Answer:
Yes, she got the predicted outcome.
Explanation:
A genetic engineer was attempting to cross a tiger and a cheetah. She predicted a phenotypic outcome of the traits she was observing to be in the following ratio: 4 (stripes only): 3 (spots only): 9 (both stipes and spots). When the cross was performed and she counted the individuals she found 50 with stripes only, 41 with spots only and 85 with both. According to the Chi-square test, did she get the predicted outcome?
Using the Chi square formula:
[tex]X^2 = \frac{(\alpha - \beta )^2}{\beta }[/tex]
Where, [tex]\alpha[/tex] = observed frequency (OF) and [tex]\beta[/tex] = expected frequency (EF).
Phenotype OF EF [tex]X^2[/tex]
Stripes only 50 4/16 x 176 = 44 [tex]\frac{(50-44)^2}{44}[/tex] = 0.8182
Spots only 41 3/16 x 176 = 33 [tex]\frac{(41-33)^2}{33}[/tex] = 1.9394
Stripe & spot 85 9/16 x 176 = 99 [tex]\frac{(85-99)^2}{99}[/tex] = 1.9798
Total [tex]X^2[/tex] = 4.74
Degree of freedom = 3 - 1 = 2
[tex]X^2[/tex] tabulated (p = 0.05, df = 2) = 5.99
The [tex]X^2[/tex] calculated is greater than the [tex]X^2[/tex] tabulated, hence, we conclude that she got the predicted outcome. It means there is no significant difference between the expected outcome and the observed outcome.
The chi-square test allows us to tell if the difference between observed and the expected numbers is statistically significant or by chance. In the exposed example, the genetic engineer got the predicted outcome.
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Available data:
4 stripes only: 3 spots only: 9 both stripes and spots
50 with stripes only, 41 with spots only and 85 with both
176 individuals
we need to know if the observed numbers coincides with the expected ratio. So we will perform a chi-square test.
Chi-square = X² = Σ (Obs-Exp)² / Exp
So first we need to get the expected numbers.
16 ------------------------ 176 individuals
4 stripes --------------X= 44 indivuals
3 spots ---------------- X = 33 individuals
9 both ----------------- X = 99 individuals
Stripes Spots Both
Observed 50 41 85
Expected 44 33 99
(Obs-Exp)² / Exp 0.82 1.94 1.98
X² = Σ (Obs-Exp)² / Exp = 0.82 + 1.94 + 1.98 = 4.74
Degrees of freedom = n - 1 = 3 - 1 = 2
Significance level, 5% = 0.05
Table value/Critical value = 5.991
X² < CV
4.74 < 5.99
There is not enough evidence to reject the null hypothesis.
The genetic engineer got the predicted outcome.
The difference between the observed individuals and the expected individuals is statistically insignificant.
The difference is by random chances
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