Non-parametric tests questions

Question 1

State what makes a hypothesis test non-parametric, and give one situation in which you would prefer such a test to a t-test.

Accepted Answer

A non-parametric, or distribution-free, test assumes no family of distributions for the population it is testing; it works from the signs or the ranks of the readings, and its hypotheses concern a median, or the identity of two populations, rather than the parameters of a named distribution. B1.
Any one reason, B1: the population is plainly not normal, for instance strongly skewed incomes or waiting times; the sample is too small for normality to be worth checking; an outlier would drag a mean about; or the data arrive as ranks, preferences or ordered categories with no scale attached.
The trade is power: where the population really is normal, the t-test detects a smaller shift from the same sample size, because it uses the readings themselves and not their order.

Question 2

A supplier claims the median mass of its bags of compost is 25 kg. Eleven bags are weighed and nine of them exceed 25 kg, with none equal to it. Carry out a sign test at the 5% level of the claim against the alternative that the median mass exceeds 25 kg.

Accepted Answer

H₀: median = 25; H₁: median > 25. One-tailed at the 5% level. B1 for both hypotheses, written about the median and not about a mean.
Under H₀ each bag is above the median with probability ½, so the number above is B(11, ½).
P(X ≥ 9) = (55 + 11 + 1)/2048 = 67/2048 = 0.0327. M1 for the binomial tail, including all three terms.
Since 0.0327 < 0.05, reject H₀. There is evidence at the 5% level that the median mass exceeds 25 kg. A1 for the conclusion in context.
No table is needed anywhere in this test: the null distribution is binomial, which is why the sign test survives without a booklet of critical values.

Question 3

State the null hypothesis of a sign test, and explain why the number of readings above the claimed value follows a binomial distribution with p = ½ when that hypothesis is true.

Accepted Answer

H₀: the population median takes the claimed value m₀. B1. Written about the median, not about a mean: the sign test says nothing about means, and for a skewed population the two are different numbers.
If m₀ really is the median, then any one reading falls above it with probability ½ by the definition of a median, and the readings of a random sample are independent of one another. A fixed number of independent trials, each with probability ½ of a success, is exactly B(n, ½). B1.

Question 4

Eight readings are 14.2, 11.9, 15.6, 12.9, 13.4, 16.1, 10.7 and 14.9, and a claimed median is 13.0. Write down the deviations from 13.0, rank their sizes, and find the two signed-rank totals T₊ and T₋.

Accepted Answer

The deviations, in the order given, are +1.2, −1.1, +2.6, −0.1, +0.4, +3.1, −2.3 and +1.9. M1 for subtracting 13.0 from each reading.
Ranking their sizes, smallest first, gives 4, 3, 7, 1, 2, 8, 6 and 5. A1.
The three negative deviations carry ranks 3, 1 and 6, so T₋ = 10 and T₊ = 26. A1.
Check: 10 + 26 = 36 = 8 × 9/2, which is what the eight ranks must total. Rank the sizes and then restore the signs; ranking the signed deviations themselves puts −2.3 at rank 1 and produces neither total.

Question 5

Name the non-parametric test you would use in each case, with a reason. (a) The house prices of 9 properties sold in one town and 9 sold in another are compared for a difference in level. (b) Twelve judges each taste two wines and say which of the two they prefer. (c) A single sample of 15 reaction times, from a population known to be skewed, is tested against a claimed median.

Accepted Answer

(a) Wilcoxon rank-sum test. Two independent groups with no partner for any reading, so the samples are unpaired; the two lists happening to be the same length settles nothing, and house prices are skewed enough that a test about means would be hard to justify. B1. (b) Sign test on the twelve pairs. Each judge supplies a pair, so the data are paired, but the record is a direction with no size attached, so there are no deviations to rank and the signed-rank test has nothing to work with. B1. (c) Single-sample Wilcoxon signed-rank test. One sample against a claimed median, with the deviations measured, so their sizes can be ranked and the test uses more of the data than a sign test would. B1. Give the reason as well as the name; the reason follows from how the data were collected.

Question 6

A machine is claimed to fill tubs to a median of 50 g. Nine tubs contain 52.4, 49.3, 55.3, 47.9, 53.8, 49.2, 51.5, 56.4 and 55.1 g. Use a Wilcoxon signed-rank test at the 5% level to test the claim against the alternative that the median exceeds 50 g. The critical value is 8.

