The chapter's outlier. Every pattern the previous four sections established breaks here: the data are quantitative rather than categorical, the null names a parameter and is written as an equation, the degrees of freedom are the familiar n minus one, and the test may be right-tailed, left-tailed or two-tailed. That last point is the genuinely new one — sections 11.2 to 11.4 had no left-tailed version because their statistics measured disagreement, which cannot be negative, whereas this statistic compares a sample variance against a claimed one and a variance can be too small as well as too large. The test assumes the underlying distribution is normal, and that assumption does not soften with sample size the way the t procedures' did. Worked through the book's exam-score and post-office examples.
Subject: Statistics · 65 slides · symbolic lesson
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Title
Statistics · Chapter 11 — The Chi-Square Distribution
Test of a Single Variance
Objectives
Six outcomes, and the fourth is what makes this section unlike the rest of the chapter.
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, pp. 582-583 — the section these objectives are drawn from
Warm-up
Chapter 8 built confidence intervals from n minus one degrees of freedom; section 11.1 gave a distribution built from squares.
Discussion prompt
A sample of 25 has a standard deviation of 3.5 when a population standard deviation of 7.2 was claimed. Is that consistent with the claim?
Hint: Ask how sample variances behave around a true variance, and what scaling would make the comparison a chi-square.
Answer:
The sample variance is 12.25 against a claimed 51.84 — less than a quarter of it. Whether that is surprising depends on how much a sample variance bounces around, which depends on the sample size.
The quantity that has a chi-square distribution is n minus one times the sample variance divided by the population variance. Here that is 24 times 12.25 over 51.84, which is 5.67, against a distribution whose mean is 24.
\[ \chi^2 = \frac{(n-1)s^2}{\sigma^2} = \frac{(24)(12.25)}{51.84} = 5.67 \]
So the statistic sits far below its mean. In sections 11.2 to 11.4 that would have meant excellent agreement and a large p-value; here it is decisive evidence that the true variance is smaller than claimed. Understanding why the same picture reads differently is most of this section.
Concept
A test of a single variance assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population variance, or the population standard deviation. The degrees of freedom are n minus one, and the test may be right-tailed, left-tailed, or two-tailed.
a test of a single variance — A chi-square test on quantitative data, comparing a sample variance against a claimed population variance. The sample standard deviation may be thought of as its random variable.
\[ \chi^2 = \frac{(n-1)s^2}{\sigma^2}, \qquad \text{df} = n - 1 \]
The book's framing is worth keeping: you may think of s as the random variable in this test. That is why the hypotheses concern sigma while the data supply s — exactly the relationship chapter 9's tests had between a claimed mean and a sample mean, transplanted to spread instead of centre.
Figure (svg): A card giving the test statistic for a single variance and naming each of its parts
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, p. 582
Section
Section 1
Concept
The statistic is n minus one times the sample variance divided by the population variance. The n is the total number of data, s squared is the sample variance and sigma squared is the population variance being claimed.
what the statistic measures — How large the sample variance is relative to the claimed one, scaled by the degrees of freedom. If the claim is right, the statistic should land near its mean of n minus one.
\[ \chi^2 = \frac{(n-1)s^2}{\sigma^2} \]
The structure repays a moment's attention. If the sample variance happened to equal the claimed variance exactly, the ratio would be one and the statistic would be n minus one — precisely the mean of the chi-square distribution being used. So the statistic is built to sit at its own mean when the null is true, which is what makes comparing it against the degrees of freedom the natural first reading.
Figure (svg): A card giving the test statistic for a single variance and naming each of its parts
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, p. 582 — the statistic and its three inputs
Picture it
Three inputs, the degrees of freedom, and the choice of tail.
Figure (svg): A card giving the test statistic for a single variance and naming each of its parts
The bottom line is what separates this test from every other in the chapter. Sections 11.2 to 11.4 were right-tailed without exception; here the tail depends on what is being claimed.
