Revision Hours Against an Exam Mark, with the Line of Best Fit Through the Mean Point
Eight students recorded the hours they spent revising for an exam and the mark they scored. The pairs of (hours, mark) were (2, 30), (3, 40), (4, 42), (4, 49), (5, 44), (6, 61), (7, 60) and (9, 74). The line of best fit cuts the mark axis at 20. (a) Find the mean point and the equation of the line of best fit. (b) Estimate the mark of a student who revises for 8 hours.
A line of best fit is drawn so that the points are balanced about it, and every such line passes through the mean point, whose coordinates are the mean of the hours and the mean of the marks. One further point on the line fixes it, and here the question gives the mark the line predicts for no revision at all. The prediction is then a substitution.
- Find the mean of the hours. 2 + 3 + 4 + 4 + 5 + 6 + 7 + 9 = 40, and there are 8 students, so the mean is 40 ÷ 8 = 5 hours.
- Find the mean of the marks. 30 + 40 + 42 + 49 + 44 + 61 + 60 + 74 = 400, so the mean is 400 ÷ 8 = 50 marks. The mean point is (5, 50), and it is marked with a cross.
- The line of best fit passes through (0, 20) and through the mean point (5, 50), so its gradient is 50 − 205 − 0 = 6. An extra hour of revision is worth about 6 marks.
- (a) The mean point is (5, 50) and the line of best fit is y = 6x + 20, where x is the hours of revision and y is the mark.
- (b) Substitute x = 8 into the line: y = 6 × 8 + 20 = 68, so the estimate is about 68 marks. Check: 8 hours lies between the 2 hours and the 9 hours recorded, so the line is being read inside the data, and 68 falls between the marks of 60 and 74 as it should.
answer(a) the mean point is (5, 50) and the line of best fit is y = 6x + 20; (b) about 68 marks
techniqueScatter Plots · Line of Best Fit
examsSAT · GCSE Higher · H2
Common pitfalls
- Joining the points one to the next instead of drawing a single straight line. A scatter diagram is not a line graph: each point is a different student, and the line of best fit is one straight line that all of them are balanced about.
- Reading the prediction off the nearest point, so a student who revises 8 hours is given the 74 marks of the student who revised 9 hours. The prediction comes from the line, which uses all eight students, and not from whichever one happens to be closest.
A Car's Age Against Its Price, Where the Gradient Is the Value Lost Each Year
A dealer recorded the age and the price of eight cars of one model. The pairs of (age in years, price in thousands of dollars) were (1, 17), (2, 14), (3, 14), (4, 14.5), (4, 9), (5, 11.5), (6, 8.5) and (7, 7.5). The line of best fit passes through (2, 15) and (6, 9). (a) Find the gradient of the line and say what it means for these cars. (b) Find the equation of the line and say what its intercept means.
The gradient of a line of best fit is a rate, and it carries the units of both axes: thousands of dollars for every year. Its sign says which way the correlation runs. The intercept is the value the line gives when the horizontal variable is zero, which here is the price of a car of no age at all.
- Plot the eight cars with the age along the horizontal axis and the price up the vertical axis. The points fall from left to right, so age and price are in negative correlation: the older a car is, the less it costs.
- Use the two points the line passes through. From (2, 15) to (6, 9) the age rises by 6 − 2 = 4 years and the price falls by 15 − 9 = 6 thousand dollars.
- (a) The gradient is 9 − 156 − 2 = −64 = −1.5, and its units are thousands of dollars for each year. A car of this model loses about $1500 of its value every year.
- Find the intercept. Going back from (2, 15) by 2 years adds 2 × 1.5 = 3 thousand dollars, so at age 0 the line gives 15 + 3 = 18 thousand dollars.
- (b) The line of best fit is p = 18 − 1.5a, where a is the age in years and p is the price in thousands of dollars. The intercept 18 is the price at age 0, so a car of this model cost about $18 000 when it was new. Check: the mean age is 32 ÷ 8 = 4 years and the mean price is 96 ÷ 8 = 12 thousand dollars, and 18 − 1.5 × 4 = 12, so the line does pass through the mean point (4, 12).
answer(a) the gradient is −1.5 thousand dollars a year, so a car of this model loses about $1500 of its value each year; (b) p = 18 − 1.5a, which passes through the mean point (4, 12), and the intercept 18 means the car cost about $18 000 when new
techniqueCorrelation · Line of Best Fit
examsSAT · GCSE Higher · H2
Common pitfalls
- Giving the gradient as 1.5 because the fall of 6 thousand dollars is a positive amount of money. The price goes down as the age goes up, so the gradient is negative, and that sign is what records the negative correlation.
