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.

## Calculating the carbon content of a tree

This assumes that you have measured

- the circumference of the tree at chest height
- the height of the tree

See the Method section for techniques

There are four steps to calculating the carbon content of a tree

- Step 1. Calculate the volume of the tree trunk
- Step 2. Calculate the biomass of the tree trunk
- Step 3. Estimate the biomass of the canopy and the root ball
- Step 4. Calculate carbon content based on the biomass of the whole tree

These steps are based on the analysis used by Jenkins et al. (2011) of Forest Research, the research agency of the Forestry Commission.

### Step 1. Calculate the volume of the tree trunk

#### (a) Calculate the radius of the tree trunk at chest height

\(\mathsf{Radius = \frac{circumference}{2\;\times\; \pi}}\)e.g. if you measured the tree circumference to be 3m

\(\mathsf{Radius = \frac{3}{2\;\times\;3.14}}\) \(\mathsf{Radius=0.48m}\)#### (b) Calculate the volume of the tree trunk above ground

The approximate volume can be calculated by assuming that the tree trunk is a cone. Use the standard formula for calculating the volume of a cone.

\(\mathsf{Volume\;of\;the\;tree\;trunk = \pi \;r^2 \; \times \frac{height}{3}}\)e.g. if the radius of the tree trunk is 0.48m and the height is 15m

\(\mathsf{Volume\;of\;the\;tree\;trunk = 3.14 \;(0.48)^2 \; \times \frac{15}{3}}\) \(\mathsf{Volume\;of\;the\;tree\;trunk = 3.62\;m^3}\)### Step 2: Calculate the biomass of the tree trunk

The biomass of the tree trunk can be calculated by

\(\mathsf{Biomass = volume \;\times\; nominal\;specific\;gravity}\)If you know what species of tree that you have measured (like oak, ash or larch), you can find precise figures for nominal specific gravity figures on page 46 of Jenkins et al. (2011). Alternatively use the figures below

- For broadleaved trees, nominal specific gravity = 0.53
- For conifers, nominal specific gravity = 0.39

e.g. if the volume of the tree trunk is \(3.62\; m^3\)and the tree is an ash tree (a broadleaved tree)

\(\mathsf{Biomass = 3.62 \;\times\; 0.53}\) \(\mathsf{Biomass = 1.92 \;oven\;dry\;tonnes}\)### Step 3. Estimate the biomass of the canopy and the root ball

As well as the tree tunk, a substantial amount of carbon is stored in the tree’s canopy (or crown) as well as below ground in the roots.

The biomass of the crown and the roots can be estimated if the radius of the trunk and the species of tree are known. This method is based on the analysis used by Jenkins et al. (2011).

#### (a) Estimate the crown biomass

First take the radius and double it to get the tree’s **diameter at breat height (DBH)**. Make sure that this is expressed in centimetres.

There are two ways to calculate the crown biomass, depending on whether the tree is large or small. A large tree has a diameter at breast height of more than 50cm. A small tree has a diameter at breast height of between 7cm and 50cm. Smaller trees cannot be assessed using this method.

If DBH is 7-50cm then \(\mathsf{Crown\;biomass = a \times DBH^b}\)

If DBH > 50cm then \(\mathsf{Crown\;biomass = c + (d\;\times\;DBH)}\)

\(a\), \(b\), \(c\) and \(d\) are species-specific constants shown in the table below.

