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 is the test statistic?
$z=-1.66$
(Round to two decimal places as needed.)
What is the P-value?
P-value $=$
(Round to four decimal places as needed.)

Step 1 ：Identify the null hypothesis and the alternative hypothesis. The null hypothesis is that the proportion of yellow peas is 27%, and the alternative hypothesis is that the proportion is not 27%.

Step 2 ：Calculate the test statistic using the formula \(z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\), where \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and \(n\) is the sample size. In this case, \(\hat{p} = \frac{135}{430 + 135}\), \(p_0 = 0.27\), and \(n = 430 + 135\).

Step 3 ：Use a Z-table or a statistical calculator to find the P-value. The P-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true.

Step 4 ：Compare the P-value with the significance level. If the P-value is less than the significance level (0.05 in this case), we reject the null hypothesis. If the P-value is greater than the significance level, we fail to reject the null hypothesis.

Step 5 ：\(\boxed{\text{Final Answer: The test statistic is } z = -1.66 \text{ and the P-value is } 0.0963}\)