- Calling the intercept the price of a car in the table that is 0 years old. No car in the table is new; 18 thousand dollars is what the line predicts at age 0, and it is worth quoting only because 0 is close to the youngest car recorded, which is 1 year old.
Cold Drinks Against the Temperature, Read Once Inside the Readings and Once Far Outside Them
A beach kiosk recorded the temperature and the cold drinks it sold on eight days. The pairs of (temperature in Celsius, drinks) were (15, 55), (17, 70), (19, 110), (21, 125), (25, 185), (27, 230), (29, 250) and (31, 295). The line of best fit is y = 15x − 180, where x is the temperature in Celsius and y is the number of drinks. (a) Estimate the sales on a day at 24 Celsius. (b) A manager reads the same line at 5 Celsius. Find what it predicts there, and say why the line must not be used at that temperature.
A line of best fit summarizes the days that were recorded. Between the lowest and the highest reading it is filling a gap that the data surrounds, which is reliable; beyond them it is a guess with nothing behind it, and the guess can turn into an impossible value. Substituting the two temperatures and then looking at where each one sits is the whole of this question.
- Find the mean point, because the line of best fit must pass through it. The temperatures total 15 + 17 + 19 + 21 + 25 + 27 + 29 + 31 = 184, so the mean temperature is 184 ÷ 8 = 23 Celsius.
- The sales total 55 + 70 + 110 + 125 + 185 + 230 + 250 + 295 = 1320, so the mean is 1320 ÷ 8 = 165 drinks. The mean point is (23, 165), and 15 × 23 − 180 = 165, so the given line does pass through it.
- (a) Substitute x = 24: y = 15 × 24 − 180 = 360 − 180 = 180, so the kiosk should expect about 180 drinks. The value 24 Celsius lies between the lowest reading, 15 Celsius, and the highest, 31 Celsius, so the line is being read inside the data and the estimate can be trusted.
- Now substitute x = 5: y = 15 × 5 − 180 = 75 − 180 = −105.
- (b) The line predicts −105 drinks, and a kiosk cannot sell a negative number of drinks, so the prediction is worthless. A temperature of 5 Celsius is far below the coldest day recorded, 15 Celsius, and nothing in the data says what the kiosk does there. Check: the line reaches zero when 15x = 180, that is at x = 12 Celsius, and it is negative at every temperature below that.
answer(a) the mean point is (23, 165) and the estimate at 24 Celsius is about 180 drinks; (b) the line gives −105 drinks, which is impossible, so it must not be used at 5 Celsius, because the readings run only from 15 Celsius to 31 Celsius
techniqueInterpolation and Extrapolation · Line of Best Fit
examsSAT · GCSE Higher · H2
Common pitfalls
- Treating the line as a law that holds at every temperature. It is a summary of eight particular days, and outside the range of those days it carries no evidence at all.
- Rounding −105 up to 0 and reporting that the kiosk sells no drinks at 5 Celsius. That hides the fault instead of reporting it: a model that returns an impossible value is showing that it is being used outside its range.
Umbrellas Against Birthday Cards, and a Prediction from Ten Days of Sales
A shop recorded, on ten days, the umbrellas and the birthday cards it sold. The pairs of (umbrellas, cards) were (2, 28), (4, 20), (5, 26), (7, 22), (8, 24), (11, 27), (13, 21), (14, 25), (16, 18) and (20, 29). (a) Find the mean point, and describe the correlation between the two sales. (b) The manager wants to predict the card sales on a day when 30 umbrellas are sold. Say what should be done, and give the best estimate the data supports.
A line of best fit follows a trend, so the first question is whether there is one. Splitting the days into the half with the fewest umbrellas and the half with the most, and comparing the mean card sales of the two halves, answers that in one step. When the two halves agree there is nothing for a line to follow, and the mean is the only honest estimate.
