Further Hypothesis Tests
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Further Hypothesis Tests
Everything you need to go from "I've seen this formula before" to "I can defend every step of this test in an exam." Confidence intervals for variance, hypothesis tests for variance, and the F-test — explained properly, not just listed.
This chapter is about testing spread, not average. Earlier in your stats course you tested whether a mean had changed. Here, you test whether a variance has changed — either on its own, or by comparing two variances against each other.
There are really only three tools in this chapter, and they build on each other in a straight line:
- The chi-squared (χ²) distribution — describes how the sample variance
S²of a single normal sample behaves. Used for confidence intervals and hypothesis tests on one variance. - The F-distribution — describes what happens when you take the ratio of two sample variances. It has two "degrees of freedom" instead of one, because two samples are involved.
- The F-test — the actual hypothesis test that uses the F-distribution to decide whether two populations really do have equal variance.
When you want to know if things have become more spread out — one sample compared to a claimed value, or two samples compared to each other — you divide variances instead of subtracting means, and you check the result against the χ² or F tables instead of the z or t tables.
Why we need a new distribution
You already know that the sample mean X̄ is an unbiased estimator of the population mean μ. In exactly the same spirit, the sample variance
Different samples from the same population give different values of s² — it's a particular observed value of the random variable S². To build a confidence interval for the true population variance σ², we need to know how S² is distributed. That's where chi-squared comes in.
Building the confidence interval
Because (n−1)S²/σ² follows a χ² distribution, we can trap it between two critical values from the χ² table with a chosen probability. Rearranging for σ² flips which critical value goes on top and bottom — this trips people up constantly, so watch closely in the example.
A fisherman catches 10 fish from a pink salmon pond. Lengths (cm): 67, 71, 62, 69, 75, 77, 63, 75, 61, 72. Find a 90% confidence interval for the population variance, assuming lengths are normally distributed.
| 1. Find x̄ and s² | x̄ = 69.2, s² = Σ(xᵢ − 69.2)² / 9 = 33.5111 |
| 2. Degrees of freedom | n = 10, so use χ²₉ (that's n − 1, not n — easy slip) |
| 3. Read the χ² table | χ²₉(0.95) = 3.325 χ²₉(0.05) = 16.919 |
| 4. Lower limit | (n−1)s² / χ²₉(0.05) = (9 × 33.5111) / 16.919 = 17.826 |
| 5. Upper limit | (n−1)s² / χ²₉(0.95) = (9 × 33.5111) / 3.325 = 90.706 |
| 6. Conclusion | 90% CI for σ²: (17.826, 90.706) |
A random sample of 15 observations from a normal distribution gives s² = 12.4. Find a 95% confidence interval for σ². (You'll need χ²₁₄(0.975) = 5.629 and χ²₁₄(0.025) = 26.119.)
Explain, in your own words, why the upper confidence limit for σ² uses the lower percentage-point value of χ² (e.g. χ²(0.95)) rather than the upper one.
Same underlying result — (n−1)S²/σ² ~ χ²ₙ₋₁ — but now we're using it to test a claim rather than build an interval. The logic is identical to every hypothesis test you've ever done: state hypotheses, find a critical region, calculate a test statistic, compare, conclude.
2. Specify α, the significance level, and ν = n − 1, the degrees of freedom
3. Find the critical region using the χ² tables
4. Calculate the test statistic: (n − 1)s² / σ² (using the claimed σ² from H₀, not the sample estimate)
5. Compare, state your conclusion, and link it back to the actual context of the question
σ² — the value stated in H₀ — not by your calculated s². It's easy to instinctively plug in s² on the bottom because that's what you just calculated, but the test statistic formula uses the population value being tested against.
A farmer suspects some chickens are eating more than their fair share, so the variance of their weights has increased above the usual 0.9kg. A sample of 8 chickens gives weights (kg): 2.5, 4.3, 2.1, 4.5, 3, 3.2, 2.7, 4.4. Test at the 2.5% significance level whether the variance has increased.
| 1. Hypotheses | H₀: σ² = 0.9 H₁: σ² > 0.9 (one-tailed — "increased") |
| 2. α and ν | α = 2.5%, ν = 8 − 1 = 7 |
| 3. Critical value | χ²₇(0.025) = 16.013 → critical region: (n−1)s²/σ² ≥ 16.013 (Note: the cheat sheet's worked source shows χ²₇(0.05)=14.067 for a 5% test — always match the tail probability to the α given in your question.) |
| 4. Calculate s² | s² = 0.8827 (from the 8 data values) |
| 5. Test statistic | (n−1)s²/σ² = (7 × 0.8827) / 0.9 = 6.865 |
| 6. Conclusion | 6.865 is not in the critical region → insufficient evidence to reject H₀. The farmer cannot conclude the variance in chicken weights has increased. |
A machine is set to fill bottles with variance 4 ml². A quality check on 12 bottles gives s² = 7.6 ml². Test at the 5% level whether the variance has changed (two-tailed). Use χ²₁₁(0.975) = 3.816 and χ²₁₁(0.025) = 21.920.
Real problems often ask you to compare two variances, not just check one against a claimed number. Think of a retailer choosing between two suppliers of the same product — both might have similar average quality, but the retailer wants the supplier whose product is more consistent (smaller variance). The F-distribution is what lets you compare two variances directly, by looking at their ratio.
