Inferences on the Population Variance

This lecture moves the inference machinery from averages to spread. Where Z and T handled claims about the population mean, the chi-square distribution takes over when the claim is about the population variance ${\sigma }^2$. The session builds the chi-square statistic, uses it to place a confidence interval on ${\sigma }^2$, tests hypotheses about ${\sigma }^2$, and closes with the chi-square test for the equality of three or more population proportions — plus a look at the two applications (independence, goodness of fit) and the F distribution that come next.

5.1 Where Variance Sits in Decision-Making and Estimation

5.1.1 The Opening Example: Old Drug versus New Drug

The very first example of this course compared an old drug with a new drug given to patients, using their heart rates per minute as the measurement. Two density tables were drawn, one for each group of patients. The comparison put us in a dilemma: the average heartbeats per minute were the same for both drugs, so with the help of averages alone we could say nothing about the efficacy of the drugs. Both drugs seemed to deliver the same information.

That is where the variance — the measure of how spread out the observations are around the average — entered the picture. Variance captures information the average cannot: which group of observations is extremely scattered and which one is less scattered. When we calculated the variance for the two groups, the decision became clear: the old drug gives better results than the new drug.

The reasoning is the lesson to keep: the lesser the variance, the more efficient the performer. When two performers tie on average, the one with the smaller spread is the dependable choice. This single example shows why variance is an important measure in decision-making scenarios — it is the measure that separates otherwise identical averages.

The same logic operates in healthcare when treatments are compared: when two treatments produce the same average outcome, the consistency of the response — measured by variance — decides which treatment performs better. It is also the core reasoning of quality control in manufacturing: a filling machine whose weights vary little around a 16-ounce target is preferable to one that overfills and underfills, because even with a correct average fill, a large process variance signals that the mechanism needs readjustment. In finance the same idea is called volatility — investors read a larger variance in returns as more risk.

5.1.2 Z and T for Means, and What We Use for Variance

So far the course has covered the Z distribution and the T distribution. Whenever the problem revolves around the population mean and validating claims about it, Z and T are the distributions to use. When the entire problem is about the population mean $\mu$ — the average of the population — and about validating it, the Z test and T test dominate the discussion.

The common rule we administered: Z is used blindly when $n>30$, and T is used when $n<30$. For large samples the Z distribution works; for small samples we fall back on T.

The test statistic is the familiar one, built from $\bar{X}-\mu$ (the sample average minus the population average) divided by a measure of the average's spread:

$$Z=\frac{\bar{X}-\mu }{\sigma /\sqrt{n}}\text{and}T=\frac{\bar{X}-\mu }{S/\sqrt{n}},$$

described in class as "X bar minus mu by sigma by root n, or X bar minus mu by S by root n". Here $\bar{X}$ is the sample average, $\mu$ is the population average (the target value in the hypotheses), $\sigma$ is the population standard deviation, $S$ is the sample standard deviation, and $n$ is the number of observations. After computing the statistic, we decided the question with the critical value approach or the p-value approach.

When the population standard deviation is known, Z is the natural tool; when $\sigma$ must be estimated from the data, T takes over. That is the reasoning packed into "$n>30$ gives Z, $n<30$ gives T".

The takeaway to carry forward: whenever the inference involves averages, the Z distribution and T distribution work well. But when the inference is about the variance instead of the average, Z and T are of no use — the very point of this session. That is the boundary the whole session builds on.

5.1.3 Replacing Unknown Population Values with Sample Values

Under the theory of estimations, the course rests on three population parameters: the population mean $\mu$, the population variance ${\sigma }^2$, and the population proportion $p$. These are the three things the text refers to, and if you understand these thoroughly, the remaining machinery — computing p-values, taking decisions depending on probability values — is very similar across all three.

One replacement principle runs through all of it. When the population mean is unknown, we replace it with the sample mean $\bar{X}$. On a similar note, when the population variance ${\sigma }^2$ or the population standard deviation $\sigma$ is unknown, we replace it with the sample variance $S^2$ (or the sample standard deviation $S$). If you do not know anything about the population, you collect sample data and let the sample stand in for it — that is the whole idea of estimation.

