A genetic experiment involving peas yielded one sample of offspring consisting of 444 green peas and 132 yellow peas. Use a 0.05 significance level to test the clai same circumstances, $26 \%$ of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypo 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.
\[
\begin{array}{l}
H_{0}: p \neq 0.26 \\
H_{1}: p=0.26
\end{array}
\]
C.
\[
\begin{array}{l}
H_{0}: p \neq 0.26 \\
H_{1}: p< 0.26
\end{array}
\]
E. H
\[
\begin{array}{l}
H_{0}: p=0.26 \\
H_{1}: p< 0.26
\end{array}
\]

What is the test statistic?
\[
z=\square
\]
(Round to two decimal places as needed.)
B.
\[
\begin{array}{l}
H_{0}: p=0.26 \\
H_{1}: p \neq 0.26
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: p=0.26 \\
H_{1}: p> 0.26
\end{array}
\]
F.
\[
\begin{array}{l}
H_{0}: p \neq 0.26 \\
H_{1}: p> 0.26
\end{array}
\]

Step 1 ：Define the null hypothesis (H0) and the alternative hypothesis (H1). The null hypothesis is a statement of no effect or no difference, while the alternative hypothesis is a statement that indicates the presence of an effect or a difference. In this case, the null hypothesis is that the proportion of yellow peas is equal to 26% (p = 0.26), and the alternative hypothesis is that the proportion of yellow peas is not equal to 26% (p ≠ 0.26). So, the null and alternative hypotheses are: \[H_{0}: p=0.26\] \[H_{1}: p \neq 0.26\]

Step 2 ：Calculate the test statistic, which is a measure of how far our sample statistic is from the null hypothesis. It is calculated using the sample data. In this case, we are dealing with proportions, so we will use the z-test for proportions. The formula for the z-test statistic is: \[z = \frac{\hat{p} - p_{0}}{\sqrt{\frac{p_{0} * (1 - p_{0})}{n}}}\] where \(\hat{p}\) is the sample proportion, \(p_{0}\) is the proportion under the null hypothesis, and n is the sample size.

Step 3 ：Substitute the given values into the formula: n = 576, \(\hat{p}\) = 0.22916666666666666, and \(p_{0}\) = 0.26. The calculation gives: \[z = -1.6870547845739476\]

Step 4 ：Round the test statistic to two decimal places to get the final answer: \[z = -1.69\]

Step 5 ：Box the final answer: \[\boxed{z = -1.69}\]