Worked example
A post office with individual lines had a waiting-time standard deviation of 7.2 minutes. With a single main line, 25 customers gave a standard deviation of 3.5 minutes.
\[ n = 25, \; s = 3.5, \; \sigma = 7.2 \]
Square the sample sd
Why: 3.5 squared.
\[ 12.25 \]
Square the claimed sd
Why: 7.2 squared.
\[ 51.84 \]
Multiply by n minus one
Why: 24 times 12.25.
\[ 294 \]
Divide
Why: 294 over 51.84.
\[ 5.67 \]
Figure (svg): The solution to Worked example Example 11.11's statistic shown as a ladder of expressions, one row per legal move
\[ \chi^2 = \frac{(24)(12.25)}{51.84} = 5.67 \]
Verify: confirm the statistic against its degrees of freedom
Why: The degrees of freedom are 24, so if the claimed variance were correct the statistic should land near 24. It came in at 5.67, less than a quarter of that — the sample varied far less than the claim predicts. Since the statistic equals df times the ratio of sample to claimed variance, and that ratio is 12.25 over 51.84, or about 0.236, the statistic is 24 times 0.236, which is the same 5.67 by a second route.
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, p. 583
Faded example
A sample of 16 gives s = 4 when sigma = 6 is claimed.
Fill in the blanks
\chi^2 = \frac16})}6.67 = ___
Why: Against 15 degrees of freedom, a statistic of 6.67 sits well below the mean — the sample varied less than the claim predicts, which points toward a left-tailed conclusion.
Worked example
Suppose the single line had given a standard deviation of exactly 7.2.
\[ s = \sigma = 7.2 \]
The ratio
Why: Variances equal.
\[ 1 \]
The statistic
Why: 24 times one.
\[ 24 \]
Compare with the mean
Why: df is 24.
Its left-tail p-value
Why: Just over half.
\[ 0.5384 \]
Figure (svg): The solution to Worked example what the statistic would be if the claim held shown as a ladder of expressions, one row per legal move
\[ \chi^2 = (24)(1) = 24 = \text{df} \]
Verify: confirm why the p-value is near but not exactly one half
Why: A right-skewed distribution has its mean above its median, so slightly MORE than half the area lies below the mean. At 24 degrees of freedom that area is 0.5384 rather than 0.5, and the excess shrinks as the degrees of freedom grow and the skew fades — 0.5940 at df 4, 0.5384 at df 24, and 0.5059 at df 1000. Section 11.1's fifth fact, showing up in an arithmetic detail.
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, pp. 582-583
Trap
\[ \chi^2 = \frac{(24)(3.5)}{7.2} = 11.67 \]
Substitute the standard deviations directly
Why: The problem states them, not the variances.
\[ \text{but the formula is written in variances} \]
Both quantities must be squared, and skipping that gives a statistic more than twice too large here.
\[ \chi^2 = \frac{(24)(3.5^2)}{7.2^2} = \frac{(24)(12.25)}{51.84} = 5.67 \]
Square both before dividing
Why: The book writes s-squared over sigma-squared for a reason.
The error is easy to make because problems usually state standard deviations — Example 11.11 gives 7.2 minutes and 3.5 minutes, never their squares. It is also easy to catch: the wrong answer here would have given a p-value of 0.0104 instead of 0.000042, still rejecting but overstating the variance by a factor of more than two.
Two truths and a lie
All three concern the statistic.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false and imports a rule from section 11.2. Here the degrees of freedom are n minus one — the familiar rule from chapter 8, since the data are quantitative and there are no categories at all.
Prediction
Commit before reasoning.
Predict first
If the claimed variance is correct, roughly where should the statistic fall?
Correct: Near n minus one.
Why: When the variances agree the ratio is one and the statistic is exactly the degrees of freedom, which is also the distribution's mean. That makes the comparison against df the fastest read available: far below points to a smaller true variance, far above to a larger one.
Estimation
Example 11.11's statistic is 5.67 on 24 degrees of freedom.
Predict first
The standard deviation of that distribution is about 6.9. How many standard deviations below the mean is the statistic?
Correct: About 2.7.
Why: The mean is 24 and the spread is the root of 48, about 6.93, so 5.67 sits roughly 2.65 standard deviations below. That is far out for a skewed distribution with a floor at zero, which is why the left-tail p-value comes to 0.000042 rather than something around 0.005.
Section
Section 2
Concept
The null and alternative hypotheses contain statements about the population variance. Even when the problem gives a standard deviation, the test can be set up using the variance — the book squares the claimed standard deviation and writes the hypotheses in those terms.
converting the claim — A claim that the standard deviation is five becomes a null that the variance is five squared. The two are equivalent statements, since a standard deviation determines a variance and the reverse.
\[ H_0: \sigma^2 = 5^2 \qquad\text{against}\qquad H_a: \sigma^2 > 5^2 \]
This is the first null in the chapter that can be written as an equation, and it restores everything chapter 9 established about hypothesis wording. The null is an equality because it has to be specific enough to generate a distribution for the statistic; the alternative carries the direction; and the direction is read from words like more, less or differs.