The sample variance is

$$S^2=\frac{{\sum }_{i=1}^n(X_i-\bar{X}{)}^2}{n-1},$$

described in class as "XI minus X bar whole square by N minus one": each observation $X_i$ minus the sample average $\bar{X}$, squared, summed over all observations from $i=1$ to $i=n$, and divided by $n-1$. Dividing by $n-1$ instead of $n$ corrects for the fact that one piece of freedom is already used up: the deviations are measured from $\bar{X}$, which is itself computed from the same data. This is the degrees-of-freedom idea, which returns in full in the next section.

Under the theory of estimations, whenever ${\sigma }^2$ is unknown, the sample variance is the best estimate for the population variance. In the same way $\bar{X}$ stands in for $\mu$, $S^2$ stands in for ${\sigma }^2$.

5.2 The Chi-Square Distribution

5.2.1 The Chi-Square Statistic and Its Degrees of Freedom

Suppose the logical statements — the null and alternative hypotheses — are written about the variance. For example, the hypotheses claim something about ${\sigma }^2$ or $\sigma$. In that situation the Z and T distributions are of no use; we will not talk about them. The fundamental distribution that addresses inferences on the population variance is the chi-square distribution, written ${\chi }^2$, with $n-1$ degrees of freedom.

The statistic is

$${\chi }^2=\frac{(n-1)S^2}{{\sigma }^2},$$

which the class described as "n minus 1 into S square by sigma square". Here $S^2$ is the sample variance, ${\sigma }^2$ is the population variance claimed in the hypotheses, and $n$ is the sample size; the quantity $(n-1)S^2/{\sigma }^2$ is the test statistic used whenever the inference concerns the population variance. By construction, this is the definition of the chi-square statistic for variance inference.

Two structural facts about the distribution matter. First, as the degrees of freedom change, the graph of the chi-square distribution transforms its shape. Second, even the chi-square distribution behaves like the normal distribution when $n$ goes large and large — a separate issue, but a useful mental anchor: many distributions converge to the normal as the sample size grows.

The Greek letter $\nu$ (pronounced "nu") is commonly used for the degrees of freedom, and generally $\nu =n-1$. The relation works in both directions: if $\nu =8$, the sample size is 9; if $\nu =12$, the sample size is 13; if $\nu =20$, the sample size is 21. Keep this relation in mind when reading tables.

There is also an important contrast with Z. For the Z distribution the sample size is immaterial, because Z is used for large samples — we never use the sample size to look up Z values. But for the T distribution and the chi-square distribution, if the sample size is $n$, you must look up the table values at $n-1$ degrees of freedom. That distinction — Z ignores the sample size, T and ${\chi }^2$ use $n-1$ — is one of the recurring mechanics of this session.

5.2.2 What Degrees of Freedom Mean

The degrees of freedom — the number of independently varying pieces of information a statistic has — is best understood through the small example given in class: choose four numbers without any specific restrictions. You freely choose 4, 7, 10 and 15 — any numbers, no constraint. All four are free.

Now impose a restriction: choose four numbers such that their average is 11. If the average of the four numbers must be 11, their total must be $4\times 11=44$. You choose the first three freely — say 4, 7 and 10 — but the fourth number is no longer free: it must be 23 so that $4+7+10+23=44$. The restriction on the average forces the last number.

So with $n$ observations under one imposed restriction, only $n-1$ of them are free. In general, the degrees of freedom count the independent observations that can be chosen freely; the constrained ones depend on the restriction. In the example, three numbers were chosen freely and the fourth depended on the average restriction, which is why we say the degrees of freedom are $n-1$. The same reasoning explains the $n-1$ in the sample variance $S^2=\sum (X_i-\bar{X}{)}^2/(n-1)$: the deviations are measured around $\bar{X}$, and the average itself is the restriction that consumes one degree of freedom. This shows up whenever the T and chi-square distributions are in play (small-sample situations). For the Z distribution it makes no impact, because Z operates on large samples where the restriction penalty is negligible.

