13.1: Introduction to Corporate Social Responsibility
What you’ll learn to do: explain corporate social responsibility (CSR) and its relationship to economic performance
In addition to profitability, consumers now expect companies to be socially responsible. The main categories of corporate social responsibility include environmental efforts, philanthropy, ethical labor practices, and volunteerism. Managers who wish to create shared value for stakeholders, and thus make their companies thrive, are socially responsible.
Contributors and Attributions
CC licensed content, Original
- Introduction to Corporate Social Responsibility. Authored by : Talia Lambarki and Lumen Learning. License : CC BY: Attribution


12.3C: Cultivation of Specimen
- Describe how direct microscope observation of a fresh or stained specimen is one of the most rapid methods of determining its characteristics
Cultivation of Specimen
The success of pathogen identification and treatment depends on how the specimen is collected, handled, and stored. It is also critical that the pathogen is isolated in a pure culture first. Direct microscope observation of a fresh or stained specimen is one of the most rapid methods of determining characteristics. Stains most often employed for bacteria are the gram stain, though they do not work on some organisms.
The direct florescence antibody (DFA) test can highlight the presence of the microbe in patient specimens by means of labeled antibodies. This test is useful for bacteria such as syphilis spirochete, which are not readily cultivated in a laboratory, or if a rapid diagnosis is essential for the survival of a patient.
In most cases, specimens are also inoculated into differential media that define such characteristics as fermentation patters (mannitol salt and MacConkey agar) and as reactions in blood (blood agar). A patient’s blood is usually cultured in a special bottle of broth that can be periodically sampled for growth. Work must be done from isolated colonies or pure cultures, as working with mixed or contaminated cultures gives misleading and inaccurate results. From such isolates, clinical microbiologists obtain information about a pathogen’s microscopic morphology and staining reactions, culture appearance, motility, oxygen requirements, and biochemical characteristics.
Serological testing uses in-vitro diagnostic testing of serum, has a high degree of specificity and sensitivity, and is based on the specificity an antibody has for its antigen. These techniques do not necessitate a cultivation step. Serum can be directly used in agglutination, precipitation, complement fixation, fluorescent microscopy, and enzyme-linked assays. Results of specimen analysis are entered in the patient’s summary chart.
Key Points
- The success of pathogen identification and treatment depends on how the specimen is collected, handled, and stored.
- Diagnostic laboratory techniques include direct testing using a microscope, and immunological or genetic methods that provide immediate clues as to the identity of the microbe or microbes in the sample.
- Following direct testing, cultivation, isolation, and identification of pathogens using a wide variety of general and specific tests is required.
Key Terms
- mannitol : A polyhydroxy alcohol, an isomer of sorbitol, used as an artificial sweetener.
- MacConkey agar : A culture medium designed to grow Gram-negative bacteria and differentiate them from lactose fermentation.


9.2: Tests in the Normal Model
Basic Theory
The Normal Model
The normal distribution is perhaps the most important distribution in the study of mathematical statistics, in part because of the central limit theorem. As a consequence of this theorem, a measured quantity that is subject to numerous small, random errors will have, at least approximately, a normal distribution. Such variables are ubiquitous in statistical experiments, in subjects varying from the physical and biological sciences to the social sciences.
So in this section, we assume that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the normal distribution with mean \(\mu\) and standard deviation \(\sigma\). Our goal in this section is to to construct hypothesis tests for \(\mu\) and \(\sigma\); these are among of the most important special cases of hypothesis testing. This section parallels the section on Estimation in the Normal Model in the chapter on Set Estimation, and in particular, the duality between interval estimation and hypothesis testing will play an important role. But first we need to review some basic facts that will be critical for our analysis.
Recall that the sample mean \( M \) and sample variance \( S^2 \) are \[ M = \frac{1}{n} \sum_{i=1}^n X_i, \quad S^2 = \frac{1}{n - 1} \sum_{i=1}^n (X_i - M)^2\]
From our study of point estimation, recall that \( M \) is an unbiased and consistent estimator of \( \mu \) while \( S^2 \) is an unbiased and consistent estimator of \( \sigma^2 \). From these basic statistics we can construct the test statistics that will be used to construct our hypothesis tests. The following results were established in the section on Special Properties of the Normal Distribution.
Define \[ Z = \frac{M - \mu}{\sigma \big/ \sqrt{n}}, \quad T = \frac{M - \mu}{S \big/ \sqrt{n}}, \quad V = \frac{n - 1}{\sigma^2} S^2 \]
- \( Z \) has the standard normal distribution.
- \( T \) has the student \( t \) distribution with \( n - 1 \) degrees of freedom.
- \( V \) has the chi-square distribution with \( n - 1 \) degrees of freedom.
- \( Z \) and \( V \) are independent.
It follows that each of these random variables is a pivot variable for \( (\mu, \sigma) \) since the distributions do not depend on the parameters, but the variables themselves functionally depend on one or both parameters. The pivot variables will lead to natural test statistics that can then be used to perform the hypothesis tests of the parameters. To construct our tests, we will need quantiles of these standard distributions. The quantiles can be computed using the special distribution calculator or from most mathematical and statistical software packages. Here is the notation we will use:
Let \( p \in (0, 1) \) and \( k \in \N_+ \).
- \( z(p) \) denotes the quantile of order \( p \) for the standard normal distribution.
- \(t_k(p)\) denotes the quantile of order \( p \) for the student \( t \) distribution with \( k \) degrees of freedom.
- \( \chi^2_k(p) \) denotes the quantile of order \( p \) for the chi-square distribution with \( k \) degrees of freedom
Since the standard normal and student \( t \) distributions are symmetric about 0, it follows that \( z(1 - p) = -z(p) \) and \( t_k(1 - p) = -t_k(p) \) for \( p \in (0, 1) \) and \( k \in \N_+ \). On the other hand, the chi-square distribution is not symmetric.
Tests for the Mean with Known Standard Deviation
For our first discussion, we assume that the distribution mean \( \mu \) is unknown but the standard deviation \( \sigma \) is known. This is not always an artificial assumption. There are often situations where \( \sigma \) is stable over time, and hence is at least approximately known, while \( \mu \) changes because of different
treatments
. Examples are given in the computational exercises below.
For a conjectured \( \mu_0 \in \R \), define the test statistic \[ Z = \frac{M - \mu_0}{\sigma \big/ \sqrt{n}} \]
- If \( \mu = \mu_0 \) then \( Z \) has the standard normal distribution.
- If \( \mu \ne \mu_0 \) then \( Z \) has the normal distribution with mean \( \frac{\mu - \mu_0}{\sigma / \sqrt{n}} \) and variance 1.
So in case (b), \( \frac{\mu - \mu_0}{\sigma / \sqrt{n}} \) can be viewed as a non-centrality parameter . The graph of the probability density function of \( Z \) is like that of the standard normal probability density function, but shifted to the right or left by the non-centrality parameter, depending on whether \( \mu \gt \mu_0 \) or \( \mu \lt \mu_0 \).
For \( \alpha \in (0, 1) \), each of the following tests has significance level \( \alpha \):
- Reject \( H_0: \mu = \mu_0 \) versus \( H_1: \mu \ne \mu_0 \) if and only if \( Z \lt -z(1 - \alpha /2) \) or \( Z \gt z(1 - \alpha / 2) \) if and only if \( M \lt \mu_0 - z(1 - \alpha / 2) \frac{\sigma}{\sqrt{n}} \) or \( M \gt \mu_0 + z(1 - \alpha / 2) \frac{\sigma}{\sqrt{n}} \).
- Reject \( H_0: \mu \le \mu_0 \) versus \( H_1: \mu \gt \mu_0 \) if and only if \( Z \gt z(1 - \alpha) \) if and only if \( M \gt \mu_0 + z(1 - \alpha) \frac{\sigma}{\sqrt{n}} \).
- Reject \( H_0: \mu \ge \mu_0 \) versus \( H_1: \mu \lt \mu_0 \) if and only if \( Z \lt -z(1 - \alpha) \) if and only if \( M \lt \mu_0 - z(1 - \alpha) \frac{\sigma}{\sqrt{n}} \).
