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Testing variances: chi-squared and the F-distribution questions
Spread can be the thing in question instead of the nuisance. One distribution tests a single variance against a claimed value; another tests two variances against each other.
7 original questions · 25 marks · the testing variances: chi-squared and the f-distribution notes · Further Statistics 2
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Write down the test statistic for a claimed population variance from a sample of size n, and its distribution.
Worked 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.A sample of 25 gives s² = 6.4. Test at 5% whether the population variance exceeds 4, given a critical value of 36.415.
Worked 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.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.
Worked 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 = 12.39. The interval is (3.90, 12.39). 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, and pairing each limit with the wrong tail is the standard slip.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.
Worked 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.Explain what determines which sample variance is placed on top in an F test, and what happens to the degrees of freedom.
Worked 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.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.
Worked 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.
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.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.
Worked 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. It is as close as a decision gets, and quoting the statistic to three significant figures is what keeps the comparison honest.
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.
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