7.1: Introduction to Hypothesis Testing
- Introduce the idea of hypothesis testing
- Define null and alternative hypotheses
- Develop the logic of identifying null and alternative hypotheses
- Define the \(p\)-value
- Explain the two possible conclusions for a hypothesis test
- Introduce the \(\alpha\) value
- Introduce type I and type II errors
- Differentiate between statistically significant and practically significant results
- Introduce one-tailed and two-tailed tests
Review and Preview
In the last chapter, we finally achieved a goal of inferential statistics: to use facts about sample data to speak confidently about the facts of the population from which it was drawn. Our method of confidence intervals provides an interval estimate of the population parameter at a certain success rate called the confidence level. When we do not know much about the population, we can utilize random sampling to build confidence intervals to learn about populations from scratch. At other times, we have claims about a certain population that we hope to test. This quest falls within the realm of inferential statistics and is the subject of this chapter.
Consider the legendary secret agent James Bond \(007,\) the central character of a series of books and action movies, who is capable of great feats of heroism and has a penchant for martinis: shaken not stirred, never stirred. With such a strong preference, one would expect that James Bond could actually tell the difference between shaken and stirred martinis simply by taste and know which was which. Such an ability may seem unlikely. It would be natural to desire some evidence to back the claim. Taste-testing martinis could serve as a method for collecting such evidence. Suppose we settled on conducting \(16\) taste tests where each martini was either stirred or shaken randomly. Upon tasting each martini, James Bond claimed the martini was either shaken or stirred. After tasting all \(16\) martinis, suppose James Bond correctly identified \(13\) of the martinis. Is this sufficient evidence to back the claim that he can distinguish between shaken and stirred martinis simply by taste? If instead he got all \(16,\) would that prove that the claim is true?
Neither result proves the claim; we cannot construct proofs of such claims. He may have been merely guessing but had great luck. Is luck a plausible explanation? If we assume that he is merely guessing, what is the probability that he actually designates \(13\) of the \(16\) martinis correctly? What is the probability that he correctly assigns at least \(13\) of the \(16\) martinis correctly? This latter question covers the case of what actually happened and anything more extreme happening and will be a frequent concern throughout this chapter. We can understand the taste testing and probability questions with binomial random variables. We have \(16\) trials with a success defined as correctly classifying the martini as shaken or stirred. If we assume James Bond is guessing, then the probability of success for a single tasting is \(p=\frac{1}{2}.\) If \(X\) is the random variable counting the number of successes in \(16\) trials, we are interested in computing \(P(X\ge13).\) The reader is encouraged to verify that \(P(X\ge13)\approx0.0106.\) With such a small probability, we would say that someone randomly guessing at least \(13\) of the \(16\) martinis correctly is quite unusual. We have two possible situations: James Bond was guessing and something quite rare occurred by chance, or James Bond actually has the ability to distinguish shaken and stirred martinis by taste. Given the evidence, the claim that James Bond was merely guessing seems dubious; the claim has not been proven false, but there is considerable doubt about its validity. We, therefore, say that there is considerable evidence that James Bond can distinguish between martinis shaken and stirred simply by taste.
We call the process described above as hypothesis testing. There is a claim or hypothesis about reality that needs to be tested (that James Bond can distinguish between shaken and stirred martinis simply by taste). A competing hypothesis is identified (that James Bond was merely guessing). Under the assumption that the competing hypothesis is true (James Bond is guessing), the probability that what happened or even something more extreme will happen is computed \(P(X\ge13)\approx0.0106.\) If this is a rare event, it casts doubt on the validity of the assumption that the competing hypothesis is true; thus, credence is lended to the original hypothesis (that James Bond can indeed distinguish between shaken and stirred martinis simply by taste). At no point in this process, just as in constructing confidence intervals, is certainty achieved. Rare things do indeed happen, but that does not mean we cannot have confidence in assessing evidence. We will now begin formalizing the process of hypothesis testing.
Hypotheses
As we interact with the world around us, we begin to notice pattens in our observations and start to form hypotheses about the world based on these patterns. Before we act as if these hypotheses are true, we want to secure reasonable and sufficient evidence in support of them. So, how do we collect evidence in support of a hypothesis that arose from the claims or observations of ourselves or others? Recall that, in our example with James Bond, we identified a competing hypothesis, a hypothesis that was opposite of the one he claimed. James Bond claimed to be able to distinguish between shaken and stirred martinis by taste. The opposite claim was that he could not distinguish by taste and, therefore, was guessing. We call these two competing hypotheses the alternative hypothesis \(H_1\) and the null hypothesis \(H_0.\) The null hypothesis is more practically or reasonably assumed to be true. While the alternative hypothesis is generally the novel or claimed hypothesis.
Within the context of the James Bond scenario, our initial disposition was that having such a refined taste palette would be abnormal. It seemed more reasonable to assume, at least initially, that one would simply be left guessing about how the martinis were mixed. Thus, the null hypothesis was that James Bond could not distinguish simply by taste and, therefore, was guessing. The alternative hypothesis was that he could distinguish simply by taste.
Consider another example: preparing to enter the shower. An important concern is the temperature of the water. Either the water is amenable to a pleasant shower or it is not. We have two hypotheses. Which of the two hypotheses do we presume to be true as we prepare to enter the shower? Not many of us turn the water on and immediately hop in the shower. Instead, we wait for it to warm up, waiting to see steam rising or periodically running a hand through the water to test the temperature. These actions speak of an initial assumption that one of the hypotheses is true. We operate with the assumption that the water is unsatisfactory until we have evidence to the contrary. Our actions reveal that the null hypothesis is that the water is not amenable to showering.\[\begin{align*}H_0&:\text{Water is not at a temperature suitable to showering}\\H_1&:\text{Water is at a temperature suitable to showering}\end{align*}\]These designations could also be thought of in terms of the potential implications of acting as if one of the hypotheses is true when indeed it is not. If we assume that the water is ready but that is not the case. What might happen? Since we assume that the water is good to go, we will hop in the shower right away. Once we are in the shower, we quickly find out that it is either too hot or too cold and immediately feel something moderately unpleasant to possibly painful. If, on the other hand, we assume that the water is not yet ready despite the water actually being perfectly suitable to us, the price is that we wait an extra minute before hopping in the shower. Which of these situations would we prefer if we were wrong with our initial assumption? We would prefer just waiting around for an extra minute over something that could potentially scald us. The hypothesis with the less drastic cost in acting as if it were true when it was false is the null hypothesis.
Within each context, determine the competing hypotheses and identify them as either the null hypothesis or the alternative hypothesis. Explain your reasoning.
- In a court of law in the United States, a prosecutor (a lawyer) argues that the defendant (a citizen) is guilty of some crime. The defendant is usually represented by a criminal defense lawyer who argues against the prosecutor (that his client is not guilty). A judge and possibly a jury follow the arguments in order to draw a final conclusion called a verdict.
- Answer
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The two competing hypotheses are related to the defendant's innocence regarding the crime charged against him. One hypothesis is that the defendant is innocent of the charge. The other is that the defendant is guilty of the charge. In the United States, our legal system is structured with the mantra "innocent until proven guilty," with sufficient evidence described as evidence beyond a reasonable doubt.\[\begin{align*}H_0&:\text{The defendant is innocent}\\H_1&:\text{The defendant is guilty}\end{align*}\]Thinking of the potential implications of acting as if one hypothesis were true when it is not can help us understand why our legal system was set up as it is. If we act as if the defendant is guilty despite the fact that he is innocent. We could send an innocent person to pay a fine, spend time in prison, or even be executed. Our founders experienced tyranny and sought in many ways to protect the citizens from the government. The potential price from the defendant's perspective is quite steep. Sending an innocent man to jail for years or to death is severe. On the other hand, if we act as if the defendant is innocent despite the fact that he is guilty the potential price from the defendant's perspective is the cost of playing the system. It may be that the defendant feels remorse and a desire to change his ways or it may be that he got away with breaking the law. He may be more or less prone to be a repeat offender. The local society may deem acting with more caution around the defendant a prudent decision. From the perspective of our legal system, the cost of an innocent man being wrongly persecuted is worse than a guilty man walking free.
"It is better that ten guilty persons escape than that one innocent suffer." - William Blackstone
"It is better 100 guilty Persons should escape than that one innocent Person should suffer." -Benjamin Franklin
- A researcher at Stine, a company that develops corn and soybean seeds, identifies a new breed of sweet corn in its breeding laboratory that he thinks will produce corn that is more tolerant to the dry conditions of northwestern Kansas than the breed that most northwestern Kansas corn farmers currently use.
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The two competing hypotheses are related to the drought tolerance of the new variety of corn. One hypothesis is that the new variety of corn has better drought tolerance than the commonly used variety of corn. The other is that the new variety of corn does not have better drought tolerance than the commonly used variety of corn. This could be that it is equally tolerant or that it is less tolerant. The Stine company would benefit tremendously from developing a variety of corn suitable for drier conditions. If the commonly used seed is of a business competitor, Stine can expand its market. Even if it is the producer of the current common variety, the new seed will likely be able to be sold at a higher price and bring new attention to the company. Making the assumption that the new seed is better than the common seed could be disastrous if false. The company would likely lose many clients and possibly face a lawsuit for false advertising. As such, the reasonable hypothesis to assume initially until there is evidence to the contrary is that the new variety of seed is not better suited to the dry conditions of northwestern Kansas.\[\begin{align*}H_0&:\text{The new variety of sweet corn seed is, at best, the same as the commonly used seed}\\H_1&:\text{The new variety of sweet corn seed is better than the commonly used seed}\end{align*}\]
The \(17^{th}\) century philosopher and mathematician Blaise Pascal engaged in a similar line of reasoning before any rigorous development of hypothesis testing in an argument called Pascal's Wager which can be found in his book \(Pens\acute{e}es.\) Dr. Peter Kreeft, a renowned philosopher and professor at Boston College, gives an exposition of the argument (which can be found here ) which we will formulate here using the framework of hypotheses that we have been developing.
Pascal lived in a time of great religious skepticism and attempted to formulate a line of reasoning that could reach a skeptic who lacked faith and did not believe that reason was sufficient to prove that God existed. There are two competing hypotheses at play: God exists and God does not exist. The skeptic knows that only one of the hypotheses is true but cannot establish intellectual certainty as to which to adopt. Pascal, just as is done within hypothesis testing, considers the potential ramifications of living as if the hypotheses were true when, indeed, they were false. Which hypothesis is it best to live by as if it were true? Pascal argues that one cannot abstain from the wager, for we are all already playing the game of life.
