Chi-Square Applications: Tests for Independence and Goodness of Fit

This session puts the chi-square distribution to work on two classic inference problems: deciding whether two categorical variables are independent, and deciding whether a sample could have come from a specific distribution. Both tests use the same chi-square statistic, and both expect you to carry out the full calculation yourself — these are the sort of problems you are likely to be asked to compute in the exam. The session works through two complete textbook examples, one for each test, and closes with a short preview of the next chapter: the design of experiments and the analysis of variance.

6.1 Chi-square Test for Independence

6.1.1 Recap: Where the Chi-square Distribution Sits

Before we look at today's problems, it helps to remember where we are. Apart from the Z distribution and the T distribution, the chi-square distribution is the third fundamental distribution our inference has been built on. Last session covered what the chi-square distribution is and why it matters for making inferences about variances — a key idea, since variance appears everywhere in statistics. We also saw the chi-square distribution at work comparing multiple proportions, where we asked whether several population proportions could all be equal. The chi-square statistic turns out to be the common engine behind all of these: it measures how far observed counts sit from the counts we would expect under a null hypothesis.

Today builds two more applications on top of that foundation. The first application is the one this section covers: judging whether two categorical variables are independent. A categorical variable is a qualitative variable — the data is collected in terms of categories rather than numbers (male/female, light/regular/dark, yes/no). If your data contains two such categorical variables, the chi-square distribution helps you answer one basic question: are these two variables independent of each other, or is there some dependency among them?

6.1.2 The Setting: Two Categorical Variables

The whole picture works like this. We take one random sample from a population and record, for every person (or item) in the sample, the values of the two categorical variables. We arrange the data in a table: the categories of one variable become the rows, and the categories of the other variable become the columns. The table has $R$ rows and $C$ columns, and every cell holds a count — the number of sample members that fell in that combination of categories. Those counts are the observed frequencies, written $F_{ij}$, where $i$ indexes the row (so $i=1,2,\dots ,R$) and $j$ indexes the column (so $j=1,2,\dots ,C$). Along the edges we keep the row totals, the column totals, and the grand total of all observations, usually written $N$. A table like this is called a contingency table, because it shows how the two variables sit together.

The test we run on this table is simply called a test for independence, and the textbook gives the whole procedure on page 525. The worked details that follow use that treatment closely.

6.1.3 The Hypotheses of the Independence Test

As with any hypothesis test, the first step is to state the two logical statements — the null and the alternative — before doing any calculation. This is a fundamental habit: whenever a problem asks for a test, you should write down $H_0$ and $H_a$ explicitly. For the test for independence, the standard pair of statements is:

$$H_0:\text{the two categorical variables are independent}$$

$$H_a:\text{the two categorical variables are not independent (an association exists between them)}$$

You can write exactly these statements in the exam whenever you encounter a test for independence. The logic behind putting independence in the null is worth understanding: the main motto here is to gather evidence against the null. The moment we reject $H_0$, we are automatically saying that a dependency exists — the variables are not independent, so there is some association between them. Claiming an association is the stronger claim, and strong claims need strong evidence; that is why the association lives in the alternative while independence sits in the null as the default we assume until proven otherwise. Think of it as the working assumption of the test: the data must convince us to give it up.

6.1.4 Expected Frequencies and the Test Statistic

Once the hypotheses are stated, the procedure runs in a small number of steps. First, assume the null hypothesis is true. Under independence, we compute for every cell the expected frequency — the count we would expect to see in that cell if the two variables were unrelated. The formula is a very simple one:

$$E_{ij}=\frac{\text{row }i\text{ total}\times \text{column }j\text{ total}}{\text{grand total}}$$

which is the same formula used in the previous method we studied. In words: to get the expected frequency for a cell, take its row total, multiply by its column total, and divide by the grand total. The instructor's verbal description of the rule was: "if I want to find out the expected frequency for this particular cell... it should be row total into each column total divided by grand total."

Why does this formula make sense? If the two variables are independent, the chance of landing in row $i$ and column $j$ at the same time is the product of the two separate chances:

$$P(\text{row }i\text{ and column }j)=P(\text{row }i)\times P(\text{column }j)$$

We estimate the two separate chances from the sample as $P(\text{row }i)\approx \text{row }i\text{ total}/N$ and $P(\text{column }j)\approx \text{column }j\text{ total}/N$. The expected count in the cell is the sample size times that probability:

$$E_{ij}=N\times \frac{\text{row }i\text{ total}}{N}\times \frac{\text{column }j\text{ total}}{N}=\frac{\text{row }i\text{ total}\times \text{column }j\text{ total}}{N}$$

so the row-total-times-column-total-over-grand-total rule is really the independence probability rule, converted into counts.

