10.6 Test of a Single Variance
SLO 4
Demonstrate an understanding of, and ability to use, basic ideas of statistical processes, including hypothesis tests and confidence interval estimation.
SLO 5
Identify appropriate statistical techniques and use technology-based statistical analysis to describe, interpret, and communicate results.
Learning Objectives
By the end of this section, you will be able to:
- state the assumption a test of a single variance requires;
- write null and alternative hypotheses in terms of a population variance;
- compute the test statistic \(\frac{(n-1)s^{2}}{\sigma^{2}}\) and its degrees of freedom;
- decide whether a claim calls for a right-tailed, left-tailed, or two-tailed test, and carry the test through to a conclusion.
Every chi-square test so far has counted things falling into categories. This one does something different: it tests a claim about spread. Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary — and to many instructors, the variance (or standard deviation) may be more important than the average. A post office cares less about the average wait than about whether the wait is predictable. The same chi-square curve that measured category mismatches turns out to measure this too, and for the same underlying reason: a variance is built out of squares.
10.6.1 The Assumption and the Statistic
A test of a single variance assumes that the underlying distribution is normal. This is a real requirement, not a formality — unlike the category tests earlier in the chapter, this one is sensitive to the shape of the population it came from, and a badly skewed population can throw the result off even with a large sample.
The null and alternative hypotheses are stated in terms of the population variance (or population standard deviation).
where
- \(n\) = the total number of data;
- \(s^{2}\) = sample variance;
- \(\sigma^{2}\) = population variance.
You may think of \(s\) as the random variable in this test. The number of degrees of freedom is
$$ df = n - 1 $$Notice this is the one test in the chapter whose degrees of freedom comes from the sample size rather than from a count of categories. There are no categories here — there is one sample and one number computed from it.
Section 10.1 defined a chi-square variable as a sum of squared standard normal variables. A sample variance is a sum of squared deviations divided by \(n-1\), so multiplying it back by \(n-1\) and dividing by \(\sigma^{2}\) rebuilds exactly that sum of squares, measured in standard-deviation units. The statistic is not merely compared to a chi-square curve by convention; when the population really is normal, it genuinely is one, with one degree of freedom spent estimating the mean.
Definition 10.6.1 — Test Statistic for a Single Variance: Here df comes from the sample size (n = 20), not from a count of categories.
A sample of 20 measurements is drawn from a normally distributed population whose standard deviation is claimed to be 4. The sample standard deviation is 5.2. Compute the test statistic and the degrees of freedom.
Solution — substituting into the formula
Step 1 — Identify the pieces. Here \(n = 20\), \(s = 5.2\) and \(\sigma = 4\), so \(s^{2} = 27.04\) and \(\sigma^{2} = 16\).
Step 2 — Find the degrees of freedom.
$$ df = n - 1 = 20 - 1 = 19 $$Step 3 — Substitute into the test statistic.
$$ \chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}} = \frac{(19)(5.2)^{2}}{4^{2}} = \frac{(19)(27.04)}{16} = \frac{513.76}{16} = 32.11 $$Answer: \(\chi^{2} \approx 32.11\) with \(df = 19\).
10.6.2 Writing Hypotheses About a Variance
The hypotheses contain statements about the population variance, even when the problem is worded in terms of the standard deviation. The move is simply to square the claimed standard deviation and leave it in that form.
A SCUBA instructor wants to record the collective depths each of his students dives during their checkout. He is interested in how the depths vary, even though everyone should have been at the same depth. He believes the standard deviation is three feet. His assistant thinks the standard deviation is less than three feet. If the instructor were to conduct a test, what would the null and alternative hypotheses be?
Solution — a claim of "less than", written as a variance
Step 1 — Identify the claimed value. The instructor believes the standard deviation is three feet, so \(\sigma = 3\).
Step 2 — Express it as a variance. Even though we are given the population standard deviation, we set the test up using the population variance: \(\sigma^{2} = 3^{2}\).
Step 3 — Write the null hypothesis. The null carries the equality and the claimed value:
$$ H_{0}: \sigma^{2} = 3^{2} $$Step 4 — Write the alternative from the challenge. The assistant thinks the standard deviation is less than three feet:
$$ H_{a}: \sigma^{2} < 3^{2} $$Answer: \(H_{0}: \sigma^{2} = 3^{2}\) and \(H_{a}: \sigma^{2} < 3^{2}\).
Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be?
Solution — squaring a claimed standard deviation
Step 1 — Identify whose claim is the status quo. The instructor's belief that the standard deviation is five points is the established position, so it becomes the null.
