Essential Maths Skills. for AS/A-level. Geography Answers. Helen Harris

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1 Essential Maths Skills for AS/A-level Geography Answers Helen Harris

2 1 Understanding data Ratios and fractions Practice question (p.10) 1 a number of semi-detached properties : total number of properties = 5 : 1 b There are four 4-bedroom properties, so number of 4-bedroom properties : total number of properties = 4 : 1 = 1 : 3 (after dividing both numbers by 4) Expressed as a fraction this is 1 3. c number of semi-detached properties : number of detached properties = 5 : 4 This ratio cannot be simplified further. d There are two terraced houses out of a total of 1 properties, so the fraction of properties that are terraced is 1 = 1 (after dividing both numbers by ). 6 Percentages Practice questions (p.11) 1 a To change a percentage to a decimal, divide by 100. So 16% = = 0.16 b 4% = = 0.04 a To change a decimal to a percentage, multiply by 100. So 0.14 = ( )% = 14% b 0.0 = ( )% = % Guided question (p.1) 1 For collection point : Step 1: the number of cars/motorcycles is 13. The total number of vehicles is 154. Step : = 0.86 Step 3: ( )% = 86% 1

3 For collection point 3: Step 1: the number of cars/motorcycles is 185. The total number of vehicles is 15. Step : = 0.86 Step 3: ( )% = 86% For collection point 4: Step 1: the number of cars/motorcycles is 16. The total number of vehicles is 178. Step : = 0.91 Step 3: ( )% = 91% Comparing the results, collection point 4 had the highest percentage of cars and motorcycles. Guided question (p.13) 1 Step 1: convert the percentage to a decimal, i.e. divide by = 0.56 Step : multiply this decimal by the initial amount of renewable energy used in = 555 Between 1993 and 011, the global consumption of renewable energy increased by 555 MTOE. Guided question (p.16) 1 Step 1: the number of tourist arrivals to Asia and the Pacific in 010 is 06 (million). Step : in 014 the value for Asia and the Pacific is 63, so calculate = 57 Step 3: = 0.8 Step 4: = 8 So between 010 and 014 there has been a 8% increase in the number of tourists to Asia and the Pacific. Practice questions (p.16) Total traffic count obtained from all four data collection points is = 695 The total amount of traffic recorded at collection point 3 was = = 31% So data collection point 3 accounts for 31% of the total traffic count.

4 3 a i Step 1: the number of tourist arrivals to the Americas in 010 is 156 (million). Step : in 014 the value for the Americas is 18, so calculate = 6 Step 3: = 0.17 Step 4: = 17 So between 010 and 014 there has been a 17% increase in the number of tourists to the Americas. ii Step 1: the number of tourist arrivals to Africa in 010 is 8 (million). Step : in 014 the value for Africa is 56, so calculate 56 8 = 8 Step 3: 8 8 = 1 Step 4: = 100 So between 010 and 014 there has been a 100% increase in the number of tourists to Africa. b Africa had the highest percentage increase in tourist arrivals between 010 and a Brazil: = = = 155 There has been a 155% increase in CO emissions in Brazil between 1981 and 011. b China: = = = 51 There has been a 51% increase in CO emissions in China between 1981 and 011. c France: = 66 ( 66) 455 = = 15 There has been a 15% decrease in CO emissions in France between 1981 and 011. d India: = = = 453 There has been a 453% increase in CO emissions in India between 1981 and

5 Densities Practice questions (p.18) 1 a Harrogate: = 10 people per km Cornwall: = 149 people per km South Tyneside: = 314 people per km b Using the Eurostat threshold of 150 people per km, Harrogate and Cornwall are rural districts, while South Tyneside is an urban district. Camden: = 46 dwellings per ha Brent: = 6 dwellings per ha Bromley: = 9 dwellings per ha 3 Australia: = 50 people per km agricultural land Kenya: = 773 people per km agricultural land India: = 84 people per km agricultural land 4