Figure (svg): Three cards showing how the alternative hypothesis selects the tail
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, p. 582 — Example 11.10's hypotheses
Picture it
Three possibilities, with the wording that signals each.
Figure (svg): Three cards showing how the alternative hypothesis selects the tail
This is chapter 9's rule, returning after three sections in which the tail was fixed by the statistic's construction. The difference is that this statistic keeps a direction, so the alternative has something to express.
Worked example
An instructor believes the standard deviation for his final exam is five points. A student claims it is more than five points.
\[ \sigma = 5 \text{ claimed} \]
Identify the parameter
Why: Spread, not centre.
Square the claim
Why: Five squared.
\[ 25 \]
Write the null
Why: An equality.
\[ \text{variance equals } 5\text{ squared} \]
Read the student's word
Why: More than.
Figure (svg): The solution to Worked example Example 11.10, the exam scores shown as a ladder of expressions, one row per legal move
\[ H_0: \sigma^2 = 5^2 \qquad H_a: \sigma^2 > 5^2 \]
Verify: confirm the conversion loses nothing
Why: Squaring is a strictly increasing operation on positive numbers, so sigma exceeds 5 exactly when sigma squared exceeds 25 — the two alternatives are the same claim. That is why the book can say that even though we are given the population standard deviation, we can set up the test using the population variance, and why the direction of the inequality is preserved.
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, p. 582
Faded example
A machine's fill volumes are claimed to have a standard deviation of 0.4 ml; an engineer suspects more.
Fill in the blanks
H_0: \sigma^2 = 0.4^2 \qquad H_a: \sigma^2 > 0.4^2
Why: Suspecting more variation gives a right-tailed test, which is Example 11.10's structure with different numbers.
Worked example
The claim is that a single line causes lower variation in waiting times.
\[ \sigma = 7.2 \text{ with individual lines} \]
The parameter
Why: The population variance.
The null
Why: The old value.
\[ 7.2\text{ squared} \]
The claim
Why: Less variation.
Read the word
Why: The book says so.
Figure (svg): The solution to Worked example Example 11.11's hypotheses shown as a ladder of expressions, one row per legal move
\[ H_0: \sigma^2 = 7.2^2 \qquad H_a: \sigma^2 < 7.2^2 \]
Verify: confirm the null carries the status-quo value
Why: The 7.2 minutes came from the post office's existing arrangement of individual lines, so the null asserts that the single line changed nothing. That is chapter 9's convention exactly: the null holds the current or claimed state, and the alternative carries what the study is trying to demonstrate. The book's own note is that the word less tells you this is a left-tailed test.
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, p. 583
Error analysis
A claim that the standard deviation is 3 feet, against an assistant's belief that it is less. Which are correct?
Annotate
On: \( \begin{aligned} &(1)\; H_0: \sigma^2 = 3^2, \; H_a: \sigma^2 < 3^2 \\ &(2)\; H_0: \sigma^2 < 3^2, \; H_a: \sigma^2 = 3^2 \\ &(3)\; H_0: s^2 = 9, \; H_a: s^2 < 9 \\ &(4)\; H_0: \sigma = 3, \; H_a: \sigma < 3 \end{aligned} \)
Error (3) is the one to guard against, because it inverts the whole logic of testing. The book's phrasing — you may think of s as the random variable — makes the roles clear: sigma is the fixed unknown being claimed about, and s is what varies from sample to sample.
Sorting
Each is an alternative hypothesis or its wording.
Sort into buckets
Sort by which tail the test uses.
Items (b) and (d) are Example 11.11 and Try It 11.10; (e) is Try It 11.11. The chapter's only left-tailed tests live here, and nowhere else.
Two truths and a lie
All three concern the hypotheses.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. Hypotheses are always statements about a population parameter; the sample variance is the evidence brought to bear on them, which is what the book means by calling s the random variable.
Prediction
Commit before reasoning.
Predict first
Why does the book restate a claim about a standard deviation as one about a variance?
Correct: The statistic is written in variances, and squaring preserves the claim.
Why: Squaring is strictly increasing on positive numbers, so a claim about sigma and the matching claim about sigma squared are the same statement — including the direction of any inequality. The variance form is used because that is what the statistic needs.
