**Answer:**

Chi-square value = **0.6**

The difference between the observed individuals and the expected individuals is **not statistically significant**. There is **not enough evidence to reject the null hypothesis.**

**Explanation:**

H₀= Individuals will be equally distributed

H₁ = Individuals will not be equally distributed.

• Chi square= ∑ ((O-E)²/E)

∑ is the sum of the terms

O are the Observed individuals

E are the Expected individuals

• Freedom degrees = 1

• Significance level, 5% = 0.05

• Table value/Critical value = 3.84

<u>The number of observed individuals</u>:

- 63 have purple flowers,
- 17 have white flowers

Total number of individuals, N = 80 = 63 + 17

<u>Observed Genotypic frequencies</u>:

- F(PF) = 63/80 =
**0.79** - F(pf) = 17/80 =
**0.21 **

p² + 2pq + q² = 1

p² + 2pq = F(PF)

0.79 + 0.21 = 1

<u>The expected genotypic frequency</u>:

F (PF)= 0.75

F (pf) = 0.25

<u>The number of expected individuals</u>:

- PF= 0.75 x 80 = 60
- wf = 0.25 x 80 = 20

<u>Chi square = sum (O-E)²/E</u>

PF =(63 - 60) ² / 60

PF = 0.15

pf= (17 - 20)²/20

pf= 0.45

Chi square= sum ((O-E)²/E) = 0.15 + 0.45 = 0.60

**Chi square = 0.6****Degrees of freedom = 1
****alpha 0.05 ****Table value/Critical value = 3.84 **

**0.6 < 3.84** meaning that **the difference between the observed individuals and the expected individuals is not statistically significant. There is not enough evidence to reject the null hypothesis.** Probably both parentals were heterozygous for the trait.