - 1590), the Kano School emerged as the dominant purveyor of Momoyama paintings, such as Chinese Guardian Lions. "Kano School" denotes a lineage of hereditary professional artists operating within the Kano familial orbit. At the start of the Momoyama period in 1573, the Kanos had been in operation for 100 years, providing affluent clients with a wide array of subject matters in various formats.
Kano Masanobu (1434 - 1530), the school's founder, was skilled in the Chinese ink-wash painting of landscapes favored in the Muromachi period. Chinese ink wash painting emphasizes the qualities of lines and brushwork; the image was constructed out of monochromatic tonalities varied by the concentration of the black ink wash. Kano Motonobu (1476 - 1559), Masanobu's son and the second patriarch of the school, was a versatile artist. He was adept in the ink wash style, also painting figural themes according to the Japanese-style convention yamato -e consisting of flat areas of bright colors without any visible brushwork. In Hosokawa Sumimoto on Horseback ( 1.6.5), the body of the horse and samurai in armor were outlined in black and filled with red, black, and white without any concept of shading. This method of painting allowed detailed patterns and decorations to be added, as in the detailing of the armor with white dots and red mosaic-like tiles.
The other Japanese convention Motonobu brought and developed in the Kano school repertoire was using delineated negative space – often in the shape of clouds – to cover and divide the composition. In Flowers and Birds of the Four Seasons (1.6.6) the sea dominates the painting's top and left-hand side and the tree branches tower above the rock outcrop. While Motonobu used very subtle, minimal shading to outline them, the sea generally remains as unpainted shapes.
In contrast, the jagged rock formation demonstrated the brushwork and different shades of ink-wash to define its craggy and sharp shapes. In this painting, Motonobu merged the two lineages -- the Chinese brushwork and the Japanese flat patterning of shapes -- into synthesis and the signature style of the Kano School. Since this painting is mainly monochromatic, the clouds are left as unpainted areas; in colored Japanese-style ( yamato-e ) paintings, the negative space/cloud formations were typically filled with gold. Motonobu's innovative synthesis and the practice of gilding large-scale surfaces were passed onto and perfected by his successor, the third patriarch of the school, Kano Eitoku.
In the context of the evolution of Kano School's visual style, Eitoku shifted Motonobu's emphasis toward bolder brushstrokes and more prominent motifs, more decorative and stylized. A comparison between Motonobu's Patriarchs of Zen Buddhism, Xiangyen Zhixian Sweeping with a Broom , and Eitoku's Chinese Guardian Lions demonstrates how the cloud formations became more stylized. The shading towards the edge of the clouds and the ground the man is standing on in Motonobu's image are gone; in contrast, Eitoku's cloud and ground have fused into a flat, ambiguous surface of gold. Brushstrokes and shading suggesting landscape features in Eitoku's painting are primarily obscured by silhouettes of clouds. The heavily stylized curls in the lions' manes and tails, and the outlined bodies of the lions, emphasize their decorative rather than naturalistic qualities.
Having struck a chord with the elite samurai clientele, Kano Eitoku received a significant number of commissions from the most powerful daimyo of the period, Oda Nobunaga and Toyotomi Hideyoshi. Eitoku involved all members of the Kano School -- from apprentices, junior artists, and himself (as the master/principal designer) – to complete the commissions on time. The apprentices prepared the pigments and papers; the junior artists painted assigned areas while Eitoku designed the overall composition and painted the most important parts. Momoyama's glittering folding screens and sliding doors of gold-leaf backgrounds were produced this way.
Colors and gold-leaf, bold brushstrokes, and monumental motifs came together in Eitoku's 8-panel byobu Cypress Tree (1.6.7). A towering, almost 2-meter-high cypress tree with a solid half-a-meter trunk anchors the composition. Despite Eitoku's palpable sweeping brushstrokes, the curving trees are highly simplified and stylized. The bold outlines of the tree branches portray undulating movements against the shimmering gold backgrounds. Complimenting the massive brown silhouette of the tree is the delicate detailing of the grey lichens on the trunk and the olive-green leaves sprouting out of the branches. Flat areas of sapphire blue peek through the silhouette of gold clouds indicating water with outcropped rocks. While bringing the painting alive and accentuating its decorativeness, colors also define this screen within Japanese-style ( yamato-e) tradition. The imposing cypress tree parallels the ferocity of the Chinese lions; both are symbols of power and dominance highly favored by the Momoyama daimyo to decorate their castle interiors.
The tea service was an important part of the Shogun ceremonial events and the room for the ceremony demonstrated the value and prestige of the shogun as the ruling regent. One of the first elaborate tea rooms was constructed for Toyotomi Hideyoshi when he was named Imperial Regent. The tearoom was portable, ready to be assembled inside of a larger room and transported with the regent wherever he traveled. Although the original tearoom was destroyed over time, replicas based on documentation were reconstructed in different castles. The gold room (1.6.8) was small, only 2.7 meters by 2.55 meters, and made from cypress wood covered with thousands of sheets of gold leaf on every surface; the walls, ceiling and shoji (sliding doors). On the floor sat crimson tatami mats made from the lush, expensive fabric.
The shimmering tea service (1.6.9) was also made from gold or gilded in gold, all meant to impress others and display the regent's power to all who came to visit him.
With the success of Eitoku's visual style, the Kano School firmly established itself as the institution in which the standard of Momoyama paintings and beyond was conceptualized. Similar to academies of art in Western Europe between the 17 th to the 19 th century, it was enormously influential. The school was not only an arbiter of taste; it was also a training ground for many young and upcoming painters between the 16 th and the 19 th centuries. "The heads of the Kano school seemed to have an affinity with the controlling figures of the day which contributed to making the school into an enormous organization of artists with a powerful system of hierarchy." [1]
[1] Jōhei, S. (1984). The Era of the Kanō School. Modern Asian Studies, 18 (4), 647-656. Retrieved from http://www.jstor.org/stable/312341


9.19: One-on-One Web Sharing Calls
- Explain when to use web sharing tools in a business context
In a planned call with a specific agenda, you will know in advance whether you want to share your screen with your colleague, so you can prepare for it by sending them connection information and making sure they have the right app at the ready. However, once you and your team are comfortable with your web sharing app of choice, it’s pretty easy to share your screen, even during calls when you hadn’t planned to.
When you’re screen sharing, there are some etiquette tips you should follow:
- Make sure you have the correct file or site open before the call starts. It’s impolite to keep your teammate waiting while your Excel or PowerPoint revs up.
- Close any unnecessary tabs in your browser and tidy up your desktop. Remember that the people sharing your screen can see your whole screen, including that sensitive email—or the fact that you have Facebook and OKCupid tabs open and your desktop image is you in a gorilla costume.
A major challenge with screensharing is that the person you’re sharing with can’t see your physical gestures. In person, you would simply point at the things on your screen that you want your colleague to focus on. When you’re screensharing, you have to do a lot more talking. Let’s look at a typical PowerPoint slide.
As you are talking through this very busy slide, you really need to narrate the navigation of the slide out loud. You might take your listener through it like this:
| Goal | Dialogue |
|---|---|
| Tell them what the slide is about overall. Make sure you have a slide title that means something. | “This slide is a summary of the second quarter, breaking out sales in a couple of different ways. |
| Help them navigate around the slide by using directional language. | “Let’s look at the pie chart on the left first. This chart shows the breakout in sales by category. As you can see, the green slice, Sportswear, is still our biggest-selling category, but Shoes, the yellow piece, is way up, and Accessories, in dark green on the left side of the chart, is holding steady.” |
| If the content is complicated, you may want to pause for questions before moving on, but be specific that you’re just taking questions on one area. | “Any questions about the pie chart or what it means before I move on?” |
| Clearly signal when you’ve moved on to another part of the slide. | “Now let’s look at the graph at the top right. This shows quarterly sales in millions by store type. As you can see, all our channels are on an uptick for Q2. A stores and B stores—the orange and gray lines—did pretty well, especially at the holidays. There was a little dip after the post-Christmas sales, but they’re back up again in Q2. Online sales—the yellow line—were reliable, but they’re really taking off with the new spring and summer offerings.” |
| Signal when you’re wrapping up. | “Let’s look at the bullet points on the bottom right where I’ve summarized some of the key factors affecting sales in Q2. . .” |
When sharing a spreadsheet, you should be sure to do the following:
- Take some time before the call to highlight key cells with color to help with navigation. That way, you can say things like, “We’re done with the yellow rows now, and if no one has any questions, we’ll turn to the green rows,” or “I can send you the full spreadsheet later, but what’s important for now are the cells outlined in red.”
- Be sure to establish what the rows and columns represent, as in, “The rows are the weeks in the fiscal year. The pink columns represent sales by category in dollars. The blue columns represent sales by category in units. Column J is the total of all categories in dollars, and Column K is the total units.”
- Use the numbers and letters for rows and columns to help with navigation.
- During the call, use your screen Zoom function to zero in on the areas that are relevant to the conversation.
When sharing a Word document, you should be sure to do the following:
- In many ways, this is the trickiest, so ask yourself whether a screen sharing call is really the way to do it. Word documents are hard because they’re usually just text with no particular landmarks to help with navigation.
- If you absolutely must go over a Word doc using screen sharing, prep the document beforehand with color or section headers to make navigation easier.
These tips may seem exhausting or elementary, but remember how distracted people can be in remote meetings. If someone zones out for even a minute, they can get seriously lost and confused about what you’re discussing. Thus whatever navigation signposts you can use will be helpful to both of you.
Additional features you can use in screen sharing apps really depend on the individual app.