- Find the mean point. The umbrellas total 2 + 4 + 5 + 7 + 8 + 11 + 13 + 14 + 16 + 20 = 100, so the mean is 100 ÷ 10 = 10 umbrellas.
- The cards total 28 + 20 + 26 + 22 + 24 + 27 + 21 + 25 + 18 + 29 = 240, so the mean is 240 ÷ 10 = 24 cards, and the mean point is (10, 24).
- Split the days at the middle. On the five days with the fewest umbrellas the cards were 28, 20, 26, 22 and 24, which total 120 and average 24. On the five days with the most umbrellas they were 27, 21, 25, 18 and 29, which also total 120 and average 24.
- (a) The two halves have the same mean, and the points are scattered over the whole diagram with no drift upward or downward, so there is no correlation between umbrella sales and card sales.
- (b) No line of best fit should be drawn. A line needs a trend to follow and there is none here, so any line drawn by eye would only be repeating the accidents of these ten days. The best estimate for the cards is the mean, 24 cards, whatever the umbrellas do. Check: 30 umbrellas is also more than any day recorded, so even a real trend could not have been followed that far.
answer(a) the mean point is (10, 24) and there is no correlation: the five days with the fewest umbrellas and the five with the most both averaged 24 cards; (b) no line of best fit should be drawn, so the best estimate is the mean, 24 cards
techniqueCorrelation · Scatter Plots
examsSAT · GCSE Higher · H2
Common pitfalls
- Drawing a line through the leftmost and the rightmost points, (2, 28) and (20, 29), and calling it the line of best fit. Two points always determine a line, so this can be done to any scatter diagram at all, and it says nothing about the other eight days.
- Reading the high value at 20 umbrellas as the start of an upward trend. One point cannot show a trend, and the day beside it, at 16 umbrellas, has the lowest card sales of all ten.
Two Years of Quarterly Sales, with a Season Riding on Top of a Trend
A garden center recorded its sales, in thousands of dollars, for eight quarters. Numbering the quarters 1 to 8, the sales were 28, 40, 50, 42, 44, 56, 66 and 58. Quarters 1 and 5 are the first quarter of a year, and quarters 3 and 7 the third. The trend line is y = 4n + 30, where n is the quarter number. (a) Say what the gradient of the trend line means, and find how far each first quarter and each third quarter lies from the trend. (b) Estimate the sales in quarter 9.
A time series has two parts. The trend is the steady movement over the whole record, and the seasonal effect is the amount by which a particular quarter sits above or below that trend, year after year. A forecast uses both: take the trend forward to the quarter wanted, then add the seasonal effect for that quarter.
- Plot the sales against the quarter number and join the points in order. The record climbs overall, but inside each year it dips and rises again, so there is a trend and a season on top of it.
- Check the trend line against the data. The quarter numbers 1 to 8 have mean 4.5, and the sales total 384, so the mean sales are 384 ÷ 8 = 48 thousand dollars. The trend line gives 4 × 4.5 + 30 = 48, so it passes through the mean point.
- Measure the first quarters. In quarter 1 the trend is 4 × 1 + 30 = 34 and the sales were 28, which is 6 below it. In quarter 5 the trend is 50 and the sales were 44, again 6 below.
- (a) The gradient 4 means the trend rises by 4 thousand dollars every quarter. In quarter 3 the trend is 42 against sales of 50, and in quarter 7 it is 58 against sales of 66, so a third quarter runs 8 thousand dollars above the trend while a first quarter runs 6 thousand below it.
- (b) Quarter 9 is the first quarter of the next year. The trend there is 4 × 9 + 30 = 66, and a first quarter is 6 thousand dollars below the trend, so the estimate is 66 − 6 = 60 thousand dollars, that is $60 000. Check: the last first quarter took 44 thousand dollars, and four quarters of growth at 4 thousand each adds 16, which gives 60 again.
answer(a) the trend rises by 4 thousand dollars a quarter, each first quarter lies 6 thousand dollars below the trend and each third quarter lies 8 thousand dollars above it; (b) about $60 000
techniqueTime Series · Line of Best Fit
examsSAT · GCSE Higher
Common pitfalls
- Taking the last figure, 58 thousand dollars, as the estimate for the next quarter. That figure is a fourth quarter and quarter 9 is a first quarter, and a quarter of growth has to be added as well, so neither part of it is right.