Accepted Answer

H₀: median = 50; H₁: median > 50. One-tailed at the 5% level, n = 9. B1.
The deviations from 50 are +2.4, −0.7, +5.3, −2.1, +3.8, −0.8, +1.5, +6.4, +5.1. M1 for subtracting the claimed median from each reading.
Ranking their sizes from 1 for the smallest gives 5, 1, 8, 4, 6, 2, 3, 9, 7. M1 A1. Rank the absolute deviations, then restore their signs.
The three negative deviations carry ranks 1, 4 and 2, so T₋ = 7 and T₊ = 38. Check: 7 + 38 = 45 = 9 × 10/2.
H₁ says the deviations run positive, so it is T₋ that should be small, and T₋ = 7 is the statistic. Since 7 ≤ 8, reject H₀. There is evidence at the 5% level that the median fill exceeds 50 g. M1 for the comparison, A1 for the conclusion in context.
A small value is the evidence here, which is the reverse of every other test in this unit. For the stated upper-tailed alternative, use the smaller negative-rank total and compare it with the supplied critical value.

Question 7

Ten operators are timed on a task before and after their workstation is rearranged. The numbers of units produced before are 42, 55, 38, 61, 47, 50, 44, 58, 36, 53, and afterwards 56, 52, 60, 68, 45, 61, 62, 53, 62, 62. State whether the samples are paired or independent, and test at the 5% level whether output has risen, using a Wilcoxon matched-pairs signed-rank test. The critical value is 10.

Accepted Answer

The samples are paired: each operator supplies a before figure and an after figure, so each reading has a partner and the two lists are in the same order. B1. Treating them as two independent samples of 10 would throw the pairing away. H₀: the median difference is 0; H₁: it is positive. One-tailed at the 5% level, n = 10 pairs. The differences, after minus before, are +14, −3, +22, +7, −2, +11, +18, −5, +26, +9. M1. Ranking their sizes gives 7, 2, 9, 4, 1, 6, 8, 3, 10, 5, so the three negative differences carry ranks 2, 1 and 3. M1 A1. T₋ = 2 + 1 + 3 = 6 and T₊ = 49; check 6 + 49 = 55 = 10 × 11/2. Since 6 ≤ 10, reject H₀. There is evidence at the 5% level that output has risen. M1 A1 for the comparison and the conclusion in context. The sign test on the same data counts 7 rises and 3 falls, giving P(X ≥ 7) = 176/1024 = 0.172 under B(10, ½) and no significant result. The three operators who slipped slipped by 2, 3 and 5 units while the risers rose by up to 26, and only the test that ranks the sizes can use that.

Question 8

Twelve tasters each try two brands of coffee and record which of the two they prefer, with no taster allowed to record a tie. Nine prefer brand A. Test at the 5% level, in two tails, whether tasters as a population have a preference between the brands.

Accepted Answer

Each taster supplies a pair and records only a direction, so this is a paired sign test. Let p be the probability that a taster prefers brand A. H₀: p = ½; H₁: p ≠ ½. B1 for the hypotheses.
Under H₀ the number preferring A is B(12, ½), and 9 is above the mean of 6, so the upper tail is the one to measure:
P(X ≥ 9) = (220 + 66 + 12 + 1)/4096 = 299/4096 = 0.0730. M1 for the binomial tail with all four terms, A1 for 0.0730.
The test is two-tailed at 5%, so the tail probability is compared with 0.025. Since 0.0730 > 0.025, do not reject H₀. There is insufficient evidence at the 5% level that tasters prefer one brand to the other. A1 for the conclusion in context.
Compare the single tail with half the significance level, or double the tail and compare with the whole of it; doing neither turns a two-tailed test at 5% into a one-tailed test at 10%, which here would have rejected. Ten preferences rather than nine would have given a tail of 0.0193 and the opposite verdict.

Question 9

Seven pairs of twins each attempt a puzzle, one twin of each pair given a hint and the other not. The times taken, in seconds, with hint minus without hint, are −12, −5, −18, +3, −9, −21 and −7. State whether the samples are paired or independent, name the test that uses the sizes of the differences, and test at the 5% level, in one tail, whether the hint reduces the time. The critical value is 3.