Most allow you to switch among the screens of the people on the call, so if Lanie needs to see Dave’s screen for a minute and then switch back to her own, just a couple of clicks can make that happen. In more sophisticated virtual environments like Amazon Chime and Slack, it’s possible to mark up or edit the screen you see, even if it isn’t your own. A touchscreen may be required for some of these functions.
Contributors and Attributions
- One-on-One Web Sharing Calls. Authored by : Barbara Egel. Provided by : Lumen Learning. License : CC BY: Attribution


1.5: Rank and Homogeneous Systems
Rank and Homogeneous Systems
There is a special type of system which requires additional study. This type of system is called a homogeneous system of equations, which we defined above in Definition 1.2.3 . Our focus in this section is to consider what types of solutions are possible for a homogeneous system of equations.
Consider the following definition.
Consider the homogeneous system of equations given by \[\begin{array}{c} a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}= 0 \\ a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}= 0 \\ \vdots \\ a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}= 0 \end{array}\nonumber \] Then, \(x_{1} = 0, x_{2} = 0, \cdots, x_{n} =0\) is always a solution to this system. We call this the trivial solution .
If the system has a solution in which not all of the \(x_1, \cdots, x_n\) are equal to zero, then we call this solution nontrivial . The trivial solution does not tell us much about the system, as it says that \(0=0\) ! Therefore, when working with homogeneous systems of equations, we want to know when the system has a nontrivial solution.
Suppose we have a homogeneous system of \(m\) equations, using \(n\) variables, and suppose that \(n > m\) . In other words, there are more variables than equations. Then, it turns out that this system always has a nontrivial solution. Not only will the system have a nontrivial solution, but it also will have infinitely many solutions. It is also possible, but not required, to have a nontrivial solution if \(n=m\) and \(n<m\) .
Consider the following example.
Find the nontrivial solutions to the following homogeneous system of equations \[\begin{array}{c} 2x + y - z = 0 \\ x + 2y - 2z = 0 \end{array}\nonumber \]
Solution
Notice that this system has \(m = 2\) equations and \(n = 3\) variables, so \(n>m\) . Therefore by our previous discussion, we expect this system to have infinitely many solutions.
The process we use to find the solutions for a homogeneous system of equations is the same process we used in the previous section. First, we construct the augmented matrix, given by \[\left[ \begin{array}{rrr|r} 2 & 1 & -1 & 0 \\ 1 & 2 & -2 & 0 \end{array} \right]\nonumber \] Then, we carry this matrix to its reduced row-echelon form, given below. \[\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \end{array} \right]\nonumber \] The corresponding system of equations is \[\begin{array}{c} x = 0 \\ y - z =0 \\ \end{array}\nonumber \] Since \(z\) is not restrained by any equation, we know that this variable will become our parameter. Let \(z=t\) where \(t\) is any number. Therefore, our solution has the form \[\begin{array}{c} x = 0 \\ y = z = t \\ z = t \end{array}\nonumber \] Hence this system has infinitely many solutions, with one parameter \(t\) .
Suppose we were to write the solution to the previous example in another form. Specifically, \[\begin{array}{c} x = 0 \\ y = 0 + t \\ z = 0 + t \end{array}\nonumber \] can be written as \[\left[ \begin{array}{r} x\\ y\\ z \end{array} \right] = \left[ \begin{array}{r} 0\\ 0\\ 0 \end{array} \right] + t \left[ \begin{array}{r} 0\\ 1\\ 1 \end{array} \right]\nonumber \] Notice that we have constructed a column from the constants in the solution (all equal to \(0\) ), as well as a column corresponding to the coefficients on \(t\) in each equation. While we will discuss this form of solution more in further chapters, for now consider the column of coefficients of the parameter \(t\) . In this case, this is the column \(\left[ \begin{array}{r} 0\\ 1\\ 1 \end{array} \right]\) .
There is a special name for this column, which is basic solution . The basic solutions of a system are columns constructed from the coefficients on parameters in the solution. We often denote basic solutions by \(X_1, X_2\) etc., depending on how many solutions occur. Therefore, Example \(\PageIndex{1}\) has the basic solution \(X_1 = \left[ \begin{array}{r} 0\\ 1\\ 1 \end{array} \right]\) .
We explore this further in the following example.
Consider the following homogeneous system of equations. \[\begin{array}{c} x + 4y + 3z = 0 \\ 3x + 12y + 9z = 0 \end{array}\nonumber \] Find the basic solutions to this system.
Solution
The augmented matrix of this system and the resulting reduced row-echelon form are \[\left[ \begin{array}{rrr|r} 1 & 4 & 3 & 0 \\ 3 & 12 & 9 & 0 \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{rrr|r} 1 & 4 & 3 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]\nonumber \] When written in equations, this system is given by \[x + 4y +3z=0\nonumber \] Notice that only \(x\) corresponds to a pivot column. In this case, we will have two parameters, one for \(y\) and one for \(z\) . Let \(y = s\) and \(z=t\) for any numbers \(s\) and \(t\) . Then, our solution becomes \[\begin{array}{c} x = -4s - 3t \\ y = s \\ z = t \end{array}\nonumber \] which can be written as \[\left[ \begin{array}{r} x\\ y\\ z \end{array} \right] = \left[ \begin{array}{r} 0\\ 0\\ 0 \end{array} \right] + s \left[ \begin{array}{r} -4 \\ 1 \\ 0 \end{array} \right] + t \left[ \begin{array}{r} -3 \\ 0 \\ 1 \end{array} \right]\nonumber \] You can see here that we have two columns of coefficients corresponding to parameters, specifically one for \(s\) and one for \(t\) . Therefore, this system has two basic solutions! These are \[X_1= \left[ \begin{array}{r} -4 \\ 1 \\ 0 \end{array} \right], X_2 = \left[ \begin{array}{r} -3 \\ 0 \\ 1 \end{array} \right]\nonumber \]
Below is a video on homogeneous systems of equations
Below is a video on solving a system of equations and determining the particular solution.
Below is a video on solving Ax = 0.
Below is another video on solving Ax = 0.
We now present a new definition.
Let \(X_1,\cdots ,X_n,V\) be column matrices. Then \(V\) is said to be a linear combination of the columns \(X_1,\cdots , X_n\) if there exist scalars, \(a_{1},\cdots ,a_{n}\) such that \[V = a_1 X_1 + \cdots + a_n X_n\nonumber \]
A remarkable result of this section is that a linear combination of the basic solutions is again a solution to the system. Even more remarkable is that every solution can be written as a linear combination of these solutions. Therefore, if we take a linear combination of the two solutions to Example \(\PageIndex{2}\) , this would also be a solution. For example, we could take the following linear combination
\[3 \left[ \begin{array}{r} -4 \\ 1 \\ 0 \end{array} \right] + 2 \left[ \begin{array}{r} -3 \\ 0\\ 1 \end{array} \right] = \left[ \begin{array}{r} -18 \\ 3 \\ 2 \end{array} \right]\nonumber \] You should take a moment to verify that \[\left[ \begin{array}{r} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{r} -18 \\ 3 \\ 2 \end{array} \right]\nonumber \]
is in fact a solution to the system in Example \(\PageIndex{2}\) .
Another way in which we can find out more information about the solutions of a homogeneous system is to consider the rank of the associated coefficient matrix. We now define what is meant by the rank of a matrix.
Let \(A\) be a matrix and consider any row-echelon form of \(A\) . Then, the number \(r\) of leading entries of \(A\) does not depend on the row-echelon form you choose, and is called the rank of \(A\) . We denote it by Rank( \(A\)) .
Similarly, we could count the number of pivot positions (or pivot columns) to determine the rank of \(A\) .
Consider the matrix \[\left[ \begin{array}{rrr} 1 & 2 & 3 \\ 1 & 5 & 9 \\ 2 & 4 & 6 \end{array} \right]\nonumber \] What is its rank?
Solution
First, we need to find the reduced row-echelon form of \(A\) . Through the usual algorithm, we find that this is \[\left[ \begin{array}{rrr} \fbox{1} & 0 & -1 \\ 0 & \fbox{1} & 2 \\ 0 & 0 & 0 \end{array} \right]\nonumber \] Here we have two leading entries, or two pivot positions, shown above in boxes.The rank of \(A\) is \(r = 2.\)
Notice that we would have achieved the same answer if we had found the row-echelon form of \(A\) instead of the reduced row-echelon form.
Suppose we have a homogeneous system of \(m\) equations in \(n\) variables, and suppose that \(n > m\) . From our above discussion, we know that this system will have infinitely many solutions. If we consider the rank of the coefficient matrix of this system, we can find out even more about the solution. Note that we are looking at just the coefficient matrix, not the entire augmented matrix.
Let \(A\) be the \(m \times n\) coefficient matrix corresponding to a homogeneous system of equations, and suppose \(A\) has rank \(r\) . Then, the solution to the corresponding system has \(n-r\) parameters.
Consider our above Example \(\PageIndex{2}\) in the context of this theorem. The system in this example has \(m = 2\) equations in \(n = 3\) variables. First, because \(n>m\) , we know that the system has a nontrivial solution, and therefore infinitely many solutions. This tells us that the solution will contain at least one parameter. The rank of the coefficient matrix can tell us even more about the solution! The rank of the coefficient matrix of the system is \(1\) , as it has one leading entry in row-echelon form. Theorem \(\PageIndex{1}\) tells us that the solution will have \(n-r = 3-1 = 2\) parameters. You can check that this is true in the solution to Example \(\PageIndex{2}\) .
Notice that if \(n=m\) or \(n<m\) , it is possible to have either a unique solution (which will be the trivial solution) or infinitely many solutions.
