12.4 Test of Two Variances

Step 1 — State the two conditions: The F test of two variances requires that (1) the populations from which the two samples are drawn are normally distributed, and (2) the two populations are independent of each other.

Step 2 — Explain why normality matters here: Every hypothesis test builds its p-value on a model of what the test statistic looks like when \(H_0\) is true. Most tests in this book tolerate a little non-normality, but the F ratio does not: squaring amplifies outliers, and the ratio of two squared quantities inherits every distortion from the underlying distributions.

Step 3 — State the consequence: If the populations are not normal, the reported p-value can be too high or too low in unpredictable ways — there is no way to tell which direction the error goes. That unpredictability is why some textbooks advise not using this test at all unless normality is reasonably satisfied.

Answer: The conditions are (1) both populations are normally distributed and (2) the populations are independent. Normality matters because squaring amplifies outliers and the F ratio inherits every distortion, so with skewed data the p-value can be wrong in either direction, making the test's conclusion untrustworthy.

Step 1 — Recall the degrees of freedom rule: For a test of two variances, the F statistic follows an F distribution with numerator degrees of freedom \(n_1 - 1\) (from sample 1) and denominator degrees of freedom \(n_2 - 1\) (from sample 2).

Step 2 — Compute each value: With \(n_1 = 14\):

With \(n_2 = 21\):

Answer: The test statistic follows an \(F_{13, 20}\) distribution — 13 degrees of freedom for the numerator and 20 for the denominator.

Step 1 — Compute the F ratio: Under \(H_0\), the F statistic is the ratio of the two sample variances:

Step 2 — Interpret the value: The neutral value for this statistic is one, since a ratio of 1 means the two sample variances match exactly. Here \(F \approx 0.94\) is very close to one.

Step 3 — Draw the conclusion about evidence: Because the two sample variances are nearly equal, the data look exactly like what equal population variances would produce. This favors \(H_0\) (the variances are equal) rather than \(H_a\) (they differ).

Answer: \(F = \dfrac{18.4}{19.6} \approx 0.9388\). Since this is close to one, the data favor \(H_0\) — there is no sign of a difference in the two population variances.

Step 1 — Put the larger variance on top: Since \(s_1^2 = 45.0 > s_2^2 = 5.0\), we use:

Step 2 — Interpret the large value: A value of one would mean the sample variances agree perfectly; values far above one mean they are far apart. Here the first sample's variance is nine times the second's.

Step 3 — State what this suggests: An F ratio as large as 9 suggests the two population variances are quite different — specifically, it suggests \(\sigma_1^2\) is substantially larger than \(\sigma_2^2\). Such a large value constitutes strong evidence against \(H_0: \sigma_1^2 = \sigma_2^2\) (a formal decision would still require comparing the p-value to \(\alpha\)).

Answer: \(F = 9.0\). A value this far above one strongly suggests the two population variances are unequal, with population 1 having much greater spread than population 2.

Step 1 — Start from the general F ratio: The general form divides each sample variance by the population variance it estimates:

Step 2 — Apply the null hypothesis: When \(H_0\) is true, \(\sigma_1^2 = \sigma_2^2\). Both unknown population variances are then the same number, so call them both \(\sigma^2\).

Step 3 — Cancel the common factor: Substituting gives

The unknown \(\sigma^2\) cancels just like any common factor in a fraction.

Step 4 — Explain why this makes the test usable: Before canceling, the formula contains \(\sigma_1^2\) and \(\sigma_2^2\) — quantities we never know (if we knew them, we wouldn't need to test anything). After canceling, only \(s_1^2\) and \(s_2^2\) remain, and those we compute directly from data. The cancellation under \(H_0\) is precisely what turns an uncomputable expression into a usable test statistic.

Answer: Under \(H_0\), both population variances are equal, so the common \(\sigma^2\) cancels, leaving \(F = \dfrac{s_1^2}{s_2^2}\). This reduction removes the unknown parameters and leaves only computable sample quantities — without it, no test would be possible.

Step 1 — Identify the claim: The claim is that her exam-score variance is smaller than the colleague's, so instructor 1's variance goes in the comparison as the smaller quantity.

Step 2 — Write the hypotheses:

The null always asserts equality (no difference); the alternative carries the claim.

Step 3 — Decide the tail: Because \(H_a\) uses a "<" sign — small values of the F ratio support the claim that \(\sigma_1^2 < \sigma_2^2\) — the test is left-tailed.

Step 4 — Explain the reasoning: The tail is chosen entirely by the alternative hypothesis, not by which sample variance happens to be larger. Here the claim puts instructor 1's variance on top of the fraction, so evidence in her favor appears as a small F value, i.e., area in the left tail of the F distribution.

Answer: \(H_0: \sigma_1^2 = \sigma_2^2\), \(H_a: \sigma_1^2 < \sigma_2^2\); the test is left-tailed because the alternative claims the first variance is smaller, making small F values the supporting evidence.

Step 1 — Compute the F statistic: Following Problem 12.4.6, instructor 1's variance goes in the numerator to match the claim:

Step 2 — Find the degrees of freedom: Numerator: \(n_1 - 1 = 25 - 1 = 24\). Denominator: \(n_2 - 1 = 25 - 1 = 24\).