In considering the ramifications, Pascal, a Christian, considers the possibility of eternal happiness because everyone seeks maximal happiness. But in considering eternal happiness, he assumes that if God exists, there is paradise (heaven), thus equivocating the existence of God such that there is no paradise (heaven) with there being no God.
Pascal continues. If one lives as if God does not exist when He really does, then that one misses out on the possibility of eternal happiness. If one lives as if God does exist when He really does not, then that one has lost nothing since there was nothing to gain or lose at the moment of death. Which of the two of these potential costs is less drastic? He argues that the cost of eternal happiness is infinitely worse than losing nothing. Then we could say, within the context of competing hypotheses, that the null hypothesis, the hypothesis initially assumed and acted upon as true, would be that God exists.\[\begin{align*}H_0&:\text{God exists}\\H_1&:\text{God does not exist}\end{align*}\]Accepting the risk assessment of the two hypotheses is a critical part of his argument, and there are reasonable grounds to object and much to consider about the context, assumptions, nuances, and ramifications of the argument. An interested reader is encouraged to ponder the argument more thoroughly and read through Kreeft's exposition linked above.
Collecting Evidence and Making Decisions
In the example about martinis and James Bond, the null hypothesis was that James Bond simply guesses whether the martinis were shaken or stirred. We collected evidence to test his claim, the alternative hypothesis (that he could distinguish between shaken and stirred martinis simply by taste) by conducting an experiment where he taste tested \(16\) martinis. We then analyzed that evidence under the assumption that the null hypothesis was true by computing the probability that a person simply guessing would get at least \(13\) identifications correct when presented with \(16\) martinis where the mixing method was chosen randomly. This probability is called the \(p\)-value.
The \(p\)-value is not the probability that James Bond was simply guessing (i.e. it is not the probability that the null hypothesis is true). Rather, it is the probability that something at least as extreme as what was observed happens assuming that the null hypothesis is true. We can understand the \(p\)-value as the probability of an event given a hypothesis. It is often misunderstood as the probability of a hypothesis given an event.
James Bond correctly identified \(13\) of the \(16\) martinis. What would be at least as extreme in the context of the taste testing? It would be that he got \(13,\) \(14,\) \(15,\) or even \(16\) martinis right. So in this context, the \(p\)-value is \(P(X\ge13).\) When the \(p\)-value is quite small, either something rare happened or there was a flaw in an assumption of our analysis, namely, that the null hypothesis is true. We cannot be certain which is the case from what we know, but if the \(p\)-value is sufficiently small, we generally consider it as significant evidence that the null hypothesis is false. When this is the case, we say that we reject the null hypothesis in favor of the alternative hypothesis . In the case of James Bond, the \(p\)-value was \(0.0106\), which is rather small, so we rejected the null hypothesis (that he was simply guessing) in favor of the alternative hypothesis (that he could indeed distinguish simply by taste). If the \(p\)-value is not sufficiently small, the event that occurred was not rare enough under the assumption that the null hypothesis is true to cast doubt on the truth of the null hypothesis. This does not prove that the null hypothesis is true; rather, we simply failed to show it was likely false, so we conclude that we fail to reject the null hypothesis when this happens. We emphasize that failing to reject is not the same thing as accepting. Hypothesis testing can only falsify, never verify, the null hypothesis, as throughout the procedure, the null hypothesis is assumed to be true.
Reports of statistical analyses outline the logical progression for the development of hypotheses, experimental design, results of the experiment, and the \(p\)-values in order to provide the readers with the full scope of the logic and evidence. This is done because there are different approaches to what is deemed a sufficiently small \(p\)-value and the decisions need to be based on more than just whether the \(p\)-value meets some given threshold. However, deciding on a threshold which considers the context and is determined independently from the evidence is a good start to measuring the weight of the evidence. Once a threshold is set, we describe the hypothesis test as having that level of significance, which is typically referred to as the \(\alpha\) value of the hypothesis test. Commonly used \(\alpha\) values are \(0.05\) and \(0.01.\)
From our discussion thus far, we note that the null hypothesis plays a pivotal role in the process of hypothesis testing. The \(p\)-value comes from a probability calculation, assuming the null hypothesis is true. The conclusions that we draw from a hypothesis test come in two forms: reject the null hypothesis or fail to reject the null hypothesis. Indeed, we can loosely understand the process of hypothesis testing as the quest for finding evidence against the null hypothesis, so that, when significant evidence (evidence that produces a \(p\)-value less than the \(\alpha\) value, the significance level) is found, we can reject the null hypothesis and favor the alternative hypothesis.
Type I and Type II Errors
Rare events are not impossible events; they do happen from time to time. As such, it is possible to conduct a hypothesis test, collecting evidence that meets our standard of significance, which leads us to reject a true null hypothesis. Perhaps, we are partially at fault for being too easily convinced that the evidence was strong. Perhaps something extraordinarily rare occurred, but we made an error either way. In rejecting a true null hypothesis, we made a statement about reality that does not match what really is happening. We call this error a t ype I error .
Recall that when deciding which of the competing hypotheses would be the null hypothesis, we were considering the potential ramifications of acting as if one of the hypotheses were true when it really was not. We could also consider the ramifications in the other direction. What are the costs of rejecting one of the hypotheses when it was true; in terms of our current discussion, what are the costs of committing a type I error if we adopt a particular hypothesis as our null hypothesis? Consider the showering example once more. If we reject that the water is suitable, when it is not, the cost is the more drastic of the two. That was the hypothesis we set as the null hypothesis. Thus, the hypothesis with a more severe cost in rejecting it when it is true is classified as the null hypothesis. We do this because we have a certain degree of control over the occurrence of type I errors; we set the thresholds regarding sufficient evidence.
We cannot completely avoid making a type I error, but we can manage the likelihood. As we have seen, evidence is collected, the \(p\)-value is computed using the evidence, and then if the \(p\)-value is less than the \(\alpha\) value \((p\)-value\(<\alpha)\), we assert that there is sufficient evidence to reject the assumption that the null hypothesis is true. The \(\alpha\) value is the upper bound regarding which \(p\)-values accounted for sufficient evidence, and recall that the \(p\)-value is the probability that something at least as extreme as what happened happens given the assumption that the null hypothesis is true. So, we can understand \(\alpha\) as the probability of making a type I error. A smaller \(\alpha\) value means a smaller rate of committing type I error.
Making a type I error is not the only way in which we may fail to recognize the reality of the world around us. It may be the case that the null hypothesis is false and would be rejected, but we fail to do so because the evidence did not meet the level of significance desired. Recall that, when the evidence gathered is not sufficient, the result of the hypothesis test is that we fail to reject the null hypothesis. Failing to reject the null hypothesis is not the same as declaring the null hypothesis is true. We are not making a strict assertion regarding which hypothesis matches reality; we are simply saying that the evidence did not cast sufficient doubt on the truthfulness of the null hypothesis. Despite the fact that, in failing to reject a false null hypothesis, we are not asserting anything false about reality, we still call it an error in the fact that we have failed to be better aligned with reality. We call this error a type II error .
Once again, consider the showering example. We set up the hypotheses as follows. \[\begin{align*}H_0&:\text{Water is not at a temperature suitable to showering}\\H_1&:\text{Water is at a temperature suitable to showering}\end{align*}\]A type I error would mean falsely believing that the water is suitable. A type II error would mean falsely believing that the water is not suitable. Notice that if we switched which hypothesis was the null hypothesis, the error types would also switch. The fact that our framing yields a type I error which is more severe than the type II error indicates the hypotheses were formulated correctly.
Our original discussion on determining which hypothesis to set as the null hypothesis centered on the ramifications of acting as if one of the hypotheses were true when that was not the case. We set the null hypothesis as the hypothesis with the less drastic costs; the alternative hypothesis would thus have the more drastic costs. This process is similar to the consideration of making a type II error. In failing to reject a null hypothesis which is false, we do not assert that the null hypothesis is true, but our initial disposition towards the hypotheses remains the same. We can connect the two ideas to formalize our hypothesis-setting process; the alternative hypothesis is set by considering the ramifications of making a type II error if the particular hypothesis is set as the alternative hypothesis. The hypothesis with the more drastic cost in making a type II error is the alternative hypothesis. We have two mechanisms for deciding how to set the hypotheses in a hypothesis test (both produce the same results).
- When considering which hypothesis to set as the null hypothesis, consider the costs of committing a type I error if the hypotheses were adopted as the null hypothesis. Set the hypothesis with the greater cost as the null hypothesis.
- When considering which hypothesis to set as the alternative hypothesis, consider the costs of committing a type II error if the hypotheses were adopted as the alternative hypothesis. Set the hypothesis with the greater cost as the alternative hypothesis.
Both methodologies produce hypotheses such that a type I error is worse than a type II error. This is because we can precisely identify the probability of a type I error, but we cannot control the likelihood of a type II error. One can verify that one has correctly set up hypotheses by checking that a type I error is worse than a type II error.
Conducting Hypothesis Testing
In inferential statistics, we are primarily interested in claims about populations and assertions about the values of parameters. We could test the claims with certainty if we could compute the actual parameter value, but alas, that is not the case in reality due to literal or practical impossibilities. We use random sampling to collect evidence and then use our knowledge of sampling distributions to compute the \(p\)-value.
Consider a seed company with a new sweet corn variety that is thought to produce greater yields in northwestern Kansas than what is commonly planted now. We can measure such a claim by looking at the average yield of the two varieties; these are facts about all seeds of the two varieties, i.e., parameters. After seasons of planting and studying the common sweet corn variety, researchers and farmers have a pretty good idea about its average yield; let us suppose that it is equal to \(70\) bushels per acre. Since the researchers at the seed company think that they have discovered a better-yielding variety of sweet corn, we can assume there is a line of scientific reasoning or some test plots that have led to such a hypothesis, but we do not know the value of the average yield for the new variety; let us simply call it \(\mu.\) Let us suppose, for pedagogical purposes, that the new sweet corn varieties from the company consistently have standard deviations of \(3.6\) bushels per acre; so, that we feel confident the new variety has a standard deviation of \(3.6\) bushels per acre too.
Just as in the case with the drought-resistant new variety of seed, we do not want to assume initially that the new seed is better than the commonly used seed. So, our null hypothesis is that the average yield of the new variety is, at best, the same as the average yield of the commonly used variety, and our alternative hypothesis is that the average yield of the new variety is better than the average yield of the commonly used variety. We can denote this symbolically as follows.\[H_0:\mu\leq 70 \text{ bushels per acre} \\ H_1:\mu>70 \text{ bushels per acre}\nonumber\]Let us suppose that the company has set a standard \(\alpha\) value for these sorts of tests at the \(0.01\) level. With the two hypotheses and \(\alpha\) value set, we now look to gather evidence by randomly sampling the yields of the new variety of sweet corn. Perhaps we randomly sample \(50\) acres across northwestern Kansas where the new variety was grown under comparable conditions to the normal farming practices. From this random sample, we compute the average yield of the sample to be \(71\) bushels per acre. The average yield of the sample is greater than the commonly grown average yield, but is it sufficient evidence to cast doubt on the truth of the null hypothesis?
