1. Quantitative and Qualitative Proportional Symbols Map
Geographical Skills
Basic Skills
Investigative Skills
Cross Profiles – Beach
Concept Maps
Risk Assessment
Histograms
Sketches
Quantitative and Qualitative
Scatter Graphs
Cartographic Skills
Data Collection
Line Graphs
Four Figure Map Skills
Graphical Skills
Statistical Skills
Six Figure Map Skills
Kite Diagrams
Mode, Middle and Range
Flow Lines
Percentages & Degrees
Chi-Squared Test
Proportional Symbols Map
Pie Charts
Mann Whitney U Test
Desire Lines
Compound Line Graphs
Student’s T-test
Choropleth Maps
Compound Bar Graphs
Spearman’s Rank
Layer Shading
Radar Graphs

2. Concept Mapping
Concept maps are graphical tools for organizing and representing knowledge. They include concepts, usually enclosed in circles or boxes Relationships between concepts are indicated by a connecting line.
Cross-referencing data collected.
Presentation is clear and neat.
They help visual demonstrate the connections between different pieces information.
They require a lot of time to create and can be made overcomplicated.
Made relevant links with arrows.
Thorough annotations that describes the evidence.
This students has a range of different pieces of information linking to a particular area of a coastline.

3. Sketches
A geographical sketch is a freehand drawing of the main elements that make up a landscape. It supports the interpretation of an actual landscape or photo of a landscape based on key geographical concepts.
Made focused on special/certain features.
Presentation is to a high standard.
They can enhance keys aspects or features of a landscapes. The are relatively easy and take little time to create.
They are not as accurate as a photograph. They can miss key details that could be relevant.
Thorough annotations that help to describe the evidence.
This students has included a number of annotations to explain the features found within their sketch.
Makes good links with other data collected.

4. Four Figure Grid Reference
Four Figure Grid References
Four Figure Grid Reference
Move ‘along the corridor’ and stop once you hit the square that the location is within. 16
Bone’s Farm
Now move ‘up the stairs’ and stop once you hit the square that the location is within again. 44
16,44 11 12 13 14 15 16 17

5. Six Figure Grid Reference
Six Figure Grid References
Six Figure Grid Reference
This time you need to move within the box. Judge which ‘mini’ grid coordinate it would be. 16 1
Bone’s Farm
Now do the same but this time go up within that square. 44 8
16 ,44 1 8 11 12 13 14 15 16 17

6. Maps lack precise interpretation unless statistical data is added.
Flow Lines
Flow lines represent movement along a given route. Variable width along a given route.
Good to show direction and size of movement. In addition they provide a visual impression of movement.
Maps lack precise interpretation unless statistical data is added.
The lines are interested only in one particular source and destination area.
Example: Internal migration within China.

7. Proportional Symbols Map
Proportional symbol maps use symbols of different sizes to represent different quantities. The symbols can be different shapes and the key shows the quantity that each different sized symbol represents.
These maps visually represent the data clearly. They also allow you to easily compare different locations quickly.
These maps may lack the precise data for a given area, making it difficult to compare locations with similar values.
Example: Traffic fatalities in the USA by state.

8. Desire Lines Example: Truck flow to St Louis, USA.
Desire lines shows individual movement with lines of proportional thickness. They are very similar to flow lines except they generalise movement, showing movement only directly from A to B whilst flow lines follow the exact path of movement.
They are effective at visually showing movement of people, goods and transports. They are also good at showing the volume and direction of movement.
In order to achieve a clear image, the real distance and direction may be distorted.
Example: Truck flow to St Louis, USA.

9. Choropleth Maps
These are maps, where areas are shaded according to a prearranged key, each shading or colour type representing a range of values.
They can give a false impression of abrupt change at the boundaries of shaded units.
Helpful for identifying relationships and comparing between different geographic location.
Example: Ethnic population of Blank Caribbean in London

10. Can visually compare and identify different areas easily.
Layer shading
Example: Land Use of Sleaford
Land use maps are visuals that provide information about how a piece of land is used, like a zoning map. Areas of land are often divided by a colour that represents a type of land use.
Key
Commercial Buildings/offices
Residential
Recreational
Businesses
Can visually compare and identify different areas easily.
Can generalise areas into categories and may not represent complete area.