Step 2 — Convert to a variance. Even though we are given the population standard deviation, we can set up the test using the population variance.
$$ H_{0}: \sigma^{2} = 5^{2} $$Step 3 — Write the student's claim as the alternative. The student says the standard deviation is more than five points:
$$ H_{a}: \sigma^{2} > 5^{2} $$Answer: \(H_{0}: \sigma^{2} = 5^{2}\) and \(H_{a}: \sigma^{2} > 5^{2}\).
10.6.3 Choosing the Tail
A test of a single variance may be right-tailed, left-tailed, or two-tailed. This is the only test in the chapter with a choice — the three category tests are always right-tailed, because any mismatch inflates their statistic. Here the direction comes from the wording of the claim.
Both solutions above write \(5^{2}\) and \(3^{2}\) rather than 25 and 9. That is deliberate and worth copying. The claimed standard deviation is the number the problem gave you and the number your reader recognizes, and keeping it visible makes the hypotheses easy to check against the wording. It also keeps you from mixing up which quantity the problem stated — squaring at the last moment, inside the test statistic, is much harder to get wrong.
| The claim says | \(H_{a}\) | Tail |
|---|---|---|
| more variation, larger, greater than | \(\sigma^{2} > \text{value}\) | right |
| less variation, smaller, lower than | \(\sigma^{2} < \text{value}\) | left |
| different, changed, not equal to | \(\sigma^{2} \neq \text{value}\) | two |
For each claim, write \(H_{a}\) and name the tail.
a. A machinist claims a lathe's part diameters vary more than the specified standard deviation of 0.02 mm.
b. A teacher claims this year's test scores have a different spread than the historical standard deviation of 8 points.
c. A bakery claims a new oven makes loaf weights vary less than the old standard deviation of 15 grams.
Solution — the wording picks the tail
Step 1 — Handle claim a. "Vary more" is a greater-than claim, so \(H_{a}: \sigma^{2} > 0.02^{2}\). This is a right-tailed test.
Step 2 — Handle claim b. "A different spread" gives no direction, so \(H_{a}: \sigma^{2} \neq 8^{2}\). This is a two-tailed test.
Step 3 — Handle claim c. "Vary less" is a less-than claim, so \(H_{a}: \sigma^{2} < 15^{2}\). This is a left-tailed test.
Answer: (a) \(\sigma^{2} > 0.02^{2}\), right-tailed; (b) \(\sigma^{2} \neq 8^{2}\), two-tailed; (c) \(\sigma^{2} < 15^{2}\), left-tailed.
10.6.4 A Full Left-Tailed Test
A quality inspector believes a filling machine's fill volumes have a standard deviation of 2.5 mL. A sample of 16 bottles from a normally distributed population has a standard deviation of 1.9 mL. The inspector wants to test the claim that the machine now varies less. State the hypotheses, the degrees of freedom, and the test statistic.
Solution — setting up a left-tailed variance test
Step 1 — Write the hypotheses. "Varies less" is a left-tailed claim.
$$ H_{0}: \sigma^{2} = 2.5^{2} \qquad H_{a}: \sigma^{2} < 2.5^{2} $$Step 2 — Find the degrees of freedom.
$$ df = n - 1 = 16 - 1 = 15 $$Step 3 — Compute the test statistic.
$$ \chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}} = \frac{(15)(1.9)^{2}}{2.5^{2}} = \frac{(15)(3.61)}{6.25} = \frac{54.15}{6.25} \approx 8.66 $$Answer: \(H_{0}: \sigma^{2} = 2.5^{2}\), \(H_{a}: \sigma^{2} < 2.5^{2}\), \(df = 15\), and \(\chi^{2} \approx 8.66\).
With individual lines at its various windows, a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times have a standard deviation of 3.5 minutes.
With a significance level of 5%, test the claim that a single line causes lower variation among waiting times for customers.
Solution — a complete left-tailed test of a single variance
Step 1 — Identify the test and the parameter. Since the claim is that a single line causes less variation, this is a test of a single variance. The parameter is the population variance, \(\sigma^{2}\), or the population standard deviation, \(\sigma\).
Step 2 — Name the random variable. The sample standard deviation, \(s\), is the random variable. Let \(s\) = standard deviation for the waiting times.
Step 3 — Write the hypotheses.
$$ H_{0}: \sigma^{2} = 7.2^{2} \qquad H_{a}: \sigma^{2} < 7.2^{2} $$The word "less" tells you this is a left-tailed test.