6 Measures of central tendency Mean, median and mode Guided question (p.) 1 a For Oban: Step 1: = 1157 mm Step : = The mean monthly rainfall for Oban is 96 mm. For Norwich: Step 1: = 596 mm Step : = The mean monthly rainfall for Norwich is 50 mm. b For Oban: Step 1: 55, 68, 75, 76, 80, 89, 98, 98, 110, 18, 130, 150 Step : the 6th value is 89; the 7th value is = = The median monthly rainfall for Oban is 94 mm. For Norwich: Step 1: 36, 39, 39, 4, 44, 45, 51, 5, 59, 61, 63, 65 Step : the 6th value is 45; the 7th value is = = 48 The median monthly rainfall for Norwich is 48 mm. c For Oban: Step 1: 55, 68, 75, 76, 80, 89, 98, 98, 110, 18, 130, 150 Step : the value 98 occurs twice; all the other values occur only once. The modal monthly rainfall for Oban is 98 mm. For Norwich: Step 1: 36, 39, 39, 4, 44, 45, 51, 5, 59, 61, 63, 65 Step : the value 39 occurs twice; all the other values occur only once. The modal monthly rainfall for Norwich is 39 mm. 5

7 d Difference in mean rainfall of Oban and Norwich = = 46 mm Difference in median rainfall of Oban and Norwich = = 46 mm Difference in modal rainfall of Oban and Norwich = = 59 mm Practice questions (p.4) a i Step 1: add the values of the exporters = 9370 Step : divide by the number of countries = 937 So the mean value of exports is US $937 billion. ii Step 1: add the values of the importers = 990 Step : divide by the number of countries = So the mean value of imports is US $990 billion. b Note that for both exports and imports, the values have already been ordered from highest to lowest. Because there are 10 countries, the median is halfway between the 5th and 6th values in each ranking. i The 5th and 6th values are 67 and = = 66 So the median value of exports is US $66 billion. ii The 5th and 6th values are 681 and = = 668 So the median value of imports is US $668 billion. c The median values for imports and exports do not accurately reflect the range of data in the table, because they use only information about the middle part of the data sets and do not take into account the higher and lower values. For instance, for both importers and exporters, there is a wide gap between the values of the top three countries and the values in the middle. 3 The modal category for site 1 is <10 mm The modal category for site is mm The modal category for site 3 is >40 mm 6

8 Frequency distributions Guided question (p.8) 1 a Number of classes = total number of data values = 160 = , so use 13 classes. b Highest data value is 60; lowest data value is 1. Range of data values is 60 1 = 59 Class width = range number of classes = = 4.5, rounded to 5. c The completed frequency distribution table is given in Table A.1. Table A.1 Frequency table for the marram grass data Class (% cover) Frequency d The vertical axis scale must cover (at least) the range of frequencies from 0 to 36. Frequency The histogram is shown in Figure A Figure A.1 Histogram of marram grass cover at Hawes sand dunes e A frequency distribution curve is shown in Figure A.1. Marram grass (% cover) The frequency distribution curve is positively skewed, meaning that the data values fall mainly below the mean. 7

9 Practice question (p.9) a Number of classes = total number of data values = 0 = , so use 4 classes. b Highest data value is 40; lowest data value is. Range of data values is 40 = 38 Class interval = range number of classes = 38 4 = 9.5, rounded to 10. c The completed frequency distribution table is given in Table A.. Table A. Frequency table for environmental quality survey data Class (score out of 40) Frequency d The vertical axis scale must span (at least) the range of frequencies from 3 to 8. The histogram is shown in Figure A.. Frequency Mean Median and Mode Score Figure A. e A frequency distribution curve is also shown in Figure A.. The frequency distribution curve is approximately symmetrical; the data can be approximately described by a normal distribution and the mean, median and mode will be similar. f Mean = ( ) 0 = = Median: arranged in order the data values are, 5, 5, 1, 18, 19, 0, 0, 0, 1, 1, 1, 1,, 3, 3, 4, 3, 34, 40 The median is halfway between the 10th and 11th values, which are both 1, so median = 1 Mode is 1, as this value occurs 4 times, more than any other value. 8

10 3 Measures of dispersion Range, interquartile range and standard deviation Guided question (p.34) 1 a The highest value in the data set is 9 (station 10) and the lowest value is 4.8 (station 6). Range = = 4. b Step 1: order the data values from high to low (Table A.3). Table A.3 Soil moisture readings ordered from high to low Soil moisture reading Step : upper quartile position = (11 + 1) 4 = 3 Upper quartile = 3rd figure from the top of the ranking = 8 Step 3: lower quartile position = (11 + 1) 4 3 = 9 Lower quartile = 9th figure from the top of the ranking = 6 Step 4: interquartile range = 8 6 = Step 5: quartile deviation = = 1 This is a small number, indicating a low degree of dispersion around the median value. c Step 1: x = ( ) 11 = = Step : see Table A.4. 9

