Questions on averages, spread and the null hypothesis

Question 1

Two heterozygous pea plants are crossed and the offspring are expected in a ratio of 3 tall to 1 dwarf. Of 400 offspring, 280 are tall and 120 are dwarf. Calculate chi-squared and use it to reach a conclusion. The critical value at p = 0.05 with 1 degree of freedom is 3.84.

Accepted Answer

M1 expected values are 400 × 3/4 = 300 tall and 400 × 1/4 = 100 dwarf, both comfortably above 5 M1 (280 − 300)² ÷ 300 = 1.33 and (120 − 100)² ÷ 100 = 4.00 A1 chi-squared = 1.33 + 4.00 = 5.33 B1 degrees of freedom = 2 categories − 1 = 1, and the calculated value of 5.33 is greater than the critical value of 3.84 at p = 0.05 B1 the null hypothesis is rejected: there is a significant difference between the observed numbers and a 3:1 ratio, so the difference is unlikely to have arisen by chance alone

Question 2

Two samples of ten daisy flower heads are measured for diameter: the population growing in full sun has a mean diameter of 28.0 mm with a standard deviation of 3.0 mm, and the population growing in shade has a mean diameter of 24.0 mm with a standard deviation of 3.5 mm. Calculate Student's t, and use it to reach a conclusion. The critical value at p = 0.05 with 18 degrees of freedom is 2.10.

Accepted Answer

M1 s1²/n1 = 3.0² ÷ 10 = 0.900 and s2²/n2 = 3.5² ÷ 10 = 1.225 M1 adds the two: 0.900 + 1.225 = 2.125, then takes the square root: √2.125 = 1.46 A1 the means differ by 28.0 − 24.0 = 4.0 mm, so t = 4.0 ÷ 1.46 = 2.74 B1 degrees of freedom = 10 + 10 − 2 = 18, and the critical value at p = 0.05 is 2.10; the calculated value of 2.74 is greater than 2.10 B1 the null hypothesis is rejected: there is a significant difference between the mean flower diameters, with daisies in full sun significantly larger than those in shade

Question 3

Five leaves taken from one shrub have widths of 12, 14, 15, 17 and 22 mm. Calculate the mean width and the standard deviation of these widths, giving the standard deviation to three significant figures.

Accepted Answer

M1 mean = (12 + 14 + 15 + 17 + 22) ÷ 5 = 80 ÷ 5 = 16 mm M1 the deviations from the mean are −4, −2, −1, +1 and +6, and the squares of these total 16 + 4 + 1 + 1 + 36 = 58 M1 divide by n − 1 rather than by n: 58 ÷ 4 = 14.5, then take the square root A1 standard deviation = 3.81 mm

Question 4

A student writes that the error bars on two mean values do not overlap, so the difference between the means is significant. Explain why this reasoning is unsound.

Accepted Answer

B1 an error bar may show one standard deviation, one standard error or a 95 per cent confidence interval, and the three have different lengths drawn on the same data, so the same means can appear to overlap or not depending on the choice B1 standard-deviation bars describe variation between individuals and barely narrow however many organisms are measured, so two genuinely different means can easily have overlapping bars B1 standard-error bars describe how precisely each mean has been estimated and shrink roughly as one over the square root of the sample size, so with a large enough sample almost any two means separate B1 significance is decided only by comparing a calculated test value with a critical value at a stated probability, so bars can suggest that a test is worth doing and no more

Question 5

Compare Student's t-test with Spearman's rank correlation, referring to the data each one needs and the question each one answers.

Accepted Answer

B1 a t-test needs two separate sets of measurements each scattered about a mean, whereas Spearman's needs two different quantities measured on each of the same individuals or quadrats B1 a t-test asks whether the two means differ, whereas Spearman's asks whether the two variables vary together B1 the t-test uses the measurements themselves and assumes they are continuous and roughly normally distributed, whereas Spearman's replaces every measurement by its rank and never sees the values B1 degrees of freedom for the t-test are n₁ + n₂ − 2, whereas Spearman's critical value is read against n, the number of pairs; for both, a calculated value at or above the critical value means the null hypothesis is rejected

Question 6

A researcher measures 2000 fish from each of two lakes and reports that the mean masses differ by 0.4 g, a difference a t-test finds significant at p = 0.05. Suggest why this significant result may still be of little biological interest.

Accepted Answer

B1 significance depends on the size of the sample as well as on the size of the difference, so a sample of 2000 from each lake can detect a difference this small B1 significant means unlikely to have arisen by chance under the null hypothesis; it does not mean the difference is large, and 0.4 g is a tiny fraction of the mass of a fish B1 the sentence reporting a significant result should give the size of the difference as well, and further work would be needed to show whether a difference this small has any effect on the fish

Question 7

Compare Pearson's linear correlation coefficient with Spearman's rank correlation coefficient.

