Apply the chi-square test to compare observed and expected frequencies, use degrees of freedom and critical values, and interpret statistical significance

What this dot point is asking

SEAB wants you to apply the chi-square test, comparing observed frequencies with expected ones, to work out degrees of freedom, to compare the statistic with a critical value, and to interpret significance. The central insight is that chi-square answers a different question from correlation: instead of asking whether two measured variables move together, it asks whether a pattern of counts across categories differs from what chance alone would produce, which is exactly the test for questions about distribution and association.

The answer

What chi-square tests

The chi-square test ($\chi^2$) compares the observed frequencies (the counts you actually recorded) with the expected frequencies (the counts you would expect if the null hypothesis were true). It works on frequency data, counts in categories, not on percentages, rates or means. Typical geographical uses are testing whether a feature is evenly distributed across categories (goodness of fit) or whether two categorical variables are associated.

The formula

The test statistic is:

where $O$ is each observed frequency and $E$ is the corresponding expected frequency. Each category contributes $\dfrac{(O-E)^2}{E}$; the larger the gaps between observed and expected, the larger $\chi^2$ becomes, signalling a bigger departure from the null.

The method, step by step

1. State the hypotheses. The null is usually "no difference from expectation" or "no association"; the alternative is that a real difference or association exists.
2. Find the expected frequencies. For an even distribution across $k$ categories, $E$ is the total divided by $k$. For an association (contingency) table, $E$ for each cell is (row total times column total) divided by the grand total.
3. Compute $\dfrac{(O-E)^2}{E}$ for every category and sum them to get $\chi^2$.
4. Find the degrees of freedom.
5. Compare with the critical value and decide on the null.

Degrees of freedom

The degrees of freedom set which critical value to use:

• For a goodness-of-fit test across $n$ categories: degrees of freedom $= n - 1$.
• For a contingency table: degrees of freedom $= (\text{rows} - 1)(\text{columns} - 1)$.

For example, four soil types in a goodness-of-fit test give $4 - 1 = 3$ degrees of freedom.

Testing significance

Compare the calculated $\chi^2$ with the critical value for those degrees of freedom at a chosen significance level (commonly 0.05):

• If $\chi^2$ is greater than or equal to the critical value, the result is significant: reject the null hypothesis, concluding the observed pattern differs from chance more than would be expected.
• If $\chi^2$ is below the critical value, fail to reject the null: the differences are within what chance could produce.

Conditions and limits

Chi-square is only valid when its conditions hold: the data must be frequencies (not percentages or means), categories must be discrete and mutually exclusive, observations must be independent, and expected frequencies should generally be at least five in each category. Small expected values make the test unreliable, so categories are sometimes combined to meet this. The test shows that a difference exists, not how strong or why.

Examples in context

Example 1. Shop types around a Singapore neighbourhood centre. A geographer testing whether the mix of shop types (food, services, retail, others) differs from an even spread counts the outlets in each category and applies a goodness-of-fit chi-square with three degrees of freedom. A statistic above the critical value would show the retail mix is significantly uneven, reflecting the centre's specialised function. It illustrates chi-square applied to categorical counts in human geography.

Example 2. Pebble orientation and a depositional process. Recording whether pebbles on a glacial or beach deposit point in particular direction classes, a geographer compares the observed counts in each direction band with an even expected spread using chi-square. A significant result indicates a preferred orientation, evidence of a directional process such as ice or current flow. The example shows the test distinguishing a real spatial pattern from random scatter, while noting it reveals difference, not strength.

Try this

Q1. State the chi-square formula and explain what $O$ and $E$ represent. [2 marks]

• Cue. $\chi^2 = \sum \dfrac{(O-E)^2}{E}$, where $O$ is the observed frequency (the count actually recorded in a category) and $E$ is the expected frequency (the count expected if the null hypothesis, such as an even distribution, were true).

Q2. A goodness-of-fit chi-square test uses five categories. State the degrees of freedom and explain how they are used. [2 marks]

• Cue. Degrees of freedom $= n - 1 = 5 - 1 = 4$; they select which critical value to read from the chi-square table at the chosen significance level, against which the calculated statistic is compared to decide whether to reject the null.

Q3. Explain why chi-square must be calculated from frequencies rather than percentages. [3 marks]

• Cue. The statistic depends on the actual counts because the size of $\dfrac{(O-E)^2}{E}$ reflects real sample sizes; percentages discard that information and would, for example, treat a pattern from 10 observations the same as one from 1000, distorting the statistic and making the significance test invalid.