Proof
In part (a), \( H_0 \) is a simple hypothesis, and under \( H_0 \), \( Z \) has the standard normal distribution. So \( \alpha \) is probability of falsely rejecting \( H_0 \) by definition of the quantiles. In parts (b) and (c), \( Z \) has a non-central normal distribution under \( H_0 \) as discussed above . So if \( H_0 \) is true, the the maximum type 1 error probability \( \alpha \) occurs when \( \mu = \mu_0 \). The decision rules in terms of \( M \) are equivalent to the corresponding ones in terms of \( Z \) by simple algebra.
Part (a) is the standard two-sided test, while (b) is the right-tailed test and (c) is the left-tailed test. Note that in each case, the hypothesis test is the dual of the corresponding interval estimate constructed in the section on Estimation in the Normal Model.
For each of the tests above , we fail to reject \(H_0\) at significance level \(\alpha\) if and only if \(\mu_0\) is in the corresponding \(1 - \alpha\) confidence interval, that is
- \( M - z(1 - \alpha / 2) \frac{\sigma}{\sqrt{n}} \le \mu_0 \le M + z(1 - \alpha / 2) \frac{\sigma}{\sqrt{n}} \)
- \( \mu_0 \le M + z(1 - \alpha) \frac{\sigma}{\sqrt{n}}\)
- \( \mu_0 \ge M - z(1 - \alpha) \frac{\sigma}{\sqrt{n}}\)
Proof
This follows from the previous result . In each case, we start with the inequality that corresponds to not rejecting \( H_0 \) and solve for \( \mu_0 \).
The two-sided test in (a) corresponds to \( \alpha / 2 \) in each tail of the distribution of the test statistic \( Z \), under \( H_0 \). This set is said to be unbiased . But of course we can construct other biased tests by partitioning the confidence level \( \alpha \) between the left and right tails in a non-symmetric way.
For every \(\alpha, \, p \in (0, 1)\), the following test has significance level \(\alpha\): Reject \(H_0: \mu = \mu_0\) versus \(H_1: \mu \ne \mu_0\) if and only if \(Z \lt z(\alpha - p \alpha)\) or \(Z \ge z(1 - p \alpha)\).
- \( p = \frac{1}{2} \) gives the symmetric, unbiased test.
- \( p \downarrow 0 \) gives the left-tailed test.
- \( p \uparrow 1 \) gives the right-tailed test.
Proof
As before \( H_0 \) is a simple hypothesis, and if \( H_0 \) is true, \( Z \) has the standard normal distribution. So the probability of falsely rejecting \( H_0 \) is \( \alpha \) by definition of the quantiles. Parts (a)–(c) follow from properties of the standard normal quantile function.
The \(P\)-value of these test can be computed in terms of the standard normal distribution function \(\Phi\).
The \(P\)-values of the standard tests above are respectively
- \( 2 \left[1 - \Phi\left(\left|Z\right|\right)\right]\)
- \( 1 - \Phi(Z) \)
- \( \Phi(Z) \)
Recall that the power function of a test of a parameter is the probability of rejecting the null hypothesis, as a function of the true value of the parameter. Our next series of results will explore the power functions of the tests above .
The power function of the general two-sided test above is given by \[ Q(\mu) = \Phi \left( z(\alpha - p \alpha) - \frac{\sqrt{n}}{\sigma} (\mu - \mu_0) \right) + \Phi \left( \frac{\sqrt{n}}{\sigma} (\mu - \mu_0) - z(1 - p \alpha) \right), \quad \mu \in \R \]
- \(Q\) is decreasing on \((-\infty, m_0)\) and increasing on \((m_0, \infty)\) where \(m_0 = \mu_0 + \left[z(\alpha - p \alpha) + z(1 - p \alpha)\right] \frac{\sqrt{n}}{2 \sigma}\).
- \(Q(\mu_0) = \alpha\).
- \(Q(\mu) \to 1\) as \(\mu \uparrow \infty\) and \(Q(\mu) \to 1\) as \(\mu \downarrow -\infty\).
- If \(p = \frac{1}{2}\) then \(Q\) is symmetric about \(\mu_0\) (and \( m_0 = \mu_0 \)).
- As \(p\) increases, \(Q(\mu)\) increases if \(\mu \gt \mu_0\) and decreases if \(\mu \lt \mu_0\).
So by varying \( p \), we can make the test more powerful for some values of \( \mu \), but only at the expense of making the test less powerful for other values of \( \mu \).
The power function of the left-tailed test above is given by
\[ Q(\mu) = \Phi \left( z(\alpha) + \frac{\sqrt{n}}{\sigma}(\mu - \mu_0) \right), \quad \mu \in \R \]- \(Q\) is increasing on \(\R\).
- \(Q(\mu_0) = \alpha\).
- \(Q(\mu) \to 1\) as \(\mu \uparrow \infty\) and \(Q(\mu) \to 0\) as \(\mu \downarrow -\infty\).
The power function of the right-tailed test above , is given by \[ Q(\mu) = \Phi \left( z(\alpha) - \frac{\sqrt{n}}{\sigma}(\mu - \mu_0) \right), \quad \mu \in \R \]
- \(Q\) is decreasing on \(\R\).
- \(Q(\mu_0) = \alpha\).
- \(Q(\mu) \to 0\) as \(\mu \uparrow \infty\) and \(Q(\mu) \to 1\) as \(\mu \downarrow -\infty\).
For any of the three tests in above , increasing the sample size \(n\) or decreasing the standard deviation \(\sigma\) results in a uniformly more powerful test.
In the mean test experiment, select the normal test statistic and select the normal sampling distribution with standard deviation \(\sigma = 2\), significance level \(\alpha = 0.1\), sample size \(n = 20\), and \(\mu_0 = 0\). Run the experiment 1000 times for several values of the true distribution mean \(\mu\). For each value of \(\mu\), note the relative frequency of the event that the null hypothesis is rejected. Sketch the empirical power function.
In the mean estimate experiment, select the normal pivot variable and select the normal distribution with \(\mu = 0\) and standard deviation \(\sigma = 2\), confidence level \(1 - \alpha = 0.90\), and sample size \(n = 10\). For each of the three types of confidence intervals, run the experiment 20 times. State the corresponding hypotheses and significance level, and for each run, give the set of \(\mu_0\) for which the null hypothesis would be rejected.
In many cases, the first step is to design the experiment so that the significance level is \(\alpha\) and so that the test has a given power \(\beta\) for a given alternative \(\mu_1\).
For either of the one-sided tests in above , the sample size \(n\) needed for a test with significance level \(\alpha\) and power \(\beta\) for the alternative \(\mu_1\) is \[ n = \left( \frac{\sigma \left[z(\beta) - z(\alpha)\right]}{\mu_1 - \mu_0} \right)^2 \]
Proof
This follows from setting the power function equal to \(\beta\) and solving for \(n\)
For the unbiased, two-sided test, the sample size \(n\) needed for a test with significance level \(\alpha\) and power \(\beta\) for the alternative \(\mu_1\) is approximately \[ n = \left( \frac{\sigma \left[z(\beta) - z(\alpha / 2)\right]}{\mu_1 - \mu_0} \right)^2 \]
Proof
In the power function for the two-sided test given above , we can neglect the first term if \(\mu_1 \lt \mu_0\) and neglect the second term if \(\mu_1 \gt \mu_0\).
Tests of the Mean with Unknown Standard Deviation
For our next discussion, we construct tests of \(\mu\) without requiring the assumption that \(\sigma\) is known. And in applications of course, \( \sigma \) is usually unknown.
For a conjectured \( \mu_0 \in \R \), define the test statistic \[ T = \frac{M - \mu_0}{S \big/ \sqrt{n}} \]
- If \( \mu = \mu_0 \), the statistic \( T \) has the student \( t \) distribution with \( n - 1 \) degrees of freedom.
- If \( \mu \ne \mu_0 \) then \( T \) has a non-central \( t \) distribution with \( n - 1 \) degrees of freedom and non-centrality parameter \( \frac{\mu - \mu_0}{\sigma / \sqrt{n}} \).
In case (b), the graph of the probability density function of \( T \) is much (but not exactly) the same as that of the ordinary \( t \) distribution with \( n - 1 \) degrees of freedom, but shifted to the right or left by the non-centrality parameter, depending on whether \( \mu \gt \mu_0 \) or \( \mu \lt \mu_0 \).
For \( \alpha \in (0, 1) \), each of the following tests has significance level \( \alpha \):
- Reject \( H_0: \mu = \mu_0 \) versus \( H_1: \mu \ne \mu_0 \) if and only if \( T \lt -t_{n-1}(1 - \alpha /2) \) or \( T \gt t_{n-1}(1 - \alpha / 2) \) if and only if \( M \lt \mu_0 - t_{n-1}(1 - \alpha / 2) \frac{S}{\sqrt{n}} \) or \( T \gt \mu_0 + t_{n-1}(1 - \alpha / 2) \frac{S}{\sqrt{n}} \).
- Reject \( H_0: \mu \le \mu_0 \) versus \( H_1: \mu \gt \mu_0 \) if and only if \( T \gt t_{n-1}(1 - \alpha) \) if and only if \( M \gt \mu_0 + t_{n-1}(1 - \alpha) \frac{S}{\sqrt{n}} \).
- Reject \( H_0: \mu \ge \mu_0 \) versus \( H_1: \mu \lt \mu_0 \) if and only if \( T \lt -t_{n-1}(1 - \alpha) \) if and only if \( M \lt \mu_0 - t_{n-1}(1 - \alpha) \frac{S}{\sqrt{n}} \).
Proof
In part (a), \( T \) has the chi-square distribution with \( n - 1 \) degrees of freedom under \( H_0 \). So if \( H_0 \) is true, the probability of falsely rejecting \( H_0 \) is \( \alpha \) by definition of the quantiles. In parts (b) and (c), \( T \) has a non-central \( t \) distribution with \( n - 1 \) degrees of freedom under \( H_0 \), as discussed above . Hence if \( H_0 \) is true, the maximum type 1 error probability \( \alpha \) occurs when \( \mu = \mu_0 \). The decision rules in terms of \( M \) are equivalent to the corresponding ones in terms of \( T \) by simple algebra.
Part (a) is the standard two-sided test, while (b) is the right-tailed test and (c) is the left-tailed test. Note that in each case, the hypothesis test is the dual of the corresponding interval estimate constructed in the section on Estimation in the Normal Model.
For each of the tests above , we fail to reject \(H_0\) at significance level \(\alpha\) if and only if \(\mu_0\) is in the corresponding \(1 - \alpha\) confidence interval.
- \( M - t_{n-1}(1 - \alpha / 2) \frac{S}{\sqrt{n}} \le \mu_0 \le M + t_{n-1}(1 - \alpha / 2) \frac{S}{\sqrt{n}} \)
- \( \mu_0 \le M + t_{n-1}(1 - \alpha) \frac{S}{\sqrt{n}}\)
- \( \mu_0 \ge M - t_{n-1}(1 - \alpha) \frac{S}{\sqrt{n}}\)
Proof
This follows from the previous result . In each case, we start with the inequality that corresponds to not rejecting \( H_0 \) and then solve for \( \mu_0 \).
The two-sided test in (a) corresponds to \( \alpha / 2 \) in each tail of the distribution of the test statistic \( T \), under \( H_0 \). This set is said to be unbiased . But of course we can construct other biased tests by partitioning the confidence level \( \alpha \) between the left and right tails in a non-symmetric way.
For every \(\alpha, \, p \in (0, 1)\), the following test has significance level \(\alpha\): Reject \(H_0: \mu = \mu_0\) versus \(H_1: \mu \ne \mu_0\) if and only if \(T \lt t_{n-1}(\alpha - p \alpha)\) or \(T \ge t_{n-1}(1 - p \alpha)\) if and only if \( M \lt \mu_0 + t_{n-1}(\alpha - p \alpha) \frac{S}{\sqrt{n}} \) or \( M \gt \mu_0 + t_{n-1}(1 - p \alpha) \frac{S}{\sqrt{n}} \).
- \( p = \frac{1}{2} \) gives the symmetric, unbiased test.
- \( p \downarrow 0 \) gives the left-tailed test.
- \( p \uparrow 1 \) gives the right-tailed test.
Proof
Once again, \( H_0 \) is a simple hypothesis, and under \( H_0 \) the test statistic \( T \) has the student \( t \) distribution with \( n - 1 \) degrees of freedom. So if \( H_0 \) is true, the probability of falsely rejecting \( H_0 \) is \( \alpha \) by definition of the quantiles. Parts (a)–(c) follow from properties of the quantile function.
The \(P\)-value of these test can be computed in terms of the distribution function \(\Phi_{n-1}\) of the \(t\)-distribution with \(n - 1\) degrees of freedom.
The \(P\)-values of the standard tests above are respectively
- \( 2 \left[1 - \Phi_{n-1}\left(\left|T\right|\right)\right]\)
- \( 1 - \Phi_{n-1}(T) \)
- \( \Phi_{n-1}(T) \)
In the mean test experiment, select the student test statistic and select the normal sampling distribution with standard deviation \(\sigma = 2\), significance level \(\alpha = 0.1\), sample size \(n = 20\), and \(\mu_0 = 1\). Run the experiment 1000 times for several values of the true distribution mean \(\mu\). For each value of \(\mu\), note the relative frequency of the event that the null hypothesis is rejected. Sketch the empirical power function.
In the mean estimate experiment, select the student pivot variable and select the normal sampling distribution with mean 0 and standard deviation 2. Select confidence level 0.90 and sample size 10. For each of the three types of intervals, run the experiment 20 times. State the corresponding hypotheses and significance level, and for each run, give the set of \(\mu_0\) for which the null hypothesis would be rejected.
The power function for the \( t \) tests above can be computed explicitly in terms of the non-central \(t\) distribution function. Qualitatively, the graphs of the power functions are similar to the case when \(\sigma\) is known, given above two-sided , left-tailed , and right-tailed cases.
If an upper bound \(\sigma_0\) on the standard deviation \(\sigma\) is known, then conservative estimates on the sample size needed for a given confidence level and a given margin of error can be obtained using the methods for the normal pivot variable, in the two-sided and one-sided cases.
Tests of the Standard Deviation
For our next discussion, we will construct hypothesis tests for the distribution standard deviation \( \sigma \). So our assumption is that \( \sigma \) is unknown, and of course almost always, \( \mu \) would be unknown as well.
For a conjectured value \( \sigma_0 \in (0, \infty)\), define the test statistic \[ V = \frac{n - 1}{\sigma_0^2} S^2 \]
- If \( \sigma = \sigma_0 \), then \( V \) has the chi-square distribution with \( n - 1 \) degrees of freedom.
- If \( \sigma \ne \sigma_0 \) then \( V \) has the gamma distribution with shape parameter \( (n - 1) / 2 \) and scale parameter \( 2 \sigma^2 \big/ \sigma_0^2 \).
Recall that the ordinary chi-square distribution with \( n - 1 \) degrees of freedom is the gamma distribution with shape parameter \( (n - 1) / 2 \) and scale parameter \( \frac{1}{2} \). So in case (b), the ordinary chi-square distribution is scaled by \( \sigma^2 \big/ \sigma_0^2 \). In particular, the scale factor is greater than 1 if \( \sigma \gt \sigma_0 \) and less than 1 if \( \sigma \lt \sigma_0 \).
For every \(\alpha \in (0, 1)\), the following test has significance level \(\alpha\):
- Reject \(H_0: \sigma = \sigma_0\) versus \(H_1: \sigma \ne \sigma_0\) if and only if \(V \lt \chi_{n-1}^2(\alpha / 2)\) or \(V \gt \chi_{n-1}^2(1 - \alpha / 2)\) if and only if \( S^2 \lt \chi_{n-1}^2(\alpha / 2) \frac{\sigma_0^2}{n - 1} \) or \( S^2 \gt \chi_{n-1}^2(1 - \alpha / 2) \frac{\sigma_0^2}{n - 1} \)
- Reject \(H_0: \sigma \ge \sigma_0\) versus \(H_1: \sigma \lt \sigma_0\) if and only if \(V \lt \chi_{n-1}^2(\alpha)\) if and only if \( S^2 \lt \chi_{n-1}^2(\alpha) \frac{\sigma_0^2}{n - 1} \)
- Reject \(H_0: \sigma \le \sigma_0\) versus \(H_1: \sigma \gt \sigma_0\) if and only if \(V \gt \chi_{n-1}^2(1 - \alpha)\) if and only if \( S^2 \gt \chi_{n-1}^2(1 - \alpha) \frac{\sigma_0^2}{n - 1} \)
Proof
The logic is largely the same as with our other hypothesis test. In part (a), \( H_0 \) is a simple hypothesis, and under \( H_0 \), the test statistic \( V \) has the chi-square distribution with \( n - 1 \) degrees of freedom. So if \( H_0 \) is true, the probability of falsely rejecting \( H_0 \) is \( \alpha \) by definition of the quantiles. In parts (b) and (c), \( V \) has the more general gamma distribution under \( H_0 \), as discussed above . If \( H_0 \) is true, the maximum type 1 error probability is \( \alpha \) and occurs when \( \sigma = \sigma_0 \).
Part (a) is the unbiased, two-sided test that corresponds to \( \alpha / 2 \) in each tail of the chi-square distribution of the test statistic \( V \), under \( H_0 \). Part (b) is the left-tailed test and part (c) is the right-tailed test. Once again, we have a duality between the hypothesis tests and the interval estimates constructed in the section on Estimation in the Normal Model.
For each of the tests in above , we fail to reject \(H_0\) at significance level \(\alpha\) if and only if \(\sigma_0^2\) is in the corresponding \(1 - \alpha\) confidence interval. That is
- \( \frac{n - 1}{\chi_{n-1}^2(1 - \alpha / 2)} S^2 \le \sigma_0^2 \le \frac{n - 1}{\chi_{n-1}^2(\alpha / 2)} S^2 \)
- \( \sigma_0^2 \le \frac{n - 1}{\chi_{n-1}^2(\alpha)} S^2 \)
- \( \sigma_0^2 \ge \frac{n - 1}{\chi_{n-1}^2(1 - \alpha)} S^2 \)
Proof
This follows from the previous result . In each case, we start with the inequality that corresponds to not rejecting \( H_0 \) and then solve for \( \sigma_0^2 \).
As before, we can construct more general two-sided tests by partitioning the significance level \( \alpha \) between the left and right tails of the chi-square distribution in an arbitrary way.
For every \(\alpha, \, p \in (0, 1)\), the following test has significance level \(\alpha\): Reject \(H_0: \sigma = \sigma_0\) versus \(H_1: \sigma \ne \sigma_0\) if and only if \(V \le \chi_{n-1}^2(\alpha - p \alpha)\) or \(V \ge \chi_{n-1}^2(1 - p \alpha)\) if and only if \( S^2 \lt \chi_{n-1}^2(\alpha - p \alpha) \frac{\sigma_0^2}{n - 1} \) or \( S^2 \gt \chi_{n-1}^2(1 - p \alpha) \frac{\sigma_0^2}{n - 1} \).
- \( p = \frac{1}{2} \) gives the equal-tail test.
- \( p \downarrow 0 \) gives the left-tail test.
- \( p \uparrow 1 \) gives the right-tail test.
Proof
As before, \( H_0 \) is a simple hypothesis, and under \( H_0 \) the test statistic \( V \) has the chi-square distribution with \( n - 1 \) degrees of freedom. So if \( H_0 \) is true, the probability of falsely rejecting \( H_0 \) is \( \alpha \) by definition of the quantiles. Parts (a)–(c) follow from properties of the quantile function.
Recall again that the power function of a test of a parameter is the probability of rejecting the null hypothesis, as a function of the true value of the parameter. The power functions of the tests for \( \sigma \) can be expressed in terms of the distribution function \( G_{n-1} \) of the chi-square distribution with \( n - 1 \) degrees of freedom.
The power function of the general two-sided test above is given by the following formula, and satisfies the given properties: \[ Q(\sigma) = 1 - G_{n-1} \left( \frac{\sigma_0^2}{\sigma^2} \chi_{n-1}^2(1 - p \, \alpha) \right) + G_{n-1} \left(\frac{\sigma_0^2}{\sigma^2} \chi_{n-1}^2(\alpha - p \, \alpha) \right)\]
- \(Q\) is decreasing on \((-\infty, \sigma_0)\) and increasing on \((\sigma_0, \infty)\).
- \(Q(\sigma_0) = \alpha\).
- \(Q(\sigma) \to 1\) as \(\sigma \uparrow \infty\) and \(Q(\sigma) \to 1\) as \(\sigma \downarrow 0\).
The power function of the left-tailed test in above is given by the following formula, and satisfies the given properties: \[ Q(\sigma) = 1 - G_{n-1} \left( \frac{\sigma_0^2}{\sigma^2} \chi_{n-1}^2(1 - \alpha) \right) \]
- \(Q\) is increasing on \((0, \infty)\).
- \(Q(\sigma_0) = \alpha\).
- \(Q(\sigma) \to 1\) as \(\sigma \uparrow \infty\) and \(Q(\sigma) \to 0\) as \(\sigma \downarrow 0\).
The power function for the right-tailed test above is given by the following formula, and satisfies the given properties: \[ Q(\sigma) = G_{n-1} \left( \frac{\sigma_0^2}{\sigma^2} \chi_{n-1}^2(\alpha) \right) \]
- \(Q\) is decreasing on \((0, \infty)\).
- \(Q(\sigma_0) =\alpha\).
- \(Q(\sigma) \to 0\) as \(\sigma \uparrow \infty)\) and \(Q(\sigma) \to 0\) as \(\sigma \uparrow \infty\) and as \(\sigma \downarrow 0\).
In the variance test experiment, select the normal distribution with mean 0, and select significance level 0.1, sample size 10, and test standard deviation 1.0. For various values of the true standard deviation, run the simulation 1000 times. Record the relative frequency of rejecting the null hypothesis and plot the empirical power curve.
- Two-sided test
- Left-tailed test
- Right-tailed test
In the variance estimate experiment, select the normal distribution with mean 0 and standard deviation 2, and select confidence level 0.90 and sample size 10. Run the experiment 20 times. State the corresponding hypotheses and significance level, and for each run, give the set of test standard deviations for which the null hypothesis would be rejected.
- Two-sided confidence interval
- Confidence lower bound
- Confidence upper bound
Exercises
Robustness
The primary assumption that we made is that the underlying sampling distribution is normal. Of course, in real statistical problems, we are unlikely to know much about the sampling distribution, let alone whether or not it is normal. Suppose in fact that the underlying distribution is not normal. When the sample size \(n\) is relatively large, the distribution of the sample mean will still be approximately normal by the central limit theorem, and thus our tests of the mean \(\mu\) should still be approximately valid. On the other hand, tests of the variance \(\sigma^2\) are less robust to deviations form the assumption of normality. The following exercises explore these ideas.
In the mean test experiment, select the gamma distribution with shape parameter 1 and scale parameter 1. For the three different tests and for various significance levels, sample sizes, and values of \(\mu_0\), run the experiment 1000 times. For each configuration, note the relative frequency of rejecting \(H_0\). When \(H_0\) is true, compare the relative frequency with the significance level.
In the mean test experiment, select the uniform distribution on \([0, 4]\). For the three different tests and for various significance levels, sample sizes, and values of \(\mu_0\), run the experiment 1000 times. For each configuration, note the relative frequency of rejecting \(H_0\). When \(H_0\) is true, compare the relative frequency with the significance level.
How large \(n\) needs to be for the testing procedure to work well depends, of course, on the underlying distribution; the more this distribution deviates from normality, the larger \(n\) must be. Fortunately, convergence to normality in the central limit theorem is rapid and hence, as you observed in the exercises, we can get away with relatively small sample sizes (30 or more) in most cases.
In the variance test experiment, select the gamma distribution with shape parameter 1 and scale parameter 1. For the three different tests and for various significance levels, sample sizes, and values of \(\sigma_0\), run the experiment 1000 times. For each configuration, note the relative frequency of rejecting \(H_0\). When \(H_0\) is true, compare the relative frequency with the significance level.