Step 4 — Name the distribution for the test. \(\chi_{24}^{2}\), where \(n\) = the number of customers sampled and
$$ df = n - 1 = 25 - 1 = 24 $$Step 5 — Calculate the test statistic, with \(n = 25\), \(s = 3.5\) and \(\sigma = 7.2\):
$$ \chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}} = \frac{(25-1)(3.5)^{2}}{7.2^{2}} = 5.67 $$Step 6 — Write the probability statement. Because the test is left-tailed, the p-value is the area to the LEFT of the statistic:
$$ p\text{-value} = P(\chi^{2} < 5.67) = 0.000042 $$Step 7 — Compare \(\alpha\) and the p-value. \(\alpha = 0.05\); p-value = 0.000042; \(\alpha >\) p-value.
Step 8 — Make a decision. Since \(\alpha >\) p-value, reject \(H_{0}\). This means that you reject \(\sigma^{2} = 7.2^{2}\). In other words, you do not think the variation in waiting times is 7.2 minutes; you think the variation in waiting times is less.
Answer: \(\chi^{2} = 5.67\) with \(df = 24\) and a p-value of 0.000042. At a 5% level of significance, from the data, there is sufficient evidence to conclude that a single line causes a lower variation among the waiting times — with a single line, customer waiting times vary less than 7.2 minutes.
Using the TI-83, 83+, 84, 84+ Calculator
In 2nd DISTR, use 7:χ²cdf. The syntax is (lower, upper, df). For Example 10.6.2, enter χ²cdf(-1E99, 5.67, 24), which gives a p-value of 0.000042. The lower bound of -1E99 stands in for negative infinity and is what makes this a left-tail area; for a right-tailed test the statistic goes in the lower slot and 1E99 in the upper.
Every other test in this chapter takes the area to the right, so reaching for the right tail here is a natural reflex — and it would have given \(1 - 0.000042\), or about 0.99996, and a decision not to reject. That is the opposite conclusion from the correct one. Let the alternative hypothesis pick the side: \(H_{a}\) points left, so the p-value is the area to the left.
10.6.5 A Right-Tailed Test
The FCC conducts broadband speed tests to measure how much data per second passes between a consumer's computer and the internet. At a certain point in time, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom and the test statistic, and draw a conclusion. Test at the 1% significance level.
Solution — a right-tailed test that does not reject
Step 1 — Write the hypotheses. The analyst claims the standard deviation is more than reported, which is a right-tailed claim.
$$ H_{0}: \sigma^{2} = 12.2^{2} \qquad H_{a}: \sigma^{2} > 12.2^{2} $$Step 2 — Find the degrees of freedom.
$$ df = n - 1 = 15 - 1 = 14 $$Step 3 — Compute the test statistic, with \(n = 15\), \(s = 13.2\) and \(\sigma = 12.2\):
$$ \chi^{2} = \frac{(15-1)(13.2)^{2}}{12.2^{2}} = \frac{(14)(174.24)}{148.84} = \frac{2439.36}{148.84} \approx 16.39 $$Step 4 — Write the probability statement. The test is right-tailed, so the p-value is the area to the right:
$$ p\text{-value} = P(\chi^{2} > 16.39) \approx 0.2902 $$Step 5 — Compare \(\alpha\) and the p-value. \(\alpha = 0.01\) and the p-value is about 0.2902, so \(\alpha <\) p-value.
Step 6 — Make a decision and conclude. Since \(\alpha <\) p-value, do not reject \(H_{0}\).
Answer: \(\chi^{2} \approx 16.39\) with \(df = 14\) and a p-value of about 0.2902. At the 1% level of significance there is insufficient evidence to conclude that the standard deviation of Internet speeds is more than 12.2 percent.
A sample standard deviation of 13.2 is larger than the claimed 12.2, yet Try It Now 10.6.5 did not reject the null hypothesis. Explain how both of those things can be true at once.
Solution — a larger sample statistic is not automatically evidence
Step 1 — Compare the two numbers directly. The sample standard deviation, 13.2, exceeds the claimed population value of 12.2 by about 8%.
Step 2 — Ask what the test actually measures. The test asks how surprising a gap that size would be if the claim were true, given only 15 ISPs. Its answer is the p-value, 0.2902.
Step 3 — Interpret that p-value. If the population standard deviation really were 12.2, a sample of 15 would produce a sample standard deviation at least this large about 29% of the time. Roughly three samples in ten. That is ordinary.
Step 4 — Identify the reason. The sample is small, so \(s\) bounces around a great deal from sample to sample. With \(df = 14\) the chi-square curve is wide and heavily right-skewed, and a statistic of 16.39 sits close to its center of 14 — nowhere near the right tail.