11 Table A.4 Sample Soil moisture Deviation from the mean, ( x x) Squared deviation from the mean, ( x x) Sum of squared deviations: ( x x) = Step 3: Σ( x x) = Step 4: = 1.3 Step 5: σ = 1.3 = 1.1 This is a low value, indicating that the data is not widely dispersed from the mean. Practice question (p.36) a For the ACs: Highest value is 73 (Finland), lowest value is 13 (UK). Range = = 60 b i For the LIDCs: Highest value is 66 (Zambia), lowest value is 8 (Sudan). Range = 66 8 = 58 The ranges of the two data sets are very similar. For the ACs: Step 1: the data values ordered from high to low are 73, 69, 47, 39, 38, 37, 35, 33, 3, 31, 13 Step : upper quartile position = (11 + 1) 4 = 3 Upper quartile = 3rd figure from the top of the ranking = 47 Step 3: lower quartile position = (11 + 1) 4 3 = 9 Lower quartile = 9th figure from the top of the ranking = 3 Step 4: interquartile range = 47 3 = 15 For the LIDCs: Step 1: the data values ordered from high to low are 66, 53, 47, 4, 41, 34, 5, 1, 11, 10, 8 Step : upper quartile = 3rd figure from the top of the ranking = 47 Step 3: lower quartile = 9th figure from the top of the ranking = 11 Step 4: interquartile range = = 36 10

12 The LIDCs have a much larger interquartile range than the ACs, which indicates that the LIDC data is more widely dispersed about the median value. ii For the ACs: Quartile deviation = 15 = For the LIDCs: Quartile deviation = 36 = 18 The LIDCs have a much larger quartile deviation than the ACs, which indicates that the LIDC data is more widely dispersed about the median value. c To calculate the standard deviation for the LIDCs: Step 1: find the mean of the data set, x. ( ) 11 = = Step : subtract the mean from each value in the data set and then square each of these differences (see Table A.5). Table A.5 Country Forest area (% total land area) Deviation from the mean, ( x x) Squared deviation from the mean, ( x x) Bangladesh = Burundi = 484 Cameroon = 9 81 Ethiopia = Ghana = Malawi = 4 Nepal = 7 49 Sudan = Tanzania = Vietnam = 15 5 Zambia = Sum of squared deviations: ( x x) = 3957 Step 3: ( x x) = 3957 Step 4: = Step 5: σ = 360 = d The standard deviation of 19 indicates that there is some variation around the mean value of 3 for the LIDCs. 11

13 4 Measures of concentration Nearest neighbour index Guided question (p.41) 1 a Step 1: A = 3 km 3 km = 9 km Step : see Figure 4., in which the outdoor shops are shown in red and numbered 1 to 5. Step 3: the distances are plotted in Figure A.3 and entered in Table A N km Figure A.3 Table A.6 Nearest neighbour distances between the outdoor shops Outdoor shop number Nearest neighbour number Distance, d (km) Sum of the distances: d = 1.5 Step 4: d = d n = = 0.5 Step 5: Rn = d n A = 0.5 = Step 6: the Rn value of 0.37 is below 1.0 and indicates that the outdoor shops have a strong tendency towards clustering. 1

14 b The outdoor shops have a smaller Rn value than the cafes, which indicates that the outdoor shops are more clustered than the cafes. c This might be because greater clustering makes comparison shopping easier and being concentrated in a smaller area will increase footfall for the shops. Practice question (p.4) a For Snowdonia: Step 1: the area mapped consists of 14 0 squares on the grid and the scale shows that the length of squares represents 1 km, so the study area is A = 7 km 10 km = 70 km. Step : see Figure 4.4, in which the corries are marked with crosses and numbered 1 to 9. Step 3: the distances are entered in Table A.7. Table A.7 Nearest neighbour distances between the corries in Snowdonia Corrie number Nearest neighbour number Distance, d (km) Sum of the distances: d = 11. Step 4: mean nearest neighbour distance is d = d n = = 1. Step 5: Rn = d n A = 1. = Step 6: the NNI for corries in Snowdonia is 0.86, which indicates a slight tendency towards clustering. b For Langdale Fell: Step 1: A = 7 km 10 km = 70 km Step : see Figure 4.5, in which the corries are marked with crosses and numbered 1 to 9. Step 3: the distances are entered in Table A.8. 13