In the variance test experiment, select the uniform distribution on \([0, 4]\). For the three different tests and for various significance levels, sample sizes, and values of \(\mu_0\), run the experiment 1000 times. For each configuration, note the relative frequency of rejecting \(H_0\). When \(H_0\) is true, compare the relative frequency with the significance level.
Computational Exercises
The length of a certain machined part is supposed to be 10 centimeters. In fact, due to imperfections in the manufacturing process, the actual length is a random variable. The standard deviation is due to inherent factors in the process, which remain fairly stable over time. From historical data, the standard deviation is known with a high degree of accuracy to be 0.3. The mean, on the other hand, may be set by adjusting various parameters in the process and hence may change to an unknown value fairly frequently. We are interested in testing \(H_0: \mu = 10\) versus \(H_1: \mu \ne 10\).
- Suppose that a sample of 100 parts has mean 10.1. Perform the test at the 0.1 level of significance.
- Compute the \(P\)-value for the data in (a).
- Compute the power of the test in (a) at \(\mu = 10.05\).
- Compute the approximate sample size needed for significance level 0.1 and power 0.8 when \(\mu = 10.05\).
Answer
- Test statistic 3.33, critical values \(\pm 1.645\). Reject \(H_0\).
- \(P = 0.0010\)
- The power of the test at 10.05 is approximately 0.0509.
- Sample size 223
A bag of potato chips of a certain brand has an advertised weight of 250 grams. Actually, the weight (in grams) is a random variable. Suppose that a sample of 75 bags has mean 248 and standard deviation 5. At the 0.05 significance level, perform the following tests:
- \(H_0: \mu \ge 250\) versus \(H_1: \mu \lt 250\)
- \(H_0: \sigma \ge 7\) versus \(H_1: \sigma \lt 7\)
Answer
- Test statistic \(-3.464\), critical value \(-1.665\). Reject \(H_0\).
- \(P \lt 0.0001\) so reject \(H_0\).
At a telemarketing firm, the length of a telephone solicitation (in seconds) is a random variable. A sample of 50 calls has mean 310 and standard deviation 25. At the 0.1 level of significance, can we conclude that
- \(\mu \gt 300\)?
- \(\sigma \gt 20\)?
Answer
- Test statistic 2.828, critical value 1.2988. Reject \(H_0\).
- \(P = 0.0071\) so reject \(H_0\).
At a certain farm the weight of a peach (in ounces) at harvest time is a random variable. A sample of 100 peaches has mean 8.2 and standard deviation 1.0. At the 0.01 level of significance, can we conclude that
- \(\mu \gt 8\)?
- \(\sigma \lt 1.5\)?
Answer
- Test statistic 2.0, critical value 2.363. Fail to reject \(H_0\).
- \(P \lt 0.0001\) so reject \(H_0\).
The hourly wage for a certain type of construction work is a random variable with standard deviation 1.25. For sample of 25 workers, the mean wage was $6.75. At the 0.01 level of significance, can we conclude that \(\mu \lt 7.00\)?
Answer
Test statistic \(-1\), critical value \(-2.328\). Fail to reject \(H_0\).
Data Analysis Exercises
Using Michelson's data, test to see if the velocity of light is greater than 730 (+299000) km/sec, at the 0.005 significance level.
Answer
Test statistic 15.49, critical value 2.6270. Reject \(H_0\).
Using Cavendish's data, test to see if the density of the earth is less than 5.5 times the density of water, at the 0.05 significance level .
Answer
Test statistic \(-1.269\), critical value \(-1.7017\). Fail to reject \(H_0\).
Using Short's data, test to see if the parallax of the sun differs from 9 seconds of a degree, at the 0.1 significance level.
Answer
Test statistic \(-3.730\), critical value \(\pm 1.6749\). Reject \(H_0\).
Using Fisher's iris data, perform the following tests, at the 0.1 level:
- The mean petal length of Setosa irises differs from 15 mm.
- The mean petal length of Verginica irises is greater than 52 mm.
- The mean petal length of Versicolor irises is less than 44 mm.
Answer
- Test statistic \(-1.563\), critical values \(\pm 1.672\). Fail to reject \(H_0\).
- Test statistic 4.556, critical value 1.2988. Reject \(H_0\).
- Test statistic \(-1.028\), critical value \(-1.2988\). Fail to Reject \(H_0\).


18.5: Structure and General Properties of the Nonmetals
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Learning Objectives
By the end of this section, you will be able to:
- Describe structure and properties of nonmetals
The nonmetals are elements located in the upper right portion of the periodic table. Their properties and behavior are quite different from those of metals on the left side. Under normal conditions, more than half of the nonmetals are gases, one is a liquid, and the rest include some of the softest and hardest of solids. The nonmetals exhibit a rich variety of chemical behaviors. They include the most reactive and least reactive of elements, and they form many different ionic and covalent compounds. This section presents an overview of the properties and chemical behaviors of the nonmetals, as well as the chemistry of specific elements. Many of these nonmetals are important in biological systems.
In many cases, trends in electronegativity enable us to predict the type of bonding and the physical states in compounds involving the nonmetals. We know that electronegativity decreases as we move down a given group and increases as we move from left to right across a period. The nonmetals have higher electronegativities than do metals, and compounds formed between metals and nonmetals are generally ionic in nature because of the large differences in electronegativity between them. The metals form cations, the nonmetals form anions, and the resulting compounds are solids under normal conditions. On the other hand, compounds formed between two or more nonmetals have small differences in electronegativity between the atoms, and covalent bonding—sharing of electrons—results. These substances tend to be molecular in nature and are gases, liquids, or volatile solids at room temperature and pressure.
In normal chemical processes, nonmetals do not form monatomic positive ions (cations) because their ionization energies are too high. All monatomic nonmetal ions are anions; examples include the chloride ion, Cl − , the nitride ion, N 3− , and the selenide ion, Se 2 − .
The common oxidation states that the nonmetals exhibit in their ionic and covalent compounds are shown in Figure \(\PageIndex{1}\). Remember that an element exhibits a positive oxidation state when combined with a more electronegative element and that it exhibits a negative oxidation state when combined with a less electronegative element.
The first member of each nonmetal group exhibits different behaviors, in many respects, from the other group members. The reasons for this include smaller size, greater ionization energy, and (most important) the fact that the first member of each group has only four valence orbitals (one 2 s and three 2 p ) available for bonding, whereas other group members have empty d orbitals in their valence shells, making possible five, six, or even more bonds around the central atom. For example, nitrogen forms only NF 3 , whereas phosphorus forms both PF 3 and PF 5 .
Another difference between the first group member and subsequent members is the greater ability of the first member to form π bonds. This is primarily a function of the smaller size of the first member of each group, which allows better overlap of atomic orbitals. Nonmetals, other than the first member of each group, rarely form π bonds to nonmetals that are the first member of a group. For example, sulfur-oxygen π bonds are well known, whereas sulfur does not normally form stable π bonds to itself.
The variety of oxidation states displayed by most of the nonmetals means that many of their chemical reactions involve changes in oxidation state through oxidation-reduction reactions. There are five general aspects of the oxidation-reduction chemistry:
- Nonmetals oxidize most metals. The oxidation state of the metal becomes positive as it undergoes oxidation and that of the nonmetal becomes negative as it undergoes reduction. For example: \[4 \overset{\large 0}{\text{Fe}}(s) + 3 \overset{\large 0}{\text{O}}_2(g) \longrightarrow 2 \overset{\large +3}{\text{Fe}}_2 \overset{\large -2}{\text{O}}_3(s) \nonumber \]
- With the exception of nitrogen and carbon, which are poor oxidizing agents, a more electronegative nonmetal oxidizes a less electronegative nonmetal or the anion of the nonmetal: \[\overset{\large 0}{\text{S}}(s) + 3 \overset{\large 0}{\text{O}}_2(g) \longrightarrow 2 \overset{\large +4}{\text{S}} \overset{\large -2}{\text{O}}_2(s) \nonumber \]\[\overset{\large 0}{\text{Cl}}_2\text{(s)} + 2 \text{I}^{-}\text{(aq)} \longrightarrow \overset{\large 0}{\text{I}_2}\text{(s)} + 2 \overset{\large }{\text{Cl}^{-}} \text{(aq)} \nonumber \]
- Fluorine and oxygen are the strongest oxidizing agents within their respective groups; each oxidizes all the elements that lie below it in the group. Within any period, the strongest oxidizing agent is in group 17. A nonmetal often oxidizes an element that lies to its left in the same period. For example: \[2 \overset{\large 0}{\text{As}}(s) + 3 \overset{\large 0}{\text{Br}}_2(l) \longrightarrow 2 \overset{\large +3}{\text{As}}_2 \overset{\large -1}{\text{Br}}_3(s) \nonumber \]
- The stronger a nonmetal is as an oxidizing agent, the more difficult it is to oxidize the anion formed by the nonmetal. This means that the most stable negative ions are formed by elements at the top of the group or in group 17 of the period.