Answer: The observed gap is real but small relative to the sampling variability of \(s\) at \(n = 15\). A sample statistic that points in the direction of \(H_{a}\) is not the same as evidence for \(H_{a}\); only the p-value settles whether the gap is bigger than chance would routinely produce.
Problem Set 10.6
Problem 1. State the assumption required by a test of a single variance.
Solution
Step 1 — Recall the requirement: A test of a single variance requires that the underlying population be normally distributed.
Step 2 — Note why it matters: Unlike the category-based chi-square tests earlier in the chapter, this test is sensitive to the shape of the population. A badly skewed population can throw off the result even with a large sample, so normality is a real assumption, not a formality.
Answer: The underlying population must be normally distributed.
Problem 2. Write the formula for the test statistic of a test of a single variance and say what \(n\), \(s^{2}\) and \(\sigma^{2}\) stand for.
Solution
Step 1 — Write the formula: The test statistic is
$$ \chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}} $$Step 2 — Define each quantity:
- \(n\) = the total number of data (sample size);
- \(s^{2}\) = the sample variance;
- \(\sigma^{2}\) = the claimed population variance.
Step 3 — Add the degrees of freedom: The degrees of freedom are \(df = n - 1\), which comes from the sample size rather than a count of categories.
Answer: \(\chi^{2} = \dfrac{(n-1)s^{2}}{\sigma^{2}}\), where \(n\) is the sample size, \(s^{2}\) is the sample variance, and \(\sigma^{2}\) is the population variance; \(df = n - 1\).
Problem 3. A sample of 30 observations is used to test a claim about a population variance. State the degrees of freedom.
Solution
Step 1 — Apply the rule: For a test of a single variance, the degrees of freedom come from the sample size: \(df = n - 1\).
Step 2 — Substitute: With \(n = 30\),
$$ df = 30 - 1 = 29 $$Answer: \(df = 29\).
Problem 4. A manager claims the standard deviation of delivery times is 6 minutes; a driver claims it is more. Write \(H_{0}\) and \(H_{a}\), and name the tail.
Solution
Step 1 — Identify the claimed value: The manager claims the standard deviation is 6 minutes, so \(\sigma = 6\). We write hypotheses in terms of the variance, keeping the square unevaluated so the original number stays visible.
Step 2 — Write the null hypothesis: The manager's claim of equality is the status quo:
$$ H_{0}: \sigma^{2} = 6^{2} $$Step 3 — Write the alternative hypothesis: The driver claims the standard deviation is more than 6 minutes:
$$ H_{a}: \sigma^{2} > 6^{2} $$Step 4 — Name the tail: Since \(H_{a}\) uses "greater than," this is a right-tailed test.
Answer: \(H_{0}: \sigma^{2} = 6^{2}\), \(H_{a}: \sigma^{2} > 6^{2}\); right-tailed.
Problem 5. A sample of 41 items from a normal population has \(s = 9\), and the claimed population standard deviation is 7. Compute the test statistic.
Solution
Step 1 — Identify the pieces: Here \(n = 41\), \(s = 9\) so \(s^{2} = 81\), and \(\sigma = 7\) so \(\sigma^{2} = 49\).
Step 2 — Find the degrees of freedom:
$$ df = n - 1 = 41 - 1 = 40 $$Step 3 — Substitute into the test statistic:
$$ \chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}} = \frac{(40)(9)^{2}}{7^{2}} = \frac{(40)(81)}{49} = \frac{3240}{49} \approx 66.12 $$Answer: \(\chi^{2} \approx 66.12\) with \(df = 40\).
Problem 6. Explain why the degrees of freedom for this test comes from the sample size rather than from a number of categories.
Solution
Step 1 — Compare with the category tests: In the goodness-of-fit, independence, and homogeneity tests, the statistic is built from counts across categories, and each category contributes one squared term. The degrees of freedom therefore count categories (minus constraints).
Step 2 — Look at what this test uses: A test of a single variance has no categories at all. There is just one sample, and one number computed from it: the sample variance \(s^{2}\).
Step 3 — Trace where the degrees go: The statistic \(\frac{(n-1)s^{2}}{\sigma^{2}}\) rebuilds a sum of \(n\) squared deviations from the sample mean. But that mean was itself estimated from the data, which uses up one degree of freedom, leaving \(n - 1\).
Answer: There are no categories in this test — only one sample. The statistic is a sum of \(n\) squared deviations measured against an estimated mean, so one degree of freedom is spent estimating the mean, leaving \(df = n - 1\).