15 Table A.8 Nearest neighbour distances between the corries in Langdale Fell Corrie number Nearest neighbour number Distance, d (km) Sum of the distances: d = 13.5 Step 4: mean nearest neighbour distance is d = d n = = 1.5 Step 5: Rn = d = 1.5 = 1.08 n A 9 70 Step 6: the NNI for corries in Langdale Fell is 1.08, which indicates a random distribution. c The corries in Snowdonia are more clustered than in Langdale Fell. d To seek an explanation, the next step would be to investigate physical factors such as the geology, aspect and height of land in the two locations. Location quotient Guided question (p.45) 1 Table A.9 Employment in three industrial sectors in seven regions of the UK, 015 Manufacturing Information & communications Health % employment LQ % employment LQ % employment LQ UK 7.8 N/A 4.0 N/A 1.4 N/A North West (NW) North East (NE) = = = 1.3 South West (SW) = = = 1.1 South East (SE) = = = 0.9 Scotland = = = 1. Wales = = = 1.1 Northern Ireland = = =

16 Practice questions (p.46) LQ = = LQ = = LQ = = LQ = = 1.3 Lorenz curves Guided question (p.51) 1 a The results of Steps 4 are given in Table A.10. Table A.10 Percentage populations of regions in England, ordered from highest to lowest Step 3: Region and rank (in descending order of % population) Step : % population of England Step 4: Cumulative % of population Step : % land area of England Step 4: Cumulative % of land area South East (SE) London North West (NW) East (E) West Midlands (WM) South West (SW) Yorkshire & Humberside (Y&H) East Midlands (EM) North East (NE) Note that the percentage population for East Midlands, = 8.49, has been rounded up to 9 rather than down to 8, so that the percentages of all the regions add up to exactly 100. Similarly, the percentage land area for the North East, = 6.5, has been rounded down instead of up. b With the ranked variable (cumulative percentage of population) plotted on the x-axis and the cumulative percentage of land area plotted on the y-axis, the curve in Figure A.4 is obtained. 15

17 Cumulative % of land area Even distribution line SW EM Y&H NE 50 WM 40 E SE London NW 0 Figure A Cumulative % of population c The even distribution line is the diagonal straight line in Figure A.4. d The Lorenz curve indicates a low degree of concentration, as it is fairly close to the even distribution line. London and the North West account for most of the concentration. The largest regions by land area do not have the highest percentage of population. Practice question (p.53) a For each country, the percentage wealth figures are already ordered (from low to high), so we only need to complete Step 4: calculating the cumulative percentages. These values are shown in Table A.11. Table A.11 Wealth distribution in the UK, Canada and Belgium Cumulative % of population (Poorest 10%) 10 % wealth held by each 10% of population from poor to rich UK Belgium Canada Cumulative % of wealth % wealth held by each 10% of population from poor to rich Cumulative % of wealth % wealth held by each 10% of population from poor to rich Cumulative % of wealth (Richest 10%)

18 b For each country, plotting the cumulative percentage of population on the x-axis and the cumulative percentage of wealth on the y-axis gives the three curves in Figure A.5. Cumulative % of wealth Even distribution line Belgium UK Canada Cumulative % of population Figure A.5 c The even distribution line is the diagonal straight line in Figure A.5. Belgium has the lowest degree of wealth concentration, as its curve is closest to the even distribution line. 17

19 5 Measures of correlation Scatter graphs and lines of best fit Guided question (p.59) 1 a,b The independent variable to plot on the x-axis is GNI per capita. The dependent variable to plot on the y-axis is percentage of territorial waters protected. The scatter graph and the line of best fit are shown in Figure A.6. Protected territorial waters (%) Australia Ghana Kenya Mexico Peru Russia Chile Bangladesh Figure A.6 Japan No relationship Norway GNI per capita (US$ 000) c From the scatter graph it appears that there is no correlation between a country s GNI per capita and the percentage of its territorial waters that are protected. Practice question (p.60) a,b The independent variable to plot on the x-axis is GNI per capita. The dependent variable to plot on the y-axis is percentage of employment in agriculture. The scatter graph and the line of best fit are shown in Figure A.7. 18

20 % employed in agriculture Negative relationship GNI per capita (US$ 000) Figure A.7 c From the scatter graph it appears that there is a negative relationship between GNI per capita and percentage employed in agriculture. As GNI increases, the proportion of the population employed in agriculture tends to decrease. Spearman rank correlation coefficient Guided question (p.63) 1 a Step 1: null hypothesis There is no relationship between time and Arctic sea ice extent. b d (Steps 5) Table A.1 Ranked average Arctic sea ice extent data Year Rank Arctic sea ice extent (million km ) Rank d d d =

