1. Chi-Square (χ 2 ) Analysis Statistical Analysis of Genetic Data

2. Statistical Analysis of Data  Statistical tests (including Chi-Square) are used to determine to test for SIGNIFICANCE in experimental results  A chi-square analysis is performed to test if the observed results are “close enough” to the expected results that they are probably the result of chance, not some other factor If the data differs greatly from what is expected and it’s NOT due to randomness, some other factor(s) must be influencing the results

3. The Null Hypothesis  The null hypothesis (H 0 )is a statement that there is no substantial STATISTICAL deviation (or difference) between observed and expected data

4. Example  If you were to toss a coin 10 times, what would your expected results be? 5 heads, 5 tails  What if you got 8 heads and 2 tails? You might ask “is this a fair coin, or is this one that is weighted to come up mostly heads?” To determine the likelihood that chance is the only thing influencing your results (not some outside factor, like a weighted coin!), you would perform a Chi-square analysis In other words, how likely is it that your results are due to chance alone?

5. The Chi-Square Equation Χ 2 = where Χ 2 = Chi-square ∑= stands for summation o = observed results e = expected results

6. Let’s Try It!  Coin Flip Example HeadsTails Expected Results (e) 55 Observed Results (o) 82

7. Let’s Try It! Χ 2 = ∑ (o-e) 2 e Χ 2 = ∑ (8 – 5) 2 + (2 – 5) 2 5 5 Χ 2 = ∑ (3) 2 + (-3) 2 5 5 Χ 2 = ∑ 9 + 9 5 Χ 2 = ∑1.8 + 1.8 Χ 2 = 3.6

8. Is that value significant?  To determine whether or not the Chi-square value is significant, we must consult the Critical Values Table Determine the degrees of freedom (df) for your experiment  It is the number of phenotypic classes (n) minus 1  Heads/tails (2 phenotypes); therefore df = 1 Find the p (probability) value  Sciences typically use the p value of 0.05 as the cutoff for accepting or rejecting the null hypothesis This means that the difference between your observed and your expected data would happen by chance alone fewer than 1 time in 20 (0.05). Therefore, there must be some significant difference between the groups! If the calculated chi-square value is greater than or equal to the critical value from the table, then the null hypothesis is REJECTED

9. Is that value significant? Probabilit y Degrees of Freedom (df) 12345 0.053.845.997.829.4911.1 0.016.649.2111.313.215.1 0.00110.813.816.318.520.5 df = 1 (2 phenotypes – 1) = 1P = 0.05 Critical Value = 3.84Our Chi-Square Value = 3.6 Since our Chi-Square value is less than the critical value, we ACCEPT the null hypothesis! This means that there is no statistical difference between the EXPECTED and OBSERVED results!

10. Another Example: Pea Plants  Two pea plants are crossed: Tt x Tt  They have 520 offspring: 338 are tall 182 are short  Is this due only to chance, or is something else going on?

11. Another Example: Pea Plants Phenotyp e Genotype# Observed (o) # Expected (e) (o – e) Tall Short Total:

12. Another Example: Pea Plants Χ 2 = ∑ (o-e) 2 e Χ 2 = ∑ (-52) 2 + (52) 2 390 130 Χ 2 = ∑ 2704 + 2704 390 130 Χ 2 = ∑6.93 + 20.8 Χ 2 = 27.7

13. Another Example: Pea Plants  Is this Χ 2 value significant? df = Consult the Chi Square critical values table!  Since our Χ 2 value of 27.7 is GREATER than the critical value of 3.84, we must REJECT the null hypothesis In other words, something else must be going on here!