Problem 7. A test of a single variance is left-tailed with \(\chi^{2} = 6.2\) and \(df = 18\). Explain which area gives the p-value and why.
Solution
Step 1 — Let the alternative pick the side: The test is left-tailed because \(H_{a}\) claims the variance is less than the stated value (\(H_{a}: \sigma^{2} < \text{value}\)).
Step 2 — Locate the p-value: The p-value is therefore the area under the \(\chi^{2}\) curve with 18 degrees of freedom to the left of the statistic:
$$ p\text{-value} = P(\chi^{2} < 6.2) \quad \text{with } df = 18 $$Step 3 — Explain why: Small values of the test statistic occur when the sample variance is much smaller than the claimed variance — exactly the direction the alternative claims. So evidence supporting \(H_{a}\) piles up on the left side of the curve, not the right as in most other chi-square tests.
Answer: The p-value is the area to the left of \(\chi^{2} = 6.2\) under the chi-square curve with \(df = 18\), because a left-tailed alternative means small statistics support \(H_{a}\).
Problem 8. A researcher runs a two-tailed test of a single variance. Describe how the p-value differs from the one-tailed case.
Solution
Step 1 — Recall the one-tailed case: In a one-tailed test, the p-value is a single tail area — either \(P(\chi^{2} > \text{statistic})\) or \(P(\chi^{2} < \text{statistic})\), depending on the direction of \(H_{a}\).
Step 2 — Describe the two-tailed case: In a two-tailed test, \(H_{a}\) allows deviation in either direction, so both extremes count as evidence. The p-value combines the two tails: it is twice the area of whichever tail is smaller.
Step 3 — State the practical effect: Doubling makes the p-value larger than either single-tail area alone, so a two-tailed test demands a larger discrepancy before rejecting \(H_{0}\).
Answer: In a two-tailed test the p-value is twice the smaller of the two tail areas (left and right of the statistic), rather than a single tail area, making rejection harder than in the one-tailed case.
Problem 9. A student computes a test statistic of \(-4.1\) for a test of a single variance. Explain how you know this is wrong.
Solution
Step 1 — Examine the formula: The test statistic is
$$ \chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}} $$Every ingredient is positive: \(n - 1 > 0\) since a sample has at least two data points, and both \(s^{2}\) and \(\sigma^{2}\) are variances, which are squares of standard deviations and hence nonnegative (positive when the data vary).
Step 2 — Connect to the distribution: Because the statistic is built entirely from squares, its distribution is a chi-square curve, which lives entirely on the nonnegative axis.
Step 3 — Conclude: A negative value cannot arise from this formula, so a computed statistic of \(-4.1\) signals an arithmetic or setup error.
Answer: The statistic is a ratio of positive quantities — a sum of squares over a squared claim — so it can never be negative; \(-4.1\) indicates a computational mistake.
Problem 10. A sample standard deviation comes out below the claimed population value, but a left-tailed test does not reject \(H_{0}\). Explain in one or two sentences how that can happen.
Solution
Step 1 — Recall what the p-value measures: The p-value asks how often chance alone would produce a result at least as extreme as the observed one if the null were true. It depends on the size of the gap between \(s\) and the claimed \(\sigma\), not merely its direction.
Step 2 — Explain the outcome: If the sample standard deviation is only slightly below the claimed value, and the sample is small, then \(s\) naturally bounces around quite a bit from sample to sample. A modest shortfall like this occurs routinely by chance, giving a large p-value.
Step 3 — Conclude: Since the p-value exceeds \(\alpha\), we do not reject \(H_{0}\): the gap points toward \(H_{a}\) but is too small to count as evidence beyond ordinary sampling variability.
Answer: The gap between \(s\) and the claimed \(\sigma\) may be real but too small relative to sampling variability (especially with a small sample) to be surprising if \(H_{0}\) were true, so the large p-value prevents rejection.
Key Terms
test of a single variance — a chi-square test of a claim about a population variance or standard deviation; it assumes the underlying distribution is normal.
test statistic for a single variance — \(\frac{(n-1)s^{2}}{\sigma^{2}}\), with \(df = n - 1\).
sample variance (\(s^{2}\)) — the variance computed from the sample; its square root \(s\) is the random variable in this test.
population variance (\(\sigma^{2}\)) — the claimed value the test is about; hypotheses are written in terms of it even when the problem states a standard deviation.
tail of the test — right, left, or two-tailed, chosen from the wording of the alternative claim; this is the only chi-square test in the chapter that is not always right-tailed.