How do we assess our evidence? We compute the \(p\)-value, the probability that something more extreme happens than what we observed given that the null hypothesis is true. We are interested in computing \(P(\bar{x}>71 \text{ bushels per acre}).\) Since we randomly sampled using a sample size of \(50,\) we expect the sampling distribution of sample means to be approximately normal with a mean \(\mu\) and standard deviation \(\frac{\sigma}{\sqrt{n}}.\) From the consistency of the company's seeds, we can feel confident that \(\sigma_{\bar{x}}=\frac{3.6}{\sqrt{50}}.\) But, from the null hypothesis, we do not know precisely what \(\mu\) is supposed to be. All we know from the assumption that the null hypothesis is true, is that \(\mu\leq70\) bushels per acre. A common adage tell us to prepare for the worst and hope for the best. We similarly want to consider the case where sufficient evidence against the null hypothesis would be hardest to obtain. Indeed, this is when we assume the new breed produces just as well as the commonly used variety, when \(\mu=70\) bushels per acre. This happens to be the \(mu\) value that produces the largest \(p\)-value (think about why this is the case). We encourage the reader to verify that the \(p\)-value is approximately \(0.0248.\)
A \(p\)-value of \(0.0248\) would constitute significant evidence at the \(\alpha=0.05\) level but would not constitute significant evidence at the \(\alpha=0.01.\) This seed company thus would fail to reject the null hypothesis in this situation. That is not to say that the new variety is not better than the common seed. There has not been sufficient evidence to doubt that the new yields are, on average, comparable to the common seed. If the research team is confident in the genetics of their newly developed variety, it may be prudent to conduct another experiment with a larger sample (recall that the standard deviation of the sampling distribution shrinks as the sample size increases) because rare things do occur.
Suppose the researchers at the seed company were very confident in the scientific reasoning behind the genetic breakthrough with this most recently developed sweet corn variety, and they decided to conduct an experiment using \(10\) times as much data, meaning, \(500\) acres were to be randomly selected across northwestern Kansas to grow this new variety. After the harvest, the researchers found that the average yield from these \(500\) acres was \(70.4\) bushels per acre.
- Using the same hypotheses as the example above, determine the \(p\)-value from this larger experiment. State and interpret the conclusion of the hypothesis test at the \(\alpha\) \(=0.01\) level.
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Since the hypotheses from the first experiment have not changed, the \(p\)-value is the probability that a random sample of \(500\) acres produces an average yield at least as large as the average yield that was observed. So we are looking to compute \(P(\bar{x}>70.4 \text{ bushels per acre}).\) The sampling distribution is approximately normal because the sample size is so large with a standard deviation of \(\sigma_{\bar{x}}\) \(=\frac{3.6}{\sqrt{500}}.\) We will again assume that \(\mu=70\) bushels per acre because that is the condition within the assumption that the null hypothesis is true that will produce the largest \(p\)-value. \(p\)-value\(=1-\text{NORM.DIST}(70.4,70,\frac{3.6}{\sqrt{500}},1)\)\( \approx 0.0065.\)
Since \(0.0065< 0.01,\) we say that at the significance level of \(0.01\) there is sufficient evidence to reject the null hypothesis that the average yield of the new variety of sweet corn yields, at best, the same as the commonly used corn seed. We conclude that the new variety of sweet corn produces, on average, a greater yield than the commonly used seed.
- At the time of the study, sweet corn was being sold at \(\$4.20\) per bushel on average. If a farmer with \(80\) acres designated for sweet corn was seeking advice about switching to the new breed of corn, what aspects of the study would be important to consider? What would your advice to the farmer be?
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Since we rejected the null hypothesis at the \(\alpha\) level of \(0.01,\), there is statistical evidence to say that the average yield of the new variety of sweet corn is larger than \(70\) bushels per acre, the average yield of the commonly used variety. The hypothesis test itself did not determine by how much the average yield would increase; it only concluded that there was an increase. If the increase was large and resulted in larger profits despite having to pay more for the seed, the farmer may be inclined to make the switch. If the increase was not large, the farmer may be less profitable despite producing more because of the increased cost associated with the new seed. If we used the average yield from the sample to make such a comparison, we would expect the farmer to have an increased yield of \(0.4\) bushels per acre on average. We could then estimate the increased revenue to be \(0.4\cdot 80\cdot 4.20\) \(=134.4\) dollars. Not knowing the actual prices of seed, we cannot estimate the profit, but it seems doubtful that the switch will lead to any increased profits worth noting.
- More technical answers for the mathematically inclined.
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Rather than just using the sample mean as a point estimate for the population mean of the new variety, we can construct a confidence interval and use the boundary points as upper and lower bounds to create an interval estimation of the increased revenue. Let us construct a \(95\%\) confidence interval; \(70.0845,70.7156.\) So, at the \(95\%\) confidence level, we expect the population mean to fall between \(70.0845\) and \(70.7156\) bushels per acre. This leads us to expect to increase the yield by something between \(0.0845\) and \(0.7156\) bushels per acre, meaning the expected increase of revenue would be between \(\$28.38\) and \(\$240.42\) which again does not seem that a switch would lead to any increased profits worth noting.
As we have just seen, there may be differences that are statistically significant that do not warrant changes in our lives. When this is the case, we say that the findings are statistically significant but not practically significant. Remember that we are trying to understand the world better so that we may better live in the world and interact with the world. The \(p\)-value does not measure the size of a difference, often referred to as the size of the effect; it measures the probability that something at least as extreme happens as what was observed to happen under the assumption that the null hypothesis is true. A small \(p\)-value does not indicate a large effect. Many people have fallen prey to misunderstanding the meaning and uses of hypothesis testing, especially regarding \(p\)-values. In fact, widespread misinterpretations of the \(p\)-value prompted the American Statistical Association published an article addressing the misuses of \(p\)-values in \(2016\) to help remedy the issue.
Types of Hypothesis Tests
Let us return to the James Bond example. What would we have concluded if, after tasting the \(16\) martinis, James Bond was only correct on \(3\) of them? If he were truly guessing, we would expect that he would be correct about half of the time, but that is not what happened. Such a low score could indicate that James Bond could taste a difference between shaken and stirred martinis, but he has confused the tastes. He consistently labels shaken as stirred and stirred as shaken. This would indicate that he can taste a difference, but his preference for shaken is mistaken!
If we were simply interested in whether James Bond's taste buds could tell the difference between shaken and stirred martinis, and not that he could distinguish between them due to his strong preference for shaken martinis, evidence against the null hypothesis that he was guessing would come in the form of either really low numbers of successes or really high numbers of successes. We are looking for evidence in two different directions. In this case, looking at the probability of something at least as extreme happening as what happened takes on a more complicated meaning.
With \(13\) out of \(16\) taste tests being successful, we easily understand more extreme as at least \(13\) martinis being correctly identified. But what would be as extreme in the other direction in which we are looking for evidence? That would be \(3\) out of \(16\) taste tests being successful. And in this direction, at least as extreme would lead us to consider at most \(3\) martinis being correctly identified. So, we are looking at the two tails of the binomial distribution to compute the \(p\)-value. We call such a test a two-tailed test . Thus the \(p\)-value is \(P(X\le 3 \text{ or } X\ge 13)\) \(\approx0.0212.\) This, again, qualifies as sufficient evidence at the \(\alpha\) value of \(0.05.\)
In our previous examples, we have looked for evidence only in one direction. The \(p\)-value came from only one of the tails of the probability distribution; we call such tests one-tailed tests . Two-tailed tests are most common in scientific research because finding any difference is generally notable, interesting, and could lead to further development. In the case of medicine and business, one-tailed tests arise with greater frequency because there are cases where there is no need to distinguish between no effect and an effect in an unpredicted or undesired direction. For example, with the varieties of corn, the seed company is interested in increasing the yield. If a new breed has the same yield or a worse yield, the new breed does not warrant further consideration.
Consider the example with James Bond without considering his preference for shaken martinis. Does he possess the taste buds necessary to note the difference between the mixing method? We identified this as a two-tailed test and could write the hypotheses as follows.\[\begin{align*}H_0&:p = 0.5\\H_1&:p \ne 0.5\end{align*}\]The evidence from the taste tests remains the same; he correctly identified \(13\) out of the \(16\) martinis which produces a \(p\)-value of \(P(X\le 3 \text{ or } X\ge 13)\) \(\approx0.0212.\) At the \(\alpha\) \(=0.05\) level, there is sufficient evidence to reject the null hypothesis that he is merely guessing. We could state the conclusion as two sided: he can distinguish (perhaps incorrectly) between shaken and stirred tastes; or, given that he correctly identified most of the drinks, we could state a stronger conclusion: he can correctly identify if a drink is shaken or stirred. We will argue that the latter conclusion is not the proper inference given how the alternative hypothesis was formulated.
The statistician Kaiser published a paper in \(1960\) arguing that we can make the claim that James Bond can correctly distinguish between shaken and stirred martinis simply by taste. Some textbooks argue this is permissible; others argue that it is not. This alternative hypothesis looks like \(H_1:\) \(p\) \(> 0.5\) rather than \(H_1:\) \(p\) \(\ne 0.5\) as originally formulated. The form of the alternative hypothesis changed to match the direction of the sample statistic. We will not adopt this practice. Since we are operating within the realm of formal hypothesis testing, we will maintain the form of the original alternative hypothesis and conclude that \(p\) \(\ne 0.5.\)
To understand our position, we further explain the process and purpose of hypothesis testing. Hypothesis testing is meant to test hypotheses formulated from previous observations, previous experimentation, or working theories. Once the hypotheses have been set, new experimentation is implemented, and the results are used as evidence in the hypothesis testing. The formulation of the hypotheses happens before the experimentation and is independent of the new data. That is not to say that the data from the new experiments cannot be studied to formulate further hypotheses or nuance the current hypotheses. This is to say that these new hypotheses are not to be tested using the same data precisely because the data led to them. That would be circular reasoning. Strictly speaking, new experiments need to be conducted to test the new hypotheses. This includes modifying the alternative hypothesis of a two-tailed test into a one-tailed alternative hypothesis.
