Printable · GCSE Foundation · ages 14-16
Statistics worksheet — GCSE Foundation
Fifteen questions across the statistics statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Statistics worksheet — GCSE Foundation
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- 1.In a scatter graph of the age, in years, and the wingspan, in cm, of 20 birds of the same species, all the points lie close to a rising line of best fit except one, which lies a long way below the line. That bird was later found to have a damaged wing. Give a reason why this point should not be used when drawing the line of best fit.
- 2.A straight line is drawn on a scatter graph to show the trend of the points. Write down the name given to this line.
- 3.Four pairs of variables are listed below. Write down the pair that you would expect to show negative correlation.
- 4.A recruitment agency in Manchester compares the weekly pay, in pounds, of seven employees at two branches. Branch A: £480, £495, £500, £505, £510, £515, £1,200. Branch B: £480, £490, £500, £510, £520, £530, £540. An advert for the agency claims 'Branch A pays more on average.' Decide whether this claim is fairly supported, using an appropriate average, and choose the correct conclusion.
- 5.The lowest temperature in a garden in Sheffield was recorded on each of seven days: 3 °C, 5 °C, 2 °C, 7 °C, 4 °C, 6 °C and 3 °C. Write down what the range of these seven temperatures describes.
- 6.A scatter graph plots the number of years, x, that 20 employees have worked at a company against their salary, y. All the plotted points lie between x = 1 and x = 15. Write down the word used to describe an estimate for y made using a value of x that lies between 1 and 15.
- 7.Two classes at a school in Coventry sit the same maths test, out of 20 marks. Class A has a mean mark of 14 and a range of 6. Class B has a mean mark of 14 and a range of 14. Write a sentence comparing the two classes, using the mean and the range.
- 8.A scatter graph of taxi journeys in Bristol shows the distance, x miles, and the fare, y pounds. The line of best fit is y = 2x + 3.50. Work out the estimated fare for a journey of 6 miles, using the line of best fit.y = 2x + 3.5
- 9.In a random sample of 40 pupils at a school, 6 are left-handed. The school has 900 pupils. Work out an estimate for the number of left-handed pupils in the school.
- 10.A two-way table records whether each of 40 people at a gym in Sheffield prefers weights or cardio, and whether they are male or female. 24 of the 40 people are female. 15 of the males prefer cardio. 10 of the females prefer weights. Work out the total number of people who prefer cardio.
- 11.A scatter graph has 50 points. Most of them lie close to a rising line of best fit, but two of them lie a long way from that line. Write down how those two points should be treated.
- 12.Four pairs of variables are listed below. Write down the pair that you would expect to show no correlation.
- 13.A school has 1,000 pupils. A student wants to estimate how many of them walk to school, so she asks 8 pupils in her own class. She says her sample is large enough to give a reliable estimate for the whole school. Is she right? Give a reason for your answer.
- 14.A council in Leeds wants to know what local people think about letting shops stay open later in the evening. It rings landline telephone numbers between 10 am and 2 pm on a Tuesday. Write down which group is most likely to be under-represented in the sample, and give a reason for your answer.
- 15.A gym in Cardiff has a scatter graph of the number of training sessions, x, attended by each member and the weight lost, y kg. The line of best fit is y = 0.4x + 1. Members want to lose at least 9 kg. Using the line of best fit, work out the least whole number of sessions needed to reach this target.y = 0.4x + 1
Answer key
- (d) An outlier from the damaged wing, not the trend. — That bird's point lies a long way from the rising trend followed by every other bird, and its low wingspan is explained by the damaged wing rather than by its age — it is an outlier caused by an unusual factor, not part of the general relationship between age and wingspan, so it should not be used when drawing the line of best fit. Saying every point must be used ignores that an outlier caused by a separate, identifiable factor can rightly be set aside. Saying it shows no correlation ignores that the other 19 points do show a clear rising trend; one outlier does not remove that. Saying it proves the line is inaccurate confuses one unusual bird with a fault in the line itself, when the line correctly describes the trend followed by the rest of the data.
- (c) A line of best fit — Method: the straight line drawn on a scatter graph is named from the job it does — it is chosen so that it follows the whole set of points as closely as possible. Working: the line passes through the middle of the points, with roughly as many points above it as below it, and it need not pass through any of the plotted points at all; the name given to the straight line chosen in that way is a line of best fit. Answer: a line of best fit. The distractors: a line of symmetry comes from confusing a trend with symmetry, which is a property of a shape rather than of a set of data; a horizontal line through the mean comes from thinking the trend is shown by an average, when a horizontal line would say that the vertical quantity does not change and so show no correlation; a line joining the first and last points comes from thinking the line must join the two extreme points, which lets two points decide a trend that all of the points should share in.