11. Data collection
Random sampling: this is the most accurate method as there is no bias involved as every person or place has an equal chance of being sampled.
Systematic sampling: this is a quick and easy method to use where a regular sample is taken, eg river depth readings may be taken every 50cm across the channel. This type of sampling does not always work well with questionnaires as it may not be possible to ask every 10th person, for example, who enters a shopping area.
Stratified sampling: this is where people or places are deliberately chosen according to the topic being investigated; for example, a questionnaire about a regeneration project might be asked to an equal number of males and females within pre-determined age groups; eg five males and five females aged 20-39, five of each gender aged and so on. This can also be used alongside random or systematic sampling. X

12. Risk Assessment
Risk assessment is the fundamental tool to ensure safety is effectively managed. The purpose of the Risk Assessment process is to identify hazards; assess who may be harmed and how; and manage the hazards through safe systems of work.
In line with Health and Safety Executive guidelines, centres should follow five steps to risk assessment:
Identify the hazards
Decide who might be harmed and how
Evaluate the risks and decide on precautions
Record your findings and implement them
Review your assessment and update if necessary
The likelihood and severity of the hazard occurring can be scored numerically (one equals low, five equals high), with resultant risk being assessed as:
More than 10 - Take immediate action to either remove or control the risk, for example a less risky option, prevent access to the hazard
Eight to 10 - Inform people of the risk and look at ways of reducing it
Less than eight - Monitor the situation more closely and aim to reduce risk over longer term.

13. Quantitative and Qualitative
Quantitative Research
Quantitative research gathers data in numerical form which can be put into categories, or in rank order, or measured in units of measurement.  This type of data can be used to construct graphs and tables of raw data.
Qualitative Research
Qualitative research gathers information that is not in numerical form.  For example, diary accounts, open-ended questionnaires, unstructured interviews and unstructured observations.
This methods mostly avoids personal basis and greater accuracy of results.
It is useful for studies at the individual level and can find out what people think.
Data is much more narrower and may not always reflect feelings or situations.
This type of data is typically descriptive data and as such is harder to analyse.

14. Cross-Section: Beach Profiles
This is a cross-section show what the landscape looks like if it’s shopped down the middle and viewed from the side.
Check the degrees of your beach profile.
Measurement
5m = 5 degrees
10m = 7 degrees
15m = 5 degrees
20m = 8 degrees
25m = 5 degrees
30m = 7 degrees
The beach profile shouldn’t be massively steep! 5 10 15 20 25 30 35 40 45 50

15. Eleven students were asked about for their opinion about geography.
Working out the Percentages and Degrees for Pie Charts
Eleven students were asked about for their opinion about geography.
Geography is the best subject in the world.
For a pie chart however you will need to turn the percentage into a degree. This can be done by multiplying by the magic 3.6.
Strongly Disagree
Disagree
Agree
Strong Agree 1 4 6 5
For each value you will need to divide it by the total amount of people. 22 90
136
112
You can check whether your result is accurate by seeing if it adds up to 100 for percentages and 360 for degrees!
1 / 16
4 / 16
6 / 16
5 / 16
= 0.06
= 0.25
= 0.38
= 0.31
x100 6%
25%
38%
31%
x 3.6
Now you need to turn the number into a percentage by multiplying by 100

16. Pie Charts
Pie charts are used to compare categories within a data set. A pie chart displays segments of data according to the share of the total values of the data.
Remember you need to include a Key!
They are visually effective as it is easy to see the relative proportions for each segment.
They can sometimes overemphasize large values and therefore smaller values are not clear.
How do you convert percentages into degrees for a pie chart?
Example:
35% x 3.6 = 126°
40% x 3.6 = 144°
25% x 3.6 = 90°

17. Kite Diagrams
A kite diagram is a chart that shows the number of or percentage of something against distance along a transect.
Kite diagrams are useful in showing changes over distance.
Can be difficult to construct and analyse.
Transect 1
Distance
Transect 1
Transect 2 2
100 4
200
300 8
400
500
600
700
800
900
1000
Transect 2
Distance

18. You can use histograms when your data can be divided into intervals.
400
Histograms are very similar to bar charts, but they have a continuous scale of numbers on the bottom and there can’t be any gaps between the bars.
You can use histograms when your data can be divided into intervals.
300
You can graph huge data sets easily with histograms.
Number of cars
200
It shows the number of values within an interval and not the actual values.
TIME
CARS
334
387
209
121 ?
100
0700
0800
0900
1000
1100
1200
Time

