Questions on choosing, carrying out and concluding a statistical test

Question 1

A test cross between a plant heterozygous for two genes and a plant homozygous recessive for both is expected, if the genes are not linked, to give offspring in a 1:1:1:1 ratio across four phenotype classes. Of 160 offspring, the four classes are observed as 48, 38, 34 and 40. Calculate chi-squared, and use it to reach a conclusion. The critical value at p = 0.05 with 3 degrees of freedom is 7.81.

Accepted Answer

M1 expected number in each of the four classes = 160 ÷ 4 = 40, all comfortably above 5 M1 (48 − 40)² ÷ 40 = 1.6 and (38 − 40)² ÷ 40 = 0.10 M1 (34 − 40)² ÷ 40 = 0.90 and (40 − 40)² ÷ 40 = 0 A1 chi-squared = 1.6 + 0.1 + 0.9 + 0 = 2.6 B1 degrees of freedom = 4 categories − 1 = 3, and the critical value at p = 0.05 is 7.81; 2.6 is less than 7.81 B1 the null hypothesis is not rejected: the observed numbers do not differ significantly from a 1:1:1:1 ratio, consistent with the two genes assorting independently

Question 2

Two varieties of tomato plant are compared for fruit mass: variety A, 12 plants, has a mean fruit mass of 85 g with a standard deviation of 8 g, and variety B, 12 plants, has a mean fruit mass of 92 g with a standard deviation of 9 g. Calculate Student's t, and use it to write a full conclusion. The critical value at p = 0.05 with 22 degrees of freedom is 2.07.

Accepted Answer

M1 s1²/n1 = 8² ÷ 12 = 5.33 and s2²/n2 = 9² ÷ 12 = 6.75 M1 adds the two: 5.33 + 6.75 = 12.08, then takes the square root: √12.08 = 3.48 M1 difference between the means = 92 − 85 = 7 g A1 t = 7 ÷ 3.48 = 2.01 B1 degrees of freedom = 12 + 12 − 2 = 22, and the critical value at p = 0.05 is 2.07; the calculated value of 2.01 is less than 2.07 B1 the null hypothesis is not rejected: there is no significant difference between the mean fruit masses of the two varieties at p = 0.05, on the evidence of these 12 plants of each

Question 3

The percentage cover of a moss species and the number of springtails per sample are recorded in eight quadrats. Ranking each variable and finding the difference in rank for each quadrat gives squared differences of 0, 1, 1, 4, 1, 0, 1 and 4. Calculate Spearman's rank correlation coefficient, and use it to reach a conclusion. The critical value at p = 0.05 for n = 8 is 0.738.

Accepted Answer

M1 Σd² = 0 + 1 + 1 + 4 + 1 + 0 + 1 + 4 = 12 M1 n(n² − 1) = 8 × (64 − 1) = 8 × 63 = 504 M1 6 × Σd² = 6 × 12 = 72 A1 rs = 1 − 72 ÷ 504 = 1 − 0.14 = 0.86 B1 the critical value for n = 8 at p = 0.05 is 0.738, and 0.86 is greater than 0.738 B1 the null hypothesis is rejected: there is a significant positive correlation between moss cover and the number of springtails

Question 4

A calculated chi-squared value of 9.49 is compared with a critical value of 7.81 at p = 0.05 with 3 degrees of freedom, and the null hypothesis is rejected. Describe the four elements a complete conclusion should include, illustrating each with this result.

Accepted Answer

B1 the comparison: state both values, for example 'the calculated value of 9.49 is greater than the critical value of 7.81' B1 the conditions: give the probability level and degrees of freedom used, for example 'at p = 0.05 with 3 degrees of freedom' B1 the decision: state plainly that the null hypothesis is rejected B1 the biology: state what the result means for the organisms or variables in question, in a sentence about the biology rather than only the numbers

Question 5

A biologist has counts of red, pink and white flowered plants from a genetic cross, and wants to compare them with a predicted 1:2:1 ratio. Identify the statistical test this data calls for, identify what is used as O and E, and identify the degrees of freedom for this design.

Accepted Answer

B1 chi-squared, because the data are counts of individuals in categories being compared with a predicted ratio B1 O is the observed number of plants counted in each flower-colour category, and E is the number expected in each category if the 1:2:1 ratio held exactly B1 degrees of freedom = number of categories − 1 = 3 − 1 = 2

Question 6

A study records whether each of 150 patients recovering from a virus is vaccinated or unvaccinated, and whether their illness was mild or severe. State a suitable null hypothesis for testing this data with chi-squared, and state the two conditions this test requires of the data.

Accepted Answer

B1 there is no significant association between vaccination status and the severity of illness B1 chi-squared needs raw counts of individuals in categories, never percentages or proportions B1 every expected value in the table should be at least 5 for the result to be trusted

Question 7

Two samples of woodlice are timed leaving a lit area: the sample kept at 15 °C has a standard deviation of 1.2 minutes and the sample kept at 25 °C has a standard deviation of 6.5 minutes. A student runs the standard t-test formula on this data regardless. Suggest why running the test in this way is a limitation, and suggest what the student should state in an evaluation.