Upper and lower critical values
F-tables (like χ² tables) only give you the upper critical values directly — the tables would be impossibly long otherwise. To find a lower critical value, you use a clever trick: flip the fraction and swap the degrees of freedom.
Find the upper critical value F₅,₈(0.95).
| 1. Table only gives upper (small p) values | Read off F₈,₅(0.05) = 4.82 (swap order, complement the probability) |
| 2. Apply the reciprocal rule | F₅,₈(0.95) = 1 / F₈,₅(0.05) = 1 / 4.82 = 0.21 |
Compare this to F₈,₅(0.95): reading F₅,₈(0.05) = 3.69 gives F₈,₅(0.95) = 1/3.69 = 0.27 — a different answer. This is exactly why order can't be swapped carelessly.
X ~ F₁₂,₁₀. Find P(1/4.30 < X < 2.91).
| 1. Upper tail | F₁₂,₁₀(5%) = 2.91, so P(X < 2.91) = 1 − 0.05 = 0.95 |
| 2. Lower tail — use the rule P(F_{ν₁,ν₂} < f) = P(1/f < F_{ν₂,ν₁}) | P(X < 1/4.3) = P(F₁₀,₁₂ > 4.3) |
| 3. Look up F₁₀,₁₂(1%) = 4.3 | P(X < 1/4.3) = 0.01 |
| 4. Subtract | P(1/4.30 < X < 2.91) = 0.95 − 0.01 = 0.94 |
Given F₆,₉(0.05) = 3.37, find the lower critical value F₉,₆(0.95).
This is where everything comes together. The F-test uses the F-distribution to formally test whether two populations have equal variance — exactly the retailer-and-suppliers situation from earlier.
2. H₀: σₓ² = σᵧ²
3. H₁: σₓ² > σᵧ² (one-tailed) or σₓ² ≠ σᵧ² (two-tailed)
4. Find the critical value F_{νₓ,νᵧ} — νₓ (larger-variance sample) goes first
5. Write down the critical region
6. Test statistic: F = sₓ² / sᵧ²
7. Compare, conclude, and relate back to the context
Samples of size 6 (X) and 13 (Y) give sₓ² = 7 and sᵧ² = 26. Test at the 5% level whether the variance of Y is greater than the variance of X.
| 1. Hypotheses | H₀: σₓ² = σᵧ² H₁: σₓ² < σᵧ² |
| 2. Degrees of freedom | νₓ = 5, νᵧ = 12 |
| 3. Critical value | Since Y's variance is the one claimed larger, put it on top: F₁₂,₅(0.05) = 4.68 |
| 4. Test statistic | The larger sample variance divides the smaller: F = sᵧ²/sₓ² = 26/7 = 3.714 |
| 5. Conclusion | 3.714 < 4.68 → not in the critical region → insufficient evidence that Y's variance is greater than X's. |
Samples of size 25 (X) and 17 (Y) give sₓ² = 8.3 and sᵧ² = 3.1. Test at the 10% level whether the variances are different.
| 1. Hypotheses | H₀: σₓ² = σᵧ² H₁: σₓ² ≠ σᵧ² |
| 2. Halve α for two tails | 10% two-tailed → look up the 5% critical value |
| 3. Critical value | νₓ = 24, νᵧ = 16 → F₂₄,₁₆(0.05) = 2.24 |
| 4. Test statistic | F = sₓ²/sᵧ² = 8.3/3.1 = 2.677 |
| 5. Conclusion | 2.677 > 2.24 → in the critical region → reject H₀. There is sufficient evidence the variances differ. |
Two factories, P and Q, produce the same component. A sample of 9 from P gives sₚ² = 12.4; a sample of 7 from Q gives s_Q² = 3.8. Test at the 5% level whether the variances differ. (You'll need F₈,₆(0.025) = 5.60.)
Why does the F-test procedure insist you always divide the larger sample variance by the smaller one, rather than just following whatever order the question states the samples in?
Unbiased variance estimator
Divide by n − 1, not n.
χ² result for one sample
Foundation for CI and hypothesis tests on a single variance.
CI for σ²
Small χ² value → upper limit. Large χ² value → lower limit.
F-distribution
Order of degrees of freedom must match order of variances.
Reciprocal rule
Swap df order + complement p + take reciprocal for lower critical values.
F-test statistic
Always ≥ 1. Compare against the upper critical value only.
n vs n − 1
Using n instead of n − 1 anywhere in this chapter — in S², in degrees of freedom, or in the test statistic — is the single most common way to lose easy marks. Every degrees-of-freedom value here comes from sample size minus 1.
Wrong table value
Mixing up χ²(α) with χ²(1−α), or F_{ν₁,ν₂} with F_{ν₂,ν₁}, gives an answer that looks plausible but is wrong. Always write out exactly which value you need before opening the table.
Forgetting to halve α
Two-tailed tests need α/2 in each tail. For the F-test specifically, this means looking up the critical value at half the stated significance level — not looking up two separate values.
No context in the conclusion
"Reject H₀" alone rarely gets full marks. Examiners want the conclusion tied back to the scenario — e.g. "there is sufficient evidence that Company B's lightbulbs have greater variance and are therefore less reliable."
Always identify which variance is larger first
Before touching the F-test, label sₓ² as the larger variance. This single habit prevents almost every ordering mistake that follows in the test.
Write the critical region before calculating
State the critical region as an inequality (e.g. "reject if F ≥ 2.24") before you compute the test statistic. It stops you from unconsciously working backwards from the answer you expect.
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