21 e Step 6: d R 1 6 s = 3 n n = = = = = f The R S value of 0.9 indicates strong negative correlation. g Degrees of freedom = n 1 = 16 1 = 15. From the table for the Spearman rank test in Appendix 1, the critical value for 15 degrees of freedom and significance level 0.01 is The size of R S (0.9) found in part e is greater than the critical value, so reject the null hypothesis. h There is a strong negative relationship between time and Arctic sea ice extent: as time increases, the Arctic sea ice extent tends to decrease. The probability that this correlation arose by chance is less than 0.01 (1%). Practice question (p.65) a c Taking GNI per capita as the independent variable (plotted on the x-axis) and average deaths per flood event as the dependent variable (plotted on the y-axis), we obtain the scatter graph and line of best fit as in Figure A.8, with anomalies circled. Average deaths per flood event China India Bangladesh Nepal Pakistan Egypt Line of best fit 100 Algeria Japan Austria Mexico Canada Brazil Italy Spain UK USA GNI per capita (US$ 000) Figure A.8 0

22 d Null hypothesis There is no relationship between GNI per capita and average number of deaths per flood event. e Steps 5 of calculating the Spearman rank correlation coefficient are in Table A.13. The values are ranked from highest to lowest. Table A.13 Ranked data on GNI per capita and average number of deaths per flood event Country GNI per capita (Step ) Rank Average deaths per flood event (Step ) Rank (Step 3) d (Step 4) d USA UK Spain Italy China Canada Austria Algeria Bangladesh India Egypt Nepal Japan Pakistan Mexico Brazil (Step 5) d = 178 d R 1 6 s = 3 n n = = = = = The R S value of 0.88 indicates negative correlation. f To test the significance of the result: Degrees of freedom = n 1 = 16 1 = 15. From the table for the Spearman rank test in Appendix 1, the critical value for 15 degrees of freedom and significance level 0.05 is and the critical value for significance level 0.01 is The size of R S (0.88) is greater than both critical values, so we can reject the null hypothesis at the 0.05 and 0.01 significance levels. 1

23 g Concluding statement: There is a negative relationship between GNI per capita and average number of deaths per flood event: as GNI per capita increases, the average number of deaths per flood event tends to decrease. The probability that this correlation arose by chance is less than 0.01 (1%).

24 6 Testing for differences between data sets Student s t test Guided question (p.71) 1 Step 1: null hypothesis Urban areas have the same level of access to improved water sources compared to rural areas. Step : n x = n y = 10 The x, y, x and y values needed for the formula are calculated in Table A.14. Table A.14 Country Urban areas (x values) x Rural areas (y values) y Tanzania Kenya Mozambique Zimbabwe Ethiopia Uganda Niger Morocco Senegal Cameroon Totals: x = 913 x = y = 566 y = x = 91.3 y = 56.6 x y = = 34.7 Step 3: t = = = x nx n x x y y x y ny + 1 n y

25 34.7 = = = = 4.59 = 7.6 Step 4: degrees of freedom = (n x 1) + (n y 1) = = 18 From the table of critical values for Student s t test in Appendix 1, the critical value for 18 degrees of freedom at the 0.01 significance level is.88. The calculated t value of 7.6 is greater than the critical value, so we reject the null hypothesis. There is a very low (lower than 1%) probability that this result was due to chance. Step 5: there is a significant difference in the level of access to improved water sources in urban and rural areas. Practice question (p.73) a (Testing if there is a significant difference between the depth of soil on the northfacing and south-facing slopes) Step 1: null hypothesis Soil on the north-facing slope has the same depth as soil on the south-facing slope. Step : n x = n y = 8 The x, y, x and y values needed for the formula are calculated in Table A.15. Table A.15 Soil sample Depth on northfacing slope (x values) x Depth on southfacing slope (y values) y Totals: x = 88 x = 108 y = 369 y =19 51 x y = = x = 11 y =

