A genetic experiment involving peas yielded one sample of offspring consisting of 423 green peas and 125 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 27% of offspring peas will be yellow. Ident 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 binomia distribution. What are the null and alternative hypotheses? OA. Ho: P 0.27 H₁: p<0.27 OC. Ho: p*0.27 H₁: p=0.27 OE. Ho: p*0.27 H₁: p>0.27 What is the test statistic? =-2.21 (Round to two decimal places as needed.) What is the P-value? P-value = 0.0272 (Round to four decimal places as needed.) What is the conclusion about the null hypothesis? OA. Fail to reject the null hypothesis because the P-value is greater than the significance level, c. OB. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a. OC. Reject the null hypothesis because the P-value is less than or equal to the significance level, a. OD. Reject the null hypothesis because the P-value is greater than the significance level, a. What is the final conclusion? OA. There is sufficient evidence to support the claim that less than 27% of offspring peas will be yellow. OB. There is not sufficient evidence to warrant rejection of the claim that 27% of offspring peas will be yellow. OC. There is sufficient evidence to warrant rejection of the claim that 27% of offspring peas will be yellow. OD. There is not sufficient evidence to support the claim that less than 27% of offspring peas will be yellow. B. Ho: p=0.27 H₁: p=0.27 OD. Ho: p=0.27 H₁: p>0.27 OF. Ho: p=0.27 H₁: p<0.27