# Geography Fieldwork

A Level

# Analysis

## 1. Statistical tests

## 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.

## Chi-squared test

Chi squared in a statistical test that is used either to test whether there is a significant difference, goodness of fit or an association between observed and expected values.

`chi^2 = ∑ (O-E)^2 / E`

The chi squared test can only be used if

- the data are in the form of frequencies in a number of categories (i.e. nominal data).
- there are more than 20 observations in total
- the observations are independent: one observation does not affect another

There are 3 steps to take when using the chi squared test

### Step 1. State the null hypothesis

There is no significant association between _______ and _______

### Step 2. Calculate the chi squared statistic

`chi^2 = ∑ (O-E)^2 / E`

`chi^2` = chi squared statistic

`O` = Observed values

`E` = Expected values

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

Compare your calculated value of `chi^2` against the critical value for `chi^2` at a confidence level of 95% / significance value of P = 0.05, and appropriate degrees of freedom.

`"Degrees of freedom" = ("number of rows" – 1) xx ("number of columns" – 1)`

If Chi Squared is equal to or greater than the critical value REJECT the null hypothesis. There is a SIGNIFICANT difference between the observed and expected values**. **

If Chi Squared is less than the critical value, ACCEPT the null hypothesis. There is NO SIGNIFICANT difference between the observed and expected values**.**

## 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 xx n_2 + 0.5 n_2 (n_2 + 1) - ∑ R_2`

`U_2 = n_1 xx 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.