Section
Section 3
Concept
A test of a single variance may be right-tailed, left-tailed or two-tailed. The chapter's other three tests are always right-tailed, because their statistics measure disagreement — and data cannot disagree with a claim in a way that makes such a statistic small.
what the statistic measures — A ratio of a sample variance to a claimed one, scaled. A ratio below one is as meaningful as a ratio above it, which is what gives the lower tail content.
\[ \text{small } \chi^2 \;\Longleftrightarrow\; s^2 \ll \sigma^2 \]
Section 11.1's fifth idea noted that a very small goodness-of-fit statistic is unusual and can indicate data that were adjusted rather than observed — but that is a forensic observation, not a test. Here the lower tail is a genuine alternative hypothesis with a genuine p-value, and Example 11.11 uses it to reach a conclusion the other tests could not have reached.
Figure (svg): A card explaining why this test can be left-tailed when the chapter's others cannot
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, pp. 582-583 — may be right-tailed, left-tailed, or two-tailed
Picture it
What each one measures, and what a small value means.
Figure (svg): A card explaining why this test can be left-tailed when the chapter's others cannot
The chi-square distribution is the same in both cases; what differs is the quantity being placed on it. A measure of disagreement has a floor that means perfect agreement, while a ratio of variances has a floor that means a much smaller spread than claimed.
Worked example
Test at 5 percent whether a single line causes lower variation than 7.2 minutes.
\[ \alpha = 0.05 \]
Hypotheses
Why: The word less.
Degrees of freedom
Why: Twenty-five minus one.
\[ 24 \]
Statistic
Why: 24 times 12.25 over 51.84.
\[ 5.67 \]
The p-value
Why: The area BELOW 5.67.
\[ 0.000042 \]
Compare and decide
Why: Alpha exceeds it.
Figure (svg): The solution to Worked example Example 11.11 in full shown as a ladder of expressions, one row per legal move
\[ p = P(\chi^2 < 5.67) = 0.000042 < 0.05 \]
Verify: confirm the direction of the probability statement
Why: The p-value is written as the probability that the statistic falls BELOW 5.67, not above it — the reverse of every other test in the chapter. Taking the right tail instead would give 0.999958, and the test would fail to reject a conclusion the data support overwhelmingly. The alternative's direction sets which tail is computed, exactly as in chapter 9.
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, p. 583
Two truths and a lie
All three concern the left tail.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The distribution has a perfectly good lower tail; sections 11.2 to 11.4 simply have no question that lives in it. Section 11.6 does, which is why the same distribution supports a left-tailed test here.
Worked example
A statistic of 5.67 on 24 degrees of freedom, read two ways.
\[ \chi^2 = 5.67, \; \text{df} = 24 \]
As a goodness-of-fit statistic
Why: Far below the mean.
Its p-value there
Why: A right tail.
\[ \text{about } 0.99996 \]
As a variance statistic
Why: Far below the claim.
Its p-value here
Why: A left tail.
\[ 0.000042 \]
Figure (svg): The solution to Worked example the same statistic in two chapters shown as a ladder of expressions, one row per legal move
\[ 0.99996 \quad\text{against}\quad 0.000042 \]
Verify: confirm the two p-values are complements
Why: They sum to one, since one is the area above the statistic and the other the area below it on the same curve. That is a clean illustration of the point: the distribution is identical and only the direction of interest differs, so identifying which test is being run has to come before any tail is read.
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, pp. 582-583
Trap
\[ p = P(\chi^2 > 5.67) = 0.99996 \]
Use the right tail, as in every earlier section of the chapter
Why: Sections 11.2 to 11.4 were right-tailed without exception.
\[ \text{but the alternative says LESS} \]
A p-value of essentially one would mean failing to reject, and the study's actual finding would be missed entirely.
\[ p = P(\chi^2 < 5.67) = 0.000042 \]
Read the tail from the alternative, as in chapter 9
Why: This is the only test in the chapter where that choice exists.
The error is particularly dangerous because it does not look like an error: 0.99996 is a perfectly ordinary p-value that leads to a perfectly ordinary do-not-reject. Nothing in the arithmetic flags it. The only defence is to notice, before computing, that the alternative points left.
Faded example
Example 11.11's, in the book's form.
Fill in the blanks
p\text< = P(\chi^2 0.000042 5.67) = ___
Why: The direction of the inequality is the whole difference between this test and every other in the chapter, and getting it backwards would give 0.999958 and the opposite decision.
Prediction
Commit before reasoning.
Predict first
Why can a goodness-of-fit test never be left-tailed?
