The chi-square (χ²) test is a crucial statistical tool in AP Biology, used to determine if observed data differs significantly from expected data. Mastering this test is vital for success in the course and the AP exam. This guide provides practice problems and explanations to help you confidently tackle chi-square analyses. We'll focus on the goodness-of-fit test, a common application in AP Biology.
Understanding the Chi-Square Goodness-of-Fit Test
The goodness-of-fit test assesses how well observed data fit a particular expected distribution. In simpler terms, it helps us determine if deviations from what we expect are due to chance or a significant factor influencing the results. The key steps involve:
-
Stating the null hypothesis (H₀): This hypothesis assumes there's no significant difference between observed and expected values. Any deviation is purely random.
-
Calculating the chi-square value (χ²): This involves comparing observed (O) and expected (E) values for each category using the formula: χ² = Σ [(O - E)² / E]
-
Determining degrees of freedom (df): This is calculated as the number of categories minus 1 (df = n - 1).
-
Finding the p-value: Using a chi-square distribution table and the calculated χ² and df, we find the p-value. This represents the probability of obtaining the observed results if the null hypothesis is true.
-
Interpreting the results: If the p-value is less than the significance level (typically 0.05), we reject the null hypothesis, concluding there's a significant difference between observed and expected values. If the p-value is greater than 0.05, we fail to reject the null hypothesis, suggesting the observed deviations are likely due to chance.
AP Bio Chi-Square Practice Problems
Let's work through some examples:
Problem 1: Mendelian Genetics
A genetics experiment involving a monohybrid cross (Tt x Tt) predicted a 3:1 phenotypic ratio (tall:dwarf). The observed results were 72 tall plants and 38 dwarf plants. Perform a chi-square test to determine if the observed results are consistent with the expected Mendelian ratio.
Solution:
-
Null Hypothesis (H₀): The observed phenotypic ratio is consistent with the expected 3:1 ratio.
-
Expected Values: Total plants = 72 + 38 = 110. Expected tall plants = (3/4) * 110 = 82.5. Expected dwarf plants = (1/4) * 110 = 27.5.
-
Chi-Square Calculation:
χ² = [(72 - 82.5)² / 82.5] + [(38 - 27.5)² / 27.5] ≈ 1.36 + 4.09 ≈ 5.45
-
Degrees of Freedom: df = 2 - 1 = 1
-
P-value: Using a chi-square distribution table with df = 1 and χ² = 5.45, the p-value is between 0.02 and 0.05.
-
Conclusion: Since the p-value is less than 0.05, we reject the null hypothesis. The observed data deviate significantly from the expected 3:1 ratio, possibly due to factors other than simple Mendelian inheritance. There might be other genes at play, environmental influence, or experimental error.
Problem 2: Hardy-Weinberg Equilibrium
A population of butterflies has two alleles for wing color: B (brown) and b (white). The expected genotype frequencies under Hardy-Weinberg equilibrium are: BB = 0.49, Bb = 0.42, bb = 0.09. A sample of 200 butterflies shows the following observed genotype frequencies: BB = 90, Bb = 80, bb = 30. Test if this population is in Hardy-Weinberg equilibrium.
Solution: (Follow the same steps as Problem 1, adapting the calculations to the Hardy-Weinberg context.) Remember to calculate expected numbers of individuals for each genotype by multiplying the expected frequencies by the total number of individuals (200).
Further Practice and Resources
These examples provide a solid foundation. For more practice, search for "AP Biology chi-square problems" online or consult your textbook. Remember to clearly define your hypotheses, show your calculations, and interpret your results in the context of the biological question. Consistent practice will build your confidence and mastery of this essential statistical tool.