19. Compound line graph
Compound graphs show layers of data and are calculated by reading off the distance in values between adjacent line. 80 70
Can visually see the proportion which makes the total. It is used for continuous data. 60 50
It takes time to read accurate numbers for each section of the total. 40
Distance
Marram
Lyme
Heather 12 14 4 16 8 18 30 10 20 40 22 58 6 24 60 26 28 54 55 25 32 52 30 20 10
12m
14m
16m
18m
20m
22m
24m
26m
28m
30m
32m

20. Compound Bar Chart
A compound bar chart is useful for when you want to express two or more quantities on one chart. The clear presentation of the bar chart allows for comparison between different values.
Number of Vehicles
100 80
This graph makes it very easy to compare the percentage of each segment. 60
They might not be necessarily easy to identify patterns and relations between different data sets. 40 20
Sweet Type
Area 1
Area 2
Area 3
Cars 40 20
Bikes 50
Lorries 30
Area 1
Area 2
Area 3

21. Can only be used for data which scale is the same.
Radar Graphs
A radar chart is a graphical method of displaying several different sets of data that are on the same scale using one focal point.
Building Quality 5 4 3
Green Space
They are a useful way of identifying patterns within a particular theme.
Shops & Services 2 1
Can only be used for data which scale is the same.
Factors
500m 0m
Building Quality 5 4
Shop & Services
Litter 3
Noise Pollution 2
Green Space
Noise Pollution
Litter

22. Scatter Graphs x x x x x x x x x x
Spearman's Rank Correlation Coefficient is a further technique for analysing this data set.
Scatter graphs let you see two sets of information at the same time. The relationship between the two sets of data is called the correlation.
A line of best fit can either be…
No correction
Positive Correction
Negative Correction
2.2
Can compare multiple sets of data and be read easily. x
2.0 x
Can only be used for continuous sets of data.
1.6 x x
Distance from museum
Price for bottled water
100m
1.80
200m
1.20
300m
2.00
400m
1.00
500m
600m
700m
0.80
800m
0.60
900m
1000m
0.85
1.2 x x x x x
0.8 x
This line of best fit shows a Negative correction between distance and price.
0.4
100
200
300
400
500
600
700
800
900
1000

23. Line Graphs
Line graphs are used to show continuous data and are useful for showing trends. This is a simple line graph. 70 60
Can compare multiple sets of data and be read easily. 50
Can only be used for continuous sets of data. 40
Population Number (millions) 30
Years UK
France
Spain
1921 40 37 21
1931  44 39 24
1941  46 41 26
1951 50 42 28
1961 53 31
1971  56 51 34
1981  54 38
1991  58 57
2001  59
2011  63 47 20 10
1921
1931
1941
1951
1961
1971
1981
1991
2000
2011

24. Mode, Median, Mean and Range
Mode, median and mean are measures of average and the range is how spread out the value is.
Sample 1 2 3 4 5 6 7
River Discharge
184 90
159
142 64 95
MODE = Most Common
MEDIAN = Middle value
MEAN = Total of items / number of items
RANGE = Difference between highest and lowest.
Calculate the mean, median, mode and range for the river discharge data shown in the table above.
The mode is the most common value = 64
To find the median, put all the numbers in order and find the middle value:
64, 64, 90, 95, 142, 159, 184. So the median is 95
Mean = total of items/ number of items = 114
Range = the difference between highest and lowest value. i.e = 120

25. Chi-squared test X² 0-45° = 2 Pebbles 46-90° = 10 Pebbles
A group of students investigated the orientation of pebbles in an exposed bed of glacial till. The glacial till was situated near the lip of a corrie in the Lake District. The students wanted to investigate whether there was a pattern to the orientation of the long axis of the till. Their hypothesis was:
There is a relation between the orientation of the glacial till and the direction of the glacier.
They measure the orientation of 40 pebbles and placed their results into four categories:
0-45° = 2 Pebbles
46-90° = 10 Pebbles
91-135° = 23 pebbles
° = 5 pebbles
The data suggests that there is a preferential direction but as this could be due to chance a chi-squared test is carried out. The test begins with the assumption that there is no preference for any direction, with the null hypothesis.
There is no significant difference between the observed orientation of pebbles and the expected random orientation.