Accepted Answer

B1 the A-level two-sample t-test formula assumes the two samples have roughly the same amount of spread, and here the two standard deviations differ by more than five times B1 the test can still be carried out and a value obtained, but the formula's assumption is clearly not well met by this data, so the result is only approximate B1 an evaluation should state that the two groups are very differently variable and that this limits confidence in the calculated value of t, even though the direction of any conclusion may still be informative

Question 8

A t-test comparing the mean fruit mass of two tomato varieties gives a calculated value below the critical value at p = 0.05. A student concludes that 'the two varieties definitely produce fruit of the same mass.' Explain why this conclusion overstates what the test has shown.

Accepted Answer

B1 failing to reach the critical value means the null hypothesis is not rejected, which shows only that the data available did not provide enough evidence of a difference B1 it does not show that no difference exists: a real but small difference could easily be missed by a small sample, so 'not rejected' is not the same as 'proved equal' B1 the safer statement is that there is no significant difference between the means at this sample size and probability level, leaving open that a larger study might detect one

Question 9

A student measures the diameter and the mass of 40 apples from one orchard. Plotted against each other the points rise in a clear straight line with no obvious outliers, and both variables are roughly normally distributed. Suggest which correlation coefficient is more appropriate here, Pearson's or Spearman's, and justify the choice.

Accepted Answer

B1 Pearson's linear correlation coefficient B1 the scatter is a straight line and both variables are continuous measurements that look roughly normally distributed, which are exactly the conditions Pearson's r assumes B1 Pearson's uses the actual measurements rather than only their order, so with data that meet its conditions it can detect a real association from a smaller sample than Spearman's would need

Question 10

Outline how the degrees of freedom, or the equivalent quantity, is found for chi-squared against a predicted ratio, for the t-test, and for Spearman's rank correlation.

Accepted Answer

B1 chi-squared against a predicted ratio: degrees of freedom = number of categories − 1 B1 the t-test: degrees of freedom = n1 + n2 − 2, the total number of individuals in both samples minus two B1 Spearman's rank: no degrees-of-freedom figure is used; instead the critical value is read against n, the number of pairs

Question 11

A student wants to test whether the number of ladybirds per plant differs between plants near a road and plants far from a road, but has only 4 plants in each group. Suggest one problem this small sample size would cause for a t-test on this data, and suggest what would need to be true of the ladybird counts for the test to be appropriate at all.

Accepted Answer

B1 with only 4 plants per group the test has very little power, so a real difference between the two locations could easily fail to be detected B1 a t-test assumes continuous, roughly normally distributed measurements, and a count of ladybirds is a discrete number that may not meet this condition, especially checked from only 4 values B1 for the test to be appropriate the ladybird counts would need to be roughly continuous or normally distributed in shape, and the two groups' spreads would need to be roughly similar

Question 12

A cross is expected to give offspring in a 9:3:3:1 ratio. Of 48 offspring, the four phenotypes are observed as 30, 10, 6 and 2. Explain why chi-squared should not be carried out on this data as it stands, and explain one way the data could be fixed so that it can be tested.

Accepted Answer

B1 the expected value for the rarest phenotype, 1 part out of 16, is 48 ÷ 16 = 3, which is below the minimum of 5 that chi-squared requires for a trustworthy result B1 with an expected value this small the calculated chi-squared value would not reliably follow the distribution the critical-value table assumes, so a conclusion drawn from it would not be trustworthy B1 the fix is to combine categories, for example the two smaller phenotype classes, or to collect more offspring so that every expected value reaches at least 5, and then repeat the test

Question 13

A biologist wants to test whether hydrogen peroxide concentration affects the rate of oxygen production by potato catalase, and records the volume of oxygen released at five different concentrations, each tested in triplicate. Explain why chi-squared would not be a suitable test for this data, and explain which test should be used instead for comparing two of the concentrations.

Accepted Answer

B1 chi-squared works on raw counts of individuals falling into categories, and a volume of oxygen produced is a continuous measurement rather than a count in a category B1 comparing two sets of continuous, roughly normally distributed measurements, each scattered about a mean, calls for a t-test rather than chi-squared

Question 14

In a Spearman's rank calculation, four quadrats have exactly the same soil moisture reading of 15%. Explain how these four readings should be ranked.

Accepted Answer

B1 the four equal readings are each given the mean of the ranks, or positions, they would otherwise have occupied between them B1 for example, if they would have occupied positions 3, 4, 5 and 6, each is given the rank 4.5, and the next distinct value still takes position 7

Question 15

A student writes 'there was a significant improvement in growth with the new fertiliser' before running any statistical test. Explain why using the word 'significant' here is premature.

Accepted Answer

B1 in everyday use 'significant' means important or noticeable, but in statistics it has a precise, technical meaning: that a calculated value has exceeded a critical value at a stated probability B1 that meaning can only be assigned once a test has actually been carried out and compared with a critical value, so calling a result 'significant' before running the test asserts something not yet established