Estimating crown biomass | ||||
---|---|---|---|---|

Species | a | b | c | d |

Larch | 0.000044 | 2.0291 | -0.129047 | 0.005039 |

Corsican pine | 0.000012 | 2.4767 | -0.299529 | 0.009949 |

Lodgepole pine | 0.000018 | 2.4767 | -0.430537 | 0.014300 |

Scots pine | 0.000016 | 2.4767 | -0.394206 | 0.013094 |

Douglas fir | 0.000017 | 2.4767 | -0.411768 | 0.013677 |

Grand Fir | 0.000015 | 2.4767 | -0.353198 | 0.011732 |

Noble Fir & other conifers | 0.000015 | 2.4767 | -0.353198 | 0.011732 |

Hemlock | 0.000015 | 2.4767 | -0.353198 | 0.011732 |

Norwegian Spruce | 0.000015 | 2.4767 | -0.353198 | 0.011732 |

Cedar | 0.000015 | 2.4767 | -0.353198 | 0.011732 |

Sitka Spruce | 0.000015 | 2.4767 | -0.353198 | 0.011732 |

Beech, Sycamore & Maple | 0.000019 | 2.4767 | -0.459519 | 0.015263 |

Oak and all other broadleaved trees | 0.000017 | 2.4767 | -0.411551 | 0.013670 |

e.g. if the DBH is 96cm and the tree is an ash tree (in “all other broadleaved trees”)

\(\mathsf{Crown\;biomass = -0.411551 + (0.013670 \times 96)}\)

\(\mathsf{Crown\;biomass = 0.901\;oven\;dry\;tonnes}\)

#### (b) Estimate the root biomass

Next you can estimate the biomass of the roots. Again the equation that you use depends on the size of the tree. Remember that the DBH must be expressed in centimetres.

If DBH is 7-50cm then…

\(\mathsf{Root\;biomass = e \times (DBH)^{2.5}}\)If DBH > 50cm then…

\(\mathsf{Root\;biomass = f + (g \times DBH)}\)\(e\), \(f\) and \(g\) are species-specific constants shown in the table below.

Species | e | f | g |
---|---|---|---|

Larch | 0.000017 | -0.133480 | 0.007296 |

Corsican pine | 0.000011 | -0.082603 | 0.004515 |

Lodgepole pine | 0.000017 | -0.133480 | 0.007296 |

Scots pine | 0.000015 | -0.118673 | 0.006487 |

Douglas fir | 0.000017 | -0.133480 | 0.007296 |

Grand Fir | 0.000015 | -0.118673 | 0.006487 |

Noble Fir | 0.000011 | -0.082603 | 0.004515 |

Hemlock | 0.000015 | -0.118673 | 0.006487 |

Norwegian Spruce & other conifers | 0.000012 | -0.091547 | 0.005004 |

Cedar | 0.000011 | -0.082603 | 0.004515 |

Sitka Spruce | 0.000021 | -0.157579 | 0.008614 |

Beech | 0.000023 | -0.174882 | 0.009559 |

Oak and all other broadleaved trees | 0.000023 | -0.174882 | 0.009559 |

e.g. if the radius is 96cm and the tree is an ash tree (in “all other broadleaved trees”)

\(\mathsf{Root\;biomass = -0.174882 + (0.009559 \times 96)}\)

\(\mathsf{Crown\;biomass = 0.743\;oven\;dry\;tonnes}\)

## Step 4. Calculate carbon content based on the biomass of the whole tree

#### (a) Find the biomass of the whole tree

\(\mathsf{Biomass\;of\;whole\;tree = Trunk\;biomass + Crown\;biomass + Root\;biomass}\)e.g. if the trunk biomass is 1.92 oven dry tonnes, the crown biomass is 0.901 oven dry tonnes, and the root biomass is 0.743 oven dry tonnes

\(\mathsf{Biomass\;of\;whole\;tree = 1.92 + 0.901 + 0.743}\)

\(\mathsf{Biomass\;of\;whole\;tree = 3.564\;oven\;dry\;tonnes}\)

#### (b) Find the carbon content of the tree

The carbon content of the tree is calculated as

\(\mathsf{Carbon\;content = \frac{Biomass\;of\;whole\;tree}{2}}\)e.g. if the biomass of the whole tree is 3.564 oven dry tonnes

\(\mathsf{Carbon\;content = \frac{3.564}{2}}\) \(\mathsf{Carbon\;content = 1.782\;tonnes\;of\;stored\;carbon}\)## Mann Whitney U test

Mann Whitney U is a statistical test that is used either to test whether there is a significant difference between the medians of two sets of data.

