Problem
A genetic experiment involving peas yielded one sample of offspring consisting of 430 green peas and 135 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. What are the null and alternative hypotheses? A. $H_{0}: P=0.27$ \[ H_{1}: p \neq 0.27 \] c. $H_{0}: p=0.27$ \[ H_{1}: p>0.27 \] E. $H_{0}: p \neq 0.27$ \[ H_{1}: p<0.27 \] B. \[ \begin{array}{l} H_{0}: p \neq 0.27 \\ H_{1}: p>0.27 \end{array} \] D. \[ \begin{array}{l} H_{0}: p=0.27 \\ H_{1}: p<0.27 \end{array} \] F. \[ \begin{array}{l} H_{0}: p \neq 0.27 \\ H_{1}: p=0.27 \end{array} \]
Solution
Step 1 ：The null hypothesis is the statement that the proportion of yellow peas is equal to 0.27, which is the claim we are testing. The alternative hypothesis is the statement that the proportion of yellow peas is not equal to 0.27. This is because we are testing the claim that the proportion is 0.27, not that it is not 0.27 or greater than 0.27. Therefore, the null and alternative hypotheses are: \(H_{0}: P=0.27\) and \(H_{1}: p \neq 0.27\). So, the answer is A.
Step 2 ：Final Answer: \(\boxed{\begin{array}{l} H_{0}: P=0.27 \\ H_{1}: p \neq 0.27 \end{array}}\)
