A genetic experiment involving peas yielded one sample of offspring consisting of 405 green peas and 138 yellow peas. Use a 0.05 significance level to test the claim that under the same circumstances, 27% of offsprinig peas will be yellow. Identify the null hypothesis, alterrnative 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.A) what are the hypotheses? alternative hypothesis?B) Identify the test statistict=?C) Identify the P-value___ (round to four decimal places as needed)D) Identify the critical value(s)The critical value(s) is (are)___(round to three decimal places as needed. Use a comma to seperate as needed)

A genetic experiment with peas resulted in 138 yellow and 405 green peas. The null hypothesis was rejected at the 0.05 level, concluding that the proportion of yellow peas is different from 27%. The test statistic was -1.7524  and the p-value was 0.0791. The critical values were -1.96 and 1.96.

The null hypothesis is that the proportion of yellow peas is equal to 0.27, and the alternative hypothesis is that the proportion of yellow peas is not equal to 0.27.

The test statistic is the z-score, which is calculated as z = (P - p) / √(p(1-p)/n), where P is the sample proportion, p is the hypothesized proportion, and n is the sample size.

The P-value for the two-tailed test is calculated as P(Z ≤ -z) + P(Z ≥ z), where z is the absolute value of the calculated z-score. Using a significance level of 0.05, the critical z-value is ±1.96.

The sample proportion of yellow peas is P = 138 / (405 + 138) ≈ 0.2543. The calculated z-score is z = (0.2543 - 0.27) / √(0.27 * 0.73 / 543) ≈ -1.7524. The P-value is P(Z ≤ -1.7524) + P(Z ≥ 1.7524) ≈ 0.0791.

The critical values for a two-tailed test with a significance level of 0.05 are ±1.96. Since the calculated z-score of -1.7524 is less than the critical value of -1.96, we fail to reject the null hypothesis.

Therefore, there is not enough evidence to conclude that the proportion of yellow peas is different from 0.27. The final conclusion is that the data do not support the claim that under the same circumstances, 27% of offspring peas will be yellow.

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