We are not limited to homogeneous systems of equations here. The rank of a matrix can be used to learn about the solutions of any system of linear equations. In the previous section, we discussed that a system of equations can have no solution, a unique solution, or infinitely many solutions. Suppose the system is consistent, whether it is homogeneous or not. The following theorem tells us how we can use the rank to learn about the type of solution we have.
Below is a video on Rank, Basic Variables, and Free Variables of a coefficient matrix.
Let \(A\) be the \(m \times \left( n+1 \right)\) augmented matrix corresponding to a consistent system of equations in \(n\) variables, and suppose \(A\) has rank \(r\) . Then
- the system has a unique solution if \(r = n\)
- the system has infinitely many solutions if \(r < n\)
We will not present a formal proof of this, but consider the following discussions.
- No Solution The above theorem assumes that the system is consistent, that is, that it has a solution. It turns out that it is possible for the augmented matrix of a system with no solution to have any rank \(r\) as long as \(r>1\) . Therefore, we must know that the system is consistent in order to use this theorem!
- Unique Solution Suppose \(r=n\) . Then, there is a pivot position in every column of the coefficient matrix of \(A\) . Hence, there is a unique solution.
- Infinitely Many Solutions Suppose \(r<n\) . Then there are infinitely many solutions. There are less pivot positions (and hence less leading entries) than columns, meaning that not every column is a pivot column. The columns which are \(not\) pivot columns correspond to parameters. In fact, in this case we have \(n-r\) parameters.
Below is a video on the geometry of Ax = 0 and Ax = b.
Below is another video on the geometry of Ax = 0 and Ax = b.
Below is a video on the geometry of Ax = 0 and Ax = b, this one with two free variables.


Chapter 9
The cutting edge of a knife that has been sharpened has a smaller surface area than a dull knife. Since pressure is force per unit area, a sharp knife will exert a higher pressure with the same amount of force and cut through material more effectively.
Lying down distributes your weight over a larger surface area, exerting less pressure on the ice compared to standing up. If you exert less pressure, you are less likely to break through thin ice.
0.809 atm; 82.0 kPa
2.2 10 2 kPa
Earth: 14.7 lb in –2 ; Venus: 1.30 × 10 3 lb in −2
(a) 101.5 kPa; (b) 51 torr drop
(a) 264 torr; (b) 35,200 Pa; (c) 0.352 bar
(a) 623 mm Hg; (b) 0.820 atm; (c) 83.1 kPa
With a closed-end manometer, no change would be observed, since the vaporized liquid would contribute equal, opposing pressures in both arms of the manometer tube. However, with an open-ended manometer, a higher pressure reading of the gas would be obtained than expected, since P gas = P atm + P vol liquid .
As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle’s law.
(a) The number of particles in the gas increases as the volume increases. (b) temperature, pressure
The curve would be farther to the right and higher up, but the same basic shape.
About 12.5 L
3.40 10 3 torr
12.1 L
217 L
8.190 10 –2 mol; 5.553 g
(a) 7.24 10 –2 g; (b) 23.1 g; (c) 1.5 10 –4 g
5561 L
46.4 g
For a gas exhibiting ideal behavior:
(a) 1.85 L CCl 2 F 2 ; (b) 4.66 L CH 3 CH 2 F
0.644 atm
The pressure decreases by a factor of 3.
4.64 g L −1
38.8 g
72.0 g mol −1
88.1 g mol −1 ; PF 3
141 atm, 107,000 torr, 14,300 kPa
CH 4 : 276 kPa; C 2 H 6 : 27 kPa; C 3 H 8 : 3.4 kPa
Yes
740 torr
(a) Determine the moles of HgO that decompose; using the chemical equation, determine the moles of O 2 produced by decomposition of this amount of HgO; and determine the volume of O 2 from the moles of O 2 , temperature, and pressure. (b) 0.308 L
(a) Determine the molar mass of CCl 2 F 2 . From the balanced equation, calculate the moles of H 2 needed for the complete reaction. From the ideal gas law, convert moles of H 2 into volume. (b) 3.72 10 3 L
(a) Balance the equation. Determine the grams of CO 2 produced and the number of moles. From the ideal gas law, determine the volume of gas. (b) 7.43 10 5 L
42.00 L
(a) 18.0 L; (b) 0.533 atm
10.57 L O 2
5.40 10 5 L
XeF 4
4.2 hours
Effusion can be defined as the process by which a gas escapes through a pinhole into a vacuum. Graham’s law states that with a mixture of two gases A and B: Both A and B are in the same container at the same temperature, and therefore will have the same kinetic energy:
Therefore,
F 2 , N 2 O, Cl 2 , H 2 S
1.4; 1.2
51.7 cm
Yes. At any given instant, there are a range of values of molecular speeds in a sample of gas. Any single molecule can speed up or slow down as it collides with other molecules. The average speed of all the molecules is constant at constant temperature.
H 2 O. Cooling slows the speeds of the He atoms, causing them to behave as though they were heavier.
(a) The number of collisions per unit area of the container wall is constant. (b) The average kinetic energy doubles. (c) The root mean square speed increases to times its initial value; u rms is proportional to
(a) equal; (b) less than; (c) 29.48 g mol −1 ; (d) 1.0966 g L −1 ; (e) 0.129 g/L; (f) 4.01 10 5 g; net lifting capacity = 384 lb; (g) 270 L; (h) 39.1 kJ min −1
Gases C, E, and F
The gas behavior most like an ideal gas will occur under the conditions that minimize the chances of significant interactions between the gaseous atoms/molecules, namely, low pressures (fewer atoms/molecules per unit volume) and high temperatures (greater kinetic energies of atoms/molecules make them less susceptible to attractive forces). The conditions described in (b), high temperature and low pressure, are therefore most likely to yield ideal gas behavior.
SF 6
(a) A straight horizontal line at 1.0; (b) When real gases are at low pressures and high temperatures, they behave close enough to ideal gases that they are approximated as such; however, in some cases, we see that at a high pressure and temperature, the ideal gas approximation breaks down and is significantly different from the pressure calculated by the ideal gas equation. (c) The greater the compressibility, the more the volume matters. At low pressures, the correction factor for intermolecular attractions is more significant, and the effect of the volume of the gas molecules on Z would be a small lowering compressibility. At higher pressures, the effect of the volume of the gas molecules themselves on Z would increase compressibility (see Figure 9.35). (d) Once again, at low pressures, the effect of intermolecular attractions on Z would be more important than the correction factor for the volume of the gas molecules themselves, though perhaps still small. At higher pressures and low temperatures, the effect of intermolecular attractions would be larger. See Figure 9.35. (e) Low temperatures


4.10: Experimental Evidence
objects large and small, slow and fast: many witnesses for the Canopus trip
Alfred Missouri has remained silent up to this point. Now he declares, "All this theory is too much for me. I won't believe a word you say unless you can show me an experimental demonstration."
“Airliner” test of twin effect
We reply, "Atomic clocks have been placed on commercial airliners and carried around Earth, some in an eastward direction, others in a westward direction. In each case the airliner clocks were compared with reference clocks at the U.S. Naval Observatory before and after their trips. These clocks disagreed. Results were consistent with the velocity-related predictions of special relativity." 1
"This verification of special relativity has two minor difficulties and a major one. Minor difficulties: (1) Each leg of a commercial airliner's trip may be at a different speed, not always accurately known and for which the time-stretching effect must be separately calculated. Also, temperature and pressure effects on airborne clocks are hard to control in a commercial airliner. (2) More fundamentally, Earth rotates, carrying the reference Naval Observatory clocks eastward around the center of Earth. Earth center can be regarded as the inertial point in free-float around Sun. With respect to this center, one airborne clock moves even faster eastward than Earth's surface, while the other one—heading west with respect to the surface—with respect to Earth's center also moves eastward, but more slowly. Taking account of these various relative velocities adds further complication to analysis of results."
The group takes a break and mills ar ound the conference room, chatting and eating refreshments. Joanne Short approaches us juggling coffee, a donut, and her notes.
"I didn't want to embarrass you in public," she says, "but isn't your plan faulty because of the turnaround? You can't be serious about leaping from one high-speed rocket to another rocket going in the opposite direction. That means certain death!  During this change of velocity you will be thrown against the front of the rocket ship, as I'm thrown when I slam on my car brakes. Release a test particle from rest and it will hurtle forward! Surely you are not in an inertial (free-float) frame. Therefore you cannot use special relativity in your analysis of this time period. What does that do to your description of the 'jump ahead' of Earth clocks as you slow down and speed up again? Don't you need general relativity to analyze events in accelerated reference frames?"
"Oh yes, general relativity can describe events in the accelerated frame," we reply, "but so can special relativity if we take it in easy steps! l like to think of a freight yard with trains moving at different speeds along parallel tracks. Each train has its own string of recording clocks along its length, each string synchronized in that particular train frame. Each adjacent train is moving at a slightly different speed from the one next to it. Now we can change frames by walking across the trains, stepping from the top of one freight car to the top of the freight car rolling next to it at a slightly different speed."
"Let these trains become rocket trains in space. Each train then has an observer passing Earth as we step on that train. Each observer, by prearrangement, reads the Earth clock at the same time that we step onto his train ('at the same time' as recorded in that frame). When you assemble all these data later on, you find that the set of observers on the sequence of trains see the Earth clock jumping forward in time much faster than would be expected.
"Notice that it takes a whole set of clocks in different frames, all reading the single Earth clock, to establish this result. So there is never any contradiction between a single clock in one frame and a single clock in any other frame. In this case special relativity can do the job just fine."
The directors reassemble and Joanne Short, smiling, takes her place with them.
