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.

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