Second, compute the test statistic. The chi-square test statistic is a double summation over all cells of the squared gap between what we observed and what we expected, relative to the expectation:

$${\chi }^2=\sum _{i=1}^R\sum _{j=1}^C\frac{(F_{ij}-E_{ij}{)}^2}{E_{ij}}$$

The verbal description that goes with it: "the chi-square test statistic is double summation $F_{ij}$ minus $E_{ij}$ whole square by $E_{ij}$," where $F_{ij}$ is given by the data and $E_{ij}$ is computed from the formula above. Three design choices in this formula matter. Squaring the gap $(F_{ij}-E_{ij}{)}^2$ makes every contribution positive, so cells where we saw too many counts and cells where we saw too few both add to the total instead of cancelling. Dividing by $E_{ij}$ weights each cell relative to its size: a deviation of 3 matters more in a cell expecting 5 than in a cell expecting 50. And the whole statistic is a sum of non-negative terms, so ${\chi }^2\ge 0$ always — it can only grow when the observed counts drift away from the expected ones. If the variables really are independent, the observed counts sit close to the expected ones and the statistic stays small; if they are not, the gaps grow and the statistic grows with them.

There is a condition attached to this formula, and it is a very, very essential condition: each expected frequency must be 5 or more ($E_{ij}\ge 5$ for every cell). If some cell comes out under 5, we cannot use the statistic as it stands; the remedy is to merge cells — combine two neighboring categories into one so the combined expected frequency climbs to 5 or above — and then recalculate. We return to this in the Q&A section, because there is a subtlety about degrees of freedom when merging happens.

Third, the degrees of freedom. For the test for independence they are

$$df=(R-1)(C-1)$$

— "rows minus 1 into columns minus 1." The intuition: once the row totals and column totals are fixed, you are free to fill only $(R-1)(C-1)$ of the cells by hand; every remaining cell is forced by the totals, so the chi-square distribution describing the statistic has exactly that many degrees of freedom. For the worked example that follows, that is $(3-1)(2-1)=2$.

One more habit from this session: instead of comparing the statistic against a critical value from a table, we use the p-value approach for the rejection rule. The instructor put it directly: "I am not interested in talking about this critical value — we will discuss the rejection rule approach with the help of the basic p-values." If the p-value comes out less than the level of significance $\alpha$, we reject the null; if it is greater, we fail to reject. The arithmetic itself is light: for an $R\times C$ table you compute $R\times C$ cell contributions, each a handful of calculator steps, so the whole test takes minutes by hand and an instant in software — the real cost is collecting a random sample large enough that every expected count reaches 5.

6.1.5 Worked Example: Beer Preference and Gender

Two practical notes from the class. First, if you are comfortable doing the work in a tabulated form like this, that is fine — presenting the calculation as a table is perfectly acceptable. Second, many statistical calculators let you enter the raw data and read off the mean and variance directly, which is handy for checking your hand calculation.

A bar chart of the two genders side by side tells the same story the test will confirm. Put beer preference (light, regular, dark) on the horizontal axis and the proportion within each gender on the vertical axis, running from 0 to about 0.7. Women's highest bar is light beer at about 57%, while men's highest bar is regular beer at about 42%; men also prefer dark beer more often than women. The two sets of bars are visibly uneven — that visual gap is what the chi-square statistic will put on a number scale.

6.1.6 Conclusion: p-value and the Decision

With ${\chi }^2=6.45$ and $df=(3-1)(2-1)=2$, we go to the chi-square table with the upper-tail area in mind. The question put to the class was: roughly what might the p-value be, and what is your conclusion — is p greater than alpha or less than alpha? The chi-square table for 2 degrees of freedom reads: an upper-tail area of 0.10 corresponds to the value 4.605, 0.05 corresponds to 5.991, 0.025 corresponds to 7.378, 0.01 corresponds to 9.210, and 0.005 corresponds to 10.597. Our statistic 6.45 sits between 5.991 and 7.378, so the p-value for this example lies somewhere between 0.05 and 0.025, which already indicates that p is strictly less than the default level of significance. The rule of thumb: if the level of significance is not given in a problem, consider it at 5% ($\alpha =0.05$) and compare with that. Since 6.45 exceeds the 5% table value 5.991, the p-value is certainly less than 0.05, and software confirms the exact value $p=0.0398$.