"We overcome these two minor difficulties by having an airplane fly round and round in circles in the vicinity of a single ground-based reference atomic clock. Then - to a high accuracy - only relative motion of these two clocks enters into the special-relativity analysis.
“Circling airplane” test of twin effect
"On November 22, 1975, a U.S. Navy P3C antisubmarine patrol plane flew back and forth for 15 hours at an altitude of 25,000 to 35,000 feet (7600 to 10,700 meters) over Chesapeake Bay in an experiment arranged by Carroll Alley and collaborators. The plane carried atomic clocks that were compared by laser pulse with identical clocks on the ground. Traveling at an average speed of \(270 \mathrm{knots}(140\) meters per second), the airborne clocks lost an average of \(5.6\) nanoseconds \(=5.6 \times\) \(10^{-9}\) seconds due to velocity-related effects in the 15-hour flight. The expected special-relativity difference in clock readings for this relative speed is \(5.7\) nanoseconds. This result is remarkably accurate, considering the low relative velocity of the two clocks: \(4.7 \times 10^{-7}\) light speed. 2
Trouble: Large frame is not inertial
"The major difficulty with all of these experiments is this: A high-flying airplane is significantly farther from Earth's center than is the ground-based clock. Think of an observer in a helicopter reading the clocks of passing airplanes and signaling these readings for comparison to a ground-based clock directly below. These two clocks the helicopter clock and the Earthbound clock - are at rest with respect to one another. Are they in the same inertial (free-float) frame? The answer is No." 3
"We know that a single inertial reference frame near Earth cannot extend far in a vertical direction (Section 2.3). Even if the two clocks - helicopter and Earthbound - were dropped in free fall, they could not both be in the same inertial frame. Released from rest 30,000 feet one above the other, they would increase this relative distance by 1 millimeter in only \(0.3\) second of free fall - too rapid a change to be ignored. But the experiment went on not for \(0.3\) second but for 15 hours!"
Solution: Use general relativity
"Since the helicopter clock and Earthbound clock are not in the same inertial frame, their behavior cannot be analyzed by special relativity. 4 Instead we must use general relativity - the theory of gravitation. General relativity predicts that during the 15-hour flight the higher-altitude clock in the Chesapeake Bay experiment will record greater elapsed time by \(52.8\) nanoseconds due to the slightly reduced gravitational field at altitudes at which the plane flew. From this must be subtracted the \(5.7\) nanoseconds by which the airborne clock is predicted to record less elapsed time due to effects of relative velocity. These velocity effects are predicted by both special relativity and general relativity and were the only results quoted above. The overall predicted result equals \(52.8-5.7=47.1\) nanoseconds net gain by the high-altitude clock compared with the clock on the ground. Contrast this with the measured value of \(47.2\) nanoseconds."
"Hence for airplanes flying at conventional speeds and conventional altitudes, tidal-gravitational effects on clocks can be greater than velocity-dependent effects to which special relativity is limited. In fact, the Chesapeake Bay experiment was conducted to verify the results of general relativity: The airplane pilot was instructed to fly as slowly as possible to reduce velocity effects! The P3C patrol plane is likely to stall below 200 knots, so a speed of 270 knots was chosen."
"In all these experiments the time-stretching effect is small because the speed of an airplane is small compared to the speed of light, but atomic clocks are now so accurate that these speed effects are routinely taken into account when such clocks are brought together for direct comparison."
“High-speed radioactive particle' test of twin effect
Professor Bright chimes in. "What the astronaut says is correct: We do not have large clocks moving fast on Earth. On the other hand, we have a great many small clocks moving very fast indeed. When particles collide in high-speed accelerators, radioactive fragments emerge that decay into other particles after an average lifetime that is well known when measured in the rest frame of the particle. When the radioactive particle moves at high speed in the laboratory, its average lifetime is significantly longer as measured on laboratory clocks than when the particle is at rest The amount of lengthening of this lifetime is easily calculated from the particle speed in the same way the astronaut calculates time stretching on the way to and from Canopus. The time-stretch factor can be as great as 10 for some of these particles: the fast-moving particles are measured to live 10 times longer, on average, than their measured lifetime when at rest! The experimental results agree with these calculations in all cases we have tried. Such time stretching is part of the everyday experience of high-energy particle physicists." 5
Earth frame: Free-float for particle experiments
"And for these increased-lifetime experiments there is no problem of principle in making observations in an inertial, free-float frame. While they are decaying, particles cover at most a few tens of meters of space. Think of the flight of each particle as a separate experiment. An individual experiment lasts as long as it takes one high-speed particle to move through the apparatus - a few tens of meters of light-travel time. Ten meters of light-travel time equals about 33 nanoseconds, or \(33 \times 10^{-9}\) seconds." 6
"Can we construct an inertial frame for such happenings? Two ball bearings released from rest say 20 meters apart do not move together very far in 33 nanoseconds! Therefore these increased-lifetime experiments could be done, in principle, in free-float frames. It follows that special relativity suffices to describe the behavior of the 'radioactive-decay clocks' employed in these experiments. We do not need the theory of gravitation provided by general relativity."
"Oscillating iron nucleus” test of twin effect
"Of course, in none of these high-speed particle experiments do particles move back and forth the way our astronaut friend proposes to do between Earth and Canopus. Even that back-and-forth result has been verified for certain radioactive iron nuclei vibrating with thermal agitation in a solid sample of iron. Atoms in a hotter sample vibrate back and forth faster, on average, and thus stay younger, on average, than atoms in a cooler sample. In this case the 'tick of the clock' carried by an iron atom is the period of electromagnetic radiation ('gamma ray') given off when its nucleus makes the transition from a radioactive state to one that is not radioactive. For detailed reasons that we need not go into here, this particular 'clock' can be read with very high accuracy. Beyond all such details, the experimental outcome is simply stated: Clocks that take one or many round trips at higher speed record a smaller elapsed time than clocks that take one or many round trips at lower speed." 7
Twin effect verified!
"These various results - plus many others we have not described - combine to give overwhelming experimental support for the predictions of the astronaut concerning the proposed trip to Canopus." 8
Dr. Bright sits back in his chair with a smile, obviously believing that he has disposed of all objections single-handedly.
"Yes," we conclude, "about the reality of the effect there is no question. Therefore if you all approve, and the Space Agency provides that new and very fast rocket, we can be on our way."
The meeting votes approval and our little story ends.
1 "Airliner" test of twin effect
2 "Circling airplane" test of twin effect
3 Trouble: Large frame is not inertial
4 Solution: Use general relativity
5 "High-speed radioactive particle" test of twin effect
6 Earth frame: Free-float for particle experiments
7 "Oscillating iron nucleus" test of twin effect
8 Twin effect verified!


15.4: Germination
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Learning Objectives
- Identify the environmental factors that stimulate germination.
- Distinguish between epigeous and hypogeous germination.
- Compare germination in eudicots versus monocots.
Many mature seeds enter a period of inactivity, or extremely low metabolic activity: a process known as dormancy , which may last for months, years or even centuries. Dormancy helps keep seeds viable during unfavorable conditions. Germination occurs when the embryo, which is dormant within a mature seed, resumes growth upon a return to favorable conditions. The embryo becomes a young seedling that is no longer confined within the seed coat.
In many seeds, the presence of a thick seed coat can inhibit germination through several mechanisms: (1) the embryo may not be able to break through the thick seed coat; (2) the seed coat may contain chemicals inhibitors; and (3) the seed coat prevents the embryo from accessing water and oxygen. Dormancy is also maintained by the relative hormone concentrations in the embryo itself.
Environmental Requirements for Germination
The requirements for germination depend on the species. Common environmental requirements include light, the proper temperature, presence of oxygen, and presence of water. Seeds of small-seeded species usually require light as a germination cue. This ensures the seeds only germinate at or near the soil surface (where the light is greatest). If they were to germinate too far underneath the surface, the developing seedling would not have enough food reserves to reach the sunlight. (Recall from 14.5 Dormancy that red light induces germination by converting the inactive form of phytochrome (Pr) to the active form (Pfr), which leads to the production of amylase. This enzyme breaks down the limited food reserves in the seed, facilitating germination.)
Not only do some species require a specific temperature to germinate, but they may also require a prolonged cold period prior to germination. In this case, cold conditions gradually break down a chemical germination inhibitor. This mechanism prevents seeds from germinating during an unseasonably warm spell in the autumn or winter in temperate climates. Similarly, plants growing in hot climates may have seeds that need a hot period in order to germinate, an adaptation to avoid germination in the hot, dry summers.
Water is always needed to allow vigorous metabolism to begin. Additionally, water can leach away inhibitors in the seed coat. This is especially common among desert annuals. Seeds that are dispersed by animals may need to pass through an animal digestive tract to remove inhibitors prior to germination. Similarly, some species require mechanical abrasion of the seed coat, which could be achieved by water dispersal. Other species are fire adapted, requiring fire to break dormancy (Figure \(\PageIndex{1}\)).
The Mechanism of Germination
The first step in germination and starts with the uptake of water, also known as imbibition . After imbibition, enzymes are activated that start to break down starch into sugars consumed by embryo. The first indication that germination has begun is a swelling in the radicle.
Depending on seed size, the time taken for a seedling to emerge may vary. Species with large seeds have enough food reserves to germinate deep below ground, and still extend their epicotyl all the way to the soil surface while the seedlings of small-seeded species emerge more quickly (and can only germinate close to the surface of the soil).
During epigeous germination , the hypocotyl elongates, and the cotyledons extend above ground. During hypogeous germination , the epicotyl elongates, and the cotyledon(s) remain belowground (Figure \(\PageIndex{2}\)). Some species (like beans and onions) have epigeous germination while others (like peas and corn) have hypogeous germination. In many epigeous species, the cotyledons not only transfer their food stores to the developing plant but also turn green and make more food by photosynthesis until they drop off.