26 Step 3: x y t = x y x y n n x y + n 1 n 1 x y = = = = = = 6.8 = 5. Step 4: degrees of freedom = (n x 1) + (n y 1) = = 14 From the table of critical values for Student s t test in Appendix 1, the critical value for 14 degrees of freedom is 1.76 at the 0.05 significance level and.98 at the 0.01 significance level. The calculated t value of 5. is greater than both critical values, so we reject the null hypothesis. There is a very low (lower than 1%) probability that this result was due to chance. Step 5: there is a significant difference in the depth of soil on the north- and southfacing slopes. b (Testing if there is a significant difference between the organic content of soil on the north-facing and south-facing slopes) Step 1: null hypothesis Soil on the north-facing slope has the same percentage of organic matter as soil on the south-facing slope. Step : n x = n y = 8 The x, y, x and y values needed for the formula are calculated in Table A.16. 5

27 Table A.16 Soil sample % organic matter on northfacing slope (x values) x % organic matter on southfacing slope (y values) y Totals: x = 315 x = y = 98 y = 1 50 x y = = 7.15 Step 3: x = y = t = = = = =.5 = 10.8 Step 4: degrees of freedom = (n x 1) + (n y 1) = = 14 The critical value for 14 degrees of freedom is 1.76 at the 0.05 significance level and.98 at the 0.01 significance level. The calculated t value of 10.8 is greater than both critical values, so we reject the null hypothesis. There is a very low probability that this result was due to chance. Step 5: there is a significant difference in the organic content of soil on the northand south-facing slopes. 6

28 Mann Whitney U test Guided question (p.76) 1 Step 1: null hypothesis The air temperature at the urban location is the same as the air temperature at the rural location. Step : n x = n y = 10. The ranking of the x and y data values is done in Table A.17. Table A.17 Ranked air temperatures at an urban location and a rural location Reading Urban location temperature (x values) Rank (r x ) Rural location temperature (y values) Rank (r y ) r x = 6.5 r y = Step 3: n ( n + 1) x x U = n n + r x x y x (10 +1) = = = = 9.5 n ( n + 1) y y U = n n + r y x y y 10 (10 +1) = = = = 7.5 Checking the calculations: U x + U y = = 100 and n x n y = = 100, so the results seem to be correct. Step 4: the critical value is the entry in row 10 and column 10 of the table in Appendix 1, which is 3. 7

29 The smaller value of U x and U y is is less than 3, so the null hypothesis should be rejected. Step 5: there is a significant difference in the air temperature at the urban location and at the rural location. Practice question (p.78) a (Testing for difference between the DO at site 1 and site ) Step 1: null hypothesis The dissolved oxygen content is the same at site 1 and at site. Step : rank the data values. Take the data for site 1 to be the x values and the data for site to be the y values. Then n x = n y = 7. The ranking of the x and y data values (from highest to lowest) is done in Table A.18. Table A.18 Ranked DO values measured from sites 1 and Site 1 DO (x values) Rank (r x ) Site DO (y values) Rank (r y ) r x = 8 r y = 77 Step 3: calculate the U values for both data sets. n ( n + 1) x x U = n n + r x x y x 7 = 7 7+ (7+1) = = = 49 n ( n + 1) y y U = n n + r y x y 7 = 7 7+ (7+1) = = = 0 y 8

30 Checking the calculations: U x + U y = = 49 and n x n y = 7 7 = 49, so the results seem to be correct. Step 4: test the significance of the result using the table of critical values for the Mann Whitney U test in Appendix 1. The critical value is the entry in row 7 and column 7 of the table, which is 8. The smaller value of U x and U y is 0. 0 is less than 8, so the null hypothesis should be rejected. Step 5: make a concluding statement. There is a significant difference between dissolved oxygen content at site 1 and site. b (Testing for difference between the DO at site 1 and site 3) Step 1: null hypothesis The level of dissolved oxygen content is the same at site 1 and at site 3. Step : rank the data values. Take the data for site 1 to be the x values and the data for site 3 to be the y values. Then n x = n y = 7. The ranking of the x and y data values (from highest to lowest) is done in Table A.19. Table A.19 Ranked DO values measured from sites 1 and 3 Site 1 DO (x values) Rank (r x ) Site 3 DO (y values) Rank (r y ) r x = 36.5 r y = 68.5 Step 3: calculate the U values for both data sets. 9