5.2.3 Properties of the Chi-Square Distribution

Two properties distinguish the chi-square distribution from everything seen before, and both came out of a class exchange:

The first property — no negative values — is the classic example of a classroom correction: the shape guess ("right skewed") is fine, but the defining property is that chi-square never goes negative, and the reason is visible in the formula itself (everything is squared).

The second property — non-symmetry — changes how you handle critical values. For a symmetric distribution you can write $Z_{\alpha /2}=-Z_{1-\alpha /2}$ and be done with one value. For chi-square, ${\chi }_{0.025}^2$ is not equal to ${\chi }_{0.975}^2$. When you compute the two critical limits of the two ends of a chi-square problem, you must be careful and read both values from the table — you cannot derive one from the other by a sign flip.

That last point is worth underlining: the empirical rule (the 68-95-99.7 pattern) is a normal-distribution property, and when a distribution is not normal, the correct tool for bounding the proportion of observations is Chebyshev's inequality.

5.2.4 Notation and Reading the Chi-Square Table

The notation mirrors the Z notation you already know. The symbol ${\chi }_{\alpha }^2$ denotes the chi-square distribution value that provides an area (probability) of $\alpha$ to the right of ${\chi }_{\alpha }^2$. All the table values are upper-tail (right-tail) probabilities: when you look for the value, you fix the degrees of freedom row, then find the column whose right-tail area equals your $\alpha$.

The worked reading in class used 19 degrees of freedom. To find ${\chi }_{0.025}^2$ at 19, freeze the 19-degrees-of-freedom row and take the column with right-tail area 0.025:

$${\chi }_{0.025}^2(19)=32.852,$$

meaning the area to the right of 32.852 is 0.025. For the other end, the value with right-tail area 0.975 is

$${\chi }_{0.975}^2(19)=8.907,$$

meaning the area to the right of 8.907 is 0.975. These two values are completely different — 32.852 versus 8.907. With the Z distribution you could write the negative of one value and be done, because of symmetry; with the chi-square distribution that does not work. The two critical limits for the two ends must each be read from the table independently.

5.3 Confidence Interval for the Population Variance

5.3.1 The Confidence Interval Formula

A single point estimate like $S^2$ will almost never hit the true population variance exactly. The natural next step is an interval: a range of plausible values for ${\sigma }^2$, built so that a chosen percentage of all such intervals (95 percent, say) will capture the true value. To build a confidence interval for the population variance we use the chi-square distribution at $n-1$ degrees of freedom.

The interval follows from the sampling fact from the last section: the statistic $(n-1)S^2/{\sigma }^2$ follows a chi-square distribution with $n-1$ degrees of freedom. Because 95 percent of the values of a chi-square distribution lie between ${\chi }_{0.975}^2$ and ${\chi }_{0.025}^2$, we can write the probability statement

$$P\left({\chi }_{1-\alpha /2}^2\le \frac{(n-1)S^2}{{\sigma }^2}\le {\chi }_{\alpha /2}^2\right)=1-\alpha ,$$

which says: with probability $1-\alpha$, the ratio $(n-1)S^2/{\sigma }^2$ falls between the small chi-square value (area $1-\alpha /2$ to its right) and the large chi-square value (area $\alpha /2$ to its right). For 95 percent, $\alpha =0.05$ and the two ends are ${\chi }_{0.975}^2$ and ${\chi }_{0.025}^2$.

