Question

if we actually did the cross and got 1150 yellow and 350 green. would this be a consistent with what was expected? chi squared calculation table category observed (o) expected (e) (o - e) (o - e)^2 (o - e)^2 / e σ x2 value σ

Explanation:

Step1: Assume a 3:1 ratio expected

Let's assume a 3:1 ratio of yellow to green as is common in simple Mendelian crosses. The total number of observed individuals is $1150 + 350=1500$.

Step2: Calculate expected values

For yellow, $E_{yellow}=\frac{3}{4}\times1500 = 1125$. For green, $E_{green}=\frac{1}{4}\times1500 = 375$.

Step3: Calculate $(O - E)$ for yellow

$O_{yellow}=1150$, so $O - E=1150 - 1125 = 25$.

Step4: Calculate $(O - E)$ for green

$O_{green}=350$, so $O - E=350 - 375=- 25$.

Step5: Calculate $(O - E)^2$ for yellow

$(O - E)^2_{yellow}=25^2 = 625$.

Step6: Calculate $(O - E)^2$ for green

$(O - E)^2_{green}=(-25)^2 = 625$.

Step7: Calculate $\frac{(O - E)^2}{E}$ for yellow

$\frac{(O - E)^2_{yellow}}{E_{yellow}}=\frac{625}{1125}\approx0.56$.

Step8: Calculate $\frac{(O - E)^2}{E}$ for green

$\frac{(O - E)^2_{green}}{E_{green}}=\frac{625}{375}\approx1.67$.

Step9: Calculate the chi - squared value

$\chi^{2}=\sum\frac{(O - E)^2}{E}=0.56 + 1.67=2.23$.

The degrees of freedom $df = n - 1$, where $n = 2$ (two categories: yellow and green), so $df=1$. Looking up the critical value for $\alpha = 0.05$ and $df = 1$ in the chi - squared distribution table, the critical value is 3.84. Since $2.23<3.84$, we fail to reject the null hypothesis.

Answer:

Yes, the results are consistent with what was expected.