Accepted Answer

Each pair of twins supplies one difference, so the samples are paired, and the test that uses the sizes as well as the signs is the Wilcoxon matched-pairs signed-rank test. B1.
H₀: the median difference is 0; H₁: it is negative, since a hint that helps makes the time with the hint the smaller. One-tailed at 5%, n = 7.
Ranking the sizes 3, 5, 7, 9, 12, 18, 21 from 1 for the smallest, the differences take ranks 5, 2, 6, 1, 4, 7 and 3. M1 for ranking the sizes, A1 for the ranks.
The one positive difference carries rank 1, so T₊ = 1 and T₋ = 27; check 1 + 27 = 28 = 7 × 8/2. M1 for T₊ as the statistic.
H₁ says the differences run negative, so it is T₊ that should be small, and small values are the evidence. Since 1 ≤ 3, reject H₀. There is evidence at the 5% level that the hint reduces the time taken. A1 for the conclusion in context.
Take the total the alternative hypothesis pushes downwards; using T₋ = 27 here and comparing it with 3 reverses the test.

Question 10

Five plants of variety P have heights 34, 41, 28, 39 and 47 cm. Six independent plants of variety Q have heights 52, 44, 38, 57, 49 and 55 cm. Test at the 5% level, in two tails, whether the two varieties have the same distribution of heights. The critical values are 18 and 42.

Accepted Answer

The two groups have no partner for any reading and are of different sizes, so they are independent and the Wilcoxon rank-sum test applies. H₀: the heights come from the same population; H₁: they do not. B1.
Rank all eleven heights together, from 1 for the shortest: 28(1), 34(2), 38(3), 39(4), 41(5), 44(6), 47(7), 49(8), 52(9), 55(10), 57(11). M1 for one ranking over the pooled data.
W is the rank total of the smaller sample, so for variety P, W = 2 + 5 + 1 + 4 + 7 = 19. A1. Check: variety Q's ranks total 47, and 19 + 47 = 66 = 11 × 12/2.
Since 18 < 19 < 42, the statistic lies between the two critical values, so do not reject H₀. There is insufficient evidence at the 5% level that the two varieties differ in height. M1 for the comparison, A1 for the conclusion in context.
Under H₀ the mean of W is 5 × 12/2 = 30, so 19 is well below what chance alone would produce, and it misses the critical value by one. Report that as a near miss rather than as evidence that the two varieties are alike.

Question 11

Name the non-parametric test that fits each of the following, with a reason. (a) A single sample of 20 waiting times is to be tested against a claimed median, but only whether each time was above or below the claim has been recorded. (b) Twelve students sit a test before and after a course, and both marks are recorded for each student. (c) Measurements from two independent processes, 7 from one and 11 from the other, are compared.

Accepted Answer

(a) Sign test. One sample against a claimed median, and the record holds directions only, so there are no deviations whose sizes could be ranked. B1. (b) Wilcoxon matched-pairs signed-rank test. Each student supplies a pair, so the data are paired and reduce to twelve differences, and the marks themselves are known, so the sizes of those differences can be ranked. B1. (c) Wilcoxon rank-sum test. No reading in one group has a partner in the other, and the unequal sizes settle that at once, so there is nothing to subtract and the pooled ranking is the only route. B1. The reason comes from how the data were collected and what each reading carries, and it is settled before any arithmetic starts.

Question 12

Six plots given fertiliser A yield 12, 15, 18, 23, 26 and 37 kg; eight independent plots given fertiliser B yield 21, 29, 31, 34, 40, 43, 46 and 52 kg. Test at the 5% level, in two tails, whether the two fertilisers give the same distribution of yields. The critical values are 29 and 61.

Accepted Answer

The samples are of different sizes, so they are not paired and the Wilcoxon rank-sum test is the one that fits. B1.
H₀: the yields come from the same population; H₁: they do not. Two-tailed at the 5% level, m = 6 and n = 8. B1. The hypotheses are about the populations being identical, not about two means.
Ranking all fourteen yields together from 1 for the smallest gives fertiliser A the ranks 1, 2, 3, 5, 6 and 10, and fertiliser B the ranks 4, 7, 8, 9, 11, 12, 13 and 14. M1 A1 for one ranking over the pooled data. Pool the observations before assigning ranks; separate rankings do not produce the Wilcoxon rank-sum statistic.
W is taken from the smaller sample, so W = 1 + 2 + 3 + 5 + 6 + 10 = 27. Check: the other total is 78, and 27 + 78 = 105 = 14 × 15/2. M1.
Under H₀ the mean of W is 6 × 15/2 = 45, so 27 lies well below what chance alone would produce.
Since 27 ≤ 29, reject H₀. There is evidence at the 5% level that the two fertilisers give different yield distributions, with A the lower. A1 A1 for the comparison and the conclusion in context.
The two critical values add to m(m + n + 1) = 6 × 15 = 90, since 29 + 61 = 90; the distribution of W is symmetric about 45, so a table printing one end gives the other.