This consideration is closely related to a problem plaguing much of modern academia, where publishing statistically significant results often takes precedence over honest intellectual inquiry, resulting in the temptation to conduct unplanned analyses of experimental data in search of a statistically significant result. Acting on this temptation is called \(p\)-hacking and is not an ethical research practice. When tests are done on patterns already seen in the data, the tests become meaningless. The proper path forward is to study the experimental data, formulate new hypotheses, and then test those hypotheses via new experimentation.
Let us consider the case of James Bond once more. Suppose that James Bond could tell the difference by taste but did not have the tastes correctly aligned with the methods. But, somehow, he still managed to get \(13\) of the \(16\) martinis labeled correctly. In this case, it is even less probable that he performed as observed than when we just assumed he was guessing. The \(p\)-value using both tails was already small enough to warrant rejection at the particular significance level; so, this is evidence that we would expect the parameter \(p\) to fall on the same side of the value in the null hypothesis \(0.5\) as the sample statistic. We agree that it is very tempting to conclude \(p > 0.5.\) Perhaps in the messiness of practical application, action may to be taken based on that conclusion, but in the rigors of hypothesis testing, another experiment is warranted to test the claim on untested data that is randomly sampled.
There is contention in the textbooks; perhaps, there is less across the field of active statisticians. Interested readers are encouraged to deepen their understanding of both sides of the argument. For an argument to the contrary of our position, see Kaiser, H. F. (1960) Directional statistical decisions. Psychological Review , 67 , 160-167.


1.1.6: End-of-Chapter Material
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Key takeaways for this Chapter
- Your health is your state of well-being.
- Wellness is the active process of achieving health through growth and change.
- Wellness is a broad concept and encompasses nine dimensions: Physical, Emotional, Social, Environmental, Financial, Intellectual, Occupational, Spiritual, and Cultural.
- A healthy population is a nations greatest resource, it means a productive population, a prosperous population, and a healthy economy.
- The top causes of death, cardiovascular disease and cancer, are highly impacted by your lifestyle choices.
- Increasing health literacy is an important skill to ensure people can find, understand, and use valid reliable information to make healthy decisions.
- Everyone is at a different readiness to change and can take steps to actively change unhealthy behaviors.
- Use SMART goal setting to change broad health goals into action oriented goals that you can achieve.
Media Attributions
- Stages-of-change © Toddatkins is licensed under a Public Domain license
- SMART-goals © Dungdm93 is licensed under a CC BY-SA (Attribution ShareAlike) license
References
- Smith, B. J., Tang, K. C., & Nutbeam, D. (2006). WHO Health Promotion Glossary: new terms. Health Promotion International , 21 (4), 340–345. https://doi.org/10.1093/heapro/dal033 , retrieved from https://www.who.int/healthpromotion/about/HPR Glossary_New Terms.pdf
- National Wellness Institute. (n.d.). Six Dimensions of Wellness, retrieved from https://nationalwellness.org/resourc...s-of-wellness/
- Heron, M. (2021). Deaths : leading causes for 2019. National Vital Statistics Reports , 70 (9). https://doi.org/10.15620/cdc:107021 , retrieved from https://www.cdc.gov/nchs/data/nvsr/n...r70-09-508.pdf
- National Center for Health Statistics. (n.d.). FastStats: Life Expectancy . Center for Disease Control and Prevention. https://www.cdc.gov/nchs/fastats/life-expectancy.htm
- Max Roser, Esteban Ortiz-Ospina and Hannah Ritchie (2013) - "Life Expectancy". Published online at OurWorldInData.org. Retrieved from: https://ourworldindata.org/life-expectancy
- Center for Disease Control and Prevention. (2011, May 20). Ten Great Public Health Achievements --- United States, 2001--2010 . https://www.cdc.gov/mmwr/preview/mmw...l/mm6019a5.htm
- U.S. Department of Health and Human Services. (n.d.). Health Literacy in Healthy People 2030 | health.gov . Health.Gov. https://health.gov/our-work/national...hy-people-2030
- Attribution CC-BY 4.0: Evaluating Information – Applying the CRAAP Test, Meriam Library, California State University, Chico, retrieved from https://library.csuchico.edu/sites/d...craap-test.pdf
- Raihan N, Cogburn M. Stages of Change Theory. [Updated 2022 Mar 9]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2022 Jan-. Available from: https://www.ncbi.nlm.nih.gov/books/NBK556005/
- Bailey R. R. (2017). Goal Setting and Action Planning for Health Behavior Change. American journal of lifestyle medicine, 13(6), 615–618. https://doi.org/10.1177/1559827617729634
- Bailey R. R. (2017). Goal Setting and Action Planning for Health Behavior Change. American journal of lifestyle medicine, 13(6), 615–618. https://doi.org/10.1177/1559827617729634
- Lorig, K., Laurent, D. D., Plant, K., Krishnan, E., & Ritter, P. L. (2013). The components of action planning and their associations with behavior and health outcomes. Chronic Illness , 10 (1), 50–59. https://doi.org/10.1177/1742395313495572
- van Cappellen, P., Rice, E. L., Catalino, L. I., & Fredrickson, B. L. (2017). Positive affective processes underlie positive health behaviour change. Psychology & Health, 33(1), 77–97. https://doi.org/10.1080/08870446.2017.1320798


5.4: Area and Triangle Trigonometry
The arc-length differential determines an area differential and the area of a region will also be an invariant of hyperbolic geometry. The area of a region will not change as it moves about the hyperbolic plane. We express the area formula in terms of polar coordinates.
Definition: Area of \(R\)
Suppose a region \(R\) in \(\mathbb{D}\) is described in polar coordinates. The area of \(R\) in \((\mathbb{D},{\cal H})\text{,}\) denoted \(A(R)\text{,}\) is given by
\[ A(R) = \iint_R \dfrac{4r}{(1-r^2)^2}dr d\theta\text{.} \]
The integral in this formula is difficult to evaluate directly in all but the simplest cases. Following is one such case.
Suppose our region is given by a circle whose hyperbolic radius is \(a\text{.}\) Since area is an invariant, we may as well assume the circle is centered at the origin. Let \(x\) be the point at which the circle intersects the positive real axis (so \(0 \lt x \lt 1\)), as pictured below. Then, by the distance formula
\[ a = \ln\left(\dfrac{1+x}{1-x}\right)\text{.} \]
Solving for \(x\text{,}\) we have
\[ x = \dfrac{e^a - 1}{e^a + 1}\text{.} \]
This circular region may be described in polar coordinates by \(0 \leq \theta \leq 2\pi\) and \(0 \leq r \leq x\text{.}\) The area of the region is then given by the following integral, which we compute with the \(u\)-substitution \(u = 1 - r^2\text{:}\)
\begin{align*} & \int_0^{2\pi} \int_0^x \dfrac{4r}{(1-r^2)^2}~drd\theta\\ & = \left(\int_0^{2\pi}~d\theta\right)\left(\int_1^{1-x^2}\dfrac{-2}{u^2}~du\right)\\ & = 2\pi\left[\dfrac{2}{u} \bigg|_1^{1-x^2}\right]\\ & = 2\pi \left[ \dfrac{2}{1-x^2} - 2\right]\\ & = 4\pi \dfrac{x^2}{1-x^2}\text{.} \end{align*}
Replace \(x\) in terms of \(a\) to obtain
\begin{align*} 4\pi \dfrac{(e^a - 1)^2}{(e^a+1)^2} \cdot \dfrac{(e^a+1)^2}{(e^a+1)^2 - (e^a- 1)^2} & = 4\pi \dfrac{(e^a - 1)^2}{4e^a}\\ & = 4\pi \left(\dfrac{e^a - 1}{2e^{a/2}}\right)^2\\ & = 4\pi \left(\dfrac{e^{a/2} - e^{-a/2}}{2} \right)^2\text{.} \end{align*}
This last expression can be rewritten using the hyperbolic sine function, evaluated at \(\dfrac{a}{2}\text{.}\) We investigate the hyperbolic sine and cosine functions in the exercises but note their definitions here.
Definition: Hyperbolic Sine Function and Hyperbolic Cosine Function
The hyperbolic sine function , denoted \(\sinh(x)\text{,}\) and the hyperbolic cosine function , denoted \(\cosh(x)\text{,}\) are functions of real numbers defined by
\[ \sinh(x) = \dfrac{e^x- e^{-x}}{2} \text{ and } \cosh(x)= \dfrac{e^x + e^{-x}}{2}\text{.} \]
The area derivation in Example \(5.4.1\) may then be summarized as follows.
Theorem \(5.4.1\)
The area of a hyperbolic circle with hyperbolic radius \(r\) is \(4\pi \sinh^2(\dfrac{r}{2})\text{.}\)
Other regions are not as simple to describe in polar coordinates. An important area for us will be the area of a \(\dfrac{2}{3}\)-ideal triangle, the figure that results if two of the three vertices of a hyperbolic triangle are moved to ideal points. See Figure \(5.4.1\).
Theorem \(5.4.2\)
The area of a \(\dfrac{2}{3}\)-ideal triangle having interior angle \(\alpha\) is equal to \(\pi - \alpha\text{.}\)
The proof of this theorem is given in the following section. The proof there makes use of a different model for hyperbolic geometry, the so-called upper half-plane model.
An ideal triangle consists of three ideal points and the three hyperbolic lines connecting them. It turns out that all ideal triangles are congruent (a fact proved in the exercises); the set of all ideal triangles is minimally invariant in \((\mathbb{D}, {\cal H})\text{.}\)
Any ideal triangle has area equal to \(\pi\text{.}\)
- Proof
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Since all ideal triangles are congruent, assume our triangle \(\Delta\) is the ideal triangle shown in Figure \(5.4.2\).
But then \(\Delta\) can be partitioned into two \(\dfrac{2}{3}\)-ideal triangles by drawing the vertical hyperbolic line from \(0\) along the imaginary axis to ideal point \(i\text{.}\) Each \(\dfrac{2}{3}\)-ideal triangle has interior angle \(\dfrac{\pi}{2}\text{,}\) so \(\Delta\) has area \(\dfrac{\pi}{2} + \dfrac{\pi}{2} = \pi\text{.}\)
It is a remarkable fact that \(\pi\) is an upper bound for the area of any triangle in \((\mathbb{D},{\cal H})\text{.}\) No triangle in \((\mathbb{D},{\cal H})\) can have area as large as \(\pi\text{,}\) even though side lengths can be arbitrarily large!