Germination in Eudicots
Upon germination in eudicot seeds, the radicle emerges from the seed coat while the seed is still buried in the soil.
For epigeous eudicots (like beans), the hypocotyl is shaped like a hook with the plumule pointing downwards. This shape is called the plumule hook, and it persists as long as germination proceeds in the dark. Therefore, as the hypocotyl pushes through the tough and abrasive soil, the plumule is protected from damage. Additionally, the two cotyledons additionally protect the from mechanical damage. Upon exposure to light, the hypocotyl hook straightens out, the young foliage leaves face the sun and expand, and the epicotyl elongates (Figure \(\PageIndex{3}\)).
In hypogeous eudicots (like peas), the epicotyl rather than the hypocotyl forms a hook, and the cotyledons and hypocotyl thus remain underground. When the epicotyl emerges from the soil, the young foliage leaves expand. The epicotyl continues to elongate (Figure \(\PageIndex{4}\)).
The radicle continues to grown downwards and ultimately produces the tap root. Lateral roots then branch off to all sides, producing the typical eudicot tap root system.
Germination in Monocots
As the seed germinates, the radicle emerges and forms the first root. In epigeous monocots (such as onion), the single cotyledon will bend, forming a hook and emerge before the coleoptile (Figure \(\PageIndex{5}\)). In hypogeous monocots (such as corn), the cotyledon remains belowground, and the coleoptile emerges first. In either case, once the coleoptile has exited the soil and is exposed to light, it stops growing. The first leaf of the plumule then pieces the coleoptile (Figure \(\PageIndex{6}\)), and additional leaves expand and unfold. At the other end of the embryonic axis, the first root soon dies while adventitious roots (roots that arise directly from the shoot system) emerge from the base of the stem (Figure \(\PageIndex{7}\)). This gives the monocot a fibrous root system.
Attributions
Curated and authored by Melissa Ha using the following sources:
- 16.4B Germination of Seeds from Biology by John W. Kimball ( CC-BY )
- 32.2 Pollination and Fertilization from Biology 2e by OpenStax (licensed CC-BY ). Access for free at openstax.org .
- 7.5 Origin of the Seed from Introduction to Botany by Alexey Shipunov (public domain)


7.18: Putting It Together- Microsoft Excel Intermediate Skills
Now that you know some of the more advanced tools available in Excel, let us take a look at our opening list of possible questions and match those with an Excel function we just covered.
- You want to see which month had your highest sales volume. You could see this by sorting your data Z to A (largest to smallest number).
- You have a total clothing inventory but you want to know how many pants you have in stock. You could see this using the COUNTIF function.
- You want to know how many months from last year you were under budget. You could see this with the IF function, just like in our learning example.
Contributors and Attributions
CC licensed content, Original
- Putting It Together: Microsoft Excel Intermediate Skills. Authored by : Shelli Carter. Provided by : Lumen Learning. License : CC BY: Attribution


17.7.1: Review Questions
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Review Questions
1 .
The laboring person is in pain and very anxious. They do not want to use traditional pain relief methods. What can the nurse offer the laboring person to help with relaxation and reducing anxiety?
-
lavender oil
-
epidural
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nalbuphine hydrochloride (Nubain)
-
cold rag
2 .
What technique, usually reserved for low-risk persons, has been shown to decrease pain, encourage movement, decrease anxiety, shorten labor, and improve labor satisfaction?
-
controlled breathing
-
aromatherapy
-
massage
-
hydrotherapy
3 .
The fetus is in an occiput posterior position. What position can the nurse assist the laboring person into that may encourage the fetus to find the occiput anterior position?
-
squatting
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shower
-
hands and knees
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semi-Fowler
4 .
What is the term for a slow, deep inhalation through the nose and exhalation through the mouth before a contraction for preparation and after a contraction for release of tension from the contraction?
-
pursed-lip breathing
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panting
-
chanting
-
cleansing breath
5 .
The laboring person wants to use open glottis breathing for pushing. How can the nurse support the person?
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Encourage the person to breathe however is most comfortable.
-
Tell the person they should hold their breath for 10 counts.
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Explain that open glottis pushing is not effective.
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Discuss that the health-care provider would not allow them to use that breathing.
6 .
The laboring person is using hypnosis for comfort measures. How can the nurse support the person?
-
Talk them through the contractions.
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Keep the lights on so that everyone can see.
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Keep the room quiet and dimly lit.
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During the contraction, ask them to rate the pain.
7 .
The nurse provides counter pressure to relieve pain and open the pelvis to help with fetal descent. What type of counter pressure is the nurse providing?
-
hip squeeze
-
perineal pressure
-
shoulder pressure
-
knee press
8 .
The laboring person is anxious and hurting. What can the nurse do to help the laboring person?
-
Offer her an epidural.
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Encourage frequent position changes.
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Explain that she has to care for two other persons.
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Tell them to call someone to come sit with them.
9 .
What drug is an anxiolytic that relieves apprehension and creates a feeling of calm?
-
hydroxyzine
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fentanyl
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codeine
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morphine
10 .
Nausea and vomiting are common discomforts in labor. When treating with promethazine, what must the nurse do when administering this medication?
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Administer subcutaneously.
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Never administer with an opioid.
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Dilute before IV administration.
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Never administer in first stage labor.
11 .
What would the nurse administer if the newborn has decreased or no respiratory effort at delivery after the person received an opiate?
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naloxone (Narcan)
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acetaminophen (Tylenol)
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oxygen
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sodium bicarbonate
12 .
What does the nurse explain to the laboring person and partner about nitrous oxide?
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It makes the newborn have respiratory depression.
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It causes the laboring person to have decreased respirations.
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It does not affect respiration as an opiate would.
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The mask is held to the person’s face by the partner.
13 .
The nurse receives orders for pharmacologic pain relief. What does the nurse explain to the laboring person after administering fentanyl?
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“I will need to monitor your oxygen saturation.”
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“It’s important for me to insert a Foley catheter.”
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“I will need to monitor for vaginal bleeding.”
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“It’s important for me to check your cervix after I administer this medication.”
14 .
What are the side effects of spinal anesthesia? Select all that apply.
- hypotension
- respiratory depression
- renal damage
- infection
15 .
What is the expected effect of pudendal anesthesia?
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loss of sensation in the perineum
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decreased ability to move one’s legs
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permanent loss of motor function
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loss of consciousness
16 .
The nurse is monitoring the person’s vital signs after the epidural is placed and notices a blood pressure of 80/50 mm Hg. What nursing intervention can be performed prior to the epidural placement to potentially reduce this side effect?
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routine use of ephedrine
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IV fluid bolus
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insertion of indwelling urinary catheter
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upright positioning of the laboring person
17 .
After insertion of the epidural catheter, what is the nurse’s responsibility regarding patient care?
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monitoring vital signs every 5 to 15 minutes
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intermittent FHR monitoring
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providing the laboring person a meal
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instructing the laboring person to get out of bed to use the restroom
18 .
What is a side effect of general anesthesia that usually limits its use to cases of emergency?
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hyperactive newborns
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newborn respiratory depression
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increase in uterine contractions
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decrease in cervical dilation
19 .
What is the disadvantage of local anesthesia?
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decreased motor function
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nausea
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hypotension
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pain during injection


19.6: Bayesian Analysis of Contingency Tables
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Time to change gears. Up to this point I’ve been talking about what Bayesian inference is and why you might consider using it. I now want to briefly describe how to do Bayesian versions of various statistical tests. The discussions in the next few sections are not as detailed as I’d like, but I hope they’re enough to help you get started. So let’s begin.
The first kind of statistical inference problem I discussed in this book appeared in Chapter 12, in which we discussed categorical data analysis problems. In that chapter I talked about several different statistical problems that you might be interested in, but the one that appears most often in real life is the analysis of
contingency tables
. In this kind of data analysis situation, we have a cross-tabulation of one variable against another one, and the goal is to find out if there is some
association
between these variables. The data set I used to illustrate this problem is found in the
chapek9.Rdata
file, and it contains a single data frame
chapek9
load("./rbook-master/data/chapek9.Rdata")
head(chapek9)
## species choice
## 1 robot flower
## 2 human data
## 3 human data
## 4 human data
## 5 robot data
## 6 human flower
In this data set, we supposedly sampled 180 beings and measured two things. First, we checked whether they were humans or robots, as captured by the
species
variable. Second, we asked them to nominate whether they most preferred flowers, puppies, or data. When we produce the cross-tabulation, we get this as the results:
crosstab <- xtabs( ~ species + choice, chapek9 )
crosstab
## choice
## species puppy flower data
## robot 13 30 44
## human 15 13 65
Surprisingly, the humans seemed to show a much stronger preference for data than the robots did. At the time we speculated that this might have been because the questioner was a large robot carrying a gun, and the humans might have been scared.
orthodox text
Just to refresh your memory, here’s how we analysed these data back in Chapter@refch:chisquare. Because we want to determine if there is some
association
between
species
and
choice
, we used the
associationTest()
function in the
lsr
package to run a chi-square test of association. The results looked like this:
library(lsr)
## Warning: package 'lsr' was built under R version 3.5.2
associationTest( ~species + choice, chapek9 )
##
## Chi-square test of categorical association
##
## Variables: species, choice
##
## Hypotheses:
## null: variables are independent of one another
## alternative: some contingency exists between variables
##
## Observed contingency table:
## choice
## species puppy flower data
## robot 13 30 44
## human 15 13 65
##
## Expected contingency table under the null hypothesis:
## choice
## species puppy flower data
## robot 13.5 20.8 52.7
## human 14.5 22.2 56.3
##
## Test results:
## X-squared statistic: 10.722
## degrees of freedom: 2
## p-value: 0.005
##
## Other information:
## estimated effect size (Cramer's v): 0.244
Because we found a small p value (in this case p<.01), we concluded that the data are inconsistent with the null hypothesis of no association, and we rejected it.