- Fluorine and oxygen are the strongest oxidizing elements known. Fluorine does not form compounds in which it exhibits positive oxidation states; oxygen exhibits a positive oxidation state only when combined with fluorine. For example: \[2 \overset{\large 0}{\text{F}}_2\text{(s)} + 2 \text{OH}^{-}\text{(aq)} \longrightarrow \overset{\large +2}{\text{O}} \text{F}_2 \text{(s)} + 2 \overset{\large -1}{\text{F}^{-}} \text{(aq)} + \text{H}_2\text{O(l)} \nonumber \]
With the exception of most of the noble gases, all nonmetals form compounds with oxygen, yielding covalent oxides. Most of these oxides are acidic, that is, they react with water to form oxyacids. Recall from the acid-base chapter that an oxyacid is an acid consisting of hydrogen, oxygen, and some other element. Notable exceptions are carbon monoxide, CO, nitrous oxide, N 2 O, and nitric oxide, NO. There are three characteristics of these acidic oxides:
- Oxides such as SO2 and N2O5, in which the nonmetal exhibits one of its common oxidation states, are acid anhydrides and react with water to form acids with no change in oxidation state. The product is an oxyacid. For example:
-
Those oxides such as NO
2
and ClO
2
, in which the nonmetal does not exhibit one of its common oxidation states, also react with water. In these reactions, the nonmetal is both oxidized and reduced. For example:
Reactions in which the same element is both oxidized and reduced are called disproportionation reactions . - The acid strength increases as the electronegativity of the central atom increases. To learn more, see the discussion in the chapter on acid-base chemistry.
The binary hydrogen compounds of the nonmetals also exhibit an acidic behavior in water, although only HCl, HBr, and HI are strong acids. The acid strength of the nonmetal hydrogen compounds increases from left to right across a period and down a group. For example, ammonia, NH 3 , is a weaker acid than is water, H 2 O, which is weaker than is hydrogen fluoride, HF. Water, H 2 O, is also a weaker acid than is hydrogen sulfide, H 2 S , which is weaker than is hydrogen selenide, H 2 Se. Weaker acidic character implies greater basic character.
Structures of the Nonmetals
The structures of the nonmetals differ dramatically from those of metals. Metals crystallize in closely packed arrays that do not contain molecules or covalent bonds. Nonmetal structures contain covalent bonds, and many nonmetals consist of individual molecules. The electrons in nonmetals are localized in covalent bonds, whereas in a metal, there is delocalization of the electrons throughout the solid.
The noble gases are all monatomic, whereas the other nonmetal gases—hydrogen, nitrogen, oxygen, fluorine, and chlorine—normally exist as the diatomic molecules H 2, N 2 , O 2 , F 2 , and Cl 2 . The other halogens are also diatomic; Br 2 is a liquid and I 2 exists as a solid under normal conditions. The changes in state as one moves down the halogen family offer excellent examples of the increasing strength of intermolecular London forces with increasing molecular mass and increasing polarizability.
Oxygen has two allotropes: O 2 , dioxygen, and O 3 , ozone. Phosphorus has three common allotropes, commonly referred to by their colors: white, red, and black. Sulfur has several allotropes. There are also many carbon allotropes. Most people know of diamond, graphite, and charcoal, but fewer people know of the recent discovery of fullerenes, carbon nanotubes, and graphene.
Descriptions of the physical properties of three nonmetals that are characteristic of molecular solids follow.
Carbon
Carbon occurs in the uncombined (elemental) state in many forms, such as diamond, graphite, charcoal, coke, carbon black, graphene, and fullerene.
Diamond, shown in Figure \(\PageIndex{2}\), is a very hard crystalline material that is colorless and transparent when pure. Each atom forms four single bonds to four other atoms at the corners of a tetrahedron ( sp 3 hybridization); this makes the diamond a giant molecule. Carbon-carbon single bonds are very strong, and, because they extend throughout the crystal to form a three-dimensional network, the crystals are very hard and have high melting points (~4400 °C).
Graphite, also shown in Figure \(\PageIndex{2}\), is a soft, slippery, grayish-black solid that conducts electricity. These properties relate to its structure, which consists of layers of carbon atoms, with each atom surrounded by three other carbon atoms in a trigonal planar arrangement. Each carbon atom in graphite forms three σ bonds, one to each of its nearest neighbors, by means of sp 2 -hybrid orbitals. The unhybridized p orbital on each carbon atom will overlap unhybridized orbitals on adjacent carbon atoms in the same layer to form π bonds. Many resonance forms are necessary to describe the electronic structure of a graphite layer; Figure \(\PageIndex{3}\): illustrates two of these forms.
Atoms within a graphite layer are bonded together tightly by the σ and π bonds; however, the forces between layers are weak. London dispersion forces hold the layers together. To learn more, see the discussion of these weak forces in the chapter on liquids and solids. The weak forces between layers give graphite the soft, flaky character that makes it useful as the so-called “lead” in pencils and the slippery character that makes it useful as a lubricant. The loosely held electrons in the resonating π bonds can move throughout the solid and are responsible for the electrical conductivity of graphite.
Other forms of elemental carbon include carbon black, charcoal, and coke. Carbon black is an amorphous form of carbon prepared by the incomplete combustion of natural gas, CH 4 . It is possible to produce charcoal and coke by heating wood and coal, respectively, at high temperatures in the absence of air.
Recently, new forms of elemental carbon molecules have been identified in the soot generated by a smoky flame and in the vapor produced when graphite is heated to very high temperatures in a vacuum or in helium. One of these new forms, first isolated by Professor Richard Smalley and coworkers at Rice University, consists of icosahedral (soccer-ball-shaped) molecules that contain 60 carbon atoms, C 60 . This is buckminsterfullerene (often called bucky balls) after the architect Buckminster Fuller , who designed domed structures, which have a similar appearance (Figure \(\PageIndex{4}\)).
Chemistry in Everyday Life: Nanotubes and Graphene
Graphene and carbon nanotubes are two recently discovered allotropes of carbon. Both of the forms bear some relationship to graphite. Graphene is a single layer of graphite (one atom thick), as illustrated in Figure \(\PageIndex{5}\), whereas carbon nanotubes roll the layer into a small tube, as illustrated in Figure \(\PageIndex{5}\).
Graphene is a very strong, lightweight, and efficient conductor of heat and electricity discovered in 2003. As in graphite, the carbon atoms form a layer of six-membered rings with sp 2 -hybridized carbon atoms at the corners. Resonance stabilizes the system and leads to its conductivity. Unlike graphite, there is no stacking of the layers to give a three-dimensional structure. Andre Geim and Kostya Novoselov at the University of Manchester won the 2010 Nobel Prize in Physics for their pioneering work characterizing graphene.
The simplest procedure for preparing graphene is to use a piece of adhesive tape to remove a single layer of graphene from the surface of a piece of graphite. This method works because there are only weak London dispersion forces between the layers in graphite. Alternative methods are to deposit a single layer of carbon atoms on the surface of some other material (ruthenium, iridium, or copper) or to synthesize it at the surface of silicon carbide via the sublimation of silicon.
There currently are no commercial applications of graphene. However, its unusual properties, such as high electron mobility and thermal conductivity, should make it suitable for the manufacture of many advanced electronic devices and for thermal management applications.
Carbon nanotubes are carbon allotropes, which have a cylindrical structure. Like graphite and graphene, nanotubes consist of rings of sp 2 -hybridized carbon atoms. Unlike graphite and graphene, which occur in layers, the layers wrap into a tube and bond together to produce a stable structure. The walls of the tube may be one atom or multiple atoms thick.