- Using the trend value 66 on its own. The trend is the average of the four seasons, so it over-estimates a first quarter and under-estimates a third; the seasonal distance has to be put back.
Braking Distance Against Speed, with a Straight Line and a Curve Both Offered
Six braking tests were made on one car. The pairs of (speed in km/h, braking distance in meters) were (20, 5), (30, 9), (40, 15), (50, 26), (60, 35) and (70, 48). Two models are offered: model A is y = 0.9x − 17 and model B is y = x2100. (a) Find the sum of the squared residuals for each model. (b) Say which model should be used, and test it against what the models give at rest.
A residual is the measured value minus the value a model gives at the same speed, so it is the vertical gap between a point and the model. Squaring the residuals removes their signs and weighs a large miss more heavily than a small one, and the model with the smaller total is the better fit. The pattern of the signs is worth reading too, and so is what each model says at a value the situation already settles.
- Work out model A at each speed: 0.9 × 20 − 17 = 1, then 10, 19, 28, 37 and 46 meters.
- Take the residuals for model A: 5 − 1 = 4, 9 − 10 = −1, 15 − 19 = −4, 26 − 28 = −2, 35 − 37 = −2 and 48 − 46 = 2. Squaring them gives 16 + 1 + 16 + 4 + 4 + 4 = 45.
- Work out model B at each speed: 202100 = 4, then 9, 16, 25, 36 and 49 meters. Its residuals are 1, 0, −1, 1, −1 and −1, and the squares total 1 + 0 + 1 + 1 + 1 + 1 = 5.
- (a) The sum of the squared residuals is 45 for model A and 5 for model B. Read the signs as well: model A is below the points at both ends and above them in the middle, and residuals that change sign in that pattern are the mark of a curve being fitted by a straight line.
- (b) Use model B. At rest the car needs no braking distance at all: model B gives 0100 = 0 meters, while model A gives 0.9 × 0 − 17 = −17 meters, and a distance cannot be negative. Check: model B says that doubling the speed from 30 to 60 km/h multiplies the braking distance by 4, from 9 to 36 meters, which is what braking distance is known to do.
answer(a) the sum of the squared residuals is 45 for model A and 5 for model B; (b) model B, y = x2100, because model A gives a braking distance of −17 meters at rest
techniqueChoosing Between Two Models · Scatter Plots
examsSAT · GCSE Higher · H2
Common pitfalls
- Adding the residuals without squaring them. The positive and the negative residuals then cancel, and a line drawn through the middle of the points totals nearly zero however badly it follows their shape.
- Choosing model A because a straight line is the simpler rule. Simplicity decides only between models that fit equally well, and this one misses by 4 meters at the slowest test and gives a negative distance at rest.
Ice Cream Sales Against Drownings, with the Temperature of Each Month
Over eight months a country recorded the ice creams sold, in thousands, and the number of drownings. The pairs of (ice creams in thousands, drownings) were (1, 4), (2, 4), (3, 8), (4, 8), (6, 12), (7, 16), (8, 16) and (9, 20). The mean temperature of those months, in the same order, was 8, 10, 13, 15, 20, 24, 26 and 28 Celsius. The line of best fit is y = 2x + 1. (a) Describe the correlation, and estimate the drownings in a month when 5 thousand ice creams are sold. (b) A councillor proposes closing the ice cream vans in order to reduce drownings. Say whether the data supports that.
A correlation says that two quantities move together. It does not say that either one moves the other, because a third quantity that moves both produces the same picture. The test is to look for such a quantity in the situation, and here the third column of the record supplies it.
- Plot the drownings against the ice creams. The points rise steadily from left to right and lie close to a straight line, so the correlation is strong and positive.
- Check the line against the mean point. The ice creams total 40, so the mean is 40 ÷ 8 = 5 thousand, and the drownings total 88, so the mean is 88 ÷ 8 = 11. The line gives 2 × 5 + 1 = 11, so it passes through (5, 11).
- (a) The correlation is strong and positive, and at 5 thousand ice creams the line estimates 2 × 5 + 1 = 11 drownings.
- Now read the temperatures. They rise month by month alongside both columns: the coldest month, at 8 Celsius, has the fewest ice creams and the fewest drownings, and the hottest, at 28 Celsius, has the most of each.