The area of a hyperbolic triangle in \((\mathbb{D},{\cal H})\) having interior angles \(\alpha, \beta\text{,}\) and \(\gamma\) is
\[ A = \pi - (\alpha + \beta + \gamma)\text{.} \]
- Proof
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Consider Figure \(5.4.3\) containing triangle \(\Delta pqr\text{.}\) We have extended segment \(qp\) to the ideal point \(t\text{,}\) \(u\) is an ideal point of line \(rq\text{,}\) and \(v\) is an ideal point of line \(pr\text{.}\) The area of the ideal triangle \(\Delta tuv\) is \(\pi\text{.}\) Notice that regions \(R_1\text{,}\) \(R_2\text{,}\) and \(R_3\) are all \(\dfrac{2}{3}\)-ideal triangles contained within the ideal triangle. Consider \(R_1\text{,}\) whose ideal points are \(u\) and \(t\text{,}\) and whose interior angle is \(\angle uqt\text{.}\) Since the line through \(q\) and \(r\) has ideal point \(u\text{,}\) the interior angle of \(R_1\) is \(\angle uqt=\pi-\beta\text{.}\) Similarly, \(R_2\) has interior angle \(\pi - \alpha\) and \(R_3\) has interior angle \(\pi - \gamma\text{.}\)
Let \(R\) denote the triangle region \(\Delta pqr\) whose area \(A(R)\) we want to compute. We then have the following relationships among areas:
\begin{align*} \pi & = A(R) + A(R_1) + A(R_2) + A(R_3)\\ & = A(R) + [\pi - (\pi -\alpha)] + [\pi - (\pi -\beta)] + [\pi - (\pi -\gamma)]\text{.} \end{align*}
Solving for \(A(R)\text{,}\)
\[ A(R) = \pi - (\alpha + \beta + \gamma)\text{,} \]
and this completes the proof.
In Euclidean geometry, trigonometric formulas relate the angles of a triangle to its side lengths. There are hyperbolic trigonometric formulas as well.
Suppose a hyperbolic triangle in \(\mathbb{D}\) has angles \(\alpha\text{,}\) \(\beta\text{,}\) and \(\gamma\) and opposite hyperbolic side lengths \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) as pictured in Figure \(5.4.4\). Then the following laws hold.
- First hyperbolic law of cosines.
\[ \cosh(c) = \cosh(a)\cosh(b)-\sinh(a)\sinh(b)\cos(\gamma). \]
- Second hyperbolic law of cosines.
\[ \cosh(c) = \dfrac{\cos(\alpha)\cos(\beta)+\cos(\gamma)}{\sin(\alpha)\sin(\beta)}\text{.} \]
- Hyperbolic law of sines.
\[ \dfrac{\sinh(a)}{\sin(\alpha)} = \dfrac{\sinh(b)}{\sin(\beta)}=\dfrac{\sinh(c)}{\sin(\gamma)}\text{.} \]
The proofs of these laws are left as exercises for the interested reader. The following result follows from the first hyperbolic law of cosines.
In a right hyperbolic triangle with hyperbolic side lengths \(a\) and \(b\text{,}\) and hypotenuse \(c\text{,}\)
\[ \cosh(c) = \cosh(a)\cosh(b)\text{.} \]
The second hyperbolic law of cosines also leads to an interesting result. In the hyperbolic plane, if we find ourselves at point \(z\text{,}\) we may infer our distance \(c\) to a point \(w\) by estimating a certain angle, called the angle of parallelism of \(z\) to the line \(L\) through \(w\) that is perpendicular to segment \(zw\text{.}\) The following is a picture of this scene:
In this setting, \(\Delta zwu\) is a \(\dfrac{1}{3}\)-ideal triangle, and the second hyperbolic law of cosines applies with \(\gamma = 0\) and \(\beta = \dfrac{\pi}{2}\) to yield the following result.
Suppose \(z\) and \(w\) are points in \(\mathbb{D}\text{,}\) and \(L\) is a hyperbolic line through \(w\) that is perpendicular to hyperbolic segment \(zw\text{.}\) Suppose further that \(u\) is an ideal point of \(L\text{.}\) Let \(\alpha = \angle wzu\) and \(c = d_H(z,w)\text{.}\) Then
\[ \cosh(c) = \dfrac{1}{\sin(\alpha)}\text{.} \]
The angle of parallelism is pursued further in Section 7.4.
Suppose a two-dimensional ship is plopped down in \(\mathbb{D}\text{.}\) What would the pilot see? How would the ship move? How would the pilot describe the world? Are all points equivalent in this world? Could the pilot figure out whether the universe adheres to hyperbolic geometry as opposed to, say, Euclidean geometry?
Recall what we know about hyperbolic geometry. First of all, any two points in the hyperbolic plane are congruent, so the geometry is homogeneous. The pilot could not distinguish between any two points, geometrically.
Second, the shortest path between two points is the hyperbolic line between them, so light would travel along these hyperbolic lines, assuming light follows geodesics. The pilot's line of sight would follow along these lines, and the ship would move along these lines to fly as quickly as possible from \(p\) to \(q\text{,}\) assuming no pesky asteroid fields block the path. To observe a galaxy at point \(q\) from the point \(p\) (as in the diagram below), the pilot would point a telescope in the direction of line \(L\text{,}\) the line along which the light from the galaxy travels to reach the telescope.
With a well-defined metric, we can say more. The pilot will view the hyperbolic plane as infinite and without boundary. In theory, the pilot can make an orbit of arbitrary radius about an asteroid located anywhere in the space.
To test for hyperbolic geometry, perhaps the pilot can turn to triangles. The angles of a triangle in the hyperbolic plane sum to less than \(180^{\circ}\text{,}\) but only noticeably so for large enough triangles. In our disk model of hyperbolic geometry, we can easily observe this angle deficiency. In the figure below, triangle \(\Delta zuw\) has angle sum about \(130^{\circ}\text{,}\) and \(\Delta pqr\) has angle sum of about \(22^{\circ}\text{!}\) Whether an intrepid \(2\)-D explorer could map out such a large triangle depends on how much ground she could cover relative to the size of her universe. We will have more to say about such things in Chapters 7 and 8.
Simply put, hyperbolic squares don't exist. In fact, no four-sided figures with four right angles exist, if we assume the sides are hyperbolic segments . If such a figure existed, its angle sum would be \(2\)\(\pi\text{.}\) But such a figure could be divided along a diagonal into two triangles whose total angle sum must then be \(2\pi\) as well. This means that one of the triangles would have angle sum at least \(\pi\text{,}\) which cannot happen.
On the other hand, there is no physical obstruction to a two-dimensional explorer making the following journey in the hyperbolic plane: Starting at a point such as \(p\) in Figure \(5.4.5\), head along a line in a certain direction for \(a\) units, turn right (\(90^{\circ}\)) and proceed in a line for \(a\) more units, then turn right again and proceed in a line \(a\) units, and then turn right one more time and proceed in a line for \(a\) units. Let \(q\) denote the point at which the explorer arrives at the end of this journey. In the Euclidean plane, \(q\) will equal \(p\text{,}\) because the journey traces out a square built from line segments. However, this is not the case in \((\mathbb{D},{\cal H})\) (though if we connect \(p\) and \(q\) with a hyperbolic line we obtain a geodesic pentagon with (at least) three right angles!). In the exercises, we investigate the distance between \(p\) and \(q\) as a function of the length \(a\text{.}\)
However, we can build a four-sided figure closely resembling a rectangle, if we drop the requirement that the legs be hyperbolic line segments.
Through any point \(0 \lt a \lt 1\) on the positive real axis, we may construct a hyperbolic line \(L_1\) through \(a\) that is perpendicular to the real axis. Also construct a hyperbolic line \(L_2\) through \(-a\) that is perpendicular to the real axis. Now pick a point \(z\) on \(L_1\) and construct the cline arc \(C_1\) through \(z, 1\) and \(-1\text{.}\) Also construct the cline arc \(C_2\) through \(\overline{z}, 1\) and \(-1\text{.}\) This creates a four-sided figure, which we call a block. We claim that each angle in the figure is right, and that opposite sides have equal length. Moreover, \(z\) and \(a\) can be chosen so that all four sides have the same length. This figure isn't a rectangle, however, in the sense that not all four sides are hyperbolic segments. The arcs \(C_1\) and \(C_2\) are not hyperbolic lines.
While squares don't exist in the hyperbolic plane, we may build right-angled regular polygons with more than four sides using hyperbolic line segments. In fact, for each triple of positive real numbers \((a,b,c)\) we may build a right-angled hexagon in the hyperbolic plane with alternate side lengths \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) We encourage the reader to work carefully through the construction of this hexagon in the proof of Theorem \(5.4.6\). We use all our hyperbolic constructions to get there.
For any triple \((a,b,c)\) of positive real numbers there exists a right-angled hexagon in \((\mathbb{D},{\cal H})\) with alternate side lengths \(a, b,\) and \(c\text{.}\) Moreover, all right-angled hexagons with alternate side lengths \(a, b,\) and \(c\) are congruent.
- Proof
-
We prove the existence of a right-angled hexagon with vertices \(v_0, v_1, \ldots, v_5\) such that \(d_H(v_0,v_1) = a, d_H(v_2,v_3) = b\text{,}\) and \(d_H(v_4,v_5) = c\text{.}\)
First, let \(v_0\) be the origin in the hyperbolic plane, and place \(v_1\) on the positive real axis so that \(d_H(v_0,v_1)= a\text{.}\) Note that
\[ v_1 = \dfrac{e^a - 1}{e^a+1}\text{.} \]
Next, construct the hyperbolic line \(A\) perpendicular to the real axis at the point \(v_1\text{.}\) This line is part of the cline that has diameter \(v_1v_1^*\text{.}\)
Pick any point \(v_2\) on the line \(A\text{.}\) For the sake of argument, assume that \(v_2\) lies above the real axis, as in Figure 5.4.20.
Next, construct the hyperbolic line \(B\) perpendicular to \(A\) at the point \(v_2\text{.}\) This line is part of the cline through \(v_2\) and \(v_2^*\) with center on the line tangent to \(A\) at \(v_2\text{.}\)
Next, construct the point \(v_3\) on line \(B\) that is a distance \(b\) away from \(v_2\text{.}\) This point is found by intersecting \(B\) with the hyperbolic circle centered at \(v_2\) with radius \(b\text{.}\) (To construct this circle, we first find the scalar \(k\) so that the hyperbolic distance between \(kv_2\) and \(v_2\) is \(b\text{.}\))
Next, draw the perpendicular \(C\) to line \(B\) at \(v_3\text{.}\)
Then construct the common perpendicular of \(C\) and the imaginary axis, call this perpendicular \(D\text{.}\) We construct this common perpendicular as follows. First find the two points \(p\) and \(q\) symmetric to both \(C\) and the imaginary axis (see Theorem \(3.2.7\)). The points \(p\) and \(q\) will live on the circle at infinity. The common perpendicular of \(C\) and the imaginary axis will be the cline through \(p\) and \(q\) that is also a hyperbolic line (i.e., orthogonal to the circle at infinity).