26. Degrees of freedom Significance Level 0.05 0.01 1 3.84 6.64 2 5.99
Step One: Calculate the Expected (E) frequencies by adding up all the observed data and dividing by the number of categories.
Step Four The chi-squared result (X²) can now be calculated by totalling all of the values in column C. Nonetheless, the result by itself is meaningless. You now need to test its significance.
Step Two: Complete column A by subtracting the expected value from the observed value. Then complete column B by squaring 0 – E.
Step Five: Work out the degrees of freedom using the formula (n – 1), where n is the number of observations – in this case the number of categories which contained observed data.
Step Three: Divide the figure in column B by the expected Value (E) to complete C.
Step Six: Using the Table below, compare your X² result with the degrees of freedom for the 95% significance level. If the X² result is the same or greater than the value given in the table, the null hypothesis can be rejected. A B C
Orientation
Observed
Expected
0 - E
(0 – E)2
𝟎−𝐄 𝟐 𝐄 E
0-45°  2
46-90°
 10
91-135°
 23 °  5
X² =
Degrees of freedom
Significance Level
0.05
0.01 1
3.84
6.64 2
5.99
9.21 3
7.82
11.34 4
9.49
13.28 5
11.07
15.09 10 -8 64
6.4 10 10 13
169
16.9 10 -5 25
2.5
25.8

27. The Mann Whitney U Test
Income deprivation scores in two areas of Lincoln
A student investigated the economic deprivation across the city of Lincoln. She has used the National Statistics website to obtain secondary data on deprivation. From her investigation she has formulated the hypothesis.
There is a greater income deprivation score for inner-city areas than outer suburbs.
She has found out the income deprivation score for eight areas in each of her two study areas—one in the inner city, and the other in the outer suburbs. The table shows her results. The income deprivation score measures the percentages of the people who are income-deprived. The higher the score, the more income deprived the area is.
Income deprivation score
Inner
City
Outer Suburb
0.53
0.05
0.40
0.09
0.24
0.06
0.25
0.12
0.08
0.16
0.10
0.20
0.13
The student decides to conduct the Mann Whitney U test to see if there is a difference between the two areas.
First she sets out the null hypothesis:
There is no difference in the income deprivation scores between the inner-city and the outer suburbs.

28. 6 is less than 13, therefore we have to REJECT the null hypothesis.
Step One: Label the data set x and y. If they are of different sizes then label the smaller one x and the larger one y.
Step Four: You now have to calculate the U values for both samples, use the formulas below:
Step Two: Rank the scores in terms of their position in both samples – insert these in column B and column D. Where the scores are the same take average of the rank values.
𝑼 𝒙 = 𝒏 𝒙 × 𝒏 𝒚 + 𝒏 𝒙 𝒏 𝒙 +𝟏 𝟐 −∑ 𝜞 𝒙
𝑈 𝑥 = × − 8 8 1
Step Three: Total the ranks in the column B and column D. 8 8 42 A B C D
Inner-city scores (x)
Rank
Outer suburb scores (y)
Total = = 64 36 42 58
𝑼 𝒚 = 𝒏 𝒙 × 𝒏 𝒚 + 𝒏 𝒚 𝒏 𝒚 +𝟏 𝟐 −∑ 𝜞 𝒚 1
0.05
0.09
0.06
0.12
0.08
0.10
0.13
15.5
0.53
0.40
0.24
0.25
0.08
0.12
0.16
0.20 2 11
𝑈 𝑦 = × × − 8 8 1 4 14 8 8 94 3
8.5 = 64 36 94 6
12.5
15.5
Step Five: Now select the smaller figure and compare it to the table of critical value to test the significance of the result. If the value is lower than or equal to the critical value, you reject the null hypothesis.
8.5
12.5 6 10 5 7
6 is less than 13, therefore we have to REJECT the null hypothesis. 42 94

