Geographical Skills and the Independent Fieldwork Investigation

A student proposes to investigate 'Is the environment getting worse?'. Explain why this is a weak enquiry question and rewrite it as a strong one.

1. 01Why it is weak

   The question is far too broad and vague, not tied to a place or a measurable variable, and cannot be answered with data a student could realistically collect - 'the environment' and 'worse' are undefined.

2. 02Make it focused and measurable

   A strong question is specific, local, grounded in theory and answerable with fieldwork - for example: 'How does river velocity change with distance downstream in the River X catchment?'

3. 03Why the rewrite works

   It names a place, a measurable variable and a testable relationship (linked to the Bradshaw model), so data can be collected, presented, analysed statistically and evaluated.

Result: The original is too broad and unmeasurable; a strong question is focused, local, theory-grounded and answerable with collectable data, e.g. how river velocity changes downstream.

A student wants to survey opinions across a town made up of distinct districts of very different wealth and age profiles. Which sampling strategy should they use, and why?

1. 01Consider the population

   The town is not uniform - it has distinct districts (strata) that differ in wealth and age, and each may hold different opinions.

2. 02Rule out the alternatives

   Simple random or systematic sampling might, by chance, over- or under-represent some districts, missing the variation that matters.

3. 03Choose and justify

   Stratified sampling divides the town into its districts and samples each in proportion to its size, ensuring every group is represented - so the sample reflects the town's real diversity.

Result: Stratified sampling is best: it divides the town into its distinct districts and samples each proportionally, so the varied population is fairly represented.

A student wants to map the total number of tourists visiting each of several resorts. Explain whether a choropleth or a proportional-symbol map is more appropriate.

1. 01The nature of the data

   The data are absolute totals (numbers of tourists) at specific points (the resorts), not rates or densities spread over areas.

2. 02Why not choropleth

   A choropleth shades whole areas and is designed for rates and densities; using it for totals would distort the impression, as large areas look more important regardless of their value.

3. 03Choose proportional symbols

   A proportional-symbol map draws a symbol at each resort sized in proportion to its total, correctly representing absolute quantities at points - so it is the appropriate choice.

Result: A proportional-symbol map is appropriate, because the data are absolute totals at points; a choropleth is for rates and densities and would distort totals.

A scatter graph plots distance from the city centre (X) against pedestrian count (Y) for eight sites, and the points rise together with one site below the trend. Describe the relationship.

1. 01State the direction

   As X increases, Y generally increases too, so there is a positive correlation between the two variables.

2. 02State the strength and scatter

   The points lie fairly close to a rising best-fit line, so the correlation is reasonably strong, though not perfect.

3. 03Note anomalies and causation

   One site lies below the trend - an anomaly worth investigating - and the correlation, however strong, shows association only, not that distance causes the change in pedestrians.

Result: The graph shows a reasonably strong positive correlation with one anomaly; the association does not by itself prove causation.

For ten paired sites, each variable is ranked, the difference in ranks (d) is found for each pair, squared, and the squared differences sum to 24. Calculate Spearman's rank correlation coefficient and interpret it.

1. 01Write the formula

   Spearman's rank is $r_s = 1 - \dfrac{6\sum d^2}{n(n^2 - 1)}$, with n = 10 and the sum of d-squared = 24.

2. 02Substitute

   Denominator = $n(n^2 - 1) = 10 \times (100 - 1) = 990$; numerator = $6 \times 24 = 144$.

   $$r_s = 1 - \dfrac{144}{990}$$
3. 03Calculate

   $144 \div 990 = 0.145$, so $r_s = 1 - 0.145 = 0.855$ (to three decimal places).

   $$r_s = 1 - 0.145 = 0.855$$
4. 04Interpret and test

   A value of +0.855 is a strong positive correlation. Comparing with the Spearman critical values for n = 10 at the 0.05 level (about 0.56 one-tailed), 0.855 exceeds it, so the correlation is statistically significant and the null hypothesis of no relationship is rejected.

Result: r_s = 0.855, a strong positive correlation that exceeds the 0.05 critical value for n = 10, so the relationship is statistically significant.

A pebble survey records 30, 20, 25 and 25 pebbles in four shape categories (total 100). Test whether the pebbles are distributed evenly among the categories using chi-squared.

1. 01State the expected values

   If the 100 pebbles were distributed evenly among the 4 categories, each expected frequency E = 100 / 4 = 25.

2. 02Apply the formula

   Chi-squared = sum of (O - E)-squared / E = (30-25)^2/25 + (20-25)^2/25 + (25-25)^2/25 + (25-25)^2/25.

   $$\chi^2 = \dfrac{25}{25} + \dfrac{25}{25} + 0 + 0 = 2$$
3. 03Test against the critical value

   Degrees of freedom = categories - 1 = 3; the critical value at the 0.05 level for 3 degrees of freedom is about 7.82. Since 2 is less than 7.82, the result is not significant.

4. 04Conclude

   We cannot reject the null hypothesis, so the observed distribution is not significantly different from an even one - the differences could have arisen by chance.

Result: Chi-squared = 2, below the 0.05 critical value of 7.82 for 3 degrees of freedom, so the pebbles are not significantly unevenly distributed - the null hypothesis stands.