- (a) Minutes a candle has burned and length remaining — As a candle burns for longer, less of it remains, so these two variables move in opposite directions as one increases — that is negative correlation. A pupil's shoe size generally increases as they get older, so age and shoe size show positive correlation, not negative, since both rise together. A football team's shirt colour is not a numerical quantity linked to how many matches it wins, so shirt colour and number of wins show no correlation at all. The number of letters in a pupil's name has no real connection to their ability in maths, so that pair also shows no correlation.
- (c) No — median £505 at A vs £510 at B. — Branch A's seven wages in order are £480, £495, £500, £505, £510, £515 and £1,200, so the median, the 4th value, is £505. Branch B's in order are £480, £490, £500, £510, £520, £530 and £540, so the median is £510. Since £505 is lower than £510, the median wage is not higher at Branch A, so the claim is not fairly supported. Choosing 'Yes — mean £600.71 at A vs £510 at B' uses the mean: 480 + 495 + 500 + 505 + 510 + 515 + 1200 = 4205, and 4205 ÷ 7 = 600.71, a figure pulled upward by the £1,200 outlier that does not represent a typical wage. Choosing 'Yes — median £515 at A vs £510 at B' miscounts the middle position, taking the 6th wage, £515, instead of the correct 4th value, £505. Choosing 'Yes — highest wage £1,200 at A vs £540 at B' compares the highest wage at each branch rather than a measure of the typical, or average, wage.
- (d) How much the temperatures varied over the seven days — Method: the range of a set of values is the largest value take away the smallest, so it is built from two values only and it measures the gap they leave between them. Working: the largest of the seven readings is 7 °C and the smallest is 2 °C, so the range is 7 − 2 = 5 °C. That figure says the week's readings covered a band 5 °C wide; it names no particular day and no particular reading. Answer: the range describes how much the temperatures varied over the seven days. The distractors: the temperature that occurred most often is the mode, which here is 3 °C, and a mode counts repeats instead of measuring a gap; the temperature typical of the week is an average, and the range is not an average, since it throws away every value lying between the two extremes; the number of different temperatures recorded is 6, a count of how many distinct values appear, while the range is a difference between two of them.
- (b) Interpolation — The salary is being estimated for a value of x between 1 and 15, which is inside the range of x-values that were actually plotted, so this is interpolation. Extrapolation would apply if the estimate used a value of x below 1 or above 15, outside the plotted range. Correlation describes the relationship between the two variables, not the reliability of an estimate, and causation describes one variable actually causing a change in the other, which is a different idea altogether — neither is the word being asked for here.
- (a) Equal means; Class A is more consistent, smaller range. — Method: when two data sets share a measure of location, compare a measure of spread to say more about consistency. Working: both classes have the same mean mark, 14, so on average they performed equally well. Class A has the smaller range, 6, so its marks are more tightly grouped around 14 than Class B's marks, which vary by as much as 14. So Class A's marks were more consistent, even though neither class did better on average. Saying Class B did better because it has the bigger range confuses a wide spread with a high score — a big range describes variability, not performance. Saying Class A did better because it has the smaller range makes the same mistake in the other direction: the two classes are tied on the mean, so neither one 'did better'. Saying the classes cannot be compared because their means are equal misses the whole point of also comparing the range. Always compare both an average AND a spread before describing two data sets — either one alone tells only half the story.
- (d) £15.50 — 2 × 6 = 12, and 12 + 3.50 = 15.50, so the estimated fare is £15.50. Choosing £12.00 stops after 2 × 6 = 12 and forgets to add the £3.50. Choosing £19.00 adds the distance and the constant first and then multiplies: 6 + 3.50 = 9.50, and 9.50 × 2 = 19.00, applying the ×2 to the whole sum instead of only to the distance. Choosing £13.00 multiplies only the constant term by 2 instead of the distance: 2 × 3.50 = 7, and 7 + 6 = 13.00.
- (b) 135 — Method: use the sample to find the PROPORTION of left-handed pupils, then apply that same proportion to the whole school population. Working: in the sample, 6 out of 40 pupils are left-handed, a proportion of 6 ÷ 40 = 0.15. Applying that proportion to the school's 900 pupils gives an estimate of 0.15 × 900 = 135 pupils. Giving 6 simply repeats the number of left-handed pupils IN THE SAMPLE, without scaling up to the whole school at all. Multiplying the population by the number of left-handed pupils in the sample without first dividing by the sample size, 900 × 6 = 5400, badly overestimates — that is more pupils than the whole school has. Dividing the population by the sample size but forgetting to multiply by the number of left-handed pupils found, 900 ÷ 40 = 22.5, finds the scale factor but stops one step short of using it. Always find the proportion in the sample first, then scale that same proportion up to the population.
