Question

A genetic experiment involving peas yielded one sample of offspring consisting of 447 green peas and 178 yellow peas. Use a 0.05 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.

A. What are the null and alternative​ hypotheses?
A. Upper H 0 : p = 0.27
Upper H 1 : p < 0.27
B. Upper H 0 : p not equals 0.27
Upper H 1 : p = 0.27
C. Upper H 0 : p not equals 0.27
Upper H 1 : p > 0.27
D. Upper H 0 : p not equals 0.27
Upper H 1 : p < 0.27
E. Upper H 0 : p = 0.27
Upper H 1 : p not equals 0.27
F. Upper H 0 : p - 0.27
Upper H 1 : p > 0.27

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

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

D. What is the conclusion about the null​ hypothesis?
A. Reject the null hypothesis because the​ P-value is less than or equal to the significance​ level, alpha.
B. Fail to reject the null hypothesis because the​ P-value is greater than the significance​ level, alpha.
C. Reject the null hypothesis because the​ P-value is greater than the significance​ level, alpha.
D. Fail to reject the null hypothesis because the​ P-value is less than or equal to the significance​ level, alpha.

E. What is the final​ conclusion?
A. There is sufficient evidence to support the claim that less than 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 not sufficient evidence to support the claim that less than 27​% of offspring peas will be yellow.
D. There is sufficient evidence to warrant rejection of the claim that 27​% of offspring peas will be yellow

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.27

H_a : p 0.27

= x / n = 178 / 447 = 0.3982

Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.3982 - 0.27 / [(0.27 * 0.73) / 447]

= 6.11

P(z > 6.11) = 1 - P(z < 6.11) = 0

P-value = 2 * 0 = 0

= 0.05

P-value <

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

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