Accepted Answer

B1 both test whether two quantities measured on the same individuals are associated, and both give a value between −1 and +1 whose sign shows the direction B1 Pearson uses the measurements themselves and tests how well a straight line fits them, whereas Spearman replaces each measurement by its rank and tests only whether the two rise or fall together B1 Pearson is parametric, so it assumes each variable is roughly normally distributed and the relationship is linear, whereas Spearman is non-parametric and assumes neither

Question 8

A student records soil moisture and the number of plant species in each of twelve quadrats. The scatter graph rises steadily but curves, and two quadrats have moisture readings far above the rest. Suggest which correlation test the student should use, and justify the choice.

Accepted Answer

B1 Spearman's rank correlation coefficient B1 Pearson's r assumes the relationship is linear and here the scatter is a curve, so a straight line does not describe the data B1 the two extreme readings would have a large effect on Pearson's r, whereas Spearman uses only the order of the values, so an extreme reading counts as one rank and nothing more

Question 9

A researcher records the blood group, A, B, AB or O, of 200 patients, and separately records the reaction time, in milliseconds, of 200 different people, a data set with one exceptionally slow outlier. Suggest which average, mean, median or mode, is most suitable for each data set, and justify each choice.

Accepted Answer

B1 blood group: the mode, because blood group is a category rather than a numerical measurement, so a mean or median has no meaning for it B1 reaction time: the median, because the one exceptionally slow outlier would pull the mean towards it while barely moving the middle value of the ordered data B1 the median is a fairer description of a typical reaction time when the data are skewed by an extreme value, whereas the mean uses every value, including the outlier

Question 10

A calculator offers two standard deviation formulae, one dividing by n and one dividing by n − 1. Explain which of the two a biologist should use when the readings are a sample from a larger population, and explain why.

Accepted Answer

B1 the n − 1 version should be used, because biological readings are almost always a sample drawn from a larger population, not a complete population B1 using the sample's own mean to measure the sample's own spread makes the calculated deviations slightly too small B1 dividing by n − 1 rather than n corrects for this, giving a less biased estimate of the population's standard deviation

Question 11

State a suitable null hypothesis for an investigation testing whether there is a correlation between the concentration of ozone in the air and the mean leaf area of a species of tree, and state what a significant result would allow a researcher to conclude about the null hypothesis.

Accepted Answer

B1 there is no significant correlation between ozone concentration and the mean leaf area of the tree B1 a significant result allows the null hypothesis to be rejected B1 rejecting it supports the conclusion that there is a genuine correlation between the two variables, though a correlation on its own does not show that ozone caused the change in leaf area

Question 12

A newspaper reports 'the probability that this difference in mean plant height is due to chance is only 0.03, so there is a 97% probability the fertiliser really works.' Explain what is wrong with turning a p-value of 0.03 into a 97% probability that the fertiliser works.

Accepted Answer

B1 a p-value is calculated by assuming the null hypothesis is true in the first place, so it cannot then be used to state how likely that starting assumption was B1 a p-value of 0.03 means that a difference this size, or larger, would occur in only 3 samples in 100 if the fertiliser had no real effect, not that there is a 97% chance the fertiliser has an effect B1 the sentence also says nothing about the size of the effect, which a complete answer should report separately from the significance

Question 13

Two samples of leaf widths have the same mean and the same range, yet a standard deviation calculation shows one sample is much more variable than the other. Explain how two samples can share the same range while differing in standard deviation, and explain why the standard deviation is generally the better statistic to report.

Accepted Answer

B1 the range depends only on the two most extreme readings and says nothing about how the values between them are distributed, so differently shaped sets of data can share the same highest and lowest value B1 the standard deviation is calculated from every reading in the set, so it is sensitive to how tightly or loosely the whole set is clustered around the mean, not only the extremes B1 the standard deviation is generally the better statistic because it uses all the information collected, whereas the range discards everything except two values and also tends to grow simply because more readings were taken

Question 14

An investigation compares the mean number of stomata per square millimetre on leaves grown in sun and on leaves grown in shade. State a suitable null hypothesis, and state the significance level conventionally used in biology.

Accepted Answer

B1 there is no significant difference between the mean number of stomata per square millimetre on sun leaves and the mean number on shade leaves B1 the conventional significance level is p = 0.05, below which the null hypothesis is rejected

Question 15

A biologist calculates Pearson's r for two variables and obtains a value of 0.04. Explain why this does not show that the two variables are unrelated.

Accepted Answer

B1 Pearson's r measures how well a straight line fits the data, so a value near zero shows only that there is no linear association B1 a variable that rises and then falls again, or that is related in any other curved way, can give a value near zero while still being strongly related, which is why the scatter graph is plotted before the test is chosen