Bayesian test
How do we run an equivalent test as a Bayesian? Well, like every other bloody thing in statistics, there’s a lot of different ways you
could
do it. However, for the sake of everyone’s sanity, throughout this chapter I’ve decided to rely on one R package to do the work. Specifically, I’m going to use the
BayesFactor
package written by Jeff Rouder and Rich Morey, which as of this writing is in version 0.9.10.
For the analysis of contingency tables, the
BayesFactor
package contains a function called
contingencyTableBF()
. The data that you need to give to this function is the contingency table itself (i.e., the
crosstab
variable above), so you might be expecting to use a command like this:
library( BayesFactor ) # ...because we have to load the package
contingencyTableBF( crosstab ) # ...because that makes sense, right?
However, if you try this you’ll get an error message.  You can specify the sampling plan using the
sampleType
argument. So I should probably tell you what your options are! The
contingencyTableBF()
function distinguishes between four different types of experiment:
-
Fixed sample size
. Suppose that in our
chapek9
example, our experiment was designed like this: we deliberately set out to test 180 people, but we didn’t try to control the number of humans or robots, nor did we try to control the choices they made. In this design, the total number of observations N is fixed, but everything else is random. This is referred to as “joint multinomial” sampling, and if that’s what you did you should specifysampleType = "jointMulti"
. In the case of thechapek9
data, that’s actually what I had in mind when I invented the data set. -
Fixed row (or column) totals
. A different kind of design might work like this. We decide ahead of time that we want 180 people, but we try to be a little more systematic about it. Specifically, the
experimenter
constrains it so that we get a predetermined number of humans and robots (e.g., 90 of each). In this design,
either
the row totals or the column totals are fixed, but not both. This is referred to as “independent multinomial” sampling, and if that’s what you did you should specify
sampleType = "indepMulti"
. -
Both row and column totals fixed
. Another logical possibility is that you designed the experiment so that
both
the row totals and the column totals are fixed. This doesn’t make any sense at all in the
chapek9
example, but there are other deisgns that can work this way. Suppose that I show you a collection of 20 toys, and then given them 10 stickers that sayboy
and another 10 that saygirl
. I then give them 10blue
stickers and 10pink
stickers. I then ask you to put the stickers on the 20 toys such that every toy has a colour and every toy has a gender. No matter how you assign the stickers, the total number of pink and blue toys will be 10, as will the number of boys and girls. In this design both the rows and columns of the contingency table are fixed. This is referred to as “hypergeometric” sampling, and if that’s what you’ve done you should specifysampleType = "hypergeom"
. -
Nothing is fixed
. Finally, it might be the case that
nothing
is fixed. Not the row columns, not the column totals, and not the total sample size either. For instance, in the
chapek9
scenario, suppose what I’d done is run the study for a fixed length of time . By chance, it turned out that I got 180 people to turn up to study, but it could easily have been something else. This is referred to as “Poisson” sampling, and if that’s what you’ve done you should specifysampleType="poisson"
.
Okay, so now we have enough knowledge to actually run a test. For the
chapek9
data, I implied that we designed the study such that the total sample size N was fixed, so we should set
sampleType = "jointMulti"
. The command that we need is,
library( BayesFactor )
## Warning: package 'BayesFactor' was built under R version 3.5.2
## Loading required package: coda
## Warning: package 'coda' was built under R version 3.5.2
## Loading required package: Matrix
## ************
## Welcome to BayesFactor 0.9.12-4.2. If you have questions, please contact Richard Morey (richarddmorey@gmail.com).
##
## Type BFManual() to open the manual.
## ************
contingencyTableBF( crosstab, sampleType = "jointMulti" )
## Bayes factor analysis
## --------------
## [1] Non-indep. (a=1) : 15.92684 ±0%
##
## Against denominator:
## Null, independence, a = 1
## ---
## Bayes factor type: BFcontingencyTable, joint multinomial
As with most R commands, the output initially looks suspiciously similar to utter gibberish. Fortunately, it’s actually pretty simple once you get past the initial impression. Firstly, note that the stuff at the top and bottom are irrelevant fluff. You already know that you’re doing a Bayes factor analysis. You already know that you’re analysing a contingency table, and you already know that you specified a joint multinomial sampling plan. So let’s strip that out and take a look at what’s left over:
[1] Non-indep. (a=1) : 15.92684 @plusorminus0%
Against denominator:
Null, independence, a = 1
Let’s also ignore those two
a=1
bits, since they’re technical details that you don’t need to know about at this stage.
268
The rest of the output is actually pretty straightforward. At the bottom, the output defines the null hypothesis for you: in this case, the null hypothesis is that there is no relationship between
species
and
choice
. Or, to put it another way, the null hypothesis is that these two variables are
independent
. Now if you look at the line above it, you might (correctly) guess that the
Non-indep.
part refers to the
alternative
hypothesis. In this case, the alternative is that there
is
a relationship between
species
and
choice
: that is, they are not independent. So the only thing left in the output is the bit that reads
15.92684 @plusorminus0%
The 15.9 part is the Bayes factor, and it’s telling you that the odds for the alternative hypothesis against the null are about 16:1. The ±0% part is not very interesting: essentially, all it’s telling you is that R has calculated an exact Bayes factor, so the uncertainty about the Bayes factor is 0%. 269 In any case, the data are telling us that we have moderate evidence for the alternative hypothesis.
Writing up the results
When writing up the results, my experience has been that there aren’t quite so many “rules” for how you “should” report Bayesian hypothesis tests. That might change in the future if Bayesian methods become standard and some task force starts writing up style guides, but in the meantime I would suggest using some common sense. For example, I would avoid writing this:
A Bayesian test of association found a significant result (BF=15.92)
To my mind, this write up is unclear. Even assuming that you’ve already reported the relevant descriptive statistics, there are a number of things I am unhappy with. First, the concept of “statistical significance” is pretty closely tied with p-values, so it reads slightly strangely. Second, the “BF=15.92” part will only make sense to people who already understand Bayesian methods, and not everyone does. Third, it is somewhat unclear exactly which test was run and what software was used to do so.
On the other hand, unless precision is extremely important, I think that this is taking things a step too far:
We ran a Bayesian test of association using version 0.9.10-1 of the BayesFactor package using default priors and a joint multinomial sampling plan. The resulting Bayes factor of 15.92 to 1 in favour of the alternative hypothesis indicates that there is moderately strong evidence for the non-independence of species and choice.
Everything about that passage is correct, of course. Morey and Rouder (2015) built their Bayesian tests of association using the paper by Gunel and Dickey (1974), the specific test we used assumes that the experiment relied on a joint multinomial sampling plan, and indeed the Bayes factor of 15.92 is moderately strong evidence. It’s just far too wordy.
In most situations you just don’t need that much information. My preference is usually to go for something a little briefer. First, if you’re reporting multiple Bayes factor analyses in your write up, then somewhere you only need to cite the software once, at the beginning of the results section. So you might have one sentence like this:
All analyses were conducted using the BayesFactor package in R , and unless otherwise stated default parameter values were used
Notice that I don’t bother including the version number? That’s because the citation itself includes that information (go check my reference list if you don’t believe me). There’s no need to clutter up your results with redundant information that almost no-one will actually need. When you get to the actual test you can get away with this:
A test of association produced a Bayes factor of 16:1 in favour of a relationship between species and choice.
Short and sweet. I’ve rounded 15.92 to 16, because there’s not really any important difference between 15.92:1 and 16:1. I spelled out “Bayes factor” rather than truncating it to “BF” because not everyone knows the abbreviation. I indicated exactly what the effect is (i.e., “a relationship between species and choice”) and how strong the evidence was. I
didn’t
bother indicating whether this was “moderate” evidence or “strong” evidence, because the odds themselves tell you! There’s nothing stopping you from including that information, and I’ve done so myself on occasions, but you don’t strictly need it. Similarly, I didn’t bother to indicate that I ran the “joint multinomial” sampling plan, because I’m assuming that the method section of my write up would make clear how the experiment was designed. (I might change my mind about that if the method section was ambiguous.) Neither did I bother indicating that this was a
Bayesian
test of association: if your reader can’t work that out from the fact that you’re reporting a Bayes factor and the fact that you’re citing the
BayesFactor
package for all your analyses, then there’s no chance they’ll understand anything you’ve written.  Bayes Bayes Bayes Bayes Bayes. See?
Other sampling plans
Up to this point all I’ve shown you is how to use the
contingencyTableBF()
function for the joint multinomial sampling plan (i.e., when the total sample size N is fixed, but nothing else is). For the Poisson sampling plan (i.e., nothing fixed), the command you need is identical except for the
sampleType
argument:
contingencyTableBF(crosstab, sampleType = "poisson" )
## Bayes factor analysis
## --------------
## [1] Non-indep. (a=1) : 28.20757 ±0%
##
## Against denominator:
## Null, independence, a = 1
## ---
## Bayes factor type: BFcontingencyTable, poisson
Notice that the Bayes factor of 28:1 here is not the identical to the Bayes factor of 16:1 that we obtained from the last test. The sampling plan actually does matter.