If \(C\) and the imaginary axis intersect, no such perpendicular exists (think triangle angles), so drag \(v_2\) away from \(v_1\) until these lines do not intersect. Then construct \(D\) as in the preceeding paragraph. Let \(v_4\) and \(v_5\) be the points of intersection of \(D\) with \(C\) and the imaginary axis, respectively.
This construction gives us a right angled hexagon such that \(d_H(v_0,v_1) = a\) and \(d_H(v_2,v_3) = b\text{.}\) We also want \(d_H(v_4,v_5) = c\text{.}\) Notice that this last distance is a function of the position of vertex \(v_2\) on line A. As \(v_2\) goes along \(A\) to \(v_1\) there is a point beyond which \(D\) no longer exists, and as \(v_2\) goes along \(A\) to the circle at infinity, the length of segment \(v_4v_5\) takes on all positive real values. So, by the intermediate value theorem, there is some point at which the segment has length \(c\text{.}\) Finally, all right-angled hexagons with alternate side lengths \(a\text{,}\) \(b\text{,}\) and \(c\) are congruent to the one just constructed because angles, hyperbolic lines, and distances are preserved under transformations in \({\cal H}\text{.}\)
We show that if one inscribes a circle in any ideal triangle, its points of tangency form an equilateral triangle with side lengths equal to \(2\ln(\varphi)\) where \(\varphi\) is the golden ratio \(\dfrac{(1+\sqrt{5})}{2}\text{.}\)
Since all ideal triangles are congruent, we choose one that is convenient to work with. Consider the ideal triangle with ideal points \(-1\), \(1\), and \(i\text{.}\)
The hyperbolic line \(L_1\) joining \(-1\) and \(i\) is part of the circle \(C_1\) with radius \(1\) centered at \(-1+i\text{.}\) The hyperbolic line \(L_2\) joining \(i\) and \(1\) is part of the circle \(C_2\) with radius \(1\) centered at \(1+i\text{.}\) Let \(C\) denote the circle with radius \(2\) centered at \(-1+2i\text{,}\) as pictured below.
Inversion about \(C\) gives a hyperbolic reflection of \(\mathbb{D}\) that maps \(L_1\) onto the real axis. Indeed, the circle \(C_1\text{,}\) since it passes through the center of \(C\text{,}\) gets mapped to a line - the real axis, in fact. Moreover, hyperbolic reflection across the imaginary axis maps \(L_1\) onto \(L_2\text{.}\) Let \(c\) be the point of intersection of these two hyperbolic lines of reflection, as pictured. The hyperbolic circle with hyperbolic center \(c\) that passes through the origin will be tangent to the real axis, and will thus inscribe the ideal triangle. Let the points of tangency on \(L_1\) and \(L_2\) be \(p\) and \(q\text{,}\) respectively.
The point \(q\) can be found analytically as the point of intersection of circles \(C\) and \(C_2\text{.}\) Working it out, one finds \(q = \dfrac{1}{5}+\dfrac{2}{5}i\text{.}\) Thus,
\begin{align*} d_H(0,q)& =\ln\bigg(\dfrac{1+|q|}{1-|q|}\bigg)\\ & =\ln\bigg(\dfrac{1+\dfrac{1}{\sqrt{5}}}{1-\dfrac{1}{\sqrt{5}}}\bigg)\\ & =\ln\bigg(\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\bigg)\\ & =\ln\bigg(\dfrac{(\sqrt{5}+1)^2}{(\sqrt{5}-1)(\sqrt{5}+1)}\bigg)\\ & =\ln\bigg(\dfrac{(\sqrt{5}+1)^2}{4}\bigg)\\ & =\ln(\varphi^2)\\ & =2\ln(\varphi)\text{.} \end{align*}
Exercises
Properties of \(\sinh(x) = (e^x-e^{-x})/2\) and \(\cosh(x)=(e^x+e^{-x})/2\).
- Verify that \(\sinh(0) = 0\) and \(\cosh(0) = 1\text{.}\)
- Verify that \(\dfrac{d}{dx}[\sinh(x)]=\cosh(x)\) and \(\dfrac{d}{dx}[\cosh(x)]=\sinh(x)\text{.}\)
- Verify that \(\cosh^2(x)-\sinh^2(x)=1\text{.}\)
- Verify that the power series expansions for \(\cosh(x)\) and \(\sinh(x)\) are \begin{align*} \cosh(x)& =1+\dfrac{x^2}{2}+\dfrac{x^4}{4!}+\cdots\\ \sinh(x)& =x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\cdots\text{.} \end{align*}
Prove that the circumference of a hyperbolic circle having hyperbolic radius \(r\) is \(C = 2\pi \sinh(r)\text{.}\)
The hyperbolic plane looks Euclidean on small scales.
-
Prove
\[ \lim_{r \to 0^+} \dfrac{4\pi\sinh^2(\dfrac{r}{2})}{\pi r^2} = 1 \text{.} \]
Thus, for small \(r\text{,}\) the Euclidean formula for the area of a circle is a good approximation to the true area of a circle in the hyperbolic plane. -
Prove
\[ \lim_{r \to 0^+}\dfrac{2\pi \sinh(r)}{2\pi r}=1\text{.} \]
Thus, for small \(r\text{,}\) the Euclidean formula for the circumference of a circle is a good approximation to the true circumference of a circle in the hyperbolic plane.
Prove that all ideal triangles are congruent in hyperbolic geometry.
- Hint
-
Prove any ideal triangle is congruent to the one whose ideal points are \(1\), \(i\text{,}\) and \(-1\) (see Figure \(5.4.2\)).
An intrepid tax collector lives in a country in the hyperbolic plane. For collection purposes, the country is divided into triangular grids. The collector is responsible for collection in a triangle having angles \(12^\circ\text{,}\) \(32^\circ\text{,}\) and \(17^\circ\text{.}\) What is the area of the collector's triangle? Can the entire space \(\mathbb{D}\) be subdivided into a finite number of triangles?
Consider the hyperbolic triangle with vertices at \(0\), \(\dfrac{1}{2}\text{,}\) and \(\dfrac{1}{2} + \dfrac{1}{2}i\text{.}\) Calculate the area of this triangle by determining the angle at each vertex.
- Hint
-
To determine the angle at a vertex it may be convenient to move it to the origin via an appropriate transformation in \(\cal H\text{.}\)
Recall the block constructed in Example \(5.4.3\). Prove that all four angles are \(90^\circ\text{,}\) and that for any choice of \(z\text{,}\) opposite sides have equal hyperbolic length.
Find a formula for the area of an \(n\)-gon, comprised of \(n\) hyperbolic line segments in terms of its \(n\) interior angles \(\alpha_1, \alpha_2, \cdots, \alpha_n\text{.}\)
- Hint
-
Decompose the \(n\)-gon into triangles.
Building a hyperbolic octagon with interior angles \(45^\circ\) . Let \(r\) be a real number such that \(0 \lt r \lt 1\text{.}\) The eight points \(v_k = re^{i\frac{\pi}{4}k}\) for \(k = 0,1,\cdots, 7\) determine a regular octagon in the hyperbolic plane, as shown in Figure \(5.4.8\). Note that \(v_0\) is the real number \(r\text{.}\) The interior angle of each corner is a function of \(r\text{.}\) We find the value of \(r\) for which the interior angle is \(45^\circ\text{.}\)
-
Prove that the center of the circle containing the hyperbolic line through \(v_0\) and \(v_1\) is
\[ z_0 = \dfrac{1+r^2}{2r}+\dfrac{1+r^2}{2r}\tan(\dfrac{\pi}{8})i \text{.} \]
- Let \(b = \dfrac{1+r^2}{2r}\) be the midpoint of segment \(v_0v_0^*\) and show that \(\angle v_0 z_0 b = \dfrac{\pi}{8}\) precisely when the interior angles of the octagon equal \(\dfrac{\pi}{4}\text{.}\)
-
Using \(\Delta v_0 b z_0\) and part (b), show that the interior angles of the octagon equal \(\dfrac{\pi}{4}\) precisely when
\[ \tan(\dfrac{\pi}{8}) = \dfrac{|b-v_0|}{|z_0 - b|} = \dfrac{\dfrac{1+r^2}{2r}-r}{\dfrac{1+r^2}{2r}\cdot \tan(\dfrac{\pi}{8})}\text{.} \]
- Solve the equation in (c) for \(r\) to obtain \(r = (\dfrac{1}{2})^{(\frac{1}{4})}\text{.}\)
- Hint
-
For part (a), note that the center \(z_0\) is on the perpendicular bisector of the Euclidean segment \(v_0v_0^*\) as well as the perpendicular bisector of the Euclidean segment \(v_0 v_1\text{.}\)
Suppose we construct a regular \(n\)-gon in the hyperbolic plane from the corner points \(r, re^{\frac{1}{n}2\pi i},\) \(re^{\frac{2}{n}2\pi i}\text{,}\) \(\cdots, re^{\frac{n-1}{n}2\pi i}\) where \(0 \lt r \lt 1\text{.}\) Calculate the hyperbolic length of any of its sides.
Prove the first hyperbolic law of cosines by completing the following steps.
-
Show that for any positive real numbers \(x\) and \(y\text{,}\)
\[ \cosh(\ln(\dfrac{x}{y})) = \dfrac{x^2+y^2}{2xy}~\text{ and }~\sinh(\ln(\dfrac{x}{y})) = \dfrac{x^2-y^2}{2xy}\text{.} \]
-
Given two points \(p\) and \(q\) in \(\mathbb{D}\text{,}\) let \(c = d_H(p,q)\text{.}\) Use the hyperbolic distance formula from Theorem \(5.3.1\) and part (a) to show
\[ \cosh(c) = \dfrac{(1+|p|^2)(1+|q|^2)-4\text{Re}(p\overline{q})}{(1-|p|^2)(1-|q|^2)}\text{.} \]
-
Now suppose our triangle has one vertex at the origin, and one point on the positive real axis. In particular, suppose \(p = r\) (\(0\lt r\lt 1\)) and \(q = ke^{i\gamma}\) (\(0\lt k\lt 1\)), with angles \(\alpha, \beta, \gamma\) and hyperbolic side lengths \(a,b,\) and \(c\) as in Figure \(5.4.9\).