Correct: A small statistic means good agreement.
Why: The statistic measures how far observations fall from expectation, so its smallest value means they coincide. There is no way for data to contradict the null in a direction that makes it small, so the lower tail contains no alternative to test.
Explain it
A classmate asks why the same statistic value can be strong evidence in one chi-square test and no evidence in another.
Discussion prompt
In two sentences or fewer, explain.
Hint: Ask what quantity each statistic is measuring.
Answer:
A goodness-of-fit statistic measures disagreement, so a small value means the data match the claim and supports the null; a single-variance statistic measures a ratio of variances, so a small value means the spread is well below the claim and contradicts it.
Same distribution, same number, opposite meaning — which is why the tail has to be chosen from the alternative before anything is computed.
Section
Section 4
Concept
A test of a single variance assumes that the underlying distribution is normal. Unlike the t procedures of chapters 8 through 10, that assumption is not rescued by a large sample — the central limit theorem describes sums and means, not sample variances.
why it matters more here — Chapter 7's theorem makes a sample MEAN approximately normal whatever the population looks like. It says nothing about the distribution of a sample VARIANCE, which depends on the population's shape however large n is.
\[ \text{underlying distribution normal} \;\Longrightarrow\; \frac{(n-1)s^2}{\sigma^2} \sim \chi^2_{n-1} \]
The practical consequence is real. A t test on a mean from a skewed population with n of 100 is close to valid; a variance test on the same data is not, and can be badly wrong in either direction. That is the main reason this test appears far less often in practice than its position in the chapter might suggest, and why Example 11.11 is careful to say the waiting times are normally distributed.
Figure (svg): A table comparing this test against the chapter's other three on five features
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, pp. 582-583 — assumes the underlying distribution is normal
Picture it
Five features, compared against sections 11.2 to 11.4.
Figure (svg): A table comparing this test against the chapter's other three on five features
Only the distribution itself is shared. Everything about the setup — the kind of data, the form of the hypotheses, the degrees of freedom and the choice of tail — comes from chapter 9 rather than from this chapter.
Worked example
The problem's own wording.
\[ \text{normally distributed waiting times} \]
Find the phrase
Why: In the problem statement.
Ask what depends on it
Why: The distribution of the statistic.
Check whether n rescues it
Why: n is 25.
Note the contrast
Why: A t test would be robust.
Figure (svg): The solution to Worked example reading the assumption in Example 11.11 shown as a ladder of expressions, one row per legal move
\[ \text{normal population} \;\Rightarrow\; \chi^2_{24} \text{ applies} \]
Verify: confirm what would go wrong without it
Why: For a population with heavier tails than a normal, sample variances vary more than the chi-square predicts, so the true tail areas are larger than the ones computed — and a p-value of 0.000042 could be substantially understated. The direction of the error depends on the population's shape, which is exactly what makes it hard to correct for and why the assumption is stated rather than checked casually.
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, p. 583
Two truths and a lie
All three concern assumptions.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false, and it misapplies a rule that is sound for means. A sample variance's distribution depends on the population's shape at every sample size, so no threshold makes the assumption unnecessary.
Worked example
Why the same assumption was gentler there.
\[ \bar{x} \text{ against } s^2 \]
A sample mean
Why: A sum, scaled.
\[ \text{chapter } 7\text{ applies} \]
Its behaviour for large n
Why: Approximately normal.
A sample variance
Why: Not a sum of the data.
Its behaviour for large n
Why: Depends on shape.
Figure (svg): The solution to Worked example comparing with chapter 8 shown as a ladder of expressions, one row per legal move
\[ \bar{X} \approx N \text{ for large } n; \quad s^2 \text{ has no such guarantee} \]
Verify: confirm this explains the different practical advice
Why: Chapter 8 could recommend a t interval for a mean from a moderately skewed population at n of 30 or more, precisely because the theorem does the work. No comparable advice exists for a variance, which is why texts that cover this test always attach the normality assumption without a sample-size escape clause. The assumption is doing genuine work rather than being a formality.
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, p. 582
Error analysis
Which are correct?
Annotate
On: \( \begin{aligned} &(1)\; \text{the underlying distribution must be normal} \\ &(2)\; \text{a large sample removes the need for it} \\ &(3)\; \text{it is the same assumption as for a } t \text{ test on a mean} \\ &(4)\; \text{Example 11.11 states it explicitly} \end{aligned} \)
Statement (2) is the dangerous one, because it is a habit carried correctly from chapters 8 through 10 and applied where it does not hold. Sample size buys a great deal for a mean and almost nothing for a variance.
