Question

A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 149 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.

What are the null and alternative hypotheses?

A. H0: p=0.24

H1: p≠0.24

B. H0: p≠0.24

H1: p=0.24

C. H0: p=0.24

H1: p>0.24

D. H0: p≠0.24

H1: p>0.24

E. H0: p≠0.24

H1: p<0.24

F. H0: p=0.24

H1 : p <0.24

What is the test statistic?

z=__?

What is the P-value?

P-value = __?

What is the conclusion about the null hypothesis?

A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α.

B. Reject the null hypothesis because the P-value is greater than the significance level, α.

C. Fail to reject

the null hypothesis because the P-value is greater than the significance level, α.

D. Reject the null hypothesis because the P-value is less than or equal to the significance level, α.

What is the final conclusion?

A. There is not sufficient evidence to support the claim that less than 24 % of offspring peas will be yellow.

B. There is sufficient evidence to warrant rejection of the claim that 24 % of offspring peas will be yellow.

C. There is

sufficient evidence to support the claim that less than 24 % of offspring peas will be yellow.

D. There is not sufficient evidence to warrant rejection of the claim that 24 % of offspring peas will be yellow.

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ : p = 0.24

H_a : p 0.24

n = 420

x = 149

= x / n = 149 / 420 = 0.3548

P₀ = 0.24

1 - P₀ = 1 - 0.24 = 0.76

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.3548 -0.24 / [(0.24 * 0.76) / 420]

= 5.507

Test statistic = 5.507

This is the two tailed test .

P(z > 5.507) = 1 - P(z < 5.507) = 1 - 1 = 0

P-value = 0

= 0.01

P-value <

D. Reject the null hypothesis because the P-value is less than or equal to the significance level, α.

B. There is sufficient evidence to warrant rejection of the claim that 24 % of offspring peas will be yellow.