- (b) No. The correlation is real, but the temperature accounts for both sides of it: in hot weather more ice creams are bought and more people swim, so more people drown. Closing the vans would change the ice cream figure and leave the swimming, and with it the drownings, exactly as they were. Check: for the proposal to work the ice creams would have to cause the drownings, and nothing in the record connects them except the month they fall in.
answer(a) strong positive correlation, and the line estimates 2 × 5 + 1 = 11 drownings; (b) no, because the temperature lies behind both columns, so closing the vans would not reduce drownings
techniqueCorrelation Is Not Causation · Correlation
examsSAT · GCSE Higher · H2
Common pitfalls
- Taking a strong correlation as proof that one quantity causes the other. A correlation says only that the two move together, and a third quantity moving both gives exactly the same diagram.
- Dismissing the correlation as a coincidence. It is not a coincidence: the two really do rise together, month after month. What is wrong is the explanation, not the pattern.
An Aquarium Gift Shop's Takings, and a Day the Shop Closed Early
An aquarium recorded, on eight days, the visitors in hundreds and the gift shop takings in hundreds of dollars. The pairs of (visitors, takings) were (1, 12), (2, 17), (3, 18), (4, 25), (5, 23), (6, 28), (7, 31) and (8, 10). On the last of those days a power cut closed the shop at midday. Drawn through all eight days the line of best fit has gradient 1; drawn through the other seven it has gradient 3. (a) Find the equation of each line. (b) Estimate the takings on an ordinary day with 600 visitors from each line, and say which estimate to use.
Every line of best fit passes through the mean point of the data it is drawn through, so a gradient and a mean point together give the equation. Leaving a point out changes both the mean point and the gradient, and a point at the edge of the diagram changes them most. Whether to leave it out is decided by the situation, not by the numbers.
- Plot the eight days. Seven of them climb steadily, while the power cut day, at 8 hundred visitors and only 10 hundred dollars, sits far below the rest: it is the outlier.
- Find the mean point of all eight days. The visitors total 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36, so the mean is 36 ÷ 8 = 4.5 hundred, and the takings total 164, so the mean is 164 ÷ 8 = 20.5 hundred dollars.
- The line through all eight days has gradient 1 and passes through (4.5, 20.5), so 20.5 = 1 × 4.5 + c and c = 16.
- Leave the power cut day out. The other seven have visitors totaling 28, a mean of 4, and takings totaling 154, a mean of 22. That line has gradient 3, so 22 = 3 × 4 + c and c = 10.
- (a) Through all eight days the line is y = x + 16, and through the seven ordinary days it is y = 3x + 10.
- (b) At 600 visitors, x = 6. The first line gives 6 + 16 = 22 hundred dollars, that is $2200, and the second gives 3 × 6 + 10 = 28 hundred dollars, that is $2800. Use $2800, because the question asks about an ordinary day and the power cut day is not one, so it should not be shaping the line. Check: the day that really had 600 visitors took 28 hundred dollars, which the second line matches exactly.
answer(a) the mean point of all eight days is (4.5, 20.5), giving y = x + 16, and without the power cut day it is (4, 22), giving y = 3x + 10; (b) $2200 from the first line and $2800 from the second, and the second is the one to use
techniqueLine of Best Fit · Scatter Plots
examsSAT · GCSE Higher
Common pitfalls
- Deleting the outlier simply because it is far from the others. A point is left out only when something outside the numbers explains it, and here there is such a reason: the shop was shut for half the day. An outlier with no explanation is real data and stays in.
- Assuming that one day in eight can hardly matter. It pulls the gradient from 3 down to 1 and moves the estimate by $600, because it lies at the far right of the diagram, where a point has the most leverage on a line.
Height Against Arm Span, Used to Recover a Measurement That Was Never Taken
A class measured the height and the arm span, in centimeters, of ten students. The pairs of (height, arm span) were (150, 150), (152, 148), (154, 153), (158, 155), (160, 158), (164, 163), (166, 162), (168, 167), (172, 168) and (176, 176). The line of best fit has gradient 1. (a) Find the mean point and the equation of the line. (b) One more student is 174 cm tall and her arm span was not measured. Estimate it.
Two facts fix a straight line: a point it passes through and its gradient. The line of best fit always passes through the mean point, so working out the two means gives that point, and the gradient the question supplies completes the equation. The missing measurement is then read off the line.