So we reject the null hypothesis. In the language of the problem: there is a dependency between the two categorical variables — beer preference is not independent of gender. If you want to write the conclusion more precisely than "reject the null," you can write it literally: "They are not independent — beer preference depends on gender." Rejecting the null is one aspect; saying what it means in words is the part that shows real understanding, and it is a skill that gets evaluated later in the course (see the Exam Guidance Summary).

6.1.7 Student Questions and Answers

The class raised five questions about this example. The first two were about the arithmetic, the next about the logic of the hypotheses, and the last two about the decision rule and its wording.

The expected-frequency arithmetic raises an immediate worry, and the next question voiced it directly.

With the mechanics settled, the discussion turned to the logic of putting independence in the null hypothesis rather than the association.

The decision rule itself drew one more question — what to conclude when p is less than alpha.

One closing question asked to see the p-value approach for this example end to end.

6.1.8 Exam Notes

The rest of the exam guidance for this concept, in summary form:

• Present the calculation tabulated if you are comfortable with that format — a clean table is a perfectly acceptable presentation and easier to check.
• Remember the condition $E_{ij}\ge 5$ for every cell, and that merging cells changes the degrees of freedom.
• Use the p-value approach with the rejection rule $p\le \alpha ⟹$ reject $H_0$, $p>\alpha ⟹$ fail to reject.
• End every problem by translating the verdict into words — "they are not independent — beer preference depends on gender" — because the writing and interpretation are part of what gets evaluated.

6.2 Chi-square Test for Goodness of Fit

6.2.1 The Idea: Does the Sample Come from a Given Distribution?

The second application of the chi-square distribution is the goodness-of-fit test. Its purpose is to determine whether a population is being sampled from a specific distribution. We come across this question all the time: is this data coming from a binomial distribution? Was this sample drawn from a Poisson distribution? Does this random sample come from a normal distribution or not? To validate whether a given sample comes from a particular distribution, we again make use of the same chi-square formula — the same statistic as in the independence test. The name says it in one line: the test asks how good the fit is between the shape of the sample data and the shape of the hypothesized distribution. If the fit is good, the sample could easily have come from that distribution; if the fit is poor, the observed data and the hypothesized distribution do not match.

6.2.2 The ChemLine Example: Hypotheses

The example is from the textbook: employee aptitude test scores for 50 randomly chosen applicants at ChemLine, a company that hires about 400 new employees annually for its four plants located throughout the United States. Define $X$ as the test score of a randomly chosen applicant — $X$ is our random variable. The personnel director asks whether a normal distribution applies to the population of test scores. If such a distribution can be used, it would be helpful in evaluating specific test scores: scores in the upper 20% or lower 40% of the population could be identified quickly, without sorting through all the individual scores.

The null and alternative follow the pattern we used before. Under $H_0$, the population test scores follow a normal distribution with mean $\mu$ and standard deviation $\sigma$. Under $H_a$, the population test scores do not follow a normal distribution — and if they do not follow a normal, we do not know what they follow, which is a complicated statement, and that is exactly why it sits in the alternative:

$$H_0:X\text{ follows a normal distribution with mean }\mu \text{ and standard deviation }\sigma$$

$$H_a:X\text{ does not follow a normal distribution}$$

As usual, we should visualize what the null is claiming: if the test scores are normal, the entire area under the distribution is spread over the test-score range in the familiar bell shape. Draw the bell curve with test score $X$ on the horizontal axis and the probability density on the vertical axis; the curve peaks at the mean, and its spread is set by the standard deviation. The null says the 50 sample scores are just draws from that single smooth bell.