To turn this probability statement about the ratio into a statement about ${\sigma }^2$, invert it algebraically. Every quantity in the ratio is positive, so multiplying by ${\sigma }^2$ and dividing by the chi-square values keeps the inequalities pointing the same way. Working the left inequality first:

$${\chi }_{1-\alpha /2}^2\le \frac{(n-1)S^2}{{\sigma }^2}\Rightarrow {\sigma }^2\le \frac{(n-1)S^2}{{\chi }_{1-\alpha /2}^2},$$

and the right inequality gives

$$\frac{(n-1)S^2}{{\sigma }^2}\le {\chi }_{\alpha /2}^2\Rightarrow \frac{(n-1)S^2}{{\chi }_{\alpha /2}^2}\le {\sigma }^2.$$

Combining the two gives the confidence interval for the population variance:

$$\frac{(n-1)S^2}{{\chi }_{\alpha /2}^2}\le {\sigma }^2\le \frac{(n-1)S^2}{{\chi }_{1-\alpha /2}^2},$$

described in class as: take $n-1$ times $S^2$, and divide it by the two chi-square values whose right-tail areas are $\alpha /2$ and $1-\alpha /2$. The larger chi-square value goes in the denominator of the lower limit, and the smaller one in the denominator of the upper limit — a consequence of the right-tail definition: ${\chi }_{\alpha /2}^2$ is the large end (small right-tail area) and ${\chi }_{1-\alpha /2}^2$ is the small end (large right-tail area).

A picture of the idea: draw the chi-square curve with 8 degrees of freedom — a right-skewed hump that starts at zero, rises to a peak, then falls off slowly to the right. Mark 2.180 on the left slope and 17.535 far out on the right tail. The area between them is 95 percent of the total; the two thin remaining slices at the ends are 2.5 percent each. The interval formula simply re-scales this band by $(n-1)S^2$. The takeaway: the band is wide because the chi-square distribution is skewed — the two ends are not symmetric mirror images.

Two cautions from class. First, keep the symbols straight: S means the standard deviation and S square means the variance — if you use $S$ where $S^2$ belongs, and $S^2$ where $S$ belongs, your entire calculation goes wrong. Second, remember that the chi-square distribution is not symmetric, so the two critical values for the two ends are different and both must come from the table.

5.3.2 Worked Example: 95% Interval from a Sample of Nine

Setup. It is observed that a sample of size 9 has sample standard deviation $S=2.81$. Develop a 95% confidence interval for the population variance ${\sigma }^2$.

Given values. The sample size is $n=9$, so the degrees of freedom are $n-1=8$. The sample standard deviation is $S=2.81$, so $S^2=2.81^2=7.8961$. A 95% confidence interval means $1-\alpha =0.95$, so $\alpha =0.05$ and $\alpha /2=0.025$. We need ${\chi }_{0.025}^2$ and ${\chi }_{0.975}^2$, both at 8 degrees of freedom.

Reading those two values from the chi-square table produced a small correction in class:

5.3.3 The 90% Interval Variant

The same problem was re-run for a 90% confidence interval in response to a question:

So for a 90% interval, $1-\alpha =0.90$, $\alpha =0.10$, $\alpha /2=0.05$, and the critical values at 8 degrees of freedom are ${\chi }_{0.05}^2(8)=15.507$ and ${\chi }_{0.95}^2(8)=2.773$. The interval becomes

$$\frac{63.1688}{15.507}\le {\sigma }^2\le \frac{63.1688}{2.773},$$

giving $4.07\le {\sigma }^2\le 22.78$ — a narrower interval than the 95% version, exactly as expected: less confidence, narrower band.

A note on the upper denominator: the lecture reads the value as 2.773, and the calculation above uses it. Standard chi-square tables list 2.733 at 8 degrees of freedom for right-tail area 0.95; with 2.733 the upper limit would be $63.1688/2.733\approx 23.11$ instead of 22.78. Both are fine on the exam — what matters is reading the two denominators off the table row you are given and using them consistently.

The lesson of the variant: when the confidence level changes from 95% to 90%, the two denominators change, and the same problem can be solved for any confidence level by swapping the critical values.

