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Testing a correlation coefficient questions
A sample correlation is almost never exactly zero, even when the population one is. Tables of critical values say how far from zero counts as evidence, and the threshold depends on how much data you have.
7 original questions · 29 marks · the testing a correlation coefficient notes · Further Statistics 2
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Write down the hypotheses for a one-tailed test of positive product moment correlation, and say why r must not appear in them.
Worked answer
H₀: ρ = 0 and H₁: ρ > 0, where ρ is the population correlation coefficient. The sample value r is what was measured, so it is the evidence rather than the claim. B1 B1 for the two hypotheses, B1 for ρ being the population coefficient. Putting r in a hypothesis asserts something already known, so it makes no testable claim.A sample of 12 pairs gives r = 0.62. The 5% one-tailed critical value is 0.4973. Carry out the test.
Worked answer
H₀: ρ = 0; H₁: ρ > 0, one-tailed at 5% with n = 12. Since 0.62 > 0.4973 the result lies in the critical region, so reject H₀: there is evidence at the 5% level of positive correlation between the two variables in the population. B1 for the hypotheses, M1 for comparing 0.62 with 0.4973, A1 for rejecting H₀ with a conclusion in context.A sample of 8 pairs gives r = 0.65. Test at the 5% level whether there is any correlation, given that the critical value for 2.5% in one tail with n = 8 is 0.7067.
Worked answer
H₀: ρ = 0; H₁: ρ ≠ 0, two-tailed at 5%, so 2.5% goes in each tail and the critical values are ±0.7067. Since 0.65 < 0.7067, the result is not in the critical region, so do not reject H₀. There is insufficient evidence at the 5% level of correlation between the two variables in the population. B1 for the hypotheses, B1 for splitting the 5% into two tails, M1 for comparing 0.65 with 0.7067, A1 for not rejecting H₀ in context. A one-tailed test at the same level would have used 0.6215 and rejected, so settle the tail from the wording of the question before looking anything up.Ten judges' rankings give rs = 0.62. The 5% one-tailed Spearman critical value for n = 10 is 0.5636. Test for positive agreement and state the advantage of this test over the product moment one.
Worked answer
H₀: ρs = 0; H₁: ρs > 0, one-tailed at 5% with n = 10. Since 0.62 > 0.5636, reject H₀: there is evidence at the 5% level that the rankings agree. Because the test uses only the orders, it makes no assumption about the shape of the underlying distributions and is unaffected by a single extreme value. B1 for the hypotheses, M1 for comparing 0.62 with 0.5636, A1 for rejecting H₀ in context, B1 for the distribution-free advantage.State the condition the product moment test requires of the data, and what should be said about it in an answer.
Worked answer
The critical values assume the pairs are drawn from a population with a bivariate normal distribution. Formal verification is not expected, but the assumption should be stated as part of the conclusion, since the test is not valid without it. Spearman's test requires no such assumption. B1 for bivariate normality, B1 for stating it in the conclusion, B1 for Spearman's test needing no such assumption.A coefficient of r = 0.35 is obtained. The 5% one-tailed critical values are 0.5494 for n = 10 and 0.3061 for n = 30. Explain what this shows about interpreting a coefficient on its own.
Worked answer
With n = 10 the value 0.35 falls short of 0.5494, so H₀ is not rejected. With n = 30 the same value exceeds 0.3061, so H₀ is rejected. The identical coefficient is significant in one case and not in the other, so a coefficient cannot be interpreted without its sample size. A small sample quite easily produces a moderate correlation by chance; a large one still can, though far less often, which is why the critical value falls with n rather than vanishing. Significance and strength are separate questions. M1 for comparing 0.35 with both critical values, A1 for the two opposite decisions, B1 for the sample size mattering, B1 for separating significance from strength.Two judges rank eight cakes. The first judge's ranks are 1 to 8 in order, and the second judge ranks the same cakes 3, 1, 2, 5, 4, 7, 6, 8. Calculate Spearman's rank correlation coefficient and test at the 5% level whether the judges agree. The 5% one-tailed critical value for n = 8 is 0.6429.
Worked answer
Differences d: −2, 1, 1, −1, 1, −1, 1, 0. Squaring gives 4, 1, 1, 1, 1, 1, 1, 0, so Σd² = 10.
rs = 1 − 6Σd²/[n(n² − 1)] = 1 − 60/(8 × 63) = 1 − 60/504 = 0.881 to three decimal places.
H₀: ρs = 0; H₁: ρs > 0, one-tailed at 5% with n = 8, since the question asks about agreement rather than any association.
Since 0.881 > 0.6429 the result is in the critical region, so reject H₀. There is evidence at the 5% level that the judges' rankings agree.
M1 for the differences, A1 for Σd² = 10, M1 for the formula, A1 for 0.881, B1 B1 for the two hypotheses, M1 for comparing with 0.6429, A1 for rejecting H₀ in context. Set the differences out in a table and show Σd² before substituting. Signs do not matter once they are squared, but a single arithmetic slip in Σd² carries through to everything after it, so total the d column and check it comes to zero.
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