Matching
Match each feature to where it belongs.
Match the pairs
Why: Section 11.6 sits opposite the rest of the chapter on both features, because it has a parameter and quantitative data while the others have neither.
Prediction
Commit before reasoning.
Predict first
For a test of a single variance on a skewed population, what does increasing n to 500 accomplish?
Correct: It narrows the distribution but does not fix the assumption.
Why: More data does make the sample variance a more precise estimate, so the test gains power. What it does not do is make the chi-square distribution correct for the statistic, since that depends on the population's shape rather than on n — so a precise answer to the wrong question is still wrong.
Faded example
A sample of 25, as in Example 11.11.
Fill in the blanks
\text25 = n - 1 = 24 - 1 = ___
Why: This is chapter 8's rule, not section 11.2's. The one constraint is that the deviations from the sample mean sum to zero, which is exactly what cost a degree of freedom there too.
Section
Section 5
Concept
The sequence is the familiar one: parameter, random variable, hypotheses, distribution, statistic, p-value, comparison with alpha, decision and conclusion. Only the distribution and statistic are new.
the random variable — The sample standard deviation. The book names it explicitly, which keeps clear that sigma is the fixed unknown and s is what varies from sample to sample.
\[ \text{parameter } \sigma^2; \quad \text{random variable } s \]
Example 11.11's write-up is worth following closely because it names the parameter and the random variable before anything else, which is a habit that pays off whenever a test concerns spread rather than centre. It is easy to slip into treating s as though it were the thing being claimed about, and naming the two separately at the start prevents it.
Figure (svg): A chi-square curve on twenty-four degrees of freedom with the small area to the left of 5.67 shaded
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, p. 583 — the full solution to Example 11.11
Picture it
The book's Figure 11.8.
Figure (svg): A chi-square curve on twenty-four degrees of freedom with the small area to the left of 5.67 shaded
The shaded region is the LEFT tail, and it is thin — 0.000042. The dashed line marks the mean at 24, and the distance between it and the statistic is what the p-value is measuring.
Worked example
Example 11.11, in the book's sequence.
\[ \alpha = 0.05 \]
Parameter and random variable
Why: Named first.
Hypotheses
Why: The word less.
Distribution for the test
Why: Chi-square.
\[ d f = 24 \]
Statistic and p-value
Why: 5.67, area below.
\[ 0.000042 \]
Decide and conclude
Why: Alpha exceeds it.
Figure (svg): The solution to Worked example the full write-up shown as a ladder of expressions, one row per legal move
\[ \text{reject } H_0: \sigma^2 = 7.2^2 \]
Verify: confirm the conclusion is about spread, not about waiting times themselves
Why: The book's wording says the variation is lower, and adds shorter waiting times only parenthetically in the problem's framing. Strictly, a smaller variance means times are more CONSISTENT, not shorter — a single line could have the same average wait with far less spread, which is in fact the usual argument for single-queue systems. Keeping that distinction clear is part of reading a variance test correctly.
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, p. 583
Matching
Match each stage of Example 11.11 to what it holds.
Match the pairs
Why: Naming the first two before computing anything is what keeps sigma and s in their proper roles — the fixed unknown being claimed about, and the quantity that varies from sample to sample.
Worked example
Internet speeds had a standard deviation of 12.2 percent. A sample of 15 providers gives 13.2, and an analyst claims the true deviation is more. Test at 1 percent.
\[ n = 15, \; s = 13.2, \; \sigma = 12.2 \]
Hypotheses
Why: The word more.
Degrees of freedom
Why: Fifteen minus one.
\[ 14 \]
Statistic
Why: 14 times 174.24 over 148.84.
\[ 16.39 \]
Compare with the mean
Why: Mean is 14.
Right tail
Why: Beyond 16.39.
\[ 0.2902 \]
Figure (svg): The solution to Worked example Try It 11.11 shown as a ladder of expressions, one row per legal move
\[ \chi^2 = 16.39, \quad p = 0.2902 > 0.01 \]
Verify: confirm why a visibly larger sample sd is not evidence
Why: The sample deviation of 13.2 exceeds the claimed 12.2 by about 8 percent, which sounds like something — but with only 15 providers a sample deviation bounces around a great deal, and the statistic of 16.39 sits barely above its mean of 14. The p-value of 0.29 says a departure this size or larger happens in nearly a third of samples when the claim is true, which is the sample-size lesson from chapter 9 arriving in a new setting.
