Questions on testing a correlation coefficient

Question 1

Write down the hypotheses for a one-tailed test of positive product moment correlation, and say why r must not appear in them.

Accepted 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.

Question 2

A sample of 12 pairs gives r = 0.62. The 5% one-tailed critical value is 0.4973. Carry out the test.

Accepted 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.

Question 3

State what makes a hypothesis test non-parametric, and name the test of correlation in this topic that is non-parametric.

Accepted Answer

A non-parametric, or distribution-free, test assumes no family of distributions for the population it is testing: it works from the ranks or the signs of the readings rather than from the readings themselves. B1.
The test on Spearman's rank correlation coefficient r_s is the non-parametric one here; the product moment test is its parametric partner, whose critical values rest on the population being bivariate normal. B1.

Question 4

A sample of 20 pairs gives r = 0.41. The one-tailed critical values for n = 20 are 0.3783 at the 5% level and 0.5155 at the 1% level. Test for positive correlation at the 5% level, and state what the same data give at the 1% level.

Accepted Answer

H₀: ρ = 0; H₁: ρ > 0, one-tailed with n = 20, where ρ is the population correlation coefficient. B1.
At 5%: 0.41 > 0.3783, so the result lies in the critical region and H₀ is rejected. There is evidence at the 5% level of positive correlation between the two variables in the population. M1 for the comparison.
At 1%: 0.41 < 0.5155, so H₀ is not rejected. There is insufficient evidence at the 1% level of positive correlation. A1 for both conclusions in context.

Question 5

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.

Accepted 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.

Question 6

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.

Accepted 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 less affected 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.

Question 7

State the condition the product moment test requires of the data, and what should be said about it in an answer.

Accepted 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.

Question 8

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.

Accepted 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.

Question 9

Explain what a non-parametric test assumes about the population, state the assumption the product moment correlation test makes that Spearman's test does not, and say what is given up by choosing the rank test when that assumption does hold.

Accepted Answer

A non-parametric test assumes no family of distributions for the population: it replaces the readings by their ranks, so its null distribution depends only on the number of readings and not on the shape they came from. B1.
The product moment correlation test assumes the pairs are drawn from a population with a bivariate normal distribution, which is what its tabulated critical values rest on. B1.
Spearman's test needs no such assumption; it still needs independent paired observations, and it tests for monotonic association rather than for a straight-line one. B1.
What is given up is power. Where the population really is bivariate normal, the product moment test detects a genuine correlation from a smaller sample, because it uses the sizes of the readings and the rank test uses only their order. B1.

Question 10

Nine paintings are ranked by two judges, and Spearman's rank correlation coefficient between the two rankings is r_s = −0.55. The one-tailed critical value for 2.5% with n = 9 is 0.7000. Test at the 5% level whether there is any association between the two rankings, and interpret the sign of the coefficient.

Accepted Answer

The question asks about any association, so the test is two-tailed: H₀: ρ_s = 0; H₁: ρ_s ≠ 0, where ρ_s is the population rank correlation coefficient. B1.
Two-tailed at 5% puts 2.5% in each tail, so the critical values are ±0.7000. B1.
Since −0.55 lies between −0.7000 and 0.7000, the result is not in the critical region, so do not reject H₀. There is insufficient evidence at the 5% level of association between the two judges' rankings. M1 for the comparison, A1 for the conclusion in context.
The negative sign says that in this sample the two judges tended to place paintings in opposite orders: what one ranked highly the other tended to rank low. The test says that a sample of nine could produce a value that size by chance, so that tendency is not established in the population.

Question 11

The table of one-tailed critical values for the product moment correlation coefficient gives, for n = 12, the entries 0.4973 at 5%, 0.5760 at 2.5% and 0.6581 at 1%. Write down the critical region for a test at the 5% level of whether two variables are negatively correlated, and the critical region for a test at the 2% level of whether they are correlated at all. A sample of 12 pairs gives r = −0.61. State the conclusion of each test.

Accepted Answer

For negative correlation the hypotheses are H₀: ρ = 0 and H₁: ρ < 0, one-tailed, so the whole 5% sits in the lower tail and the critical region is r < −0.4973: the 5% column, with a minus sign attached because the alternative points that way. B1 for the critical region.
For any correlation the hypotheses are H₀: ρ = 0 and H₁: ρ ≠ 0, two-tailed, so 2% splits into 1% in each tail and the column to read is the 1% one: the critical region is r < −0.6581 or r > 0.6581. B1 for reading the 1% column and giving both tails.
Now place r = −0.61. Since −0.61 < −0.4973, the first test rejects H₀: there is evidence at the 5% level of negative correlation between the two variables in the population. Since −0.61 lies between −0.6581 and 0.6581, the second test does not reject H₀: there is insufficient evidence at the 2% level of correlation between them. M1 for comparing −0.61 with both regions, A1 for the two conclusions in context.