5.4 Hypothesis Testing for the Population Variance

5.4.1 The Testing Setup and Decision Rules

Hypothesis testing for the variance follows exactly the logic used for the population mean, with two differences: the hypotheses are about ${\sigma }^2$, and the test statistic is different. The test statistic is the chi-square quantity from before:

$${\chi }^2=\frac{(n-1)S^2}{{\sigma }^2},$$

where ${\sigma }^2$ in the denominator is the value claimed by the null hypothesis — written ${\sigma }_0^2$ to make that explicit. Everything else is the familiar machinery: we reject the null hypothesis when the p-value is less than or equal to $\alpha$, and we fail to reject (accept) when the p-value is greater than $\alpha$.

The critical value approach mirrors the p-value logic. For a right-tailed test (alternative of the form ${\sigma }^2>{\sigma }_0^2$), reject the null hypothesis when the observed chi-square is greater than or equal to ${\chi }_{\alpha }^2$ at $n-1$ degrees of freedom. For a left-tailed test the condition flips to the left end. The two-tailed case involves both ends.

One recurring clarification concerned the level of the test:

5.4.2 Worked Example: One-Sided Test (p-Value and Critical Value Approaches)

Setup. A sample of 16 items gave a sample standard deviation of $S=9.5$. Test the hypotheses $H_0:{\sigma }^2=50$ against $H_A:{\sigma }^2>50$, using both the p-value approach and the critical value approach.

Where the claimed value 50 comes from drew a question, and the answer is the same as in the Z test:

A table check confirmed the critical value:

How to report such p-values drew a clarifying exchange:

5.4.3 Worked Example: Two-Sided Test from a Sample of Fifteen

Setup. The textbook problem (Chapter 11, page 496, Problem 10) is a consumer example in which the population standard deviation is claimed to be $\sigma =12$. A sample of 15 observations is collected. Test whether the claim $\sigma =12$ should be rejected at the 0.95 level of significance, i.e. $\alpha =0.05$.

A student checked the p-value magnitude:

A second student checked the degree-of-freedom reading:

A note on exact p-values. For the Z distribution we can compute the exact p-value easily. For the T and chi-square distributions, exact p-values are cumbersome by hand — the tables only bracket them — so we report a range ("p-value greater than 0.10", "p-value between 0.025 and 0.05") or get the exact value from software such as Excel.

5.5 Chi-Square Test for Equality of Population Proportions

5.5.1 The Setting: Comparing Three Population Proportions

Beyond general inference on the variance, the chi-square distribution has several further applications — the equality of proportions (Section 12.1 of the text), the test of independence between two categorical variables, and the goodness of fit of a claimed distribution. The first of these is the subject of this section; the repurchase-intention survey of car owners used as the worked example lives here too.

The situation: until now the analysis was single-population — one population, sample observations, hypotheses of the form $H_0:\mu ={\mu }_0$ against $H_A:\mu \ne {\mu }_0$. The same analysis extends to two populations — is population 1 better than population 2? — and for proportions the same pattern holds: with a single sample we test $H_0:p=p_0$, and with two populations we take two sample sizes and test whether the population proportions $p_1$ and $p_2$ differ. The Z and T distributions serve those comparisons well.

But what happens with three or more populations? The hypotheses generalize to

$$H_0:p_1=p_2=p_3\text{against}{H}_A:\text{not all equal}.$$

This is the situation where the chi-square distribution helps rather than Z or T. The observed frequencies — the counts actually recorded from the survey or questionnaire — are written $F_{ij}$, where $i$ indexes the rows and $j$ the columns. The observed frequencies may come from anything: a questionnaire collecting feedback from individual customers, tabulated by the categories under study.

The expected frequencies $E_{ij}$ — the counts we would expect if the null hypothesis were true — come from a simple formula:

$$E_{ij}=\frac{R_i\times C_j}{N},$$

described in class as "row i total into column j total by the grand total": the total of row $i$, times the total of column $j$, divided by the total sample size $N$.

