# Chi-Squared Test.

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Chi-Squared Test

Statistical Testing Compare expected and observed Mendel’s crosses
Theoretical (3:1, 9:3:3:1) Did Mendel get exactly 75 tall plants and 25 short plants? Close – deviation due to chance Natural variability of living systems (vs an incorrect hypothesis) Mistakes in counting (vs inappropriate testing) How much deviation is acceptable? Are my underlying assumptions accurate? Is my sampling reliable?

Hypothesis Testing Cross two pea plants with green seeds
Yields a population of 880 plants 639 have green seeds 241 have yellow seeds Propose the genotypes of the parents

Scientific Hypothesis
Allele for green is dominant to allele for yellow Parent plants were both heterozygous for this trait If this hypothesis is true Predicted ratio of offspring from this cross is 3:1 (based on Mendel’s laws) G g GG Gg gG gg Green (G) is dominant Yellow (g) Parents are heterozygous (Gg x Gg) Phenotypic ratio is 3:1

Chi-Square Test Compare observed data with data you would expect to obtain according to a specific hypothesis How much deviation can occur before you conclude something other than chance at work? p = 0.05 Χ2 = Σ(O-E)2 / E O = observed value E = expected value Tests the null hypothesis States that there is no significant difference between the expected and observed result

Null Hypothesis There is no significant difference between your observed pea offspring and offspring produced according to Mendel’s laws. Goal: determine if null hypothesis should be rejected or not rejected Use chi-squared

Step 1 Determine number of expected in each category Ratio is 3:1
Total number of observed individuals = 880 Expected numerical values 660 green (3/4 x 880 = 660) 220 yellow (1/4 x 880 = 220)

Step 2 Calculate X2 Green Yellow Observed (o) 639 241 Expected (e) 660
220 Deviation (o-e) -21 21 Deviation2 (d2) 441 d2/e 0.668 2.005 X2 = Σd2/e = 2.673

Step 3 Determine degrees of freedom (df)
Number of categories in the problem minus 1 Two categories (green and yellow) 1 df

Step 4 Determine a relative standard to serve as the basis for rejecting the hypothesis Commonly used in biological research: p<0.05 P value – probability of rejecting the null hypothesis (no difference between observed and expected) when it is actually true Probability of making an error of falsely stating that there is a significant difference between e and o when there is not a significant difference. p < 0.05 Stating that there is a less than 5% chance of error of stating there is a difference when there actually isn’t a significant difference

Step 5 Refer to chi-square distribution table
Locate appropriate df (1) Locate the value corresponding to the p value of 0.05

Degrees of Freedom (df)
Conclusions If your X2 > distribution table # Reject null hypothesis Conclude observed numbers significantly different from the expected X2 (2.673) is not greater than 3.84 (df=1, p=0.05) Therefore your observed distribution of plants with green and yellow seeds is not significantly different from the distribution that would be expected under Mendel’s Laws. Any minor difference -> due to chance Degrees of Freedom (df) Probability (p) 0.95 0.90 0.80 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.001 1 0.004 0.02 0.06 0.15 0.46 1.07 1.64 2.71 3.84 6.64 10.83 2 0.21 0.45 0.71 1.39 2.41 3.22 4.60 5.99 9.21 13.82 3 0.35 0.58 1.01 1.42 2.37 3.66 4.64 6.25 7.82 11.34 16.27 4 1.06 1.65 2.20 3.36 4.88 7.78 9.49 13.28 18.47 5 1.14 1.61 2.34 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52 Nonsignificant Significant