Question

A genetic experiment involving peas yielded one sample of offspring consisting of 442 green peas and 123 yellow peas. Use a 0.01 significance level to test the claim that under the same​ circumstances, 26​% 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

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.26

H_a : p 0.26

= x / n = 123 / 442 = 0.2783

P₀ = 0.26

1 - P₀ = 0.74

Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.2783 - 0.26 / [(0.26 * 0.74) / 442]

= 0.876

P(z > 0.876) = 1 - P(z < 0.876) = 0.1905

P-value = 0.381

= 0.01

P-value >

Fail to reject the null hypothesis .

There is not sufficient evidence that the  0.01 significance level to test the claim that under the same​ circumstances, 26​% of offspring peas will be yellow .