The Mann Whitney U test can only be used if there are at least 6 pairs of data. It does not require a normal distribution.

There are 3 steps to take when using the Mann Whitney U test

### Step 1. State the null hypothesis

There is no significant difference between _______ and _______

### Step 2. Calculate the Mann Whitney U statistic

\(U_1= n_1 \times n_2 + 0.5 n_2 (n_2 + 1) \;- ∑ R_2\) \(U_2 = n_1 \times n_2 + 0.5 n_1 (n_1 + 1) \;- ∑ R_1\)- \(n_1\) is the number of values of \(x_1\)
- \(n_2\) is the number of values of \(x_2\)
- \(R_1\) is the ranks given to \(x_1\)
- \(R_2\) is the ranks given to \(x_2\)

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

Compare the value of U against the critical value for U at a confidence level of 95% / significance value of P = 0.05.

If U is equal to or smaller than the critical value (p=0.05) the REJECT the null hypothesis. There is a SIGNIFICANT difference between the 2 data sets.

If U is greater than the critical value, then ACCEPT the null hypothesis. There is NOT a significant difference between the 2 data sets.

### Worked example

A geographer was interested in the impact of footpath erosion on peat moorland at Kinder Scout in the Peak District. A point frame quadrat was used to measure vegetation cover at 10 randomly chosen points in each of two sites: one close to an unpaved section of the Pennine Way footpath, and the other site 250m from the edge of the footpath.

Here are the results.

Number of times out of 10 the pin hits vegetation | |
---|---|

Site 1 (on footpath) | Site 2 (Away from footpath) |

1 | 2 |

1 | 8 |

4 | 10 |

2 | 4 |

4 | 3 |

3 | 9 |

1 | 9 |

0 | 7 |

5 | 3 |

2 | 10 |

### Step 1. State the null hypothesis

There is no significant difference in vegetation cover between a site on the footpath and a site away from the footpath on peat moorland at Kinder Scout.

### Step 2. Calculate Mann Whitney U statistic

(a) Give each result a rank. Calculate the sum of the ranks for the two columns.

Site 1 (on footpath) | Site 2 (Away from footpath) | ||
---|---|---|---|

Hits | Rank | Hits | Rank |

1 | 3 | 2 | 6 |

1 | 3 | 8 | 16 |

4 | 12 | 10 | 19.5 |

2 | 6 | 4 | 12 |

4 | 12 | 3 | 9 |

3 | 9 | 9 | 17.5 |

1 | 3 | 9 | 17.5 |

0 | 1 | 7 | 15 |

5 | 14 | 3 | 9 |

2 | 6 | 10 | 19.5 |

SUM | 69 | SUM | 141 |

(b) Calculate \(∑R_1\)and \(∑R_2\)

\(∑R_1\) is the sum of the ranks in the first column (deciduous woodland) = \(69\)

\(∑R_2\) is the sum of the ranks in the first column (evergreen woodland) = \(141\)

\(n_1 = 10\) and \(n_2 = 10\)

(c) Calculate \(U_1\) and \(U_2\)

\(U_1 = 10 \times 10 + 0.5 \times 10 \times (10 + 1) \;– 141 = 14\) \(U_2 = 10 \times 10 + 0.5 \times 10 \times (10 + 1) \;– 69 = 86\)### Step 3. Test the significance of the result

In this example, \(U_1 = 14\) and \(U_2 = 86\)

Select the smaller of the values. \(U_14\) is the smallest, so \(U=14\)

The critical value at \(p=0.05\) significance level for \(n_1=10\) and \(n_2=10\) is \(\)23\(\). Since our calculated value of \(\)14 < 23\(\), the null hypothesis can be rejected.

In conclusion, there is a significant difference in vegetation cover between a site on the footpath and a site away from the footpath on peat moorland at Kinder Scout.