1. Scatter Graphs Teach GCSE Maths x x x x x x x x x x Weight and Length of Broad Beans Length (cm) 3 1·5 Weight (g) 0·5 1

2. "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages" © Christine Crisp Scatter Graphs

3. The data set shown in the table gives the weight and length of a sample of 10 broad beans. Bean Number Weight (g) Length (cm) 10·71·7 21·22·2 30·92·0 41·42·3 51·22·4 61·12·2 71·02·0 80·91·9 91·02·1 100·81·6 We can see if there is a link between the variables by plotting a scatter graph. Weight and length both vary: they are called variables. Source: Statistics for Biology by O N Bishop published by Pearson Education

4. 1·60·8 2·11·0 1·90·9 2·01·0 2·21·1 2·41·2 2·31·4 2·00·9 2·21·2 1·70·7 Length (cm) Wt. (g) x x x x x x x x x x Weight and Length of Broad Beans Weight (g) Length (cm) Tell your partner what you notice about the lengths as the weights increase. Ans:Generally the heavier beans are longer. 0·7 1·7 1·2 2·2

5. 1·60·8 2·11·0 1·90·9 2·01·0 2·21·1 2·41·2 2·31·4 2·00·9 2·21·2 1·70·7 Length (cm) Wt. (g) We say that, for the beans, weight and length are positively correlated. “Correlated” means “there is a relationship”. “Positive” means “as one variable increases, the other also increases”. x x x x x x x x x x Weight and Length of Broad Beans Length (cm) Weight (g)

6. x x x x x x x x x x Weight and Length of Broad Beans Length (cm) 1·60·8 2·11·0 1·90·9 2·01·0 2·21·1 2·41·2 2·31·4 2·00·9 2·21·2 1·70·7 Length (cm) Wt. (g) Joining the points has no meaning but we can draw a line, the line of best fit, through the middle of the points. Tip: We draw the line “by eye” with about the same number of points above it as below. Weight (g)

7. x x x x x x x x x x Weight and Length of Broad Beans Length (cm) 1·60·8 2·11·0 1·90·9 2·01·0 2·21·1 2·41·2 2·31·4 2·00·9 2·21·2 1·70·7 Length (cm) Wt. (g) The line can be used to estimate the length of a bean if we are given its weight. e.g.Estimate the length of a bean weighing 1·3 g. Ans: 2·35 cm. 1·3 Weight (g)

8. x x x x x x x x x x Weight and Length of Broad Beans Length (cm) 1·60·8 2·11·0 1·90·9 2·01·0 2·21·1 2·41·2 2·31·4 2·00·9 2·21·2 1·70·7 Length (cm) Wt. (g) We must always use the graph even if a bean of the given weight is in the table. e.g.Estimate the length of a bean weighing 0·8 g. Ans: 1·8 cm. 0·8 Weight (g) This value must not be used as the bean may not be typical.

9. Can you and your partner see why it doesn’t make sense to extend the line to the y -axis ? x x x x x x x x x x Weight and Length of Broad Beans Length (cm) 1·60·8 2·11·0 1·90·9 2·01·0 2·21·1 2·41·2 2·31·4 2·00·9 2·21·2 1·70·7 Length (cm) Wt. (g) Ans:We would estimate that a bean weighing nothing is 0·9 cm long ! The line must not be used far beyond the data points as the relationship between the variables may not hold. Weight (g)

10. 1·60·8 2·11·0 1·90·9 2·01·0 2·21·1 2·41·2 2·31·4 2·00·9 2·21·2 1·70·7 Length (cm) Wt. (g) In this example, the points all lie close to the line of best fit. x x x x x x x x x x Weight and Length of Broad Beans Length (cm) We say the correlation is strong. Correlation can be perfect, strong or weak, or there can be no correlation. Decide with your partner what a set of points with perfect correlation would look like. Ans: All the points would lie on a straight line. Weight (g)

11. 1·60·8 2·11·0 1·90·9 2·01·0 2·21·1 2·41·2 2·31·4 2·00·9 2·21·2 1·70·7 Length (cm) Wt. (g) x x x x x x x x x x Weight and Length of Broad Beans Length (cm) With weak correlation, the points are more scattered than here. With no correlation they are all over the graph. Weight (g)

12. If one variable decreases as the other increases, we say the correlation is negative. The scatter graph shows how, as the percentage of the population with access to clean water in Peru increases, the proportion of deaths of young children decreases. Infant Mortality (per 1000 births) Access to Clean Water (%) Infant Mortality and Access to Clean Water in Regions of Peru Source: PAHO

13. Infant Mortality (per 1000 births) Source: PAHO Infant Mortality and Access to Clean Water in Regions of Peru Access to Clean Water (%) The line of best fit slopes down to the right. The original data actually had one more point. We would say again that the correlation is strong, even though it is not as strong as before.

14. Infant Mortality and Access to Clean Water in Regions of Peru Infant Mortality (per 1000 births) Source: PAHO A point lying well off the line of best fit is called an outlier. It may have arisen because of an error in collecting or entering the data so it is sometimes missed out. Access to Clean Water (%) Extra point

15. The points on the following scatter graph do not lie on, or near, a straight line. However, the 2 variables are related as they lie close to a smooth curve. We say the relationship is non-linear. Zero correlation means there is no linear relationship but there may be a non-linear one.

16. SUMMARY  A scatter graph plots values of 2 variables to show any relationship between them.  The relationship is called correlation. Positive correlation means as one variable increases, the other also increases. Negative correlation means as one variable increases, the other decreases.  Correlation can be perfect, strong or weak.  Zero correlation means there is no linear relationship between the variables.  A line of best fit through the centre of the points can be used to estimate the value of one variable from a value of the other. The line should only be used within, or close to, the range of the points.  Outliers lie well away from the other points.

17. Exercise 1.Pick a word from each list ( or choose None ) to describe the correlation between the variables in each of the following: (a) (b) (c) List B: Positive, Negative List A:Perfect, Strong, Weak, None

18. Exercise (b) Perfect, Negative Answers: (a) Weak, Positive (c) None

19. Exercise 2.The table shows the number of accidents to children as a percentage of those to adults, y, in 9 areas of London together with the percentage of open space in those areas, x. ABCDEFGHI Open Spaces (%), x 51·31·474·55·26·314·614·8 Children’s Accidents (%), y 46·342·94038·23733·630·823·817·1 (a)Plot the data on a scatter diagram and draw, by eye, the line of best fit. ( Suitable scales for squared paper are given on the next slide. A version suitable for photocopying is available at the end. ) (b)Estimate the percentage of accidents to children if the open space is 10%.

20. Exercise Open Spaces (%) Children’s Accidents (%) Childrens’ Accidents and Open Spaces

21. Exercise Open Spaces (%) Children’s Accidents (%) Childrens’ Accidents and Open Spaces Solution: With 10 % open spaces, the percentage of accidents that happen to children is estimated as 28 %. (a) (b) We won’t all have drawn the line in exactly the same place. You are not wrong if your line cuts the y -axis a bit higher up (e.g. at y = 46 ). ( Your graph may give a slightly different answer. )

23. Children’s Accidents (%) Scatter Graph Showing Accidents to Children in 9 Areas of London ( as a percentage of all accidents ) and Percentage of Open Spaces Open Spaces (%)