# Geography Fieldwork

A Level

# Data analysis

## 4. Data analysis

Sophisticated data analysis will help you spot patterns, trends and relationships in your results. Data analysis can be qualitative and/or quantitative, and may include statistical tests. An example of a statistical test is outlined below.

## Sediment analysis

### (a) Size of coarse sediments: mean size

Coarse sediments are pebbles and cobbles. If you have measured the a, b and c axes using calipers or a ruler, you could calculate the mean pebble size for each sample site on the beach.

`"Mean size" = (a + b + c)/3` for each pebble

### (b) Size of fine sediments: phi sizes

Fine sediments are clay, silt and sand. Once you have sieved the sediment, you can calculate phi sizes. Use the conversion table if you do not have the phi sizes already.

Sediment size | |
---|---|

mm | phi |

1.00 | 0 |

0.50 | 1 |

0.25 | 2 |

0.13 | 3 |

0.06 | 4 |

0.03 | 5 |

0.01 | 6 |

Calculate the percentage mass of sediment in each phi size category. For example, if total mass=100g and the mass of material at 5-10mm = 20g, then 20% of the total mass of sediment is 5-10mm in diameter. This can be presented in a number of ways

- using a histogram with % mass on the y axis and sediment size on the x-axis
- pie charts to show changes along the transect, which might be overlaid on a map or aerial photograph
- plot a scattergraph to show how mean sediment size varies with distance along the beach (see below).

Alternatively, use semi-logarithmic graph paper to plot a cumulative frequency graph of phi against mass. Plot phi size on the linear x-axis. Plot the cumulative mass of sediment on the logarithmic y-axis.

On your finished graph, find the phi size values at 16% and 84% cumulative mass. Use these figures in the following formula

`("phi at 84% mass" - "phi at 16% mass")/2`

Use the following table to interpret the result

result | interpretation |
---|---|

0.35 | very well sorted |

0.35 - 0.5 | well sorted |

0.5 - 0.7 | moderately well sorted |

0.7 - 1.0 | moderately sorted |

1.0 - 2.0 | poorly sorted |

2.0 - 4.0 | very poorly sorted |

4.0 | extremely poorly sorted |

### (c) Zingg's shape classification

The analysis of the shape of coarse sediments can be divided into 4 categories: shape, sphericity, flatness and roundness.

The raw data needed for each pebble are the lengths of the a, b and c axes.

Calculate the ratio `b-:a`

Calculate the ratio `c-:b`

Now classify each pebble into one of the four groups shown in the table

Type of pebble | `b-:a` | `c-:b` |
---|---|---|

Sphere | > 0.67 | > 0.67 |

Disc | > 0.67 | < 0.67 |

Rod | < 0.67 | > 0.67 |

Blade | < 0.67 | < 0.67 |

### (d) Krumbein's Index of Sphericity

The raw data needed for each pebble are the lengths of the a, b and c axes.

For each stone, calculate Krumbein's Index as follows

`"Krumbein's Index" = ((bc)/a^2)^(1/3)`

Krumbein's Index (K) K must be between 0 and 1. K = 1 for a perfectly spherical pebble. The lower that K is, the less spherical the pebble.

### (e) Cailleux’s Flatness Index

The raw data needed for each pebble are the lengths of the a, b and c axes.

For each stone, calculate Cailleux’s Flatness Index as follows

`"Flatness Index" = (a + b) / (2c) xx 100`

A perfectly equidimensional particle will have a Flatness Index of 100 and will increase infinitely as it become flatter.

### (f) Cailleux’s Roundness Index

The raw data needed for each pebble are:

- the length of the longest axis (l)
- the radius of curvature of the sharpest angle (r)

For each stone, calculate Cailleux Index as follows

`"Roundness Index" = (2r)/lxx1000`

Roundness Index =1000 for a perfectly spherical pebble. The lower the Roundness Index is, the more angular the pebble.

Cailleux's Roundness Index may be presented using box and whisker plots.

## Spearman’s Rank Correlation Test

Spearman’s Rank Correlation is a statistical test to test whether there is a significant relationship between two sets of data.

The Spearman’s Rank Correlation test can only be used if there are at least 10 (ideally at least 15-15) pairs of data.

