1. Chi-Squared Test

2. 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?

3. 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

4. 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

5. 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

6. 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

7. 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)

8. 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

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

10. 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

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

12. 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