6.2.3 Estimating the Parameters from the Sample

Before any further analysis, we develop estimates of the mean and standard deviation of the normal distribution, because at the population level we do not know $\mu$ and $\sigma$. This is the theory of estimation at work — the very fundamental idea of inference: we do not know the population mean, but we hold a random sample in our hands, so we work at the sample level and then generalize to the population. If the population mean $\mu$ is unknown, the best replacement is the sample mean; wherever ${\sigma }^2$ appears we use the sample variance, and wherever $\sigma$ appears we use the sample standard deviation:

$$\mu \to \bar{x},\sigma \to s$$

With these estimates, the null hypothesis can be restated in concrete numbers: the population of test scores has a normal distribution with mean 68.42 and standard deviation 10.41. So under the null, $X\sim \mathcal{N}(68.42,10.41^2)$. Remember: for a goodness-of-fit problem you explicitly calculate $\bar{x}$ and $s$, or if they are supplied you just use those given observations directly.

6.2.4 Building the Intervals with the Z-table

Because the normal distribution is a continuous probability distribution, we must use a different procedure for finding the observed and expected frequencies than we did with the categorical table. We hold 50 test scores; to get the observed and expected frequencies we follow two points.

First, to find the observed frequencies, we divide the entire data into different categories — call them intervals — each with a lower bound and an upper bound. The second point is the rule of thumb that controls how many intervals we can make: the expected frequency in each category or interval must be at least 5. With 50 observations, how many intervals should we divide the data into? The answer the class gave: 10. Very good — with 50 observations and 10 equal intervals, each interval holds an expected frequency of 5, which maintains the condition.

The area under the normal curve tells us the probabilities: the whole area is 1, and dividing it into 10 equal parts means each interval carries probability $1/10=0.1$. Cumulatively, the areas build up 0.1, 0.2, 0.3, and so on, until the whole interval is 1. These are the probabilities we use to find the cutoff points between intervals.

Since $X$ follows a normal distribution, we can standardize it using the relation between $X$ and the standard normal variable $Z$:

$$Z=\frac{X-\mu }{\sigma }$$

and when $\mu$ and $\sigma$ are unknown, we replace them with the estimates from the sample:

$$Z=\frac{X-\bar{x}}{s}$$

This is exactly how the interval bounds in the worked problem (55.10, 59.68, and the rest) are found. On the bell curve, draw vertical lines at each of these cutoffs; they slice the area under the curve into ten equal strips, each holding probability 0.1.

6.2.5 Observed and Expected Frequencies

With the intervals fixed, the two frequency columns are filled in different ways. The observed frequencies $F_{ij}$ come from counting: tally how many of the 50 test scores actually fall inside each interval. The expected frequencies $E_{ij}$ come from the null hypothesis: if the scores really are normal with mean 68.42 and standard deviation 10.41, then each of the 10 intervals should hold an expected frequency of $50\times 0.1=5$ — the sample size times the interval probability. Doing the tally for the ChemLine data gives:

The expected column is uniform — exactly 5 in every row — because the intervals were designed to be equal-probability. The observed column shows where the real data bunches: 9 scores landed between 59.68 and 63.01 (more than the 5 the normal would predict), while only 2 scores landed in each of the bands 65.82–68.42 and 71.02–73.83 (fewer than predicted). Those bumps and dips are the "misfit" the chi-square statistic will measure.

The comparison with the independence test is worth making explicit:

When to pick which: if your data is two categorical variables already summarized in a table, use the independence test; if your data is one set of raw values and you want to check its distribution family, use goodness of fit.

6.2.6 The Test Statistic, Degrees of Freedom, and p-value

The test statistic is the same chi-square expression as before:

$${\chi }^2=\sum _{i=1}^k\frac{(f_i-e_i{)}^2}{e_i}$$

with the sum taken over the intervals. For the ChemLine data, the cell-by-cell computation is:

So the computed value of the test statistic is ${\chi }^2=7.2$. The only nonzero contributions come from the intervals where the observed count actually differed from 5 — the two overcrowded bands (3.2 and 0.2 from the +4 and +1 surplus) and the two undercrowded bands (1.8 and 1.8 from the two −3 gaps) — which is the whole logic of the statistic in miniature: no difference, no contribution.

The degrees of freedom, however, follow a different rule. For the independence test it was $(R-1)(C-1)$; for the goodness-of-fit test it is

$$df=k-p-1$$

where $k$ stands for the number of intervals or categories — here $k=10$ — and $p$ stands for the number of parameters of the distribution estimated from the sample. The distribution here is the normal distribution, whose parameters are $\mu$ and $\sigma$; both were unknown, and we estimated both from the sample observations (that is what we did with $\bar{x}$ and $s$), so $p=2$. This gives

$$df=10-2-1=7$$

This $k-p-1$ convention is the one people generally follow whenever a goodness-of-fit problem comes into the picture. Why subtract $p$? Because each estimated parameter eats one degree of freedom: the sample mean and sample standard deviation were not specified by the theory, they were fitted to the data, so the test must be less strict — and $df$ drops by one for each.

