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Stage 4: Data analysis

Stage 4: Data presentation | Statistics

Data presentation

(a) Maps

It is most straightforward to produce one map of building age and another map of building function. Although you may be tempted to photocopy a published map, there are copyright restrictions, and it will inevitably contain too much unnecessary information. Instead draw your own street plan of the study settlement, with scale and north arrow. Use colour codes to indicate the age of each building and its functions.

(b) Traffic survey results

One way of displaying all the data that you have collected is to draw divided bar charts, showing the number of vehicles at each survey time, with the proportion of each type (cars, buses, lorries, etc) indicated on each bar. See example below.

Statistics

The chi-squared test can be used to compare two villages. Need more information about this test?

If you have compared two villages, say at different distances from a city, you can use the chi-squared test to see if there is a significant difference between them. The data you have collected is observed data and will look as follows

	Number of buildings of each age
	Village x	Village y	TOTAL
Pre-Victorian	45	12	57
Victorian	287	113	400
Early 20th century	140	141	281
Post-war	50	102	152
Recent	34	118	152
TOTAL	556	486	1042

Test the null hypothesis: There is no significant difference in the age of houses in village X and village Y.

Expected values are shaded in grey.

	Number of buildings of each age
	Village x		Village y		TOTAL
Pre-Victorian	45	30.4	12	26.6	57
Victorian	287	213.4	113	186.6	400
Early 20th century	140	149.9	141	131.1	281
Post-war	50	81.1	102	70.9	152
Recent	34	81.1	118	70.9	152
TOTAL	556	556	486	486	1042

Now calculate the sum of (observed - expected)2 ÷ (expected)

chi squared = (45-30.4)2 ÷ 30.4 + (287-213.4)2 ÷ 213.4 + (140-149.9)2 ÷ 149.9 + (50-81.1)2 ÷ 81.1 + (34-81.1)2 ÷81.1 + (26.6-12)2 ÷12 + (113-186.6)2 ÷ 186.6 + (141-131.1)2 ÷131.1 + (102-70.9)2 ÷70.9 + (118-70.9)2 ÷70.9

chi squared = 7.0 + 25.4 + 0.7 + 27.4 + 8.0 + 29.0 + 0.7 + 13.6 + 31.3

chi squared = 155.1

Now compare the calculated value of chi squared against the critical values for (columns of observed values - 1) x (rows of observed values-1) degrees of freedom.

In our example, at the p=0.05 probability level where degrees of freedom = 4, the critical value of chi squared = 9.49.

Since 155.1 > 9.49, the null hypothesis is rejected at the p=0.05 level.

Therefore, we are 95% certain that there is a significant difference in the age of houses in village X and village Y.

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