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, p. 583
Trap
\[ \text{reject} \;\Rightarrow\; \text{a single line means SHORTER waits} \]
Read a variance conclusion as one about the mean
Why: The problem's framing mentions shorter waiting times.
\[ \text{but the test compared } \sigma^2, \text{ not } \mu \]
A smaller variance means more consistent waits, which is compatible with the same average or even a longer one.
\[ \text{with a single line, waiting times VARY less} \]
State the conclusion about the parameter actually tested
Why: Spread, not centre.
The distinction matters practically. Single-queue systems are usually adopted precisely because they reduce variability rather than average wait — everyone waits about the same time instead of some customers being unlucky in a slow line. Testing the mean would need a different test entirely, from chapter 9.
Faded example
n = 15, s = 13.2, sigma = 12.2.
Fill in the blanks
\chi^2 = \frac148.8416.39 = \frac______} = ___
Why: Against a mean of 14, a statistic of 16.39 is unremarkable — the right-tail p-value is 0.2902, far above the 1 percent level.
Two truths and a lie
All three concern the conclusion.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. A smaller variance means waits are more consistent; the average could be unchanged or even higher. Testing the mean would require a separate test on a different parameter.
Estimation
Try It 11.11's sample sd of 13.2 against a claimed 12.2 gave p = 0.2902 at n = 15.
Predict first
With the same ratio of deviations but n = 100, would the p-value be larger or smaller?
Correct: Much smaller.
Why: The statistic is n minus one times a fixed ratio, so it grows with the sample size while the distribution's mean grows at the same rate — but its spread grows only like the square root, so the same proportional departure sits further out. At n = 100 the statistic would be 115.9 against a mean of 99 with spread 14.1, giving a p-value near 0.11 rather than 0.29, and by n = 400 it would be decisive.
Comparison
Fill the blanks. Only one row agrees.
Comparison matrix
| Sections 11.2 to 11.4 | Section 11.6 | |
|---|---|---|
| Data | categorical counts | quantitative |
| Hypotheses | in words | as equations about a parameter |
| Degrees of freedom | cells, or a product | n minus one |
| Distribution used | chi-square | chi-square |
The last row is the only thing shared, and even there the reason differs: the earlier tests reach a chi-square because they sum squared departures across cells, while this one reaches it because a scaled sample variance from a normal population has that distribution exactly.
Pattern
Six steps, and the third is the one this chapter has not needed before.
Compare the statistic with n minus one first: at it means the claim looks right, well below means less spread than claimed, well above means more.
OpenStax Introductory Business Statistics 2e, §11.2 Test of a Single Variance §11.2 Test of a Single Variance
Check
The statistic.
Check your understanding
A sample of 20 has a standard deviation of 6 when 8 was claimed. What is the test statistic?
Answer: A
Why: Nineteen times 36 over 64 is 10.69, which sits below the mean of 19 and points toward less spread than claimed.
Check
The tail.
Check your understanding
An engineer claims a process now varies LESS than the historical standard deviation. Which tail?
Answer: A
Why: Less variation means a smaller variance, so the alternative points below the claimed value and the p-value is a left-tail area.
Check
The assumption.
Check your understanding
A sample of 200 is drawn from a strongly skewed population. May a test of a single variance be used?
Answer: A
Why: The central limit theorem describes sample means, not sample variances, so the distribution of the statistic still depends on the population's shape however large n is.
Real world
A factory's bottle-filling machine is specified to have a standard deviation of at most 1.5 ml. A quality engineer samples 30 bottles, finds a standard deviation of 2.1 ml, computes a chi-square statistic of 56.84 on 29 degrees of freedom with a right-tail p-value of 0.0014, and recommends shutting the line down.
Discussion prompt
Assess the calculation, then say what should be checked before acting.
Hint: Ask what the statistic assumes about the fill volumes, and what a single sample can establish.
Answer:
The arithmetic is right. Twenty-nine times 4.41 over 2.25 is 56.84, and against a mean of 29 with a standard deviation of about 7.6, that sits nearly four standard deviations out — so a p-value near 0.0014 is exactly what the numbers give. The sample genuinely varies more than the specification allows.
\[ \chi^2 = \frac{(29)(2.1^2)}{1.5^2} = \frac{(29)(4.41)}{2.25} = 56.84 \]
The normality assumption is the first thing to check, and it is not automatic here. Fill volumes are often close to normal, but a machine that is drifting, or that occasionally double-fills, produces a skewed or heavy-tailed distribution — and those are precisely the failure modes that would inflate a sample variance. If the data are heavy-tailed, the chi-square tail area understates the true probability and the evidence is weaker than 0.0014 suggests. A histogram or normal quantile plot of the 30 values costs nothing and settles it.