There are 3 steps to take when using the Spearman’s Rank Correlation Test

### Step 1. State the null hypothesis

There is no significant relationship between _______ and _______

### Step 2. Calculate the Spearman’s Rank Correlation Coefficient

`r_s = 1-(6∑D^2) / (n(n^2-1))`

- `r_s` = Spearman's Rank correlation coefficient
- `D` = differences between ranks
- `n` = number of pairs of measurements

Step 3. Test the significance of the result

Compare the value of `r_s` that you have calculated *against* the critical value for `r_s` at a confidence level of 95% / significance value of p = 0.05.

If `r_s` is equal to or above the critical value (p=0.05) the REJECT the null hypothesis. There is a SIGNIFICANT relationship between the 2 variables.

A positive sign for `r_s` indicates a significant positive relationship and a negative sign indicates a significant negative relationship.

If `r_s` (ignoring any sign) is less than the critical value, ACCEPT the null hypothesis. There is NO SIGNIFICANT relationship between the 2 variables.

### Worked example

A geographer is interested in whether longshore drift is operating along a coastal spit at Abermenai Point in Anglesey. Random samples of pebbles were taken at the base of the storm beach at 15 sites every 50m along a spit. Here are the results.

Distance along spit (m) | Mean pebble size (cm) |
---|---|

0 | 11.3 |

50 | 12.2 |

100 | 12.8 |

150 | 13.6 |

200 | 14.5 |

250 | 15.3 |

300 | 13.5 |

350 | 11.2 |

400 | 10.0 |

450 | 9.3 |

500 | 8.0 |

550 | 10.2 |

600 | 9.8 |

650 | 10.2 |

700 | 7.4 |

### Step 1. State the null hypothesis

There is no significant relationship between distance along the spit and mean pebble size at Abermenai Point.

### Step 2. Calculate the Spearman’s Rank Correlation Coefficient

(a) Rank the measurements

Distance along spit (m) | Mean pebble size (cm) | ||
---|---|---|---|

Data | Rank | Data | Rank |

0 | 1 | 11.3 | 8 |

50 | 2 | 12.2 | 9 |

100 | 3 | 12.8 | 10 |

150 | 4 | 13.6 | 12 |

200 | 5 | 14.5 | 13 |

250 | 6 | 15.3 | 14 |

300 | 7 | 13.5 | 11 |

350 | 8 | 11.2 | 7 |

400 | 9 | 10.0 | 5 |

450 | 10 | 9.3 | 3 |

500 | 11 | 8.0 | 2 |

550 | 12 | 10.2 | 6.5 |

600 | 13 | 9.8 | 4 |

650 | 14 | 10.2 | 6.5 |

700 | 15 | 7.4 | 1 |

(b) Calculate `D` and `D^2`

Distance along spit (m) | Mean pebble size (cm) | ||||
---|---|---|---|---|---|

Data | Rank | Data | Rank | `D` | `D^2` |

0 | 1 | 11.3 | 8 | 7 | 49 |

50 | 2 | 12.2 | 9 | 7 | 49 |

100 | 3 | 12.8 | 10 | 7 | 49 |

150 | 4 | 13.6 | 12 | 8 | 64 |

200 | 5 | 14.5 | 13 | 8 | 64 |

250 | 6 | 15.3 | 14 | 8 | 64 |

300 | 7 | 13.5 | 11 | 4 | 16 |

350 | 8 | 11.2 | 7 | -1 | 1 |

400 | 9 | 10.0 | 5 | -4 | 32 |

450 | 10 | 9.3 | 3 | -7 | 49 |

500 | 11 | 8.0 | 2 | -9 | 81 |

550 | 12 | 10.2 | 6.5 | -5.5 | 30.25 |

600 | 13 | 9.8 | 4 | -9 | 81 |

650 | 14 | 10.2 | 6.5 | -7.5 | 56.25 |

700 | 15 | 7.4 | 1 | -14 | 196 |

(c) Calculate `∑D^2`i.e. the sum of the `D^2`column`= 865.6`

(d) Calculate `r_s`

`r_s = 1-(6∑D^2) / (n(n^2-1))`

`r_s = 1-(6xx865.5) / (15xx(225-1))`

`r_s = -0.55`

### Step 3. Test the significance of the result

The critical value at `p=0.05` significance level for `15` pairs of measurements is `0.521`

Since our calculated value of `0.55> 0.521` (ignore the minus sign), the null hypothesis can be rejected.

In conclusion, there is a significant relationship between distance along the spit and mean pebble size at Abermenai Point. The negative sign for `r_s`indicates a significant negative relationship.