Carbon nanotubes are extremely strong materials that are harder than diamond. Depending upon the shape of the nanotube, it may be a conductor or semiconductor. For some applications, the conducting form is preferable, whereas other applications utilize the semiconducting form.
The basis for the synthesis of carbon nanotubes is the generation of carbon atoms in a vacuum. It is possible to produce carbon atoms by an electrical discharge through graphite, vaporization of graphite with a laser, and the decomposition of a carbon compound.
The strength of carbon nanotubes will eventually lead to some of their most exciting applications, as a thread produced from several nanotubes will support enormous weight. However, the current applications only employ bulk nanotubes. The addition of nanotubes to polymers improves the mechanical, thermal, and electrical properties of the bulk material. There are currently nanotubes in some bicycle parts, skis, baseball bats, fishing rods, and surfboards.
Phosphorus
The name phosphorus comes from the Greek words meaning light bringing. When phosphorus was first isolated, scientists noted that it glowed in the dark and burned when exposed to air. Phosphorus is the only member of its group that does not occur in the uncombined state in nature; it exists in many allotropic forms. We will consider two of those forms: white phosphorus and red phosphorus.
White phosphorus is a white, waxy solid that melts at 44.2 °C and boils at 280 °C. It is insoluble in water (in which it is stored—see Figure \(\PageIndex{6}\)), is very soluble in carbon disulfide, and bursts into flame in air. As a solid, as a liquid, as a gas, and in solution, white phosphorus exists as P 4 molecules with four phosphorus atoms at the corners of a regular tetrahedron, as illustrated in Figure \(\PageIndex{6}\). Each phosphorus atom covalently bonds to the other three atoms in the molecule by single covalent bonds. White phosphorus is the most reactive allotrope and is very toxic.
Heating white phosphorus to 270–300 °C in the absence of air yields red phosphorus. Red phosphorus (shown in Figure \(\PageIndex{6}\)) is denser, has a higher melting point (~600 °C), is much less reactive, is essentially nontoxic, and is easier and safer to handle than is white phosphorus. Its structure is highly polymeric and appears to contain three-dimensional networks of P 4 tetrahedra joined by P-P single bonds. Red phosphorus is insoluble in solvents that dissolve white phosphorus. When red phosphorus is heated, P 4 molecules sublime from the solid.
Sulfur
The allotropy of sulfur is far greater and more complex than that of any other element. Sulfur is the brimstone referred to in the Bible and other places, and references to sulfur occur throughout recorded history—right up to the relatively recent discovery that it is a component of the atmospheres of Venus and of Io, a moon of Jupiter. The most common and most stable allotrope of sulfur is yellow, rhombic sulfur, so named because of the shape of its crystals. Rhombic sulfur is the form to which all other allotropes revert at room temperature. Crystals of rhombic sulfur melt at 113 °C. Cooling this liquid gives long needles of monoclinic sulfur. This form is stable from 96 °C to the melting point, 119 °C. At room temperature, it gradually reverts to the rhombic form.
Both rhombic sulfur and monoclinic sulfur contain S 8 molecules in which atoms form eight-membered, puckered rings that resemble crowns, as illustrated in Figure \(\PageIndex{7}\). Each sulfur atom is bonded to each of its two neighbors in the ring by covalent S-S single bonds.
When rhombic sulfur melts, the straw-colored liquid is quite mobile; its viscosity is low because S 8 molecules are essentially spherical and offer relatively little resistance as they move past each other. As the temperature rises, S-S bonds in the rings break, and polymeric chains of sulfur atoms result. These chains combine end to end, forming still longer chains that tangle with one another. The liquid gradually darkens in color and becomes so viscous that finally (at about 230 °C) it does not pour easily. The dangling atoms at the ends of the chains of sulfur atoms are responsible for the dark red color because their electronic structure differs from those of sulfur atoms that have bonds to two adjacent sulfur atoms. This causes them to absorb light differently and results in a different visible color. Cooling the liquid rapidly produces a rubberlike amorphous mass, called plastic sulfur.
Sulfur boils at 445 °C and forms a vapor consisting of S 2 , S 6 , and S 8 molecules; at about 1000 °C, the vapor density corresponds to the formula S 2 , which is a paramagnetic molecule like O 2 with a similar electronic structure and a weak sulfur-sulfur double bond.
As seen in this discussion, an important feature of the structural behavior of the nonmetals is that the elements usually occur with eight electrons in their valence shells. If necessary, the elements form enough covalent bonds to supplement the electrons already present to possess an octet. For example, members of group 15 have five valence electrons and require only three additional electrons to fill their valence shells. These elements form three covalent bonds in their free state: triple bonds in the N 2 molecule or single bonds to three different atoms in arsenic and phosphorus. The elements of group 16 require only two additional electrons. Oxygen forms a double bond in the O 2 molecule, and sulfur, selenium, and tellurium form two single bonds in various rings and chains. The halogens form diatomic molecules in which each atom is involved in only one bond. This provides the electron required necessary to complete the octet on the halogen atom. The noble gases do not form covalent bonds to other noble gas atoms because they already have a filled outer shell.


2.6: Careful Use of Language in Mathematics
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This section might seem like a bit of a sidetrack from the idea of problem solving, but in fact it is not. Mathematics is a social endeavor. We do not just solve problems and then put them aside. Problem solving has (at least) three components:
- Solving the problem. This involves a lot of scratch paper and careful thinking.
- Convincing yourself that your solution is complete and correct. This involves a lot of self-check and asking yourself questions.
- Convincing someone else that your solution is complete and correct. This usually involves writing the problem up carefully or explaining your work in a presentation.
If you are not able to do that last step, then you have not really solved the problem. We will talk more about how to write up a solution soon. Before we do that, we have to think about how mathematicians use language (which is, it turns out, a bit different from how language is used in the rest of life).
Mathematical Statements
Definition
A mathematical statement is a complete sentence that is either true or false, but not both at once.
So a “statement” in mathematics cannot be a question, a command, or a matter of opinion. It is a complete, grammatically correct sentence (with a subject, verb, and usually an object). It is important that the statement is either true or false, though you may not know which! (Part of the work of a mathematician is figuring out which sentences are true and which are false.)
Think / Pair / Share
For each English sentence below, decide if it is a mathematical statement or not. If it is, is the statement true or false (or are you unsure)? If it is not a mathematical statement, in what way does it fail?
- Blue is the prettiest color.
- 60 is an even number.
- Is your dog friendly?
- Honolulu is the capital of Hawaii.
- This sentence is false.
- All roses are red.
- UH Manoa is the best college in the world.
- 1/2 = 2/4.
- Go to bed.
- There are a total of 204 squares on an 8 × 8 chess board.
Now write three mathematical statements and three English sentences that fail to be mathematical statements.
Notice that “1/2 = 2/4” is a perfectly good mathematical statement. It does not look like an English sentence, but read it out loud. The subject is “1/2.” The verb is “equals.” And the object is “2/4.” This is a very good test when you write mathematics: try to read it out loud. Even the equations should read naturally, like English sentences.
Statement (5) is different from the others. It is called a paradox : a statement that is self-contradictory. If it is true, then we conclude that it is false. (Why?) If it is false, then we conclude that it is true. (Why?) Paradoxes are no good as mathematical statements, because it cannot be true and it cannot be false.
And / or
Consider this sentence:
After work, I will go to the beach, or I will do my grocery shopping.
In everyday English, that probably means that if I go to the beach, I will not go shopping. I will do one or the other, but not both activities. This is called an “exclusive or.”
We can usually tell from context whether a speaker means “either one or the other or both,” or whether he means “either one or the other but not both.” (Some people use the awkward phrase “and/or” to describe the first option.)
Remember that in mathematical communication, though, we have to be very precise. We cannot rely on context or assumptions about what is implied or understood.
Definition
In mathematics, the word “or” always means “one or the other or both.”
The word “and” always means “both are true.”
Think / Pair / Share
For each sentence below:
- Decide if the choice x = 3 makes the statement true or false.
- Choose a different value of that makes the statement true (or say why that is not possible).
- Choose a different value of that makes the statement false (or say why that is not possible).
- x is odd or x is even.
