Questions on testing variances: chi-squared and the F-distribution

Question 1

Write down the test statistic for a claimed population variance from a sample of size n, and its distribution.

Accepted Answer

The statistic is (n − 1)S²/σ², where S² is the sample variance and σ² the value claimed under the null hypothesis. It follows a chi-squared distribution on n − 1 degrees of freedom, provided the population is normal. B1 for the statistic, B1 for the chi-squared distribution on n − 1 degrees of freedom.

Question 2

A sample of 25 gives s² = 6.4. Test at 5% whether the population variance exceeds 4, given a critical value of 36.415.

Accepted Answer

H₀: σ² = 4; H₁: σ² > 4, one-tailed at 5% with 24 degrees of freedom. The statistic is 24 × 6.4/4 = 38.4. Since 38.4 > 36.415, reject H₀: there is evidence at the 5% level that the variance exceeds 4. B1 for the hypotheses, M1 for the statistic, A1 for 38.4 with the conclusion in context.

Question 3

Two independent random samples, of sizes n₁ and n₂, are taken from normal populations with equal variances. Write down the distribution of the ratio S₁²/S₂² of the two sample variances, and state the two conditions it requires.

Accepted Answer

S₁²/S₂² follows an F distribution on n₁ − 1 and n₂ − 1 degrees of freedom, the numerator's first. B1. It requires both populations to be normal, and the two samples to be independent of each other. B1. The order of the two parameters matters, because the F distribution is not symmetric in them: F on 10 and 15 is a different distribution from F on 15 and 10, with different critical values.

Question 4

A random sample of 10 observations from a normal population gives s² = 3.6. Write down the number of degrees of freedom and find the value of the chi-squared test statistic for H₀: σ² = 2.5.

Accepted Answer

The degrees of freedom are n − 1 = 9. B1.
The statistic is (n − 1)s²/σ² = 9 × 3.6/2.5 = 32.4/2.5 = 12.96. M1 for the statistic, A1 for 12.96.
The claimed variance 2.5 goes in the denominator, not the sample variance: dividing the other way up gives 6.25 and answers nothing. Note also that the statistic is larger than the 9 degrees of freedom, which is where the mean of the chi-squared distribution sits, so this sample points upwards from the claim.

Question 5

A sample of 25 taken from a normal population gives s² = 6.4. Find a 95% confidence interval for σ², given chi-squared values of 39.364 and 12.401 on 24 degrees of freedom.

Accepted Answer

The numerator is (n − 1)s² = 24 × 6.4 = 153.6. Dividing by the larger chi-squared value gives the lower limit, 153.6/39.364 = 3.90, and dividing by the smaller gives the upper limit, 153.6/12.401 = 12.4. The interval is (3.90, 12.4). M1 for (n − 1)s², A1 for 153.6, M1 for dividing by both chi-squared values, A1 for the interval. The two ends look swapped because the chi-squared value sits in the denominator, so each limit pairs with the opposite tail.

Question 6

A sample of 13 gives s² = 24.5 and an independent sample of 9 gives s² = 8.2. Test at 5% whether the first population is more variable, given a critical value of 3.284.

Accepted Answer

H₀: σ₁² = σ₂²; H₁: σ₁² > σ₂². The alternative names the first population, so the first sample variance goes on top: F = 24.5/8.2 = 2.99, on 12 and 8 degrees of freedom, numerator first.
Since 2.99 < 3.284, do not reject H₀. There is insufficient evidence at the 5% level that the first population is more variable, even though one sample variance is about three times the other. B1 for the hypotheses, M1 for F = 24.5/8.2, A1 for 2.99 on 12 and 8 degrees of freedom, A1 for not rejecting H₀ in context.

Question 7

Explain what determines which sample variance is placed on top in an F test, and what happens to the degrees of freedom.

Accepted Answer

For a two-tailed test, put the larger sample variance on top, so that F is at least 1 and only the printed upper tail is needed. For a one-tailed test, the alternative hypothesis decides it: the population claimed to be more variable supplies the numerator. Either way the degrees of freedom must follow the variances, numerator first, since the F distribution is not symmetric in its two parameters. B1 for the larger variance on top in a two-tailed test, B1 for the degrees of freedom following the variances.

Question 8

A sample of 16 gives s² = 2.5. Test at 5% in two tails whether the population variance is 5, given critical values of 6.262 and 27.488.