Question 12

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.

Accepted Answer

Differences d: −2, 1, 1, −1, 1, −1, 1, 0. Squaring gives 4, 1, 1, 1, 1, 1, 1, 0, so Σd² = 10.
r_s = 1 − 6Σd²/[n(n² − 1)] = 1 − 60/(8 × 63) = 1 − 60/504 = 0.881 to three decimal places.
H₀: ρ_s = 0; H₁: ρ_s > 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.

Question 13

A coach ranks eight athletes 1 to 8 in order of expected performance, and the eight are then timed over a fixed distance, in seconds: 12.4, 12.1, 12.9, 12.6, 13.1, 12.8, 13.4 and 13.2, listed in the coach's order. Calculate Spearman's rank correlation coefficient between the coach's ranking and the times, and test at the 5% level whether the coach's ranking agrees with the times. The 5% one-tailed critical value for n = 8 is 0.6429.

Accepted Answer

Rank the times, fastest first, so that rank 1 means the best performance and the two rankings run the same way: 12.1(1), 12.4(2), 12.6(3), 12.8(4), 12.9(5), 13.1(6), 13.2(7), 13.4(8). In the coach's order the time ranks are 2, 1, 5, 3, 6, 4, 8, 7. M1 for ranking the times, A1 for the ranks.
The differences d are −1, 1, −2, 1, −1, 2, −1, 1, so d² gives 1, 1, 4, 1, 1, 4, 1, 1 and Σd² = 14. M1 for the differences and their sum.
r_s = 1 − 6(14)/[8(8² − 1)] = 1 − 84/504 = 0.833. A1.
H₀: ρ_s = 0; H₁: ρ_s > 0, one-tailed at 5% with n = 8, since the question asks about agreement rather than about any association. B1.
Since 0.833 > 0.6429, reject H₀. There is evidence at the 5% level that the coach's ranking agrees with the times recorded. A1 for the conclusion in context.
Ranking the times slowest first would reverse the sign and give −0.833, which the same one-tailed test would then read as no evidence of agreement.

Question 14

A student tests r = 0.88 from a sample of 10 pairs against a 5% one-tailed critical value of 0.5494, rejects H₀, and writes: 'there is a strong positive correlation between x and y, so increasing x will increase y.' Identify three separate faults in that conclusion, state the condition on the population that the critical value assumes, and write a correct conclusion.

Accepted Answer

First fault. The test establishes evidence of correlation in the population; it does not establish that the correlation is strong. Significance and strength are separate questions, and with n = 10 even 0.55 would have been significant. B1. Second fault. Correlation does not establish causation. The association could run the other way, or a third variable could be driving both. B1. Third fault. The significance level and the population are missing. A conclusion has to name the level the test was carried out at and say that the claim is about ρ, the population coefficient, rather than about the ten pairs in hand. B1. The condition. The tabulated critical values for the product moment coefficient assume the pairs come from a population with a bivariate normal distribution. B1. A correct conclusion. Since 0.88 > 0.5494, reject H₀: there is evidence at the 5% level of positive correlation between x and y in the population from which the sample was drawn, assuming that population is bivariate normal. B1.

Question 15

Six pairs of readings give Σx = 30, Σy = 42, Σx² = 190, Σy² = 350 and Σxy = 244. Calculate the product moment correlation coefficient and test at the 5% level, in two tails, whether the two variables are correlated in the population. The one-tailed critical value for 2.5% with n = 6 is 0.8114.

Accepted Answer

Build the three summary statistics, each correction term dividing by n = 6.
S_xx = 190 − 30²/6 = 190 − 150 = 40.
S_yy = 350 − 42²/6 = 350 − 294 = 56.
S_xy = 244 − (30)(42)/6 = 244 − 210 = 34.
M1 for the three formulas, A1 for all three values.
r = 34/√(40 × 56) = 34/√2240 = 34/47.33 = 0.718. M1 for the formula, A1 for 0.718.
H₀: ρ = 0; H₁: ρ ≠ 0, two-tailed at 5% with n = 6, so the critical values are ±0.8114. M1 for the comparison.
Since 0.718 < 0.8114, do not reject H₀. There is insufficient evidence at the 5% level of correlation between the two variables in the population. A1 for the conclusion in context.
A coefficient over 0.7 failing to reach significance is what six pairs cost: with n = 30 the two-tailed 5% value is 0.3610 and the same coefficient would have been significant several times over.