Question

A genetic experiment involving peas yielded one sample of offspring consisting of 437 green peas and 121 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

Null and alternative hypotheses

Ho : p = 0.24

Ha : p = 0.24

Test statistic Z

Z = ( p^ - p)/sqrt[ p*q/n]

Where p^ = X/n = number of yellow offspring peas/total number of offspring peas

p^ = 121/558 = 0.22

Z = (0.22 - 0.24)/sqrt[0.24*0.76/558]

Z = -1.11 this is test statistic value

p-value for Z = -1.11 and two tailed test

p-value = 2* P( Z < -1.11)

p-value = 0.2670

Here p-value = 0.2670 > 0.01

Decision : we fail to reject the null hypothesis Ho

Conclusion : There is sufficient evidence to support the claim that

under the same​ circumstances, 24​% of offspring peas will be yellow