Accepted Answer

H₀: σ² = 5; H₁: σ² ≠ 5, with 15 degrees of freedom and 2.5% in each tail. The statistic is 15 × 2.5/5 = 7.5. Since 6.262 < 7.5 < 27.488 the value lies between the two critical values, so do not reject H₀. There is insufficient evidence at the 5% level that the population variance differs from 5.
B1 for the hypotheses, M1 for the statistic, A1 for 7.5, A1 for not rejecting H₀ in context. Note how far apart the critical values are. The chi-squared distribution is strongly skewed, so the acceptance region is nothing like symmetric about the mean of 15.

Question 9

Two independent random samples are taken from normal populations. A sample of 11 gives s² = 18.5 and a sample of 16 gives s² = 6.2. Test at the 5% level whether the two population variances differ, given that the 2.5% upper critical value of F on 10 and 15 degrees of freedom is 3.060.

Accepted Answer

H₀: σ₁² = σ₂²; H₁: σ₁² ≠ σ₂², two-tailed. A two-tailed test at 10% puts 5% in each tail, so the 5% upper critical value is the one to use. B1.
No direction is named, so the larger sample variance goes on top: F = 20.4/5.1 = 4.00. That variance came from the sample of 8, so the numerator carries 7 degrees of freedom and the denominator 12. M1 for the ratio, A1 for 4.00 on 7 and 12 degrees of freedom.
Since 4.00 > 2.913, reject H₀. There is evidence at the 10% level that the two population variances differ. A1 for the conclusion in context.
That verdict rules out the plan. The two-sample t test with a pooled estimate of variance assumes a single common population variance, and the F test has just found evidence against it, so pooling these two estimates is not justified on this data. B1.

Question 10

A process is claimed to have variance 0.5. A random sample of 20 items from a normal population gives s² = 0.28. Test at the 5% level whether the variance is less than 0.5, given a lower critical value of 10.117 on 19 degrees of freedom.

Accepted Answer

H₀: σ² = 0.5; H₁: σ² < 0.5, one-tailed at the 5% level with 19 degrees of freedom, so the whole 5% sits in the lower tail. B1.
The statistic is (n − 1)s²/σ² = 19 × 0.28/0.5 = 5.32/0.5 = 10.64. M1 for the statistic, A1 for 10.64.
Small values are the evidence here, because a small sample variance drives the statistic down. Since 10.64 > 10.117, the statistic is not in the critical region, so do not reject H₀. There is insufficient evidence at the 5% level that the process variance is below 0.5. A1 for the conclusion in context.
The direction of the tail has to be settled from the alternative hypothesis, and comparing 10.64 with an upper critical value such as 30.144 would answer a different question.

Question 11

A random sample of 16 observations from a normal population gives s² = 9.61. Find a 90% confidence interval for the population standard deviation, given chi-squared values of 24.996 and 7.261 on 15 degrees of freedom.

Accepted Answer

The numerator is (n − 1)s² = 15 × 9.61 = 144.15. M1.
For the variance, divide by the larger chi-squared value for the lower limit and by the smaller for the upper: 144.15/24.996 = 5.767 and 144.15/7.261 = 19.853, so σ² lies in (5.767, 19.853). A1.
The question asks about the standard deviation, so take square roots of both limits. M1.
√5.767 = 2.401 and √19.853 = 4.456, so the interval for σ is (2.40, 4.46). A1.
Square-rooting the two limits is legitimate because taking square roots preserves order across positive numbers, so the interval maps onto the interval. The sample standard deviation √9.61 = 3.1 sits inside it, though not at its centre: the interval for σ is not symmetric about 3.1, because the chi-squared distribution is skewed.

Question 12

Two independent random samples are taken from normal populations. A sample of 10 from the first gives s² = 4.2, and a sample of 16 from the second gives s² = 12.6. Test at the 10% level whether the two population variances differ, given that the 5% upper critical value of F on 15 and 9 degrees of freedom is 3.006. Explain what goes wrong if the ratio is taken the other way up.

Accepted Answer

H₀: σ₁² = σ₂²; H₁: σ₁² ≠ σ₂², two-tailed.
A two-tailed test at 10% puts 5% in each tail, so the 5% critical value is the one to use. Comparing with the 10% value would double the size of the test.
The alternative names no direction, so put the larger sample variance on top: F = 12.6/4.2 = 3.00. That variance came from the sample of 16, so the numerator has 15 degrees of freedom and the denominator has 9. The degrees of freedom follow the variances, in that order.
Since 3.00 < 3.006, do not reject H₀. There is insufficient evidence at the 10% level that the population variances differ.
Taking the ratio the other way up gives 4.2/12.6 = 0.333 on 9 and 15 degrees of freedom. A value below 1 must be judged against the lower 5% point of F on 9 and 15, and the tables print only upper points, so an extra step is needed: the lower point is the reciprocal of the printed 3.006. Putting the larger variance on top avoids that step entirely. B1 for the hypotheses, B1 for 5% in each tail, M1 for F = 12.6/4.2, A1 for 3.00 on 15 and 9 degrees of freedom, A1 for not rejecting H₀ in context, B1 for the effect of inverting the ratio.