31 n ( n + 1) x x U = n n + r x x y x = (7+1) 36.5 = = = 40.5 n ( n + 1) y y U = n n + r y x y = (7+1) 68.5 = = = 8.5 y Checking the calculations: U x + U y = = 49 and n x n y = 7 7 = 49, so the results seem to be correct. Step 4: test the significance of the result using the table of critical values for the Mann Whitney U test in Appendix 1. The critical value is the entry in row 7 and column 7 of the table, which is 8. The smaller value of U x and U y is is greater than 8, so the null hypothesis is accepted. Step 5: make a concluding statement. There is no significant difference between dissolved oxygen content at site 1 and site 3. c Since there is a significant difference between dissolved oxygen content at site 1 (upstream of factory) and site (1 km downstream), but no significant difference between site 1 and site 3 (5 km downstream), this indicates that 5 km past the factory the dissolved oxygen content had nearly returned to the levels before the factory s influence. Chi-squared test Guided question (p.8) 1 Step 1: null hypothesis There is no significant difference (or change) in the number of fast food outlets with increasing distance from the CBD. Step : E = 8 4 = 7 for each category, so we have Table A.0. 30

32 Table A.0 Observed frequency Expected frequency 0 1 km 3 km 4 5 km >5 km Total Step 3: the calculations of O E, (O E) and O Table A.1 Distance from CBD Observed frequency (O) ( ) E E are shown in Table A.1. Expected frequency ( O E) (E) O E (O E) E 0 1 km km km >5 km Step 4: χ = = Step 5: degrees of freedom = number of categories 1 = 4 1 = 3 From the table for the chi-squared test in Appendix 1, the critical value is: 7.8 at the 0.05 significance level at the 0.01 significance level The χ value is greater than both critical values so the null hypothesis can be rejected with 99% certainty. Step 6: there is a significant difference in the number of fast food outlets with increasing distance from the CBD. Practice question (p.83) First calculate the expected frequencies for all of the categories (i.e. fill in a copy of Table 6.17). This is Step of both parts a and b. Table A. Expected frequencies Town centre Rural idyll Positive Negative Positive Negative Female < = = = = 11.0 Male < = = = = 10.4 Female = = = = 5. Male = = = =.3 a Step 1: null hypothesis There is no significant difference between the age/gender groups in positive perception of the town centre. Step : the expected frequency of positive perception of the town centre in each age/gender group is given by the first column of values in Table A.. Step 3: the calculations of O E, (O E) and (O E) E are shown in Table A.3. 31

33 Table A.3 Age/gender group Observed frequency (O) Expected frequency ( O E) (E) O E (O E) E Female < Male < Female Male Step 4: χ = =. Step 5: degrees of freedom = number of categories 1 = 4 1 = 3 From the table for the chi-squared test in Appendix 1, the critical value is: 7.8 at the 0.05 significance level at the 0.01 significance level The χ value. is lower than both these critical values so the null hypothesis is accepted. Step 6: there is no significant difference between the age/gender groups in positive perception of the town centre. b Step 1: null hypothesis There is no significant difference between the age/gender groups in negative perception of the rural idyll. Step : the expected frequency of negative perception of the rural idyll in each age/gender group is given by the fourth column of values in Table A.. Step 3: the calculations of O E, (O E) and Table A.4 Age/gender group Observed frequency (O) (O E) E are shown in Table A.4. Expected frequency ( O E) (E) O E ( O E) E Female < Male < Female Male Step 4: χ = = 10.3 Step 5: degrees of freedom = number of categories 1 = 4 1 = 3 From the table for the chi-squared test in Appendix 1, the critical value is: 7.8 at the 0.05 significance level at the 0.01 significance level The χ value 10.3 is higher than the 0.05-level critical value but lower than the 0.01-level critical value so the null hypothesis can be rejected at the 0.05 significance level. Step 6: there is a significant difference between the age/gender groups in negative perception of the rural idyll. We can be 95% certain that these results were not due to chance. 3

34 Exam-style questions AS level questions 1 a i The independent variable is the one plotted on the x-axis, beach width. (1 mark) ii Average beach slope angle ( ) (1 mark) Figure A.9 Beach width (m) A line of best fit in shown in Figure A.9. (Any appropriate line that goes through the middle of the set of plotted points, with about the same number of points above as below the line, will be accepted.) There is a negative relationship between beach width and beach slope angle: as the beach width increases, the slope angle tends to decrease. (1 mark) b Wider beaches are generally made of sand, while narrower beaches tend to be composed of pebbles, which can support a steeper angle than sand. (1 mark for a correct reason and marks for explanation) a i Figure A.10 shows Figure E. with the Lorenz curve for the USA added. ( marks a mark will be deducted for an inaccuracy at any point) 33