29. Health expenditure per capita
Spearman's Rank
Calculate the Spearman rank correlation coefficient for the strength of relationship between the GNI per capita and health expenditure per capita of 10 selected countries in 2014.
The null hypothesis of this statistical test is written as…
There is no relationship between a country’s GNI per capita and health expenditure.
Country
GNI per capita
Rank
Health expenditure per capita d d²
Australia
65000
6000
Norway
104000
10000
Japan
40000
4000
Switzerland
88000
9000
Spain
29000
3000
Chile
15000
1000
Mexico
700
Russia
13000
Peru
400
China
7000
Step 1: Rank both sets of data values for each sets of data. 3 3
Step 2: For each pair of data values, calculate the difference, d, between the two ranks; this can be a positive or negative value. 1 1 4 4 2 2 5 5
Step 3: Now square the d values to eliminate negative signs. Make an extra column in the table to enter these d² values. 6
6.5
-0.5
0.25 8 8 7
6.5
0.5
0.25
Step 4: Add up all the d² values. 10
9.5
0.5
0.25
𝒅² = 1 9
9.5
-0.5
0.25

30. Spearman's Rank 𝑹 𝒔 =1− 6𝛴d² n³−n =1− 6 𝑥 1 10³−10 =1− 6 1000−10
Step 7: But is this result statistically significant? In other words, is it possible that it has occurred by chance?
Now test the significance of the result using the Spearman Rank test.
To apply the test you need the degrees of freedom of the data and the significance level you want to test at
Step 5: Insert the values for 𝒅² and n into the formula and calculate the 𝑹 𝒔 value.
Degrees of freedom
Significance Level
0.05
0.01 4
1.000 5
0.900 6
0.829
0.943 7
0.714
0.893 8
0.643
0.833 9
0.600
0.783 10
0.564
0.745
𝑹 𝒔 =1− 6𝛴d² n³−n
r = relationship
 = sum of
d = differences between the ranks
n = number of variables
=1− 6 𝑥 1 10³−10
Step 8: Looking at the critical values table. The entry for 10 degrees of freedom at for significance level 0.01 is
As is well above 0.745, there is less than 1 in 100 chance of the result having occurred randomly.
Step 6: If the value of r is very close to 1, there is a strong positive correlation.
If the value of r is very close to -1, there is a strong negative correlation.
This value suggests a strong correlation.
=1− −10
=1−
As 𝑹 𝒔 value you have calculated is greater than the critical value, you can reject the null hypothesis. This means that there is a significant relationship between the GNI and Health expenditure.
=1−0.006
𝑹 𝒔 =0.994

31. The richness of species is the same at location A as at location B.
Student’s t test
As part of an ecosystem study at Hunstanton Dunes, students conducted a quadrat survey at two location, A (20m from the sea) and B (25m from the sea). At each location there were ten sample sites and at each sites the richness of species was represented by total number of individual species found in a single quadrat.
Based on the data within the table, use the student’s t test to determine of there is a significant difference between the number of plant species at locations A and B.
Site Number
Species richness at A
Species richness at B 1 3 2 6 13 4 8 15 5 14 7 9 11 10 16
The null hypothesis of this statistical test is written as…
The richness of species is the same at location A as at location B.

32. Species richness at A (x values) Species richness at B (y values)
Step One: First calculate the values needed for the formula using the table below.
Step Three: Next add up the values in the third and fourth column to find x² and y² .
x = =𝟖𝟒
Site Number
Species richness at A (x values) x²
Species richness at B (y values) y² 1 3 2 6 13 4 8 15 5 14 7 9 11 10 16
Totals
x 84
y 113
𝑥 = 8.4
𝑦 = 11.3
𝒙 = 84 / 10 = 8.4
y = =𝟏𝟏𝟑
𝒚 = 113 / 10 = 11.3
Step Three: Now calculate the absolute difference of 𝑥 and 𝑦 .
= 2.9
Step Four: Insert the various values you have found out into the t test formula to calculate the t value.
Step Two: Now calculate the mean by adding the data together for each set of values (x & y). Then divide the total by the total amount of sites (i.e. 10) to find x and y .

33. Student’s t test 𝑛 𝑥 −1 + 𝑛 𝑦 −1 = 9 + 9 + 18 1 4 Degrees of freedom
Step Five: Now test the significance of the results using the table of critical values for Student’s t test.
For Student’s t test the degrees of freedom is: 1 4
𝑛 𝑥 −1 + 𝑛 𝑦 −1 =
Degrees of freedom
Significance Level
0.05
0.01 17
1.74
2.90 18
1.73
2.88 19
2.85 5 2 6
Step Six: As 1.6 is less the both of the critical values, we can then accept the null hypothesis.
i.e. The richness of species is the same at location A as at location B. 3 7