Where does this formula come from? Suppose the null of equal proportions is true. Then the single best estimate of the repurchase proportion for every group is the pooled proportion from all the data: the total number of "yes" answers divided by the total sample size. Multiplying that pooled proportion by the size of a given group gives the expected count for that group — which, after rearrangement, is exactly the row-total-times-column-total rule: the expected frequency for the cell in row $i$ and column $j$ equals the fraction of the grand total that the row represents, scaled up by the column's size. This formula generalizes the intuitive approach of applying one pooled proportion to every group.

The test statistic compares each observed count with its expected count under $H_0$. For every cell, the difference $F_{ij}-E_{ij}$ is squared and divided by $E_{ij}$, and all the cells are summed:

$${\chi }^2=\sum _i\sum _j\frac{(F_{ij}-E_{ij}{)}^2}{E_{ij}}.$$

Think of it as a disagreement meter. If the observations match the expectations closely, every $(F_{ij}-E_{ij}{)}^2$ term is small and the total stays small — the data behave as if the null were true. If some cells deviate a lot, the squared differences grow and the total grows — evidence that the proportions are not all equal. Squaring removes the sign of the deviation, and dividing by $E_{ij}$ gives big deviations in small cells more weight per unit. That is why the test is always right-tailed: only large values of ${\chi }^2$ argue against the null.

Two operating conditions matter. First, the chi-square test works well only when each expected frequency is 5 or more; if a cell's expected frequency falls below 5, we combine that cell with a neighboring one to push the value above 5. Second, the test is right-tailed, and the degrees of freedom are

$$\text{degrees of freedom}=(r-1)(c-1),$$

where $r$ is the number of rows and $c$ the number of columns. For an equality-of-proportions table with two rows and $k$ columns (one column per population), this gives $(2-1)(k-1)=k-1$ degrees of freedom — the same $k-1$ as "three proportions minus one". And as always, when no level of significance is given, the default is the 5 percent level.

5.5.2 Worked Example: Car Owners and Repurchase Intention

Setup. A survey of car owners asked whether they are likely to repurchase the same model. Three models were compared — Swift, i10 and Figo — and the owners' answers fell into two categories: "yes" (likely to repurchase the same model) and "no" (want to change the model). The observed frequencies $F_{ij}$ are given in the table.

The conclusion follows by inspection of the table — no exact p-value needed:

5.6 Two More Chi-Square Applications and What Comes Next

5.6.1 Test of Independence and Goodness of Fit

Two more chi-square methods were previewed for the coming sessions.

The chi-square test of independence checks whether two categorical variables behave independently or whether there is a dependency between them. The data are again tabulated into observed and expected frequencies, and the same ${\chi }^2=\sum \sum (F_{ij}-E_{ij}{)}^2/E_{ij}$ machinery applies. The hypotheses ask directly about the relationship: the null claims the two variables are independent (knowing one gives no information about the other), and the alternative claims they are not. The degrees of freedom for this test are $(r-1)(c-1)$ — the same count as in the equality-of-proportions test.

The goodness of fit test answers a different question: given some sample observations, do they come from a specific named distribution — a binomial distribution, a Poisson distribution, or, very commonly, the normal distribution? People routinely ask whether observations are from a normal distribution; the chi-square distribution validates whether a given random sample comes from the claimed distribution. The sample categories are compared against the frequencies the claimed distribution predicts, and a large chi-square value says the claim does not fit the data.

5.6.2 Beyond Two Populations: ANOVA and the Road Ahead

Where does this session sit in the map of the course? So far the analysis was single-population: hypotheses of the form $H_0:\mu ={\mu }_0$ against $H_A:\mu \ne {\mu }_0$, validated by the Z or T test. The text's Sections 9 and 10 cover the methods of statistical inference for population means and proportions with one and two populations — hypothesis tests and conclusions in both situations, where the Z and T distributions are used very well.

The comparison of two population variances was deliberately stopped here. When it comes to the design of experiments and the analysis of variance, another distribution appears at that point — the F distribution — and the comparison of two population variances belongs there rather than here.