35 Cumulative % of wealth Even distribution line Belgium USA Figure A Cumulative % of population ii Of the two countries, Belgium has the more equitable wealth distribution, as its Lorenz curve lies closer to the even distribution line. (1 mark) b Answers could include: The rich have more money to invest and can therefore become even richer. People from wealthy families can spend more on gaining education and qualifications and hence access higher-paid jobs. Low-skilled workers are paid less and often become trapped in low-paid jobs. (1 mark for an accurate cause and up to marks for an explanation of how it affects income inequality) 3 a i Possible graphical presentations of the data include: Scatter graph with GNI plotted on the x-axis and number of refugees on the y-axis (e.g. one graph for each year, or the data for both years plotted on the same graph in different colours). World map showing the named countries with proportional symbols to represent the numbers of refugees and GNI (e.g. one map for each year, or a single map with the data of the two years shown in different colours). Raw data figures converted to percentages and presented in pie charts. Bar chart showing bars for the number of refugees and GNI of each country, in each of 005 and 014. (1 mark for acceptable graphical method and marks for accurate sketch) ii Possible quantitative methods for analysing the data include: Measures of central tendency mean, median, mode. Measures of dispersion range, interquartile range, standard deviation. (1 mark for identification of an appropriate quantitative technique plus marks available for explanation of a correct application with regards to the data in Table E.) 34

36 b Push factors affecting the flow of refugees might include: natural disasters government persecution environmental deterioration food insecurity war civil unrest, persecution by a different sector of society 4 Mean (6 marks 1 mark for each relevant push factor and marks for each explanation) Total values = (number of values) = 3.9 (1 mark) Interquartile range Ranked data: 6., 5., 4.8, 4.7, 4.6, 4., 3.3, 3.1,.5,., 1.9 (1 mark) UQ = = 3 = third value, 4.8 (1 mark) LQ = = 9 = ninth value,.5 (1 mark) =.3 (1 mark) A small difference between the mean and interquartile range means a small spread of data. (1 mark) A-level questions 1 a i The completed rows for Botswana and Vietnam in Table E.4 are as follows. Table A.5 Country Internet users per 100 people Rank Exports of goods and services (% GDP) Rank d d Botswana Vietnam Adding up all the d values: d = 6 Insert the values for d and n into the formula and calculate the R S value: d R 1 6 s = 3 n n = = = = (4 marks 1 mark for correct values in each row (Botswana and Vietnam), 1 mark for substituting the correct values in the formula and 1 mark for the correct final answer) ii df = number of paired values in the data set = 14 (1 mark) For 13 degrees of freedom, the critical value at the 0.05 or 95% level is Since > 0.457, the result is significant at the 0.05 level. (1 mark) With 95% certainty, we can reject the null hypothesis and conclude that there is a significant positive correlation/relationship between the levels of internet use and exports of goods and services. (1 mark) 35

37 b Explanations may include: Greater internet use leads to more exchange and sharing of ideas, especially in R&D (research and development). Internet use leads to a more informed and better-educated workforce in the industrial sector. Use of and other online messaging tools enable more efficient communication. The internet allows increased access to information regarding competitors and market activity. ( 3 = 6 marks awarded for two reasons that are well explained, or 3 = 6 marks awarded for three reasons with partial explanations) a i The completed column of Table A.6 (O E) E (O E) E values is as follows. So χ (O E) = = = 10.5 E (1 mark for correct completion of column, 1 mark for correct chi-squared value) ii df = 3 1 = (1 mark) For degrees of freedom, the critical value is 5.99 at the 0.05 level and 9.1 at the 0.01 level. Since 10.5 > 5.99 and 10.5 > 9.1, the result is significant at both the 0.05 and the 0.01 levels, so the null hypothesis can be rejected (with 95% and 99% confidence). (1 mark) We conclude that there is a significant difference between infiltration rate and location on a slope profile. (1 mark) b Explanations may include the following points: Infiltration is the process whereby water soaks into soil. The rate of movement of water into the soil is called the infiltration rate. The infiltration rate is controlled by gravity, capillary action and soil porosity. Soil porosity is affected by texture, structure and organic content of the soil. Soils with coarse texture have high infiltration rates. Heavy rain can break up clumps of soil into smaller particles, thus reducing the movement of water. Saturated soils and those with a high organic content will have lower infiltration rates. (6 marks explanation must centre on different factors affecting the infiltration rate. 1 mark awarded for each correct factor and up to marks for an accurate explanation linking it to the infiltration rate) 36