Chi-square hypothesis testing

What the exam asks

• State the null hypothesis ($H_0$) for a given genetic cross or investigation.
• Calculate expected counts ($e$) and the test statistic ($\chi^2$) from a table of observed counts ($o$).
• Determine the degrees of freedom ($df$) and look up the critical value on the AP Biology Equations and Formulas sheet.
• Evaluate or justify a claim by comparing $\chi^2$ with the critical value at $p = 0.05$ (or $p = 0.01$ if specified) and explaining the biological meaning of rejecting or failing to reject $H_0$.

Core ideas

The null hypothesis ($H_0$)

A null hypothesis is a testable statement that there is no statistically significant difference between the observed counts and the counts expected from a specific model (any deviation is due to random chance alone).

• Genetics example: “The two genes assort independently, and any difference between the observed offspring numbers and the expected $9:3:3:1$ ratio is due to chance.”
• Behavior/ecology example: “Pillbugs show no preference between wet and dry chambers; any difference from a $1:1$ distribution is due to chance.”

Calculating expected values and $\chi^2$

The AP Biology Equations and Formulas sheet provides both the formula and the critical-value table:

• $o$ = observed count in a category (must be raw counts, never percentages or proportions).
• $e$ = expected count in that category, calculated by multiplying the total sample size ($N$) by the expected fraction or probability for that category.
• $\sum$ means you calculate $\frac{(o - e)^2}{e}$ for and add the results together.

Degrees of freedom and the critical value

Degrees of freedom ($df$) equal the number of distinct phenotypic or outcome categories ($n$) minus one: