A wellconstructed 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.
\(\mathsf{Mean\;size = \frac{(a + b + c)}{3}}\) 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  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 510mm = 20g, then 20% of the total mass of sediment is 510mm 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 xaxis
 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 semilogarithmic graph paper to plot a cumulative frequency graph of phi against mass. Plot phi size on the linear xaxis. Plot the cumulative mass of sediment on the logarithmic yaxis.
On your finished graph, find the phi size values at 16% and 84% cumulative mass. Use these figures in the following formula
\(\frac{phi\;at\;84\%\;mass\;\;phi\;at\;16\%\;mass}{2}\)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 \(b \div a\)
Calculate the ratio \(c \div b\)
Now classify each pebble into one of the four groups shown in the table
Type of pebble  \(b \div a\)  \(c \div b\) 

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
\(\mathsf{Krumbein’s\;Index} = \sqrt[3]\frac{bc}{a^2}\)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
\(\mathsf{Flatness\;Index} = \frac{(a + b)}{2} \times 100\)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
\(\mathsf{Roundness\;Index} = \frac{2r}{l}\times 1000\)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 1515) 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
\(r_s = 1\frac{(6∑D^2)}{n(n^21)}\) \(r_s\) = 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  16 
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 \(r_s\)
\(r_s = 1\frac{(6∑D^2)}{n(n^21)}\) \(r_s = 1\frac{(6\;\times\;865.5)}{(15\;\times\;(2251)}\) \(r_s = 0.55\)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.