Question 13

Twenty-five paired differences are ranked by size and the negative ranks total 95. The table of critical values does not extend to n = 25. Use a normal approximation, with a continuity correction, to test at the 5% level in one tail whether the median difference is positive. State also what difference the continuity correction makes here.

Accepted Answer

H₀: the median difference is 0; H₁: it is positive. One-tailed at the 5% level, so the critical value is −1.645. B1.
For n = 25 the signed-rank statistic has mean n(n + 1)/4 = 25 × 26/4 = 162.5 and variance n(n + 1)(2n + 1)/24 = 25 × 26 × 51/24 = 33150/24 = 1381.25, so the standard deviation is √1381.25 = 37.17. M1 A1 for the mean and the variance.
H₁ says the differences run positive, so T₋ = 95 is the statistic and it is the lower tail that matters. T takes whole values only, so P(T ≤ 95) becomes the area below 95.5.
z = (95.5 − 162.5)/37.17 = −67/37.17 = −1.803. M1 A1 for the standardising with the half included.
Since −1.803 < −1.645, reject H₀. There is evidence at the 5% level that the median difference is positive. A1.
Without the correction z would be −67.5/37.17 = −1.816, the same verdict from a slightly overstated statistic. The correction moves z towards zero, so leaving it out tilts the test towards rejecting; here it changes the third decimal place, not the conclusion. Note also that a two-tailed test at 5% would compare 1.803 with 1.96 and would not reject, so the tail has to be settled before the critical value is quoted.

Question 14

Two independent samples, of sizes 15 and 18, are compared with a Wilcoxon rank-sum test, and the rank total for the sample of 15 is 200. The tables of critical values do not reach these sample sizes. Use a normal approximation, with a continuity correction, to test at the 5% level, in two tails, whether the two populations are identical.

Accepted Answer

H₀: the two samples come from the same population; H₁: they do not. Two-tailed at 5%, so the critical values are ±1.96. B1.
Take the smaller sample first, so m = 15 and n = 18. Under H₀, W has mean m(m + n + 1)/2 = 15 × 34/2 = 255 and variance mn(m + n + 1)/12 = 15 × 18 × 34/12 = 9180/12 = 765, so the standard deviation is √765 = 27.66. M1 A1 for the mean and the variance.
W = 200 lies below the mean, so it is the lower tail that matters. W takes whole values only, so P(W ≤ 200) becomes the area below 200.5.
z = (200.5 − 255)/27.66 = −54.5/27.66 = −1.970. M1 A1 for standardising with the half included.
Since −1.970 < −1.96, reject H₀. There is evidence at the 5% level that the two populations differ, with the sample of 15 taking the lower ranks. A1.
Without the correction z would be −55/27.66 = −1.989: the same verdict from a statistic that overstates the case. The correction always moves z towards zero, so leaving it out tilts a test towards rejecting, and here the margin over 1.96 is thin enough for that to matter to how the result is reported.

Question 15

Two independent samples, of sizes 7 and 9, are compared with a Wilcoxon rank-sum test, and the rank total for the sample of 7 is W = 42. Write down the smallest and the largest values W could take. The lower critical value for a two-tailed test at the 5% level is 40; find the upper one and carry out the test. The Mann-Whitney statistic is U = W − ½m(m + 1), where m is the smaller sample size; find U and state what it counts.

Accepted Answer

The smallest total the sample of 7 could take is 1 + 2 + … + 7 = 28, when its readings are the seven lowest, and the largest is 10 + 11 + … + 16 = 91, when they are the seven highest. B1 for both.
Those two extremes add to 119 = m(m + n + 1) = 7 × 17, and the distribution of W is symmetric about its mean of 119/2 = 59.5, so the two critical values add to 119 as well. M1 for m(m + n + 1).
The upper critical value is therefore 119 − 40 = 79. A1.
H₀: the two samples come from the same population; H₁: they do not. Since 40 < 42 < 79, the statistic lies between the critical values, so do not reject H₀. There is insufficient evidence at the 5% level that the two populations differ. A1 for the conclusion in context.
U = 42 − ½(7)(8) = 42 − 28 = 14. M1 for the subtraction, A1 for 14 with what it counts: taking the smallest possible total off W shifts the origin to zero, and what is left counts the pairs, one reading from each sample, in which the reading from the sample of 7 is the larger. Out of 7 × 9 = 63 such pairs, 14 go that way.
U and W are the same test written from different origins, so a table for one converts to a table for the other by the same shift of 28.