What about the design in which the row columns (or column totals) are fixed? As I mentioned earlier, this corresponds to the “independent multinomial” sampling plan. Again, you need to specify the
sampleType
argument, but this time you need to specify whether you fixed the rows or the columns. For example, suppose I deliberately sampled 87 humans and 93 robots, then I would need to indicate that the
fixedMargin
of the contingency table is the
"rows"
. So the command I would use is:
contingencyTableBF(crosstab, sampleType = "indepMulti", fixedMargin="rows")
## Bayes factor analysis
## --------------
## [1] Non-indep. (a=1) : 8.605897 ±0%
##
## Against denominator:
## Null, independence, a = 1
## ---
## Bayes factor type: BFcontingencyTable, independent multinomial
Again, the Bayes factor is different, with the evidence for the alternative dropping to a mere 9:1. As you might expect, the answers would be diffrent again if it were the columns of the contingency table that the experimental design fixed.
Finally, if we turn to hypergeometric sampling in which everything is fixed, we get…
contingencyTableBF(crosstab, sampleType = "hypergeom")
#Error in contingencyHypergeometric(as.matrix(data2), a) :
# hypergeometric contingency tables restricted to 2 x 2 tables; see help for contingencyTableBF()
… an error message. Okay, some quick reading through the help files hints that support for larger contingency tables is coming, but it’s not been implemented yet. In the meantime, let’s imagine we have data from the “toy labelling” experiment I described earlier in this section. Specifically, let’s say our data look like this:
toys <- data.frame(stringsAsFactors=FALSE,
gender = c("girl", "boy"),
pink = c(8, 2),
blue = c(2, 8)
)
The Bayesian test with hypergeometric sampling gives us this:
contingencyTableBF(toys, sampleType = "hypergeom")
#Bayes factor analysis
#--------------
#[1] Non-indep. (a=1) : 8.294321 @plusorminus0%
#
#Against denominator:
# Null, independence, a = 1
#---
#Bayes factor type: BFcontingencyTable, hypergeometric
The Bayes factor of 8:1 provides modest evidence that the labels were being assigned in a way that correlates gender with colour, but it’s not conclusive.


1.6: Types of Statistical Studies (4 of 4)
Learning Objectives
- Based on the study design, determine what types of conclusions are appropriate.
Example
Multitasking
Do you constantly text-message while in class? Do you jump from one website to another while doing homework?  In a study of high-tech multitasking at Stanford University, researchers put 100 students into two groups: those who regularly do a lot of media multitasking and those who don’t. The two groups performed a series of three tasks:
(1) A task to measure the ability to pay attention:
- Students view two images of red and blue rectangles flashed one after the other on a computer screen. They try to tell if the red rectangles are in a different position in the second frame.
(2) A task to measure control of memory:
- Students view a sequence of letters flashed onto a computer screen, then recall which letters occurred more than once.
(3) A task to measure the ability to switch from one job to another:
- Students view numbers and letters together with the instructions to pay attention to the numbers, then recall if the numbers were even or odd. Then the instructions switch. Students are to pay attention to the letters and recall if the letters were vowels or consonants.
On every task, the multitaskers did worse than the non-multitaskers.
The researchers concluded that “people who are regularly bombarded with several streams of electronic information do not pay attention, control their memory, or switch from one job to another as well as those who prefer to complete one task at a time” (as reported in Stanford News in 2009).
“When they’re [high-tech multitaskers] in situations where there are multiple sources of information coming from the external world or emerging out of memory, they’re not able to filter out what’s not relevant to their current goal,” said Wagner, an associate professor of psychology at Stanford.
In general, we should not make cause-and-effect statements from observational studies, but in reality, researchers do it all the time. This does not mean that researchers are drawing incorrect conclusions from observational studies. Instead, they have developed techniques that go a long way toward decreasing the impact of confounding variables. These techniques are beyond the scope of this course, but we briefly discuss a simplified example to illustrate the idea.
Example
Smoking and Cancer
Consider this excerpt from the National Cancer Institute website:
- Smoking is a leading cause of cancer and of death from cancer. Millions of Americans have health problems caused by smoking. Cigarette smoking and exposure to tobacco smoke cause an estimated average of 438,000 premature deaths each year in the United States.
Notice that the National Cancer Institute clearly states a cause-and-effect relationship between smoking and cancer. Now let’s think about the evidence that is required to establish this causal link. Researchers would need to conduct experiments similar to the hormone replacement therapy experiments done by the Women’s Health Initiative. Such experiments would be very difficult to do. The researchers cannot manipulate the smoking variable.  Obviously, this is impossible. So how can we say that smoking causes cancer?
In practice, researchers approach this challenge in a variety of ways. They may use advanced techniques for making statistical adjustments within an observational study to control the effects of confounding variables that could influence the results. A simple example is the cell phone and brain cancer study.
- In this observational study, researchers identified a group of 469 people with brain cancer. They paired each person who had brain cancer with a person of the same sex, of similar age, and of the same race who did not have brain cancer. Then they compared the cell phone use for each pair of people. This matching attempts to control the confounding effects of sex, age, and race on the response variable, cancer. With these adjustments, the study will provide stronger evidence for (or against) a casual link.
However, even with such adjustments, we should be cautious about using evidence from an observational study to establish a cause-and-effect relationship. Researchers used these types of adjustments in the observational studies with hormone replacement therapy. We saw in that research that the results were still misleading when compared to those of an experiment.
So how can the National Cancer Institute state as a fact that smoking causes cancer?
They used other nonstatistical guidelines to build evidence for a cause-and-effect relationship from observational studies. In this approach, researchers review a large number of observational studies with criteria that, if met, provide stronger evidence of a possible cause-and-effect relationship. Here are some simplified examples of the criteria they use:
(1) There is a reasonable explanation for how one variable might cause the other.
- For example, experiments with rats show that chemicals found in cigarettes cause cancer in rats. It is therefor reasonable to infer that these same chemicals may cause cancer in humans.
- Consider these experiments together with the observational studies showing the association between smoking and cancer in humans. We now have more convincing evidence of a possible cause-and-effect relationship between smoking and cancer in humans.
(2) The observational studies vary in design so that factors that confound one study are not present in another.
- For example, one observational study shows an association between smoking and lung cancer, but the people in the study all live in a large city. Air pollution in a large city may contribute to the lung cancer, so we cannot be sure that smoking is the cause of cancer in this study.
- Another observational study looks only at nonsmokers. This study shows no difference in lung cancer rates for nonsmokers living in rural areas compared to nonsmokers living in cities.
- Consider these two studies together. The second study suggests that air pollution does not contribute to lung cancer, so we now have more convincing evidence that smoking (not air pollution) is the cause of higher cancer rates in the first study.
Let’s Summarize
-
There are four steps in a statistical investigation:
- Ask a question that can be answered by collecting data.
- Decide what to measure, and then collect data.
- Summarize and analyze.
- Draw a conclusion, and communicate the results.
-
There are two types of statistical research questions:
- Questions about a population
- Questions about cause-and-effect
- To answer a question about a population, we select a sample and conduct an observational study. To answer a question about cause-and-effect we conduct an experiment.
-
There are two types of statistical studies:
- Observational studies: An observational study observes individuals and measures variables of interest. We conduct observational studies to investigate questions about a population or about an association between two variables. An observational study alone does not provide convincing evidence of a cause-and-effect relationship.
- Experiments: An experiment intentionally manipulates one variable in an attempt to cause an effect on another variable. The primary goal of an experiment is to provide evidence for a cause-and-effect relationship between two variables.
- In statistics, a variable is information we gather about individuals or objects.
- When we investigate a relationship between two variables, we identify an explanatory variable and a response variable. To establish a cause-and-effect relationship, we want to make sure the explanatory variable is the only thing that impacts the response variable. Other factors, however, may also influence the response. These other factors are called confounding variables.
- The influence of confounding variables on the response variable is one of the reasons that an observational study gives weak, and potentially misleading, evidence of a cause-and-effect relationship. A well-designed experiment takes steps to eliminate the effects of confounding variables, such as random assignment of people to treatment groups, use of a placebo, and blind conditions. For this reason, a well-designed experiment provides convincing evidence of cause-and-effect.
Contributors and Attributions
- Concepts in Statistics. Provided by : Open Learning Initiative. Located at : oli.cmu.edu. License : CC BY: Attribution


13.3: Lab Report
Exercise 1 Data:
Make a sketch of your chromatogram using the template that follows. Using colored pencils, note the color of the bands. Use Table 1 to help you identify each pigment. On your sketch, label each band with the name of the pigment. Mark the distance (in cm) from the initial pigment band to each colored band as well as the total distance from the initial pigment band to the solvent front.
Sketch of Resulting Chromatogram
- Use the data from your sketch to complete the data table below.
- Calculate the value and include that in your data table.
|
A |
B |
C |
D |
E |
|---|---|---|---|---|
|
Name of Pigment |
Description of color |
Distance solvent front traveled from initial pigment band |
Distance pigment traveled from initial pigment band |
Rf value of pigment (column D / column C) |
Exercise 1 Review Questions:
- Which plant pigments are most polar?
- Which are least polar?
- How do these differences in polarity affect the movement of the pigments up the chromatography paper? Explain why this is observed.
4. What does the abbreviation Rf stand for? How is it calculated? What does it tell us?
- If Solution A moves 4 cm and Solution B moves 4.5 cm on a piece of chromatography paper, when the solvent moves 10 cm, which is the most polar solution? Explain your answer.