Step 3 — State the distribution: The test statistic follows \(F \sim F(24, 24)\).

Answer: \(F = \dfrac{31.2}{58.7} \approx 0.5315\), distributed as \(F_{24, 24}\).

Step 1 — Compute the F ratio: Putting machine A's variance on top:

Step 2 — Interpret the value: One is the neutral value for an F ratio — it means the two sample variances agree perfectly. Here \(F \approx 1.03\) is extremely close to one: the machines' fill variabilities differ by only about 2.5%.

Step 3 — Explain why no table lookup is needed: Evidence against \(H_0\) comes from F drifting far from one. A statistic essentially equal to one lies near the center of the null distribution, where almost all of the probability mass sits when the variances truly are equal. The p-value would be close to its maximum possible size (near 1 for a two-tailed test), far above any conventional significance level, so rejection is impossible regardless of the exact cutoff.

Answer: \(F = \dfrac{0.81}{0.79} \approx 1.0253\). Because the ratio is nearly one, the data look exactly like what equal variances predicts; the p-value would be very large, so no formal lookup is needed to conclude there is no evidence of a difference in variability between the two filling machines.

Step 1 — Explain why the calculator asks for standard deviations: The D:2-SampFTest routine is built around the same machinery as the two-sample t procedures, whose inputs Sx1 and Sx2 are sample standard deviations. Internally the calculator squares these values to get the variances it needs for the F ratio, so the input convention matches the rest of the STAT TESTS menu rather than the variance-based formula in the textbook.

Step 2 — Convert before entering: The problem supplies variances, so take square roots first:

Step 3 — Note the danger of skipping this step: Entering 52.3 and 89.9 directly makes the calculator treat them as standard deviations, squaring them into variances of about 2735 and 8082. The resulting F statistic is completely wrong, yet the calculator reports it without any error message — the wrong answer looks exactly as trustworthy as the right one.

Answer: The calculator expects standard deviations because Sx1/Sx2 follow the menu-wide convention and it squares them internally. You must enter \(\sqrt{52.3} \approx 7.2319\) and \(\sqrt{89.9} \approx 9.4816\), not the variances themselves.

Step 1 — Recall what the p-value model assumes: A p-value is computed under a model of how the test statistic behaves when \(H_0\) is true. For the F test, that model assumes both populations are normally distributed.

Step 2 — Apply the sensitivity discussion: The F ratio is unusually fragile to violations of this assumption. Squaring amplifies outliers, and the ratio of two squared quantities inherits every distortion from the underlying distributions. With clearly skewed data, the actual distribution of the F statistic under \(H_0\) can be very different from the theoretical F distribution the p-value was read from.

Step 3 — Explain the consequence for the student's result: Because the skew distorts the null distribution in unpredictable ways — sometimes pushing p-values too high, sometimes too low, with no way to tell which — the reported p-value of 0.02 may be far from the true probability of seeing data this extreme if the variances were equal. It could correspond to a true p-value well above 0.05 (in which case the "significant" finding is spurious) or even further below it. Either way, the number cannot be taken at face value.

Answer: The p-value of 0.02 rests on the normality assumption, and the F test is notoriously sensitive to its violation: with clearly skewed data, the true null distribution of F may differ greatly from the theoretical one, so the reported p-value may be too high or too low by an unknowable amount. The student should check normality (or use a more robust procedure) before trusting that result.

Step 1 — Give a scenario: Consider two pharmacies filling prescriptions. Management wants to know whether the two pharmacies' waiting times have the same variability — not whether their average waits are equal.

Step 2 — State why variances are the natural question: Two pharmacies could both average 15 minutes per prescription, but if one's wait swings between 5 and 30 minutes while the other stays between 12 and 18, customers experience them very differently. Predictability is what matters for staffing and for service guarantees, and predictability is measured by variance.

Step 3 — Interpret unequal variances practically: If the test showed pharmacy A's wait-time variance significantly larger than pharmacy B's, it would mean A's service is unreliable even if its average looks fine: customers at A face unpredictable waits, so management might shift staff toward A, investigate its workflow, or quote wider time windows to A's customers. Equal variances would mean both locations deliver equally dependable service.

Answer: Example: comparing the variability of customer waiting times at two pharmacies. Unequal variances would mean one location's waits are far less predictable than the other's — a practical reliability problem even if the average waits match — while equal variances would indicate both provide equally dependable service.

Step 1 — Evaluate the statement: The statement is false — in fact it has things backwards.

Step 2 — Explain why: One is the neutral value of the F ratio: if the two population variances really are equal, the sample variances tend to be close together and F lands near one. So an F statistic very close to one is exactly what the null hypothesis predicts.

Step 3 — State the correct reading: Evidence against \(H_0\) comes from F drifting far from one — typically far above one when the larger variance is placed in the numerator. An F near one therefore supports the null hypothesis (equal variances), not the alternative.

Answer: False. An F statistic close to one means the two sample variances nearly match, which is precisely what equal population variances would produce — so it is strong evidence for the null hypothesis, not for the alternative.