Show
\[ \cosh(a) = \dfrac{1+r^2}{1-r^2}~~;~~\sinh(a) = \dfrac{2r}{1-r^2}\text{;} \]
\[ \cosh(b) = \dfrac{1+k^2}{1-k^2}~~;~~\sinh(b) = \dfrac{2k}{1-k^2}\text{.} \]
-
Show that for the triangle in part (c),
\[ \cosh(c) = \cosh(a)\cosh(b)-\sinh(a)\sinh(b)\cos(\gamma)\text{.} \]
- Explain why this formula works for any triangle in \(\mathbb{D}\text{.}\)
In this exercise we prove the hyperbolic law of sines. We assume our triangle is as in Figure \(5.4.10\). Thus, \(q = ke^{i\gamma}\) for some \(0 \lt k \lt 1\) and \(p = r\) for some real number \(0 \lt r \lt 1\text{.}\) Suppose further that the circle containing side \(c\) has center \(z_0\) and Euclidean radius \(R\text{,}\) shown in the figure, and that \(m_q\) is the midpoint of segment \(qq^*\) and \(m_p\) is the midpoint of segment \(pp^*\text{,}\) so that \(\Delta z_0m_qq\) and \(\Delta pm_pz_0\) are right triangles.
- Verify that the triangle angles \(\alpha\) and \(\beta\) correspond to angles \(\angle m_qz_0q\) and \(\angle pz_0m_p\text{,}\) respectively.
- Notice that \(\sin(\alpha) = \dfrac{|m_q-q|}{R}\) and \(\sin(\beta) = \dfrac{|m_p-p|}{R}.\) Verify that \(|m_q-q| = \dfrac{\frac{1}{k} - k}{2} = \dfrac{1 - k^2}{2k}\) and that \(|m_p-p| = \dfrac{1-r^2}{2r}\text{.}\)
-
Verify that
\[ \dfrac{\sinh(a)}{\sin(\alpha)} = \dfrac{\sinh(b)}{\sin(\beta)}\text{.} \]
-
Explain why we may conclude that for any hyperbolic triangle in \(\mathbb{D}\text{,}\)
\[ \dfrac{\sinh(a)}{\sin(\alpha)} = \dfrac{\sinh(b)}{\sin(\beta)}=\dfrac{\sinh(c)}{\sin(\gamma)}\text{.} \]
Prove the second law of hyperbolic cosines.
- Answer
-
This result follows from repeated applications of the first law and judicial use of the two identities \(\cos^2(x) + \sin^2(x) = 1\) and \(\cosh^2(x) - \sinh^2(x) = 1\text{.}\)
Recall the journey in Example \(5.4.3\) in which a bug travels a path that would trace a square in the Euclidean plane. For convenience, we assume the starting point \(p\) is such that the first right turn of \(90^{\circ}\) occurs at the point \(x\) on the positive real axis and the second turn occurs at the origin. (This means that \(x = \dfrac{(e^a-1)}{(e^a+1)}\text{.}\)) The third corner must then occur at \(xi\text{.}\) We have reproduced the journey with some more detail in Figure \(5.4.11\). In this exercise, we make use of hyperbolic triangle trig to measure some features of this journey from \(p\) to \(q\) in terms of the length \(a\) of each leg.
- Determine the hyperbolic distance between \(p\) and the origin (corner two of the journey). In particular, show that \(\cosh(d_H(0,p)) = \cosh^2(a)\text{.}\) Note that if this journey had been done in the Euclidean plane, the corresponding distance would be \(\sqrt{2}a\text{.}\) How do these distances compare for small positive values of \(a\text{?}\) Is \(\cosh^2(a)\) close to \(\cosh(\sqrt{2}a)\text{?}\)
- Let \(\theta = \angle x0p\text{.}\) Show that \(\tan(\theta) = \dfrac{1}{\cosh(a)}\text{.}\) What is the corresponding angle if this journey is done in the Euclidean plane? What does \(\theta\) approach as \(a \to 0^+\text{?}\)
- Show that \(\angle x0p = \angle 0px\text{.}\)
- Show that \(\cosh(d_H(p,q)) = \cosh^4(a)[1-\sin(2\theta)]+\sin(2\theta)\text{.}\)
-
Let \(\alpha = \angle q p 0\text{,}\) \(b = d_H(0,p)\) and \(c = d_H(p,q)\text{.}\) Show that
\[ \sin(\alpha) = \dfrac{\sinh(b)}{\sinh(c)}\cos(2\theta). \]
- Determine the area of the pentagon enclosed by the journey if, after reaching \(q\) we return to \(p\) along the geodesic. In particular, show that the area of this pentagon equals \(\dfrac{3\pi}{2}-2(\theta+\alpha)\text{.}\) What is the corresponding area if the journey had been done in the Euclidean plane?
- Would any of these measurements change if we began at a different point in the hyperbolic plane and/or headed off in a different direction initially than the ones in Figure \(5.4.11\)?


9.6: The Hybrid Orbital Model
- Explain why the sharing of atomic orbitals (as implied in the Lewis model) cannot adequately account for the observed bonding patterns in simple molecules.
- Sketch out a diagram illustrating how the plots of atomic s - and p - orbital wave functions give rise to a pair of hybrid orbitals.
- Draw "orbital box" diagrams showing how combinations of an atomic s orbital and various numbers of p orbitals create sp , sp 2 , and sp 3 hybrid orbitals.
- Show how hybrid orbitals are involved in the molecules methane, water, and ammonia.
As useful and appealing as the concept of the shared-electron pair bond is, it raises a somewhat troubling question that we must sooner or later face: what is the nature of the orbitals in which the shared electrons are contained? Up until now, we have been tacitly assuming that each valence electron occupies the same kind of atomic orbital as it did in the isolated atom. As we shall see below, his assumption very quickly leads us into difficulties.
Atomic orbitals alone do not work for Molecules
Consider how we might explain the bonding in a compound of divalent beryllium, such as beryllium hydride, BeH 2 . The beryllium atom, with only four electrons, has a configuration of 1 s 2 2 s 2 . Note that the two electrons in the 2 s orbital have opposite spins and constitute a stable pair that has no tendency to interact with unpaired electrons on other atoms.
The only way that we can obtain two unpaired electrons for bonding in beryllium is to promote one of the 2 s electrons to the 2 p level. However, the energy required to produce this excited-state atom would be sufficiently great to discourage bond formation. It is observed that Be does form reasonably stable bonds with other atoms. Moreover, the two bonds in BeH 2 and similar molecules are completely equivalent; this would not be the case if the electrons in the two bonds shared Be orbitals of different types, as in the "excited state" diagram above.
These facts suggest that it is incorrect to assume that the distribution of valence electrons that are shared with other atoms can be described by atomic-type s, p , and d orbitals at all.
Remember that these different orbitals arise in the first place from the interaction of the electron with the single central electrostatic force field associated with the positive nucleus. An outer-shell electron in a bonded atom will be under the influence of a force field emanating from two positive nuclei, so we would expect the orbitals in the bonded atoms to have a somewhat different character from those in free atoms. In fact, as far as valence electrons are concerned, we can throw out the concept of atomic orbital altogether and reassign the electrons to a new set of molecular orbitals that are characteristic of each molecular configuration. This approach is indeed valid, but we will defer a discussion of it until a later unit.
For now, we will look at a less-radical model that starts out with the familiar valence-shell atomic orbitals, and allows them to combine to form hybrid orbitals whose shapes conform quite well to the bonding geometry that we observe in a wide variety of molecules.
What are hybrid orbitals?
First, recall that the electron, being a quantum particle, cannot have a distinct location; the most we can do is define the region of space around the nucleus in which the probability of finding the electron exceeds some arbitrary value, such as 90% or 99%. This region of space is the orbital . Because of the wavelike character of matter, the orbital corresponds to a standing wave pattern in 3-dimensional space which we can often represent more clearly in 2-dimensional cross section. The quantity that is varying (“waving”) is a number denoted by ψ ( psi ) whose value varies from point to point according to the wave function for that particular orbital.
Orbitals of all types are simply mathematical functions that describe particular standing-wave patterns that can be plotted on a graph but have no physical reality of their own. Because of their wavelike nature, two or more orbitals (i.e., two or more functions ψ) can be combined both in-phase and out-of-phase to yield a pair of resultant orbitals which, to be useful, must have squares that describe actual electron distributions in the atom or molecule.
The s,p,d and f orbitals that you are familiar with are the most convenient ones for describing the electron distribution in isolated atoms because assignment of electrons to them according to the usual rules always yields an overall function Ψ 2 that predicts a spherically symmetric electron distribution, consistent with all physical evidence that atoms are in fact spherical. For atoms having more than one electron, however, the s,p,d , f basis set is only one of many possible ways of arriving at the same observed electron distribution. We use it not because it is unique, but because it is the simplest.
In the case of a molecule such as BeH 2 , we know from experimental evidence that the molecule is linear and therefore the electron density surrounding the central atom is no longer spherical, but must be concentrated along two directions 180° apart, and we need to construct a function Ψ 2 having these geometrical properties. There are any number of ways of doing this, but it is convenient is to use a particular set of functions ψ (which we call hybrid orbitals ) that are constructed by combining the atomic s,p,d, and f functions that are already familiar to us.
You should understand that hybridization is not a physical phenomenon ; it is merely a mathematical operation that combines the atomic orbitals we are familiar with in such a way that the new (hybrid) orbitals possess the geometric and other properties that are reasonably consistent with what we observe in a wide range (but certainly not in all) molecules. In other words, hybrid orbitals are abstractions that describe reality fairly well in certain classes of molecules (and fortunately, in much of the very large class of organic substances) and are therefore a useful means of organizing a large body of chemical knowledge... but they are far from infallible.
Hybridization is not a physical phenomenon ; it is merely a mathematical operation that combines the atomic orbitals we are familiar with in such a way that the new (hybrid) orbitals possess the geometric and other properties that are reasonably consistent with what we observe in a wide range (but certainly not in all) molecules.
This approach, which assumes that the orbitals remain more or less localized on one central atom, is the basis of the theory which was developed in the early 1930s, mainly by Linus Pauling .
Linus Pauling (1901-1994) was the most famous American chemist of the 20th century and the author of the classic book The Nature of the Chemical Bond . His early work pioneered the application of X-ray diffraction to determine the structure of complex molecules; he then went on to apply quantum theory to explain these observations and predict the bonding patterns and energies of new molecules. Pauling, who spent most of his career at Cal Tech, won the Nobel Prize for Chemistry in 1954 and the Peace Prize in 1962.