Now for the p-value. Consider a portion of the chi-square table at 7 degrees of freedom: the value 2.833 has an area to its right of 0.90, and the value 12.017 has an area to its right of 0.10. Our statistic 7.2 lies somewhere between 2.833 and 12.017, so the p-value — the area to the right of the statistic — lies between 0.1 and 0.9, and by mere inspection of the diagram it is certainly greater than 0.1. There is no doubt: the p-value is greater than alpha, where alpha is 0.1 as given in the problem. From software, the exact p-value comes out to about 0.4084 — a value we cannot read from the tables, but we do not need to: writing "it lies between 0.1 and 0.9," or "p is greater than 0.1," or "p is less than 0.9" is all acceptable, as long as you can anticipate where the p-value falls. The decision rule is the same good old story as for the Z test, the T test, and the chi-square distribution alike:

$$p\le \alpha ⟹\text{reject }{H}_0,p>\alpha ⟹\text{fail to reject }{H}_0$$

6.2.7 Conclusion: Accepting the Normal Distribution

Since the p-value is greater than alpha, we fail to reject the null hypothesis — we accept that the population test scores follow a normal distribution with mean 68.42 and standard deviation 10.41. The conclusion can be carried back to the person who asked the question: you can recommend to the personnel director that the normal distribution may be applied to assist the interpretations he wants — finding the lowest 20%, the lowest 40%, the lowest 60%, and so on, quickly. That is the whole payoff of the test: once we establish that the sample plausibly came from a normal population, the normal distribution becomes a working tool for evaluating individual scores.

Think about what this unlocks in practice. With the normal model in hand, the personnel director no longer needs to sort 400 applicants' files by hand: a score's distance from the mean in standard-deviation units gives its percentile immediately, so "upper 20%" and "lower 40%" cutoffs fall straight out of the Z table. The same logic runs through grade assignment in classes, performance-award decisions in companies, and quality screening in manufacturing — anywhere a bell-shaped assumption lets you turn one raw score into a position in the population.

6.2.8 Student Questions and Answers

The class asked seven questions about the goodness-of-fit example, from the setup of the intervals to the fine print of the decision.

With the intervals settled, the first cutoff value came under scrutiny.

Next, a student read a number from the chi-square table and wondered whether it was the statistic.

The discussion then broadened to what belongs in the course and which normality checks exist beyond chi-square.

One student asked how far the method reaches beyond the normal case.

Another question tested the significance-level default.

Finally, a question about the normal table itself — which convention to follow.

6.2.9 Exam Notes

The rest of the exam guidance for this concept, in summary form:

• No binomial or Poisson goodness-of-fit questions will be asked; normal-distribution problems are acceptable in the exam.
• The p-value decision rule (p ≤ alpha rejects, p > alpha fails to reject) is the same for the Z test, the T test, and the chi-square distribution.
• The observed frequencies come from tallying the raw data; the expected frequencies are $50\times 0.1=5$ per interval — build the two columns the goodness-of-fit way, not the independence-test way.
• The ten equal-probability intervals come from the Z table cutoffs (55.10, 59.68, 63.01, 65.82, 68.42, 71.02, 73.83, 77.16, 81.74); remember the expected frequency of at least five in every interval.

6.3 Preview: Design of Experiments and Analysis of Variance

6.3.1 Design of Experiments

The next chapter is Chapter 13, and it opens a new area: the design of experiments. The plan is to cover, in the best possible way, what exactly experimental designs are, why designing an experiment is necessary, and what its relevance is. There are many types of designs; depending on the practical situation you are in, you will design an experiment and recommend the type that should be performed so that you get the best possible output or response.

This matters most for people who will work in the pharmaceutical industry or in biological and life sciences. They should not perform experiments throughout the year — that is hugely time-consuming and cost-consuming. Instead, they should run a clever combination of experiments and still reach sound conclusions. Finding the best possible combination of runs to perform, and getting the best output for the least effort, is what design of experiments is all about. The chapter also covers what types of designs are available and how to address those situations.

