A genetic experiment involving peas yielded one sample of offspring consisting of 450 green peas and 175 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 25% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesi: 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? O A. Ho: p= 0.25 H,:p>0.25 о В. Но: р3D0.25 H,:p#0.25 Ос. Но: р#0.25 O D. Ho: p+0.25 H1: p<0.25 H1:p>0.25 ОЕ. Но: р#0.25 H1:p=0.25 O F. Ho:p= 0.25 H1: p<0.25 What is the test statistic? (Round to two decimal places as needed.) What is the P-value? P-value = (Round to four decimal places as needed.) What is the conclusion about the null hypothesis? O A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a. B. Reject the null hypothesis because the P-value is greater than the significance level, a. OC. Fail to reject the null hypothesis because the P-value is greater than the significance level, a. O D. Reject the null hypothesis because the P-value is less than or equal to the significance level, a. What is the final conclusion? O A. There is sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. O B. There is sufficient evidence to support the claim that less than 25% of offspring peas will be yellow. OC. There is not sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. D. There is not sufficient evidence to support the claim that less than 25% of offspring peas will be yellow.

Let p is defined as the proportion of offspring peas will be yellow.

The claim of the test is 25% of offspring peas will be yellow.

The hypothesis is,

Null hypothesis:

$H_0:p=0.25$

Alternative hypothesis:

$H_1:p\ne 0.25$

The proportion of offspring peas will be yellow is,

$\begin{array}{rcl}\hat{p}& =& \frac{175}{450+175}\\ & =& \frac{175}{625}\\ & =& 0.28\end{array}$

The test statistic is,

$\begin{array}{rcl}z& =& \frac{\hat{p}-p}{\sqrt{{\displaystyle \frac{pq}{n}}}}\\ & =& \frac{0.28-0.25}{\sqrt{{\displaystyle \frac{0.25\left(1-0.25\right)}{625}}}}\\ & =& \frac{0.03}{0.0173}\\ & =& 1.73\end{array}$

Thus, the test statistic is 1.73.