The generalization pattern is worth stating explicitly. One population means tests like $\mu ={\mu }_0$; two populations compare ${\mu }_1$ and ${\mu }_2$; and when there are three or more, the hypotheses become ${\mu }_1={\mu }_2={\mu }_3$, ${\sigma }_1^2={\sigma }_2^2={\sigma }_3^2$, or $p_1=p_2=p_3$. This session showed how the chi-square distribution handles the equality of three or more population proportions; the equality of three or more population means is the analysis of variance (ANOVA) problem, which will connect these two questions when it arrives.

Exam Guidance Summary

• Mid semester: expect a chi-square problem of the confidence-interval type for the population variance, worth five or six marks. A related question on the three chi-square applications (equality of proportions, independence, goodness of fit) is the usual practice in the mid — spend time on these applications.
• Quiz: the quiz covered normal distribution problems and the chapters taught in the previous sessions — nothing else. There is no linear programming in the syllabus (it is an optimization sort of thing). If the exam is open book, you can use the material.
• Empirical rule: the 68-95-99 percent pattern applies only to the normal distribution. For other distributions use Chebyshev's inequality; you cannot quote 68, 95 or 99 percent for them.
• p-values: for the Z distribution the p-value is exact. For the T and chi-square distributions, give a range ("p-value between 0.025 and 0.05", "p-value greater than 0.10") or state that the p-value is less than alpha; exact values come from software such as Excel (e.g., 0.0281 in the one-sided example).
• Decision rule: reject the null hypothesis when p-value $\le \alpha$; fail to reject (accept) when p-value $>\alpha$.
• Default level: when alpha is not given, assume the 5 percent level of significance.
• One-sided vs two-sided: one-sided tests keep alpha whole; only two-sided tests split alpha by 2. The alternative hypothesis decides which case applies.
• Degrees of freedom: for the T and chi-square distributions, always read the table at $n-1$, never at $n$.
• Chi-square symmetry: the chi-square distribution is not symmetric; ${\chi }_{0.025}^2\ne {\chi }_{0.975}^2$, so both critical limits must be read from the table separately — you cannot take the negative of one.
• Notation: ${\chi }_{\alpha }^2$ is the value with area $\alpha$ to the right; the tables give upper-tail areas.
• Expected frequencies: the chi-square test for proportions requires each expected frequency to be at least 5; otherwise combine cells.
• Text references: Section 12.1 (equality of proportions, with a different worked example), Sections 9 and 10 (means and proportions, one and two populations), and Chapter 11 (page 496, Problem 10 — the two-sided variance test).
• Symbol discipline: S is the standard deviation, S square is the variance; mixing them sends the entire calculation wrong.

Key Industry Applications

• Drug efficacy trials in healthcare. When two treatments produce the same average outcome, the variance (consistency of response) decides which treatment is better. This was the motivating example for variance as a decision-making measure: the treatment whose effect is tightly clustered around the average is the dependable performer. The same logic appears in pharmaceutical quality control, where the variance in drug weights across manufactured units is a regulated quality measure.
• Automobile owner surveys and market research. The repurchase-intention survey across Swift, i10 and Figo owners is a market research application of the chi-square test for equality of proportions: survey responses tabulated as observed frequencies, expected frequencies computed under equal-proportion null hypotheses, and a chi-square conclusion about whether models genuinely differ. Car makers use exactly such loyalty surveys to decide where marketing effort pays.
• Quality and process consistency. "The lesser the variance, the more efficient it is" transfers directly to quality control and process consistency: the performer with smaller variance is the more efficient one whenever averages tie. In manufacturing, a filling process whose weight variance exceeds its specification triggers readjustment even when the average weight is correct; control of the variance, not just the mean, is what keeps production within tolerance.
• Statistical software. Excel (and similar software) is the practical route to exact p-values for the T and chi-square distributions, where hand computation from tables only brackets the p-value — for example, the exact 0.0281 behind the bracket "between 0.025 and 0.05".