- Find the mean height. The heights total 150 + 152 + 154 + 158 + 160 + 164 + 166 + 168 + 172 + 176 = 1620, so the mean is 1620 ÷ 10 = 162 cm.
- Find the mean arm span. The spans total 150 + 148 + 153 + 155 + 158 + 163 + 162 + 167 + 168 + 176 = 1600, so the mean is 1600 ÷ 10 = 160 cm, and the mean point is (162, 160).
- The line passes through the mean point with gradient 1, so 160 = 1 × 162 + c, which gives c = −2.
- (a) The mean point is (162, 160) and the line of best fit is s = h − 2, where h is the height and s is the arm span, both in centimeters. A gradient of 1 says that two students whose heights differ by 1 cm differ by about 1 cm in arm span as well.
- (b) Substitute h = 174: s = 174 − 2 = 172 cm. Check: 174 cm lies between the shortest student at 150 cm and the tallest at 176 cm, so the line is read inside the data, and 172 cm falls between the arm spans of the students who are 172 cm and 176 cm tall.
answer(a) the mean point is (162, 160) and the line of best fit is s = h − 2; (b) about 172 cm
techniqueLine of Best Fit · Scatter Plots
examsSAT · GCSE Higher
Common pitfalls
- Assuming that the arm span equals the height and answering 174 cm. The gradient is 1 but the intercept is not zero, and the measurements bear that out: eight of the ten students have an arm span shorter than their height and the other two have the two measurements equal.
- Estimating from the nearest student instead, and giving the 168 cm span of the student who is 172 cm tall. One student is one measurement; the line uses all ten and is far less affected by whichever happens to be nearest.
Oven Temperature Against Baking Time, with the Line of Best Fit Read Backwards
A baker baked one loaf at eight oven temperatures and recorded the time it needed. The pairs of (temperature in Celsius, time in minutes) were (180, 50), (190, 44), (200, 43), (210, 43), (220, 39), (230, 40), (240, 34) and (250, 35). The line of best fit has gradient −0.2. (a) Find the mean point and the equation of the line, and say what the gradient means. (b) The baker wants the loaf to take 40 minutes. Find the oven temperature the line suggests.
A line of best fit can be read either way round. Reading forward means putting in a temperature and getting a time; reading backwards means putting in the time that is wanted and solving the equation for the temperature. The answer is then checked against the range of the readings, because a temperature outside them would be a guess.
- Find the mean temperature. The temperatures total 180 + 190 + 200 + 210 + 220 + 230 + 240 + 250 = 1720, so the mean is 1720 ÷ 8 = 215 Celsius.
- Find the mean time. The times total 50 + 44 + 43 + 43 + 39 + 40 + 34 + 35 = 328, so the mean is 328 ÷ 8 = 41 minutes, and the mean point is (215, 41).
- The line passes through the mean point with gradient −0.2, so 41 = −0.2 × 215 + c, that is 41 = −43 + c, which gives c = 84.
- (a) The line of best fit is m = 84 − 0.2x, where x is the temperature in Celsius and m is the time in minutes. The gradient −0.2 is −0.2 minutes for each degree, so every 10 Celsius hotter takes about 2 minutes off the baking time.
- (b) Put m = 40 into the equation: 40 = 84 − 0.2x, so 0.2x = 84 − 40 = 44 and x = 44 ÷ 0.2 = 220 Celsius. Check: 84 − 0.2 × 220 = 84 − 44 = 40, and 220 Celsius lies between the 180 and the 250 the baker tested, so the answer is inside the readings.
answer(a) the mean point is (215, 41) and the line of best fit is m = 84 − 0.2x, with gradient −0.2, so every 10 Celsius hotter takes about 2 minutes off; (b) 220 Celsius
techniqueLine of Best Fit · Interpolation and Extrapolation
examsSAT · GCSE Higher
Common pitfalls
- Reading across from 40 minutes to the nearest point and answering 230 Celsius. That loaf took 40 minutes on one trial; the line averages all eight trials and puts the temperature at 220 Celsius.
- Solving 40 = 84 − 0.2x by taking 84 from both sides and then dividing by 0.2 without minding the sign, which gives −220. Move the 0.2x to the left and the 40 to the right instead, so that 0.2x = 44.