- (b) 29 — There are 40 − 24 = 16 males, and 15 of them prefer cardio, so 16 − 15 = 1 male prefers weights. There are 24 females, and 10 prefer weights, so 24 − 10 = 14 females prefer cardio. Altogether, 15 + 14 = 29 people prefer cardio. Choosing 15 only counts the males who prefer cardio and forgets the females. Choosing 11 adds the two weights figures, 1 + 10 = 11, instead of the two cardio figures. Choosing 30 comes from 40 − 10, subtracting only the number of females who prefer weights from the grand total, rather than finding both cardio sub-totals separately.
- (b) Treat them as outliers and check them before deciding — Method: a point lying a long way from the pattern the rest of the data make is called an outlier, and an outlier is investigated before anything is done with it, because it may be an error in the data or it may be a genuine but unusual case. Working: 48 of the 50 points lie close to the rising line of best fit, so the trend is set by those 48; the two remaining points do not follow it, so they are identified as outliers and checked — a mistake in measuring or recording would be corrected, while a genuine reading would be kept and reported. Answer: treat them as outliers and check them before deciding what to do with them. The distractors: deleting them at once assumes that every point far from the line must be an error, which throws away real data; moving the line so that it passes through them assumes a line of best fit must touch particular points, when it is drawn to follow all 50; taking them as proof that there is no correlation lets two points overturn the pattern that the other 48 agree on.
- (b) A person's shoe size and their favourite colour — A person's shoe size is not linked to which colour they prefer, so these two show no correlation. The other three pairs are all genuinely correlated: distance travelled and fuel used rise together, which is positive correlation; hours of revision and test score generally rise together, which is also positive correlation; and as outdoor temperature rises, fewer woolly hats are sold, which is negative correlation. Negative correlation is still a real relationship between two variables — it is not the same thing as no relationship at all, so the temperature and hats pair is not the answer to this question.
- (c) No — 8 from one class is too small to represent the school. — Method: judge reliability by asking whether the sample is both large enough, and spread across the population, relative to what it is meant to represent. Working: 8 pupils is a tiny fraction of the school's 1,000 pupils, and all 8 come from a single class rather than a range of year groups, so the sample is both too small and too narrow to represent the whole school reliably. She is not right. Saying any sample size gives an equally reliable estimate ignores that reliability generally improves with a larger, more representative sample. Saying the method is unreliable because it was not done online is not a reason connected to sample size or representativeness at all. Saying 8 is reliable because it is more than half her class compares the sample to the wrong population — the school has 1,000 pupils, not one class. Always judge a sample's size against the population it is meant to represent, not against a smaller group within it.
- (d) Full-time workers, as most are at work at that time — Method: a sample is biased when the method of contact makes part of the population much less likely to be reached, so test each group against where its members actually are between 10 am and 2 pm on a weekday, and test each stated reason against the facts. Working: those hours are the middle of the working day, so people in full-time employment are at work and not beside a landline telephone, while people who are retired and people who are unemployed are far more likely to be at home and are reached at the usual rate; the method therefore collects far fewer replies from full-time workers than their share of the adult population the council is consulting. Answer: full-time workers, as most are at work at that time. The distractors: the reply naming retired people rests on the false claim that most retired people are at work in the daytime, when in fact a daytime call reaches them more easily than anyone; the reply naming unemployed people rests on the false claim that they are out during the day, when they too are among the easiest people to reach by a daytime call; the reply naming children rests on the false claim that children are at home at 11 am on a Tuesday, when they are at school and so are not reached by the call at all, and school-age children are in any case not the adults whose views the council is collecting.
- (d) 20 — 0.4x + 1 = 9, so 0.4x = 9 − 1 = 8, and 8 ÷ 0.4 = 20, so 20 sessions are needed. Choosing 23 divides 9 by 0.4 without first subtracting the 1: 9 ÷ 0.4 = 22.5, rounded up to 23. Choosing 25 subtracts the wrong way, adding the 1 instead of taking it away: 9 + 1 = 10, and 10 ÷ 0.4 = 25. Choosing 2 misplaces the decimal point in the gradient, dividing by 4 instead of by 0.4: 8 ÷ 4 = 2.
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