"In December 1930 Pauling had his famous 'breakthrough' where, in a rush of inspiration, he ' stayed up all night, making, writing out, solving the equations, which were so simple that I could solve them in a few minutes '. This flurry of calculations would eventually become the first of Pauling's germinal series of papers on the nature of the chemical bond. ' I just kept getting more and more euphorious as time went by ', Pauling would recall. "
Although the hybrid orbital approach has proven very powerful (especially in organic chemistry), it does have its limitations. For example, it predicts that both H 2 O and H 2 S will be tetrahedrally coordinated bent molecules with bond angles slightly smaller than the tetrahedral angle of 109.5° owing to greater repulsion by the nonbonding pair. This description fits water (104.5°) quite well, but the bond angle in hydrogen sulfide is only 92°, suggesting that atomic p orbitals (which are 90° apart) provide a better description of the electron distribution about the sulfur atom than do sp 3 hybrid orbitals.
The hybrid orbital model is fairly simple to apply and understand, but it is best regarded as one special way of looking at a molecule that can often be misleading. Another viewpoint, called the molecular orbital theory , offers us a complementary perspective that it is important to have if we wish to develop a really thorough understanding of chemical bonding in a wider range of molecules.
Constructing hybrid orbitals
Below: "Constructive" and "destructive" combinations of 2 p and 2 s wave functions (line plots) give rise to the sp hybrid function shown at the right. The solid figures depict the corresponding probability functions ψ 2 .
Hybrid orbitals are constructed by combining the ψ functions for atomic orbitals. Because wave patterns can combine both constructively and destructively, a pair of atomic wave functions such as the s - and p - orbitals shown at the left can combine in two ways, yielding the sp hybrids shown.
From an energy standpoint, we can represent the transition from atomic s - and p -orbitals to an sp hybrid orbital in this way:
Notice here that 1) the total number of occupied orbitals is conserved, and 2) the two sp hybrid orbitals are intermediate in energy between their parent atomic orbitals. In terms of plots of the actual orbital functions ψ we can represent the process as follows:
The probability of finding the electron at any location is given not by ψ , but by ψ 2 , whose form is roughly conveyed by the solid figures in this illustration.
Hybrids derived from atomic s - and p orbitals
Digonal bonding: sp-hybrid orbitals
Returning to the example of BeH 2 , we can compare the valence orbitals in the free atoms with those in the beryllium hydride molecule as shown here. It is, of course, the overlap between the hydrogen-1 s orbitals and the two lobes of the beryllium sp -hybrid orbitals that constitutes the two Be—H "bonds" in this molecule. Notice that whereas a single p -orbital has lobes on both sides of the atom, a single sp -hybrid has most of its electron density on one side, with a minor and more spherical lobe on the other side. This minor lobe is centered on the central atom (some textbook illustrations don't get this right.)
As far as the shape of the molecule is concerned, the result is exactly the same as predicted by the VSEPR model (although hybrid orbital theory predicts the same result in a more fundamental way.) We can expect any central atom that uses sp -hybridization in bonding to exhibit linear geometry when incorporated into a molecule.
Trigonal (sp 2 ) hybridization
We can now go on to apply the same ideas to some other simple molecules. In boron trifluoride , for example, we start with the boron atom, which has three outer-shell electrons in its normal or ground state, and three fluorine atoms, each with seven outer electrons. As is shown in this configuration diagram, one of the three boron electrons is unpaired in the ground state. In order to explain the trivalent bonding of boron, we postulate that the atomic s - and p - orbitals in the outer shell of boron mix to form three equivalent hybrid orbitals. These particular orbitals are called sp 2 hybrids , meaning that this set of orbitals is derived from one s- orbital and two p-orbitals of the free atom.
This illustration shows how an s -orbital mixes with two p orbitals to form a set of three sp 2 hybrid orbitals. Notice again how the three atomic orbitals yield the same number of hybrid orbitals.
Boron Trifluoride BF 3 is a common example of sp 2 hybridization. The molecule has plane trigonal geometry.
Tetrahedral (sp 3 ) hybridization
Let us now look at several tetravalent molecules, and see what kind of hybridization might be involved when four outer atoms are bonded to a central atom. Perhaps the commonest and most important example of this bond type is methane, CH 4 .
orbitals of carbon mix into four sp 3 hybrid orbitals which are chemically and geometrically identical; the latter condition implies that the four hybrid orbitals extend toward the corners of a tetrahedron centered on the carbon atom.
Methane is the simplest hydrocarbon; the molecule is approximately spherical, as is shown in the space-filling model:
By replacing one or more of the hydrogen atoms in CH 4 with another sp 3 hybridized carbon fragments, hydrocarbon chains of any degree of complexity can be built up. The simplest of these is ethane:
This shows how an sp 3 orbital on each of two two carbon atoms join (overlap) to form a carbon-carbon bond, and then the remaining carbon sp 3 orbital overlaps with six hydrogen 1s orbitals to form the ethane molecule.
Lone pair electrons in hybrid orbitals
If lone pair electrons are present on the central atom, these can occupy one or more of the sp 3 orbitals. This causes the molecular geometry to be different from the coordination geometry, which remains tetrahedral. In the ammonia molecule, for example, the nitrogen atom normally has three unpaired p electrons, but by mixing the 2 s and 2 p orbitals, we can create four sp 3 -hybrid orbitals just as in carbon. Three of these can form shared-electron bonds with hydrogen, resulting in ammonia, NH 3 . The fourth of the sp 3 hybrid orbitals contains the two remaining outer-shell electrons of nitrogen which form a non-bonding lone pair. In acidic solutions these can coordinate with a hydrogen ion, forming the ammonium ion NH 4 + .
Although no bonds are formed by the lone pair in NH 3 , these electrons do give rise to a charge cloud that takes up space just like any other orbital.
In the water molecule, the oxygen atom can form four sp 3 orbitals. Two of these are occupied by the two lone pairs on the oxygen atom, while the other two are used for bonding. The observed H-O-H bond angle in water (104.5°) is less than the tetrahedral angle (109.5°); one explanation for this is that the non-bonding electrons tend to remain closer to the central atom and thus exert greater repulsion on the other orbitals, thus pushing the two bonding orbitals closer together.
Molecular ions
Hybridization can also help explain the existence and structure of many inorganic molecular ions . Consider, for example, electron configurations of zinc in the compounds in the illustrations below. The tetrachlorozinc ion (top row) is another structure derived from zinc and chlorine. As we might expect, this ion is tetrahedral; there are four chloride ions surrounding the central zinc ion. The zinc ion has a charge of +2, and each chloride ion is –1, so the net charge of the complex ion is –2.
At the bottom is shown the electron configuration of atomic zinc, and just above it, of the divalent zinc ion. Notice that this ion has no electrons at all in its 4-shell. In zinc chloride, shown in the next row up, there are two equivalent chlorine atoms bonded to the zinc. The bonding orbitals are of sp character; that is, they are hybrids of the 4s and one 4p orbital of the zinc atom. Since these orbitals are empty in the isolated zinc ion, the bonding electrons themselves are all contributed by the chlorine atoms, or rather, the chloride ions, for it is these that are the bonded species here. Each chloride ion possesses a complete octet of electrons, and two of these electrons occupy each sp bond orbital in the zinc chloride complex ion. This is an example of a coordinate covalent bond , in which the bonded atom contributes both of the electrons that make up the shared pair.


8.17F: Glomeromycota
- Describe the ecology and reproduction of Glomeromycetes
In the kingdom Fungi, the Glomeromycota is a newly-established phylum comprised of about 230 species that live in close association with the roots of trees and plants. Fossil records indicate that trees and their root symbionts share a long evolutionary history. It appears that most members of this family form arbuscular mycorrhizae: the hyphae interact with the root cells forming a mutually-beneficial association where the plants supply the carbon source and energy in the form of carbohydrates to the fungus while the fungus supplies essential minerals from the soil to the plant. This association is termed biotrophic. The Glomeromycota species that have arbuscular mycorrhizal are terrestrial and widely distributed in soils worldwide where they form symbioses with the roots of the majority of plant species. They can also be found in wetlands, including salt-marshes, and are associated with epiphytic plants.
The glomeromycetes do not reproduce sexually and cannot survive without the presence of plant roots. They have coenocytic hyphae and reproduce asexually, producing glomerospores. The biochemical and genetic characterization of the Glomeromycota has been hindered by their biotrophic nature, which impedes laboratory culturing. This obstacle was eventually surpassed with the use of root cultures. With the advent of molecular techniques, such as gene sequencing, the phylogenetic classification of Glomeromycota has become clearer. The first mycorrhizal gene to be sequenced was the small-subunit ribosomal RNA (SSU rRNA). This gene is highly conserved and commonly used in phylogenetic studies so it was isolated from spores of each taxonomic group. Using a molecular clock approach based on the substitution rates of SSU sequences, scientists were able to estimate the time of divergence of the fungi. This analysis shows that all glomeromycetes probably descended from a common ancestor 462 and 353 million years ago, making them a monophyletic lineage. A long-held theory is that Glomeromycota were instrumental in the colonization of land by plants.
Key Points
- Most glomeromycetes form arbuscular mycorrhizae, a type of symbiotic relationship between a fungus and plant roots; the plants supply a source of energy to the fungus while the fungus supplies essential minerals to the plant.
- Glomeromycota that have arbuscular mycorrhizal are mostly terrestrial, but can also be found in wetlands.
- The glomeromycetes reproduce asexually by producing glomerospores and cannot survive without the presence of plant roots.
- DNA analysis shows that all glomeromycetes probably descended from a common ancestor 462 and 353 million years ago.
- The classification of fungi as Glomeromycota has been redefined with adoption of molecular techniques.
Key Terms
- biotrophic : describing a parasite that needs its host to stay alive
- arbuscular mycorrhizae : a type of symbiotic relationship between a fungus and the roots of a plant where the plants supply a source of energy to the fungus while the fungus supplies essential minerals to the plant
- glomeromycete : an organism of the phylum Glomeromycota


2.2: Case Study - HACC Resources
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Juan is in his second semester at HACC. He has been doing well academically and currently has a 3.0 GPA. This semester has been extremely challenging financially for Juan. He lost his job, and he just received notice that his electricity may be shut off for lack of payment. His computer also died, and he does not have the funds for replacement. He is contemplating quitting school at this point.
Jennifer is in her first year at HACC. She originally thought she wanted to be an elementary school teacher. After getting a job at a local daycare, she has realized that working with younger children is not the best fit for her. Now she feels stressed and confused because she doesn't know what major she should choose.
Maria is taking English this semester. She was assigned a research project and must find articles in a HACC library database for her project. She has no idea how to find articles because she never had to find research in the library before. In addition, she's been struggling with her writing skills. Maria is feeling very overwhelmed and is considering withdrawing from the class.
Reflections Questions: