Question

genetic experiment involving peas yielded one sample of offspring consisting of

422

green peas and

134

yellow peas. Use a

0.01

significance level to test the claim that under the same​ circumstances,

24​%

of offspring peas will be yellow. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method and the normal distribution as an approximation to the binomial distribution.

Answer 1

To Test :-

H0 :- P = 0.24
H1 :- P ≠ 0.24

P = X / n = 134/556 = 0.241

Test Statistic :-
Z = ( P - P0 ) / ( √((P0 * q0)/n)
Z = ( 0.241 - 0.24 ) / ( √(( 0.24 * 0.76) /556))
Z = 0.0556

Test Criteria :-
Reject null hypothesis if Z > Z(α/2)
Z(α/2) = Z(0.01/2) = 2.576
Z < Z(α/2) = 0.0556 < 2.576, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

Decision based on P value
P value = 2 * P ( Z > 0.0556 )  = 0.9557
Reject null hypothesis if P value < α = 0.01
Since P value = 0.9557 > 0.01, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

There is sufficient evidence to support the claim that 24​% of offspring peas is yellow.