Question 13

Two machines are compared. A random sample of 9 items from machine A gives Σx = 450 and Σx² = 22 600; an independent random sample of 7 items from machine B gives Σy = 350 and Σy² = 17 550. Both populations are normal. Test at the 5% level whether machine A is the more variable, given a critical value of 4.147 on 8 and 6 degrees of freedom.

Accepted Answer

Build each unbiased variance estimate from its own totals.
Machine A: S_xx = 22 600 − 450²/9 = 22 600 − 22 500 = 100, so s_A² = 100/8 = 12.5. M1 for S_xx, A1 for 12.5.
Machine B: S_yy = 17 550 − 350²/7 = 17 550 − 17 500 = 50, so s_B² = 50/6 = 8.33. A1.
H₀: σ_A² = σ_B²; H₁: σ_A² > σ_B², one-tailed at 5%. B1.
The alternative names machine A, so machine A's estimate goes on top: F = 12.5/8.33 = 1.50, on 8 and 6 degrees of freedom. M1.
Since 1.50 < 4.147, do not reject H₀. There is insufficient evidence at the 5% level that machine A is the more variable. A1 for the conclusion in context.
Divide each sum of squares by n − 1 rather than by n: the divisors 8 and 6 differ from the sample sizes and from each other, and using 9 and 7 would give 11.1 and 7.14 and a different ratio.

Question 14

A random sample of 21 observations from a normal population gives s² = 12.4. Test at the 5% level, in two tails, whether σ² = 8, given critical values of 9.591 and 34.170 on 20 degrees of freedom. Find also the 95% confidence interval for σ² from the same sample, and explain why the two results were bound to agree.

Accepted Answer

H₀: σ² = 8; H₁: σ² ≠ 8, on 20 degrees of freedom with 2.5% in each tail.
The statistic is (n − 1)s²/σ² = 20 × 12.4/8 = 248/8 = 31. M1 for the statistic, A1 for 31.
Since 9.591 < 31 < 34.170, the statistic lies between the critical values, so do not reject H₀. There is insufficient evidence at the 5% level that the population variance differs from 8. A1 for the conclusion in context.
For the interval, the numerator is (n − 1)s² = 248. Dividing by the larger chi-squared value gives the lower limit, 248/34.170 = 7.26, and by the smaller gives the upper, 248/9.591 = 25.9, so the interval is (7.26, 25.9). M1 for both divisions, A1 for the interval.
The value 8 lies inside, which is the same verdict the test reached. B1 for the explanation: the interval is built from the same statistic and the same two critical values, and it holds exactly the values of σ² that would not be rejected, so a claimed variance falls inside the interval precisely when the two-tailed test at the matching level does not reject it.

Question 15

Before pooling two sample variances for a two-sample t test, a researcher tests them for equality. Independent random samples from normal populations give s² = 20.4 from 8 observations and s² = 5.1 from 13 observations. Test at the 10% level whether the two population variances differ, given that the 5% upper critical value of F on 7 and 12 degrees of freedom is 2.913, and state what the result means for the t test that was to follow.

Accepted Answer

H₀: σ₁² = σ₂²; H₁: σ₁² ≠ σ₂², two-tailed.
A two-tailed test at 10% puts 5% in each tail, so the 5% critical value is the one to use. Comparing with the 10% value would double the size of the test.
The alternative names no direction, so put the larger sample variance on top: F = 12.6/4.2 = 3.00. That variance came from the sample of 16, so the numerator has 15 degrees of freedom and the denominator has 9. The degrees of freedom follow the variances, in that order.
Since 3.00 < 3.006, do not reject H₀. There is insufficient evidence at the 10% level that the population variances differ.
Taking the ratio the other way up gives 4.2/12.6 = 0.333 on 9 and 15 degrees of freedom. A value below 1 must be judged against the lower 5% point of F on 9 and 15, and the tables print only upper points, so an extra step is needed: the lower point is the reciprocal of the printed 3.006. Putting the larger variance on top avoids that step entirely. B1 for the hypotheses, B1 for 5% in each tail, M1 for F = 12.6/4.2, A1 for 3.00 on 15 and 9 degrees of freedom, A1 for not rejecting H₀ in context, B1 for the effect of inverting the ratio.