6.3.2 Analysis of Variance

Alongside the designs, the next session will take up another very interesting procedure conceptually: the analysis of variance, or ANOVA. We already know the basic significance of variance. Analysis of variance asks what we can analyze further in terms of variance, especially when it comes to the design of experiments. The plan is to work through Chapter 13 with examples, showing how the calculation part is carried out — and how the designs connect to the conceptual understanding of analysis of variance. These topics are all in Chapter 13, which is the focus of the coming session.

The connection between the two halves of the preview is direct: an experiment is designed so that the variance in the measured response can be split into parts — variance due to the treatments you changed on purpose, and variance due to chance. ANOVA is the tool that reads that split. Where today's chi-square tests asked about frequencies in categories, ANOVA asks about differences between group means — a new question, a new distribution (the F distribution), and the same overall habit: state the hypotheses, compute a test statistic, read the p-value, and state the conclusion in words.

Exam Guidance Summary

• The two chi-square applications are exam-required. Both the test for independence and the goodness-of-fit test were flagged as "very, very required," with the expectation that you will calculate them in the exam. Practice both calculation styles: the table-based independence computation and the interval-based goodness-of-fit computation.
• Open-book exam. No extra tables are needed — check the table values given and write your own comment.
• Always write the hypotheses. State $H_0$ and $H_a$ explicitly for every problem. The standard independence statements can be written as-is; justify whatever variant you choose, and always give the conclusion in words, not just "reject."
• Default alpha. When the level of significance is not given, use 5% by default — that is the common rule.
• p-value approach. The rejection rule runs on p-values: $p\le \alpha$ rejects the null, $p>\alpha$ fails to reject. Approximate p-values ("lies between", "greater than 0.1") are acceptable; exact values come only from software.
• Degrees of freedom. Independence test: $(R-1)(C-1)$. Goodness of fit: $k-p-1$, where $k$ is the number of intervals and $p$ is the number of parameters estimated from the sample.
• The $E\ge 5$ condition. Every expected frequency must be at least 5; if not, merge cells — and merge the observed frequencies in the same way — then adjust the degrees of freedom.
• Goodness of fit specifics. Alpha is 0.1 in the worked example; checking at 5% gives the same conclusion. No binomial or Poisson goodness-of-fit questions will be asked. Normal-distribution problems are acceptable in the exam.
• Presentation. Tabulating your calculation is fine and makes the work easier to follow.
• Revision support. The next session is a planned revision of everything discussed so far for the mid-semester, with small questions, and practice problems with solutions will be shared before the mid-semester so you can practice the calculations.
• Where to find it. The chi-square applications live in Chapter 12 of the textbook: the test for independence is on page 525, and the goodness-of-fit example runs through pages 532 to 536.
• Quizzes and assignment. There will be two quizzes, and Quiz 2 covers these chapters; the assignment is a 15-mark group data-analysis report (see the Key Industry Applications section) that includes an interpretation component.

Key Industry Applications

• Real-world: beer industry association survey — a consumer-preference study where the test for independence decides whether beer type (light, regular, dark) is associated with the gender of the drinker. Market-research studies like this are a standard setting for the test for independence, and the verdict feeds straight into promotion and advertising decisions: when preference depends on gender, a manufacturer can customize campaigns for each target market.
• Real-world: ChemLine — a company hiring roughly 400 new employees each year across four plants in the United States. Its personnel director uses the goodness-of-fit test to check whether aptitude test scores follow a normal distribution, which then allows quick evaluation of specific scores (upper 20%, lower 40%, and so on) without sorting through individual files.
• Real-world: normal-curve grading — if an entire class follows a normal (empirical) distribution, then roughly 10% of the students should get A grades, 10% should get E grades, and the middle bands split 20, 20, and so on. The same logic turns a percentile from the Z table into a grade band.
• Real-world: performance awards — companies may lean on the Gaussian distribution when deciding hikes or awards, so that extremely talented performers get recognized; the same goodness-of-fit check that validates the aptitude-test data legitimizes that practice.
• Real-world: pharmaceutical and life-science industries — designed experiments (Chapter 13) let researchers run smart combinations of experiments instead of performing experiments throughout the year, saving huge amounts of time and cost, with ANOVA reading the variance split.
• Real-world: data-analysis software — the assignment invites you to use whatever software is available around you: Power BI, Minitab, SAS, SPSS, RStudio, or Excel, depending on the context; statistical calculators can also compute the mean and variance directly from entered data.