A single sample also cannot separate a persistent problem from a transient one. Thirty consecutive bottles sampled over a few minutes describe the machine's behaviour in those minutes; a run of bottles taken during one jam would produce this result even if the process is well-behaved the rest of the day. Sampling at several separated times, or plotting the values in production order, distinguishes the two — and a control chart is the tool actually used for this in practice.
What the test supports: at the time of sampling, fill variability exceeded specification, assuming near-normal fills. What should happen before a shutdown: look at the 30 values plotted in order and as a histogram, take a second sample at a different time, and check whether the excess variance comes from a few outliers or from all thirty. The statistical conclusion is sound; the operational decision needs more than one sample to rest on.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why can a test of a single variance be left-tailed when the chapter's other chi-square tests cannot?
Correct: Because it measures a ratio, and a variance can be too small.
\[ \text{small } \chi^2: \quad \text{good fit (no evidence)} \quad\text{against}\quad s^2 \ll \sigma^2 \text{ (evidence)} \]
Why: The chapter's other statistics measure disagreement between observed and expected counts, and their smallest value means perfect agreement — a state that never contradicts the null, so the lower tail holds no alternative. This statistic is proportional to the sample variance, so a small value means the spread is well below the claim, which is a genuine and testable alternative. Same distribution, different quantity placed on it.
Explain it
They computed Example 11.11's p-value as 0.999958 and concluded there was no evidence.
Discussion prompt
In two sentences or fewer, correct them.
Hint: Ask which direction the alternative points.
Answer:
The alternative says the variance is LESS than 7.2 squared, so the p-value is the area to the left of 5.67 rather than to the right — 0.000042, not 0.999958.
This is the one test in the chapter where the tail has to be read from the alternative, because it is the only one whose statistic can be small for a reason that contradicts the null.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the first, square the claimed value and keep the inequality's direction. For the second, read more, less or differs from the alternative. For the third, this statistic measures a ratio while the others measure disagreement. For the fourth, the central limit theorem covers means, not variances. Do five problems of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the test statistic as n minus one times s squared over sigma squared, label its three inputs, and write the degrees of freedom as n minus one beside it. Under that, draw three boxes for the three alternatives — variance greater than, less than, and not equal to the claimed value — and write beneath each the wording that signals it and which tail it uses. In the middle of the page, work Example 11.11 completely: n of 25, s of 3.5, sigma of 7.2, the hypotheses in variance form, the statistic as 24 times 12.25 over 51.84, which is 5.67, and the degrees of freedom of 24. Beside it draw a chi-square curve for 24 degrees of freedom with a dashed line at the mean of 24, a solid line at 5.67, and the area to the LEFT of 5.67 shaded and labelled 0.000042. At the bottom left, write a two-column table contrasting this test with sections 11.2 to 11.4 on data type, hypothesis form, degrees of freedom and tail. At the bottom right, write one sentence saying why a small statistic means good agreement in a goodness-of-fit test and strong evidence here.
Check your statistic by a second route: the ratio of variances is 12.25 over 51.84, about 0.236, and 24 times that is 5.67 again. Check your shaded curve by confirming the shading is on the LEFT — the single most common error in this section, and one the arithmetic will not catch for you.
Recap
Six things, and most of them come from chapter 9 rather than from this chapter.
| If you see | Then |
|---|---|
| A claim about a standard deviation | Square it and write the hypotheses in variances |
| The word more or greater | A right-tailed test |
| The word less or lower | A left-tailed test, the chapter's only kind |
| The word differs or changed | A two-tailed test |
| A statistic near n minus one | The sample is consistent with the claim |
| A statistic far below n minus one | Less spread than claimed |
| A skewed population, any n | The assumption fails; the test is not licensed |
| A rejection | A conclusion about spread, never about the average |
That completes chapter 11. Chapter 12 turns from counts and spread to relationships between two quantitative variables — scatter plots, a regression line, and a test of whether the correlation between them is real.
OpenStax Introductory Statistics 2e, §11.6 Test of a Single Variance §11.6, pp. 582-583 — everything on these slides traces back here
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