Question

A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 120 yellow peas. Use a 0.01 significance level to test the claim that under the same​ circumstances, 27​% 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. Upper H 0 : p not equals 0.27 Upper H 1 : p less than 0.27

B. Upper H 0 : p equals 0.27 Upper H 1 : p less than 0.27

C. Upper H 0 : p not equals 0.27 Upper H 1 : p equals 0.27

D. Upper H 0 : p equals 0.27 Upper H 1 : p not equals 0.27

E. Upper H 0 : p equals 0.27 Upper H 1 : p greater than 0.27

F. Upper H 0 : p not equals 0.27 Upper H 1 : p greater than 0.27

What is the test​ statistic? z=___________​(Round to two decimal places as​ needed.)

What is the​ P-value? ​P-=_____________ ​(Round to four decimal places as​ needed.)

What is the conclusion about the null​ hypothesis?

A. Fail to reject the null hypothesis because the​ P-value is greater than the significance​ level, alpha.

B. Fail to reject the null hypothesis because the​ P-value is less than or equal to the significance​ level, alpha.

C. Reject the null hypothesis because the​ P-value is less than or equal to the significance​ level, alpha.

D. Reject the null hypothesis because the​ P-value is greater than the significance​ level, alpha.

What is the final​ conclusion?

A. There is sufficient evidence to warrant rejection of the claim that 27​% of offspring peas will be yellow.

B. There is not sufficient evidence to warrant rejection of the claim that 27​% of offspring peas will be yellow.

C. There is sufficient evidence to support the claim that less than 27​% of offspring peas will be yellow.

D. There is not sufficient evidence to support the claim that less than 27​% of offspring peas will be yellow

Answer 1

Solution:

Given: A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 120 yellow peas.

thus n = 420 + 120 = 540

x = number of yellow peas = 120

Claim: under the same​circumstances, 27​% of offspring peas will be yellow.

Level of significance = 0.01

Part a)  What are the null and alternative​hypotheses?

Part b) What is the test​ statistic?

Part c) What is the​ P-value?

For two tailed test , p-value is:

p-value = 2* P(Z > z test statistic) if z is positive

p-value = 2* P(Z < z test statistic) if z is negative

thus

p-value = 2* P(Z < z test statistic)

p-value = 2* P(Z < -2.50 )

Look in z table for z = -2.5 and 0.00 and find corresponding area.

P( Z< -2.50) = 0.0062

thus

p-value = 2* P(Z < -2.50 )

p-value = 2*  0.0062

p-value = 0.0124

Part d) What is the conclusion about the null​ hypothesis?

Decision Rule:
Reject null hypothesis H0, if P-value < 0.01 level of significance, otherwise we fail to reject H0

Since p-value = 0.0124 > 0.01 level of significance, we fail to reject H0.

Thus

A. Fail to reject the null hypothesis because the​ P-value is greater than the significance​ level, alpha

Part e) What is the final​ conclusion?

B. There is not sufficient evidence to warrant rejection of the claim that 27​% of offspring peas will be yellow.