Exercise 2 Data and Review Questions:
|
Pigment |
Peak Absorption Wavelengths (nm) |
Sample 1 |
Sample 2 |
Sample 3 |
|---|---|---|---|---|
|
Chlorophyll A |
430 and 662 |
|||
|
Chlorophyll B |
453-642 |
|||
|
Carotenoids |
460-550 |
|||
|
Anthocyanins |
520* |
|||
|
Xanthophyll |
494 |
|||
|
Betalains |
535 or 480* |
*peaks may vary and are pH-dependent.
- How do you think the knowledge obtained from your chromatogram and the spectrograms relate to our understanding of plant pigments?
- Why do leaves change color in the fall?
Exercise 3 Employing Steps in the Scientific Method:
- Record the Question that is being investigated in this experiment. ________________________________________________________________
- Record a Hypothesis for the question stated above. ________________________________________________________________
- Predict the results of the experiment based on your hypothesis (if/then). ________________________________________________________________
|
Minute |
In Water/Detergent |
In Bicarbonate/Detergent |
|---|---|---|
|
1 |
||
|
2 |
||
|
3 |
||
|
4 |
||
|
5 |
||
|
6 |
||
|
7 |
||
|
8 |
||
|
9 |
||
|
10 |
||
|
11 |
||
|
12 |
||
|
13 |
||
|
14 |
||
|
15 |
||
|
16 |
Exercise 3 Review Questions:
- Complete the following sentence.
Photosynthesis is a set of ____________________________ in which ____________energy is converted to __________________energy.
- Why do some trees appear green in the summer but change colors in the fall?
3. What optimal wavelengths (or peaks) did you observe for both chlorophyll a and b ?
4. Do your leaf disks float? Use the information in this diagram of a cross-section of a leaf to explain why a leaf disk would float.
- Where does photosynthesis occur in a leaf? State which organelles carry out photosynthesis and which type or types of leaf cells have this organelle.
- Explain why it is useful to the plant to have air spaces around the spongy mesophyll cells in the leaves. (Hint: Recall the chemical equation for photosynthesis.)
- What was the purpose of the sodium bicarbonate in this experiment?
- Leaf disks normally float. What caused the leaf disks to sink?
- To measure the rate of photosynthesis, you replaced the air in the spongy mesophyll in your leaf disks with a liquid. This caused the leaf disks to sink. Then you put these leaf disks in water with dissolved CO2 and measured the amount of time it took for the leaf disks to float. Which product of photosynthesis accumulated in the spongy mesophyll and caused the leaf disks to float?
- Suppose that a leaf disk that has had the air sucked out is placed in a bicarbonate solution under a dim light that results in a low rate of photosynthesis that just equals the rate of cellular respiration. Would you expect this leaf disk to float? Explain why or why not.
Exercise 4: Observing and Quantifying Stomata (Optional)
- Record the Question that is being investigated in this experiment. ________________________________________________________________
- Develop a hypothesis about the number of open stomata found on the upper side of a leaf as compared to the lower side of the leaf. ___________________________________________________________
- Develop a hypothesis about the number of open stomata found on the upper side of a leaf as compared to the lower side of the leaf. Write your hypothesis in the space below. ____________________________________________________________
- Predict the results of the experiment based on your hypothesis (if/then). ________________________________________________________________
|
Surface of Leaf |
# of Stomata: View 1 |
# of Stomata: View 2 |
# of Stomata: View 3 |
Average # of Stomata |
|---|---|---|---|---|
|
Upper |
||||
|
Lower |
Extension Activity: (Optional)
The results of this experiment can be presented graphically. The presentation of your data in a graph will assist you in interpreting your results. Based on your results, you can complete the final step of scientific investigation, in which you must be able to propose a logical argument that either allows you to support or reject your initial hypothesis.
- Graph your results using the data from Table 3.
- What is the dependent variable? Which axis is used to graph this data? ______________________________________________________________________
- What is your independent variable? Which axis is used to graph this data? _____________________________________________________________________
Exercise 4 Review Questions:
- What are stomata?
2. What is the importance of stomata in photosynthesis?
3. What is the function of the guard cells?
4. During the lab activity, were more stomata observed on the upper surface of the leaf or the lower surface of the leaf? Explain why you think this distribution exists.


9.3: Current Signal Systems
It is possible through the use of electronic amplifiers to design a circuit outputting a constant amount of current rather than a constant amount of voltage. This collection of components is collectively known as a current source , and its symbol looks like this:
A current source generates as much or as little voltage as needed across its leads to produce a constant amount of current through it. This is just the opposite of a voltage source (an ideal battery), which will output as much or as little current as demanded by the external circuit in maintaining its output voltage constant. Following the “conventional flow” symbology typical of electronic devices, the arrow points against the direction of electron motion. Apologies for this confusing notation: another legacy of Benjamin Franklin’s false assumption of electron flow!
Current sources can be built as variable devices, just like voltage sources, and they can be designed to produce very precise amounts of current. If a transmitter device were to be constructed with a variable current source instead of a variable voltage source, we could design an instrumentation signal system based on current instead of voltage:
The internal workings of the transmitter’s current source need not be a concern at this point, only the fact that its output varies in response to changes in the float position, just like the potentiometer setup in the voltage signal system varied voltage output according to float position.
Notice now how the indicator is an ammeter rather than a voltmeter (the scale calibrated in inches, feet, or meters of water in the tank, as always). Because the circuit is a series configuration (accounting for the cable resistances), current will be precisely equal through all components. With or without cable resistance, the current at the indicator is exactly the same as the current at the transmitter, and therefore there is no error incurred as there might be with a voltage signal system. This assurance of zero signal degradation is a decided advantage of current signal systems over voltage signal systems.
The most common current signal standard in modern use is the 4 to 20 milliamp (4-20 mA) loop, with 4 milliamps representing 0 percent of measurement, 20 milliamps representing 100 percent, 12 milliamps representing 50 percent, and so on. A convenient feature of the 4-20 mA standard is its ease of signal conversion to 1-5 volt indicating instruments. A simple 250 ohm precision resistor connected in series with the circuit will produce 1 volt of drop at 4 milliamps, 5 volts of drop at 20 milliamps, etc:
The current loop scale of 4-20 milliamps has not always been the standard for current instruments: for a while there was also a 10-50 milliamp standard, but that standard has since been obsoleted. One reason for the eventual supremacy of the 4-20 milliamp loop was safety: with lower circuit voltages and lower current levels than in 10-50 mA system designs, there was less chance for personal shock injury and/or the generation of sparks capable of igniting flammable atmospheres in certain industrial environments.
Review
- A current source is a device (usually constructed of several electronic components) that outputs a constant amount of current through a circuit, much like a voltage source (ideal battery) outputting a constant amount of voltage to a circuit.
- A current “loop” instrumentation circuit relies on the series circuit principle of current being equal through all components to insure no signal error due to wiring resistance.
- The most common analog current signal standard in modern use is the “4 to 20 milliamp current loop.”


11.6: Hypothesis Testing for Two Means and Two Proportions (Worksheet)
-
- Last updated
- Save as PDF
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
Student Learning Outcomes
- The student will select the appropriate distributions to use in each case.
- The student will conduct hypothesis tests and interpret the results.
Supplies:
- the business section from two consecutive days’ newspapers
- three small packages of M&Ms®
- five small packages of Reese's Pieces®
Increasing Stocks Survey
Look at yesterday’s newspaper business section. Conduct a hypothesis test to determine if the proportion of New York Stock Exchange (NYSE) stocks that increased is greater than the proportion of NASDAQ stocks that increased. As randomly as possible, choose 40 NYSE stocks, and 32 NASDAQ stocks and complete the following statements.- \(H_{0}\): _________
- \(H_{a}\) : _________
- In words, define the random variable.
- The distribution to use for the test is _____________.
- Calculate the test statistic using your data.
-
Draw a graph and label it appropriately. Shade the actual level of significance.
- Graph: Figure 10.6.1.
- Calculate the \(p\text{-value}\).
- Do you reject or not reject the null hypothesis? Why?
- Write a clear conclusion using a complete sentence.
Decreasing Stocks Survey
Randomly pick eight stocks from the newspaper. Using two consecutive days’ business sections, test whether the stocks went down, on average, for the second day.- \(H_{0}\): ________
- \(H_{a}\) : ________
- In words, define the random variable.
- The distribution to use for the test is _____________.
- Calculate the test statistic using your data.
-
Draw a graph and label it appropriately. Shade the actual level of significance.
- Graph: Figure 10.6.2.
- Calculate the \(p\text{-value}\):
- Do you reject or not reject the null hypothesis? Why?
- Write a clear conclusion using a complete sentence.
Candy Survey
Buy three small packages of M&Ms and five small packages of Reese's Pieces (same net weight as the M&Ms). Test whether or not the mean number of candy pieces per package is the same for the two brands.
- \(H_{0}\): ________
- \(H_{a}\): ________
- In words, define the random variable.
- What distribution should be used for this test?
- Calculate the test statistic using your data.
-
Draw a graph and label it appropriately. Shade the actual level of significance.
- Graph: Figure 10.6.3.
- Calculate the \(p\text{-value}\).
- Do you reject or not reject the null hypothesis? Why?
- Write a clear conclusion using a complete sentence.
Shoe Survey
Test whether women have, on average, more pairs of shoes than men. Include all forms of sneakers, shoes, sandals, and boots. Use your class as the sample.
- \(H_{0}\): ________
- \(H_{a}\): ________
- In words, define the random variable.
- The distribution to use for the test is ________________.
- Calculate the test statistic using your data.
-
Draw a graph and label it appropriately. Shade the actual level of significance.
- Graph: Figure 10.6.4.
- Calculate the \(p\text{-value}\).
- Do you reject or not reject the null hypothesis? Why?
- Write a clear conclusion using a complete sentence.