- x is odd and x is even.
- x is prime or x is odd.
- x > 5 or x < 5.
- x > 5 and x < 5.
- x + 1 = 7 or x – 1 = 7.
- x ·1 = x or x ·0 = x.
- x ·1 = x and x ·0 = x.
- x ·1 = x or x ·0 = 0.

- What would convince you beyond any doubt that the sentence is true? How could you convince someone else that the sentence is true?
- What would convince you beyond any doubt that the sentence is false? How could you convince someone else that the sentence is false?
Suppose you were given a different sentence: “There is a $100 bill in this envelope.”
- What would convince you beyond any doubt that the sentence is true? How could you convince someone else that the sentence is true?
- What would convince you beyond any doubt that the sentence is false? How could you convince someone else that the sentence is false?
Think / Pair / Share
What is the difference between the two sentences? How does that difference affect your method to decide if the statement is true or false?
Some mathematical statements have this form:
- “Every time...”
- “For all numbers. . . ”
- “For every choice. . . ”
- “It’s always true that. . . ”
These are universal statements. Such statements claim that something is always true, no matter what.
- To prove a universal statement is false, you must find an example where it fails. This is called a counterexample to the statement.
- To prove a universal statement is true, you must either check every single case, or you must find a logical reason why it would be true. (Sometimes the first option is impossible, because there might be infinitely many cases to check. You would never finish!)
Some mathematical statements have this form:
- “Sometimes...”
- “There is some number. . . ”
- “For some choice. . . ”
- “At least once...”
These are existential statements. Such statements claim there is some example where the statement is true, but it may not always be true.
- To prove an existential statement is true, you may just find the example where it works.
- To prove an existential statement is false, you must either show it fails in every single case, or you must find a logical reason why it cannot be true. (Sometimes the first option is impossible!)
Think / Pair / Share
For each statement below, do the following:
- Decide if it is a universal statement or an existential statement. (This can be tricky because in some statements the quantifier is “hidden” in the meaning of the words.)
- Decide if the statement is true or false, and do your best to justify your decision.
- Every odd number is prime.
- Every prime number is odd.
- For all positive numbers \(x, x^{3} > x\).
- There is some number \(x\) such that \(x^{3} = x\).
- The points (1, 1), (2, 1), and (3, 0) all lie on the same line.
- Addition (of real numbers) is commutative.
- Division (of real numbers) is commutative.
Look back over your work. you will probably find that some of your arguments are sound and convincing while others are less so. In some cases you may “know” the answer but be unable to justify it. That is okay for now! Divide your answers into four categories:
- I am confident that the justification I gave is good.
- I am not confident in the justification I gave.
- I am confident that the justification I gave is not good, or I could not give a justification.
- I could not decide if the statement was true or false.
Conditional Statements
Problem 24 (Card Logic)
You have a deck of cards where each card has a letter on one side and a number on the other side. Your friend claims: “If a card has a vowel on one side, then it has an even number on the other side.”
These cards are on a table.
Which cards must you flip over to be certain that your friend is telling the truth?
Think / Pair / Share
After you have thought about the problem on your own for a while, discuss your ideas with a partner. Do you agree on which cards you must check? Try to come to agreement on an answer you both believe.
Here is another very similar problem, yet people seem to have an easier time solving this one:
Problem 25 (IDs at a Party)
You are in charge of a party where there are young people. Some are drinking alcohol, others soft drinks. Some are old enough to drink alcohol legally, others are under age. You are responsible for ensuring that the drinking laws are not broken, so you have asked each person to put his or her photo ID on the table. At one table, there are four young people:
- One person has a can of beer, another has a bottle of Coke, but their IDs happen to be face down so you cannot see their ages.
- You can, however, see the IDs of the other two people. One is under the drinking age, the other is above it. They both have fizzy clear drinks in glasses, and you are not sure if they are drinking soda water or gin and tonic.
Which IDs and/or drinks do you need to check to make sure that no one is breaking the law?
Think / Pair / Share
After you have thought about the problem on your own for a while, discuss your ideas with a partner. Do you agree on which cards you must check? Compare these two problems. Which question is easier and why?
Definition
A conditional statement can be written in the form
If some statement then some statement.
Where the first statement is the hypothesis and the second statement is the conclusion .
Think / Pair / Share
These are each conditional statements, though they are not all stated in “if/then” form. Identify the hypothesis of each statement. (You may want to rewrite the sentence as an equivalent “if/then” statement.)
-
If the tomatoes are red, then they are ready to eat.
The tomatoes are red. / The tomatoes are ready to eat. -
An integer
n
is even if it is a multiple of 2.
n is even. / n is a multiple of 2. -
If
n
is odd, then
n
is prime.
n is odd. / n is prime. -
The team wins when JJ plays.
The team wins. / JJ plays.
Remember that a mathematical statement must have a definite truth value. It is either true or false, with no gray area (even though we may not be sure which is the case). How can you tell if a conditional statement is true or false? Surely, it depends on whether the hypothesis and the conclusion are true or false. But how, exactly, can you decide?
The key is to think of a conditional statement like a promise, and ask yourself: under what condition(s) will I have broken my promise?
Example \(\PageIndex{1}\):
Here is a conditional statement:
If I win the lottery, then I’ll give each of my students $1,000.
There are four things that can happen:
- True hypothesis, true conclusion: I do win the lottery, and I do give everyone in class $1,000. I kept my promise, so the conditional statement is TRUE.
- True hypothesis, false conclusion: I do win the lottery, but I decide not to give everyone in class $1,000. I broke my promise, so the conditional statement is FALSE.
- False hypothesis, true conclusion: I do not win the lottery, but I am exceedingly generous, so I go ahead and give everyone in class $1,000. I did not break my promise! (Do you see why?) So the conditional statement is TRUE.
- False hypothesis, false conclusion: I do not win the lottery, so I do not give everyone in class $1,000. I did not break my promise! (Do you see why?) So the conditional statement is TRUE.
What can we conclude from this? A conditional statement is false only when the hypothesis is true and the conclusion is false. In every other instance, the promise (as it were) has not been broken. If a mathematical statement is not false, it must be true.
Example \(\PageIndex{2}\):
Here is another conditional statement:
If you live in Honolulu, then you live in Hawaii.
Is this statement true or false? It seems like it should depend on who the pronoun “you” refers to, and whether that person lives in Honolulu or not. Let us think it through:
- Sookim lives in Honolulu, so the hypothesis is true. Since Honolulu is in Hawaii, she does live in Hawaii. The statement is true about Sookim, since both the hypothesis and conclusion are true.
- DeeDee lives in Los Angeles. The statement is true about DeeDee since the hypothesis is false.
So in fact it does not matter! The statement is true either way. The right way to understand such a statement is as a universal statement: “Everyone who lives in Honolulu lives in Hawaii.”
This statement is true, and here is how you might justify it: “Pick a random person who lives in Honolulu. That person lives in Hawaii (since Honolulu is in Hawaii), so the statement is true for that person. I do not need to consider people who do not live in Honolulu. The statement is automatically true for those people, because the hypothesis is false!”
Example \(\PageIndex{3}\):
How do we show a (universal) conditional statement is false?
You need to give a specific instance where the hypothesis is true and the conclusion is false. For example:
If you are a good swimmer, then you are a good surfer.
Do you know someone for whom the hypothesis is true (that person is a good swimmer) but the conclusion is false (the person is not a good surfer)? Then the statement is false!
Think / Pair / Share
For each conditional statement, decide if it is true or false. Justify your answer.
- If \(2 \times 2 = 4\) then \(1 + 1 = 3\).
- If \(2 \times 2 = 5\) then \(1 + 1 = 3\).
- If \(\pi > 3\) then all odd numbers are prime.
- If \(\pi < 3\) then all odd numbers are prime.
- If a number has a 4 in the one’s place, then the number is even.
- If a number is even, then the number has a 4 in the one’s place.
- If the product of two numbers is 0, then one of the numbers is 0.
- If the sum of two numbers is 0, then one of the numbers is 0.
Think / Pair / Share (Two truths and a lie)
On your own, come up with two conditional statements that are true and one that is false. Share your three statements with a partner, but do not say which are true and which is false. See if your partner can figure it out!